Chapter 13 INTERNAL FORCED CONVECTION - …mbahrami/ENSC 388/Solution manual/IntroTHT_2e...13-2 General Flow Analysis 13-1C Liquids are usually transported in circular pipes because

13-1

Solutions Manual for

Introduction to Thermodynamics and Heat Transfer Yunus A. Cengel 2nd Edition, 2008

Chapter 13 INTERNAL FORCED CONVECTION

PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill.

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

13-2

General Flow Analysis 13-1C Liquids are usually transported in circular pipes because pipes with a circular cross-section can withstand large pressure differences between the inside and the outside without undergoing any distortion. 13-2C Reynolds number for flow in a circular tube of diameter D is expressed as

ν

DVavgRe = where ρμν

ρππρρ====∞ and 4

)4/( 22 Dm

Dm

AmV

c

&&&

Substituting,

m, Vavg μπρμρπν D

mD

DmDV && 4)/(

4Re2

avg ===

13-3C Engine oil requires a larger pump because of its much larger density. 13-4C The generally accepted value of the Reynolds number above which the flow in a smooth pipe is turbulent is 4000. 13-5C For flow through non-circular tubes, the Reynolds number as well as the Nusselt number and the

friction factor are based on the hydraulic diameter Dh defined as pA

D ch

4= where Ac is the cross-sectional

area of the tube and p is its perimeter. The hydraulic diameter is defined such that it reduces to ordinary

diameter D for circular tubes since DD

DpA

D ch ===

ππ 4/44 2

.

13-6C The region from the tube inlet to the point at which the boundary layer merges at the centerline is called the hydrodynamic entry region, and the length of this region is called hydrodynamic entry length. The entry length is much longer in laminar flow than it is in turbulent flow. But at very low Reynolds numbers, Lh is very small (Lh = 1.2D at Re = 20). 13-7C The friction factor is highest at the tube inlet where the thickness of the boundary layer is zero, and decreases gradually to the fully developed value. The same is true for turbulent flow. 13-8C In turbulent flow, the tubes with rough surfaces have much higher friction factors than the tubes with smooth surfaces. In the case of laminar flow, the effect of surface roughness on the friction factor is negligible. 13-9C The friction factor f remains constant along the flow direction in the fully developed region in both laminar and turbulent flow. 13-10C The fluid viscosity is responsible for the development of the velocity boundary layer. For the idealized inviscid fluids (fluids with zero viscosity), there will be no velocity boundary layer.


13-3

13-11C The number of transfer units NTU is a measure of the heat transfer area and effectiveness of a heat transfer system. A small value of NTU (NTU < 5) indicates more opportunities for heat transfer whereas a large NTU value (NTU >5) indicates that heat transfer will not increase no matter how much we extend the length of the tube. 13-12C The logarithmic mean temperature difference lnTΔ is an exact representation of the average temperature difference between the fluid and the surface for the entire tube. It truly reflects the exponential decay of the local temperature difference. The error in using the arithmetic mean temperature increases to undesirable levels when differs from eTΔ iTΔ by great amounts. Therefore we should always use the logarithmic mean temperature. 13-13C The region of flow over which the thermal boundary layer develops and reaches the tube center is called the thermal entry region, and the length of this region is called the thermal entry length. The region in which the flow is both hydrodynamically (the velocity profile is fully developed and remains unchanged) and thermally (the dimensionless temperature profile remains unchanged) developed is called the fully developed region. 13-14C The heat flux will be higher near the inlet because the heat transfer coefficient is highest at the tube inlet where the thickness of thermal boundary layer is zero, and decreases gradually to the fully developed value. 13-15C The heat flux will be higher near the inlet because the heat transfer coefficient is highest at the tube inlet where the thickness of thermal boundary layer is zero, and decreases gradually to the fully developed value. 13-16C In the fully developed region of flow in a circular tube, the velocity profile will not change in the flow direction but the temperature profile may. 13-17C The hydrodynamic and thermal entry lengths are given as DLh Re05.0= and

for laminar flow, and Lt = 0 05. Re Pr D DLL th 10≈≈ in turbulent flow. Noting that Pr >> 1 for oils, the thermal entry length is larger than the hydrodynamic entry length in laminar flow. In turbulent, the hydrodynamic and thermal entry lengths are independent of Re or Pr numbers, and are comparable in magnitude. 13-18C The hydrodynamic and thermal entry lengths are given as DLh Re05.0= and

for laminar flow, and DLt PrRe05.0= DLL th 10≈≈ in turbulent flow. Noting that Pr << 1 for liquid metals, the thermal entry length is smaller than the hydrodynamic entry length in laminar flow. In turbulent, the hydrodynamic and thermal entry lengths are independent of Re or Pr numbers, and are comparable in magnitude. 13-19C In fluid flow, it is convenient to work with an average or mean velocity Vavg and an average or mean temperature Tm which remain constant in incompressible flow when the cross-sectional area of the tube is constant. The Vavg and Tm represent the velocity and temperature, respectively, at a cross section if all the particles were at the same velocity and temperature. 13-20C When the surface temperature of tube is constant, the appropriate temperature difference for use in the Newton's law of cooling is logarithmic mean temperature difference that can be expressed as

)/ln(ln

ie

ie

TTTT

TΔΔΔ−Δ

=Δ


13-4

13-21 Air flows inside a duct and it is cooled by water outside. The exit temperature of air and the rate of heat transfer are to be determined. Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the duct is constant. 3 The thermal resistance of the duct is negligible.

10°C

L = 12 m

D = 25 cm

Air 50°C 7 m/s

Properties The properties of air at the anticipated average temperature of 30°C are (Table A-22)

CJ/kg. 1007

kg/m 164.1 3

°==

pcρ

Analysis The mass flow rate of water is

kg/s 0.400=m/s) (74

m) (0.25)kg/m 164.1(4

23

avg

2

avgππρρ =⎟

⎟⎠

⎞⎜⎜⎝

⎛== VDVAm c&

2m 9.425=m) m)(12 25.0(ππ == DLAs

The exit temperature of air is determined from

C 15.47 °=−−=−−=−− )1007)(400.0(

)425.9)(85( )/( )5010(10)( eeTTTT ps cmhA

isse&

The logarithmic mean temperature difference and the rate of heat transfer are

kW 13.9==°°=Δ=

°=⎟⎠⎞

⎜⎝⎛

−−−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=Δ

W13,900)C36.17)(m 425.9)(C. W/m85(

C36.17

501047.1510ln

5047.15

ln

22ln

ln

ThAQ

TTTT

TTT

s

is

es

ie

&


13-5

13-22 Steam is condensed by cooling water flowing inside copper tubes. The average heat transfer coefficient and the number of tubes needed are to be determined. Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the pipe is constant. 3 The thermal resistance of the pipe is negligible. Properties The properties of water at the average temperature of (10+24)/2=17°C are (Table A-15)

Steam, 30°C

L = 5 m

D = 1.2 cm

Water 10°C 4 m/s

24°C CJ/kg. 8.4183

kg/m 7.998 3

°==

pcρ

Also, the heat of vaporization of water at 30°C is . kJ/kg 2431=fgh

Analysis The mass flow rate of water and the surface area are

kg/s 0.4518=m/s) (44

m) (0.012)kg/m 7.998(4

23

avg

2

avgππρρ =⎟

⎟⎠

⎞⎜⎜⎝

⎛== VDVAm c&

The rate of heat transfer for one tube is

W460,26)C1024)(CJ/kg. 8.4183)(kg/s 4518.0()( =°−°=−= iep TTcmQ &&

The logarithmic mean temperature difference and the surface area are

C63.11

10302430ln

1024

lnln °=

⎟⎠⎞

⎜⎝⎛

−−−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=Δ

is

es

ie

TTTT

TTT

2m 0.1885=m) m)(5 012.0(ππ == DLAs

The average heat transfer coefficient is determined from

C.kW/m 12.1 2 °=⎟⎠⎞

⎜⎝⎛

°=

Δ=⎯→⎯Δ=

W1000kW 1

)C63.11)(m 1885.0( W460,26

2ln

ln TAQ

hThAQs

s

&&

The total rate of heat transfer is determined from

kW 65.364kJ/kg) 2431)(kg/s 15.0( === fgcondtotal hmQ &&

Then the number of tubes becomes

13.8=== W460,26 W650,364

QQ

N totaltube &

&


13-6

13-23 Steam is condensed by cooling water flowing inside copper tubes. The average heat transfer coefficient and the number of tubes needed are to be determined. Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the pipe is constant. 3 The thermal resistance of the pipe is negligible. Properties The properties of water at the average temperature of (10+24)/2=17°C are (Table A-15)

Steam, 30°C

L = 5 m

D = 1.2 cm

Water 10°C 4 m/s

24°C CJ/kg. 8.4183

kg/m 7.998 3

°==

pcρ

Also, the heat of vaporization of water at 30°C is . kJ/kg 2431=fgh

Analysis The mass flow rate of water is

kg/s 0.4518=m/s) (44

m) (0.012)kg/m 7.998(4

23

avg

2

avgππρρ =⎟

⎟⎠

⎞⎜⎜⎝

⎛== VDVAm c&

The rate of heat transfer for one tube is

W460,26)C1024)(CJ/kg. 8.4183)(kg/s 4518.0()( =°−°=−= iep TTcmQ &&


C63.11

10302430ln

1024

lnln °=

⎟⎠⎞

⎜⎝⎛

−−−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=Δ

is

es

ie

TTTT

TTT

2m 0.1885=m) m)(5 012.0(ππ == DLAs

The average heat transfer coefficient is determined from

C.kW/m 12.1 2 °=⎟⎠⎞

⎜⎝⎛

°=

Δ=⎯→⎯Δ=

W1000kW 1

)C63.11)(m 1885.0( W460,26

2ln

ln TAQ

hThAQs

s

&&

The total rate of heat transfer is determined from

kW 6.1458kJ/kg) 2431)(kg/s 60.0( === fgcondtotal hmQ &&

Then the number of tubes becomes

55.1=== W460,26

W600,458,1Q

QN total

tube &

&


13-7

13-24 Combustion gases passing through a tube are used to vaporize waste water. The tube length and the rate of evaporation of water are to be determined. Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the pipe is constant. 3 The thermal resistance of the pipe is negligible. 4 Air properties are to be used for exhaust gases. Properties The properties of air at the average temperature of (250+150)/2=200°C are (Table A-22)

kJ/kg.K 287.0

CJ/kg. 1023

=

°=

R

c p

Also, the heat of vaporization of water at 1 atm or 100°C is (Table A-15). kJ/kg 2257=fgh

Analysis The density of air at the inlet and the mass flow rate of exhaust gases are

3kg/m 7662.0K) 273250(kJ/kg.K) 287.0(

kPa 115=

+==

RTPρ

Ts=110°C

L

D =3 cm

Exh. gases250°C 5 m/s

150°C

kg/s 0.002708=m/s) (54

m) (0.03)kg/m 7662.0(4

23

avg

2

avgππρρ =⎟

⎟⎠

⎞⎜⎜⎝

⎛== VDVAm c&

The rate of heat transfer is

W0.277)C150250)(CJ/kg. 1023)(kg/s 002708.0()( =°−°=−= eip TTcmQ &&


C82.79

250110150110ln

250150

lnln °=

⎟⎠⎞

⎜⎝⎛

−−−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=Δ

is

es

ie

TTTT

TTT

22

lnln m 0.02891

)C82.79)(C. W/m120( W0.277

=°°

=Δ

=⎯→⎯Δ=Th

QAThAQ ss

&&

Then the tube length becomes

cm 30.7====⎯→⎯= m 3067.0m) 03.0(m 0.02891 2

πππ

DA

LDLA ss

The rate of evaporation of water is determined from

kg/h 0.442=kg/s 0001227.0kJ/kg) 2257(

kW) 2770.0(===⎯→⎯=

fgevapfgevap h

QmhmQ

&&&&


13-8

13-25 Combustion gases passing through a tube are used to vaporize waste water. The tube length and the rate of evaporation of water are to be determined. Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the pipe is constant. 3 The thermal resistance of the pipe is negligible. 4 Air properties are to be used for exhaust gases. Properties The properties of air at the average temperature of (250+150)/2=200°C are (Table A-22)

kJ/kg.K 287.0

CJ/kg. 1023

=

°=

R

c p

Also, the heat of vaporization of water at 1 atm or 100°C is (Table A-15). kJ/kg 2257=fgh

Analysis The density of air at the inlet and the mass flow rate of exhaust gases are

3kg/m 7662.0K) 273250(kJ/kg.K) 287.0(

kPa 115=

+==

RTPρ

Ts =110°C

L

D =3 cm

Exh. gases250°C 5 m/s

150°C

kg/s 0.002708=m/s) (54

m) (0.03)kg/m 7662.0(4

23

avg

2

avgππρρ =⎟

⎟⎠

⎞⎜⎜⎝

⎛== VDVAm c&

The rate of heat transfer is

W0.277)C150250)(CJ/kg. 1023)(kg/s 002708.0()( =°−°=−= eip TTcmQ &&


C82.79

250110150110ln

250150

lnln °=

⎟⎠⎞

⎜⎝⎛

−−−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=Δ

is

es

ie

TTTT

TTT

22

lnln m 0.08673

)C82.79)(C. W/m40( W0.277

=°°

=Δ

=⎯→⎯Δ=Th

QAThAQ ss

&&

Then the tube length becomes

cm 92.0====⎯→⎯= m 920.0m) 03.0(m 0.08673 2

πππ

DA

LDLA ss

The rate of evaporation of water is determined from

kg/h 0.442=kg/s 0001227.0kJ/kg) 2257(

kW) 2770.0(===⎯→⎯=

fgevapfgevap h

QmhmQ

&&&&


13-9

Laminar and Turbulent Flow in Tubes 13-26C The friction factor for flow in a tube is proportional to the pressure drop. Since the pressure drop along the flow is directly related to the power requirements of the pump to maintain flow, the friction factor is also proportional to the power requirements. The applicable relations are

ρ

ρ PmWVDLfP Δ

==Δ&&

pump

2 and

2

13-27C The shear stress at the center of a circular tube during fully developed laminar flow is zero since the shear stress is proportional to the velocity gradient, which is zero at the tube center. 13-28C Yes, the shear stress at the surface of a tube during fully developed turbulent flow is maximum since the shear stress is proportional to the velocity gradient, which is maximum at the tube surface. 13-29C In fully developed flow in a circular pipe with negligible entrance effects, if the length of the pipe is doubled, the pressure drop will also double (the pressure drop is proportional to length). 13-30C Yes, the volume flow rate in a circular pipe with laminar flow can be determined by measuring the velocity at the centerline in the fully developed region, multiplying it by the cross-sectional area, and dividing the result by 2 since . cc AVAV )2/( maxavg ==V&

13-31C No, the average velocity in a circular pipe in fully developed laminar flow cannot be determined by simply measuring the velocity at R/2 (midway between the wall surface and the centerline). The mean velocity is Vmax/2, but the velocity at R/2 is

43

1)2/( max

2/2

2

maxV

RrVRV

Rr

=⎟⎟⎠

⎞⎜⎜⎝

⎛−=

=

13-32C In fully developed laminar flow in a circular pipe, the pressure drop is given by

2avg

2avg 328

D

LV

R

LVP

μμ==Δ

The mean velocity can be expressed in terms of the flow rate as 4/2avg

DAV

c πVV &&

== . Substituting,

4222avg

2avg 128

4/32328

DL

DDL

D

LV

R

LVP

πμ

πμμμ VV &&

====Δ

Therefore, at constant flow rate and pipe length, the pressure drop is inversely proportional to the 4th power of diameter, and thus reducing the pipe diameter by half will increase the pressure drop by a factor of 16 . 13-33C In fully developed laminar flow in a circular pipe, the pressure drop is given by

2avg

2avg 328

D

LV

R

LVP

μμ==Δ

When the flow rate and thus mean velocity are held constant, the pressure drop becomes proportional to viscosity. Therefore, pressure drop will be reduced by half when the viscosity is reduced by half. 13-34C The tubes with rough surfaces have much higher heat transfer coefficients than the tubes with smooth surfaces. In the case of laminar flow, the effect of surface roughness on the heat transfer coefficient is negligible.


13-10

13-35 The flow rate through a specified water pipe is given. The pressure drop and the pumping power requirements are to be determined. Assumptions 1 The flow is steady and incompressible. 2 The entrance effects are negligible, and thus the flow is fully developed. 3 The pipe involves no components such as bends, valves, and connectors. 4 The piping section involves no work devices such as pumps and turbines. Properties The density and dynamic viscosity of water are given to be ρ = 999.1 kg/m3 and μ = 1.138×10-3 kg/m⋅s, respectively. The roughness of stainless steel is 0.002 mm (Table 13-3). Analysis First, we calculate the mean velocity and the Reynolds number to determine the flow regime:

L = 30 m

D = 4 cm

Water 5 L/s

53

3avg

2

3

2avg

1040.1skg/m 10138.1

m) m/s)(0.04 98.3)(kg/m 1.999(Re

/m 98.34/m) (0.04

/m 0.0054/

×=⋅×

==

====

−μ

ρππ

DV

ssDA

Vc

VV &&

which is greater than 10,000. Therefore, the flow is turbulent. The relative roughness of the pipe is

105m 04.0

m 102/ 56

−−

×=×

=Dε

The friction factor can be determined from the Moody chart, but to avoid the reading error, we determine it from the Colebrook equation using an equation solver (or an iterative scheme),

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

×+

×−=→

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+−=

−

fffD

f 5

5

1040.151.2

7.3105log0.21

Re51.2

7.3/log0.21 ε

It gives f = 0.0171. Then the pressure drop and the required power input become

kPa 5.101kN/m 1

kPa 1m/skg 1000

kN 12

m/s) 98.3)(kg/m 1.999(m 0.04

m 300171.02 22

232avg =⎟

⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅==Δ

VDLfPρ

kW 0.508=⎟⎠

⎞⎜⎝

⎛

⋅=Δ=

/smkPa 1kW 1)kPa 5.101)(/m 005.0(

33

upump, sPW V&&

Therefore, useful power input in the amount of 0.508 kW is needed to overcome the frictional losses in the pipe. Discussion The friction factor could also be determined easily from the explicit Haaland relation. It would give f = 0.0169, which is sufficiently close to 0.0171. Also, the friction factor corresponding to ε = 0 in this case is 0.0168, which indicates that stainless steel pipes can be assumed to be smooth with an error of about 2%. Also, the power input determined is the mechanical power that needs to be imparted to the fluid. The shaft power will be more than this due to pump inefficiency; the electrical power input will be even more due to motor inefficiency.


13-11

13-36 In fully developed laminar flow in a circular pipe, the velocity at r = R/2 is measured. The velocity at the center of the pipe (r = 0) is to be determined. Assumptions The flow is steady, laminar, and fully developed. Analysis The velocity profile in fully developed laminar flow in a circular pipe is given by

R r

0

Vmax

V(r)=Vmax(1-r2/R2)

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

2

2

max 1)(RrVrV

where Vmax is the maximum velocity which occurs at pipe center, r = 0. At r =R/2,

4

3411)2/(1)2/( max

max2

2

maxV

VR

RVRV =⎟⎠⎞

⎜⎝⎛ −=⎟

⎟⎠

⎞⎜⎜⎝

⎛−=

Solving for Vmax and substituting,

m/s 8===3m/s) 6(4

3)2/(4

maxRV

V

which is the velocity at the pipe center. 13-37 The velocity profile in fully developed laminar flow in a circular pipe is given. The mean and maximum velocities are to be determined. Assumptions The flow is steady, laminar, and fully developed.

R r

0

Vmax

V(r)=Vmax(1-r2/R2) Analysis The velocity profile in fully developed laminar flow in a circular pipe is given by

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

2

2

max 1)(RrVrV

The velocity profile in this case is given by

)/1(4)( 22 RrrV −=

Comparing the two relations above gives the maximum velocity to be Vmax = 4 m/s. Then the mean velocity and volume flow rate become

m/s 2===2m/s 4

2max

avgV

V

/sm 0.0628 3==== ]m) (0.10m/s)[ 2()( 22avgavg ππRVAV cV&


13-12

13-38 The velocity profile in fully developed laminar flow in a circular pipe is given. The mean and maximum velocities are to be determined. Assumptions The flow is steady, laminar, and fully developed. Analysis The velocity profile in fully developed laminar flow in a circular pipe is given by

R r

0

Vmax

V(r)=Vmax(1-r2/R2)

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

2

2

max 1)(RrVrV


)/1(4)( 22 RrrV −=

Comparing the two relations above gives the maximum velocity to be Vmax = 4 m/s. Then the mean velocity and volume flow rate become

m/s 2===2m/s 4

2max

avgV

V

/sm 0.0157 3==== ]m) (0.05m/s)[ (2)( 22avgavg ππRVAV cV&


13-13

13-39 The convection heat transfer coefficients for the flow of air and water are to be determined under similar conditions. Assumptions 1 Steady flow conditions exist. 2 The surface heat flux is uniform. 3 The inner surfaces of the tube are smooth. Properties The properties of air at 25°C are (Table A-22)

7296.0Pr

/sm 10562.1

C W/m.02551.025-

=×=

°=

ν

k

Water or Air 2 m/s

L = 7 m

D = 8 cm

The properties of water at 25°C are (Table A-15)

14.6Pr/sm 10937.8997/10891.0/

C W/m.607.0kg/m 997

27-3

3

=×=×==

°==

−ρμν

ρk

Analysis The Reynolds number is

243,10/sm 10562.1

m) m/s)(0.08 (2Re

25=

×==

−νVD

which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly m 8.0m) 08.0(1010 ==≈≈ DLL th

which is much shorter than the total length of the tube. Therefore, we can assume fully developed turbulent flow in the entire duct, and determine the Nusselt number from

76.32)7296.0()243,10(023.0PrRe023.0 4.08.04.08.0 ====k

hDNu

Heat transfer coefficient is

C. W/m10.45 2 °=°

== )76.32(m 08.0

C W/m.02551.0NuDkh

Repeating calculations for water:

035,179/sm 10937.8

m) m/s)(0.08 (2Re

27=

×==

−νVD

4.757)14.6()035,179(023.0PrRe023.0 4.08.04.08.0 ====k

hDNu

C. W/m5747 2 °=°

== )4.757(m 08.0

C W/m.607.0NuDkh

Discussion The heat transfer coefficient for water is 550 times that of air.


13-14

13-40 Air flows in a pipe whose inner surface is not smooth. The rate of heat transfer is to be determined using two different Nusselt number relations. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The pressure of air is 1 atm. Properties Assuming a bulk mean fluid temperature of 20°C, the properties of air are (Table A-22)

7309.0Pr

CJ/kg. 1007/sm 10516.1

C W/m.02514.0kg/m 204.1

25-

3

=

°=×=

°==

pc

k

ν

ρ

Analysis The mean velocity of air and the Reynolds number are

Air 10ºC

0.065 kg/s

L = 5 m

D = 12 cm

m/s 773.44/m) 12.0()kg/m 204.1(

kg/s 065.023avg ===

πρ cAmV&

785,37/sm 10516.1m) m/s)(0.12 (4.773Re

25avg =

×==

−ν

DV


which is much shorter than the total length of the pipe. Therefore, we can assume fully developed turbulent flow in the entire duct. The friction factor may be determined from Colebrook equation using EES to be

02695.0785,37

51.27.3

12.0/00022.0log21Re

51.27.3

/log21=⎯→⎯

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+−=⎯→⎯

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+−= f

fffD

fε

The Nusselt number from Eq. 13-66 is 7.114)7309.0)(785,37)(02695.0(125.0PrRe125.0 3/13/1 === fNuHeat transfer coefficient is

C. W/m02.24)7.114(m 12.0

C W/m.02514.0 2 °=°

== NuDkh

Next we determine the exit temperature of air 2m 1.885=m) m)(5 12.0(ππ == DLA

C0.30)1050(50)( )1007)(065.0()885.1)(02.24(

)/( °=−−=−−=−− eeTTTT pcmhA

isse&

This result verifies our assumption of bulk mean fluid temperature that we used for property evaluation. Then the rate of heat transfer becomes W1307=°−°=−= C)100.30)(CJ/kg. 1007)(kg/s 065.0()( iep TTcmQ &&

Repeating the calculations using the Nusselt number from Eq. 13-70:

2.105)17309.0()8/02695.0(7.121)7309.0)(1000785,37)(8/02695.0(

)1(Pr)8/(7.121Pr)1000)(Re8/(

3/25.03/25.0=

−+

−=

−+

−=

ffNu

C. W/m04.22)2.105(m 12.0

C W/m.02514.0 2 °=°

== NuDkh

C8.28)1050(50)( )1007)(065.0()885.1)(04.22(


isse&

W1230=°−°=−= C)108.28)(CJ/kg. 1007)(kg/s 065.0()( iep TTcmQ &&

The result by Eq. 13-66 is about 6 percent greater than that by Eq. 13-70.


13-15

13-41 Air flows in a square cross section pipe. The rate of heat loss and the pressure difference between the inlet and outlet sections of the duct are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The pressure of air is 1 atm. Properties Taking a bulk mean fluid temperature of 80°C assuming that the air does not loose much heat to the attic, the properties of air are (Table A-22)

7154.0Pr

CJ/kg. 1008/sm 10097.2

C W/m.02953.0kg/m 9994.0

25-

3

=

°=×=

°==

pc

k

ν

ρAir

80ºC 0.15 m3/s

L = 8 m

a = 0.2 m

Analysis The mean velocity of air, the hydraulic diameter, and the Reynolds number are

m/s 75.3)m 2.0(/sm 15.02

3===

AV V&

m 2.04

44 2==== a

aa

PADh

765,3510097.2

)m 2.0)(m/s 75.3(Re5

=×

==−ν

hVD

which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly m 2m) 2.0(1010 ==≈≈ hth DLL

which is much shorter than the total length of the pipe. Therefore, we can assume fully developed turbulent flow in the entire duct, and determine the Nusselt number from

4.91)7154.0()765,35(023.0PrRe023.0 3.08.03.08.0 ====k

hDNu h


C. W/m13.5)4.91(m 2.0

C W/m.02953.0 2 °=°

== NuDkh

Next we determine the exit temperature of air

2m 6.4=m) m)(8 2.0(44 == aLA

C3.71)8060(60)( )1008)(15.0)(9994.0()4.6)(5.13(


isse&

Then the rate of heat transfer becomes

W1315=°−°=−= C)3.7180)(CJ/kg. 1008)(/sm 15.0)(kg/m 9994.0()( 33iep TTcmQ &&

From Moody chart: Re = 35,765 and ε/D = 0.001 → f = 0.026

Then the pressure drop is determined to be

Pa 7.3===Δ )m 8()m 2.0(2

)m/s 75.3)(kg/m 9994.0()026.0(2

232L

DVfP ρ


13-16

13-42 A liquid is heated as it flows in a pipe that is wrapped by electric resistance heaters. The required surface heat flux, the surface temperature at the exit, and the pressure loss through the pipe and the minimum power required to overcome the resistance to flow are to be determined. Assumptions 1 Steady flow conditions exist. 2 The surface heat flux is uniform. 3 The inner surfaces of the tube are smooth. 4 Heat transfer to the surroundings is negligible. Properties The properties of the fluid are given to be ρ = 1000 kg/m3, cp = 4000 J/kg⋅K, μ = 2x10-3 kg/s⋅m, k = 0.48 W/m⋅K, and Pr = 10 Analysis (a) The mass flow rate of the liquid is

( ) kg/s 0.0628m/s) (0.84/m) (0.01)kg/m 1000( 23 === πρAVm&

The rate of heat transfer and the heat flux are

W560,12C)2575)(CJ/kg. 4000)(kg/s 0628.0()( =°−°=−= iep TTcmQ &&

2 W/m40,000===)m 10)(m 01.0(

W560,12πsA

Qq

&&

Liquid0.8 m/s

L = 10 m

D = 1 cm

(b) The Reynolds number is

4000skg/m 002.0

m) m/s)(0.01 )(0.8kg/m (1000Re3

=⋅

==μ

ρVD

which is greater than 2300 and smaller than 10,000. Therefore, we have transitional flow. However, we use turbulent flow relation. The entry lengths in this case are roughly m 1.0m) 01.0(1010 ==≈≈ DLL th

which is much shorter than the total length of the tube. Therefore, we can assume fully developed turbulent flow in the entire duct, and determine the Nusselt number from

44)10()4000(023.0PrRe023.0 4.08.04.08.0 ====k

hDNu


C. W/m2112)44(m 01.0

C W/m.48.0 2 °=°

== NuDkh

The surface temperature at the exit is

C93.9°=⎯→⎯°−°⋅=⎯→⎯−= sses TTTThq C)75)(C W/m2112( W000,40)( 2&

(c) From Moody chart: Re = 4000, ε = 0.046 mm, ε/D = 0.045/10 = 0.0045 → f = 0.044

Then the pressure drop and the minimum power required to overcome this pressure drop are determined to be

Pa 14,080===Δ )m 10()m 01.0(2

)m/s 8.0)(kg/m 1000()044.0(2

232L

DVfP ρ

( ) W0.88==Δ= Pa) 0m/s)(14,08 8.0(4/)m 10.0( 2πPW V&&


13-17

13-43 The average flow velocity in a pipe is given. The pressure drop and the pumping power are to be determined. Assumptions 1 The flow is steady and incompressible. 2 The entrance effects are negligible, and thus the flow is fully developed. 3 The pipe involves no components such as bends, valves, and connectors. 4 The piping section involves no work devices such as pumps and turbines. Properties The density and dynamic viscosity of water are given to be ρ = 999.7 kg/m3 and μ = 1.307×10-3 kg/m⋅s, respectively. Analysis (a) First we need to determine the flow regime. The Reynolds number of the flow is

Water 1.2 m/s

L = 15 m

D = 0.2 cm 1836

skg/m 10307.1m) 10m/s)(2 2.1)(kg/m 7.999(

Re3-

-33avg =

⋅×

×==

μ

ρ DV

which is less than 2300. Therefore, the flow is laminar. Then the friction factor and the pressure drop become

kPa 188=⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅==Δ

===

22

232avg

kN/m 1kPa 1

m/skg 1000kN 1

2m/s) 2.1)(kg/m 7.999(

m 0.002m 150349.0

2

0349.01836

64Re64

VDLfP

f

ρ

(b) The volume flow rate and the pumping power requirements are

W0.71=⎟

⎠

⎞⎜⎝

⎛

⋅×=Δ=

×====

−

−

/smkPa 1 W1000)kPa 188)(/m 1077.3(

/m 1077.3]4/m) (0.002m/s)[ 2.1()4/(

336

pump

3622avgavg

sPW

sDVAV c

V

V

&&

& ππ

Therefore, power input in the amount of 0.71 W is needed to overcome the frictional losses in the flow due to viscosity.


13-18

13-44 Water is to be heated in a tube equipped with an electric resistance heater on its surface. The power rating of the heater and the inner surface temperature are to be determined. Assumptions 1 Steady flow conditions exist. 2 The surface heat flux is uniform. 3 The inner surfaces of the tube are smooth. Properties The properties of water at the average temperature of (80+10) / 2 = 45°C are (Table A-15)

Water 10°C

5 L/min

80°C

(Resistance heater)

L

D = 2 cm

91.3Pr

CJ/kg. 4180/sm 10602.0/

C W/m.637.0kg/m 1.990

26-

3

=

°=×==

°==

pc

k

ρμν

ρ

Analysis The power rating of the resistance heater is

kg/s 0825.0kg/min 951.4)/minm 005.0)(kg/m 1.990( 33 ==== V&& ρm

W24,140=°−°=−= C)1080)(CJ/kg. 4180)(kg/s 0825.0()( iep TTcmQ &&

The velocity of water and the Reynolds number are

m/s 2653.04/m) 02.0(

/sm )60/105(2

33

avg =×

==−

πcAV V&

8813/sm 10602.0m) m/s)(0.02 (0.2653

Re26

avg =×

==−ν

hDV

which is less than 10,000 but much greater than 2300. We assume the flow to be turbulent. The entry lengths in this case are roughly m 20.0m) 02.0(1010 ==≈≈ DLL th

which is much shorter than the total length of the duct. Therefore, we can assume fully developed turbulent flow in the entire duct, and determine the Nusselt number from

85.56)91.3()8813(023.0PrRe023.0 4.08.04.08.0 ====k

hDNu h


C. W/m1811)85.56(m 02.0

C W/m.637.0 2 °=°

== NuDkh

h

Then the inner surface temperature of the pipe at the exit becomes

C96.3°=

°−°=

−=

es

s

eess

TT

TThAQ

,

2

,

C)80)](m 13)(m 02.0()[C. W/m1811( W140,24

)(

π

&


13-19

13-45 Flow of hot air through uninsulated square ducts of a heating system in the attic is considered. The exit temperature and the rate of heat loss are to be determined. Assumptions 1 Steady operating conditions exist. 2 The inner surfaces of the duct are smooth. 3 Air is an ideal gas with constant properties. 4 The pressure of air is 1 atm. Properties We assume the bulk mean temperature for air to be 80°C since the mean temperature of air at the inlet will drop somewhat as a result of heat loss through the duct whose surface is at a lower temperature. The properties of air at 1 atm and this temperature are (Table A-22)

7154.0Pr

CJ/kg. 1008/sm 10097.2

C W/m.02953.0kg/m 9994.0

25-

3

=

°=×=

°==

pc

k

ν

ρ

Air 85°C 0.1 m3/min

L = 10 m Ts = 70°C

Te

Analysis The characteristic length that is the hydraulic diameter, the mean velocity of air, and the Reynolds number are

m 15.04

44 2==== a

aa

PA

D ch

m/s 444.4m) 15.0(

/sm 10.02

3

avg ===cA

V V&

791,31/sm 10097.2m) m/s)(0.15 (4.444Re

25avg =

×==

−νhDV

which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly m 5.1m) 15.0(1010 ==≈≈ hth DLL


16.83)7154.0()791,31(023.0PrRe023.0 3.08.03.08.0 ====k

hDNu h


C. W/m37.16)16.83(m 15.0

C W/m.02953.0 2 °=°

== NuDkh

h

Next we determine the exit temperature of air,

kg/s 0.09994=/s)m )(0.10kg/m 9994.0(

m 6=m) m)(10 15.0(4433

2

==

==

V&& ρm

aLAs

C75.7°=−−=−−=−− )1008)(09994.0(

)6)(37.16()/( )8570(70)( eeTTTT pcmhA

isse&

Then the logarithmic mean temperature difference and the rate of heat loss from the air becomes

W941=°°=Δ=

°=⎟⎠⎞

⎜⎝⎛

−−−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=Δ

)C58.9)(m 6)(C. W/m37.16(

C58.9

85707.7570ln

857.75

ln

22ln

ln

ThAQ

TTTT

TTT

s

is

es

ie

&

Note that the temperature of air drops by almost 10°C as it flows in the duct as a result of heat loss.


13-20

13-46 EES Prob. 13-45 is reconsidered. The effect of the volume flow rate of air on the exit temperature of air and the rate of heat loss is to be investigated.

Analysis The problem is solved using EES, and the solution is given below. "GIVEN" T_i=85 [C] L=10 [m] side=0.15 [m] V_dot=0.10 [m^3/s] T_s=70 [C] "PROPERTIES" Fluid$='air' C_p=CP(Fluid$, T=T_ave)*Convert(kJ/kg-C, J/kg-C) k=Conductivity(Fluid$, T=T_ave) Pr=Prandtl(Fluid$, T=T_ave) rho=Density(Fluid$, T=T_ave, P=101.3) mu=Viscosity(Fluid$, T=T_ave) nu=mu/rho T_ave=1/2*(T_i+T_e) "ANALYSIS" D_h=(4*A_c)/p A_c=side^2 p=4*side Vel=V_dot/A_c Re=(Vel*D_h)/nu "The flow is turbulent" L_t=10*D_h "The entry length is much shorter than the total length of the duct." Nusselt=0.023*Re^0.8*Pr^0.3 h=k/D_h*Nusselt A=4*side*L m_dot=rho*V_dot T_e=T_s-(T_s-T_i)*exp((-h*A)/(m_dot*C_p)) DELTAT_ln=(T_e-T_i)/ln((T_s-T_e)/(T_s-T_i)) Q_dot=h*A*DELTAT_ln

V [m3/s] Te [C] Q [W] 0.05 74.89 509

0.055 75 554.1 0.06 75.09 598.6

0.065 75.18 642.7 0.07 75.26 686.3

0.075 75.34 729.5 0.08 75.41 772.4

0.085 75.48 814.8 0.09 75.54 857

0.095 75.6 898.9 0.1 75.66 940.4

0.105 75.71 981.7 0.11 75.76 1023

0.115 75.81 1063 0.12 75.86 1104


13-21

0.125 75.9 1144 0.13 75.94 1184

0.135 75.98 1224 0.14 76.02 1264

0.145 76.06 1303 0.15 76.1 1343

0.04 0.06 0.08 0.1 0.12 0.14 0.1674.8

75.1

75.3

75.7

75.9

76.2

500

600

700

800

900

1000

1100

1200

1300

1400

V [m3/s]

T e [

C]

Q [

W]

Te

Q


13-22

13-47 Air enters the constant spacing between the glass cover and the plate of a solar collector. The net rate of heat transfer and the temperature rise of air are to be determined. Assumptions 1 Steady operating conditions exist. 2 The inner surfaces of the spacing are smooth. 3 Air is an ideal gas with constant properties. 4 The local atmospheric pressure is 1 atm. Properties The properties of air at 1 atm and estimated average temperature of 35°C are (Table A-22)

7268.0Pr C,J/kg. 1007 /s,m 10655.1

C W/m.02625.0 ,kg/m145.125-

3

=°=×=

°==

pc

k

ν

ρ

Analysis Mass flow rate, cross sectional area, hydraulic diameter, mean velocity of air and the Reynolds number are

Air 30°C

0.15 m3/min

60°C Collector plate

(insulated)

Glass cover 20°C

kg/s 1718.0)/sm 15.0)(kg/m 145.1( 33 === V&& ρm

2m 03.0m) m)(0.03 (1 ==cA

m 05825.0m) 03.0m 1(2)m 03.0(44 2

=+

==PA

D ch

m/s 5m 0.03

/sm 15.02

3

avg ===cA

V V&

600,17/sm 10655.1m) 25m/s)(0.058 (5

Re25

avg =×

==−ν

hDV


which are much shorter than the total length of the collector. Therefore, we can assume fully developed turbulent flow in the entire collector, and determine the Nusselt number from

43.50)7268.0()600,17(023.0PrRe023.0 4.08.04.08.0 ====k

hDNu h

and C. W/m73.22)43.50(m 05825.0

C W/m.02625.0 2 °=°

== NuDkh

h

The exit temperature of air can be calculated using the “average” surface temperature as

, 2m 10m) m)(1 (52 ==sA C402

2060, °=

+=avgsT

C31.3710071718.01073.22exp)3040(40exp)( ,, °=⎟

⎠⎞

⎜⎝⎛

××

−−−=⎟⎟⎠

⎞⎜⎜⎝

⎛−−−=

p

siavgsavgse cm

hATTTT

&

The temperature rise of air is C7.3°=°−°=Δ C30C3.37TThe logarithmic mean temperature difference and the heat loss from the glass are

C32.13

302031.3720ln

3031.37

lnln, °=

−−−

=

−−−

=Δ

is

es

ieglass

TTTT

TTT

W1514=C))(13.32m C)(5. W/m(22.73 22ln °°=Δ= ThAQ sglass

&

The logarithmic mean temperature difference and the heat gain of the absorber are

C17.26

306031.3760ln

3031.37

lnln, °=

−−−

=

−−−

=Δ

is

es

ieabsorber

TTTT

TTT

W2975=C))(26.17m C)(5. W/m(22.73 22ln °°=Δ= ThAQabsorber

&

Then the net rate of heat transfer becomes W1461=−= 15142975netQ&


13-23

13-48 Oil flows through a pipeline that passes through icy waters of a lake. The exit temperature of the oil and the rate of heat loss are to be determined. Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the pipe is very nearly 0°C. 3 The thermal resistance of the pipe is negligible. 4 The inner surfaces of the pipeline are smooth. 5 The flow is hydrodynamically developed when the pipeline reaches the lake. Properties The properties of oil at 10°C are (Table A-19)

Oil 10°C

0.5 m/s

(Icy lake, 0°C)

L = 1500

D = 0.4 m

750,28Pr C,J/kg. 1839

/sm 10592.2 kg/m.s, 326.2

C W/m.1460.0 ,kg/m 6.89323-

3

=°=×==

°==

pc

k

νμ

ρ

Analysis (a) The Reynolds number in this case is

16.77/sm 10592.2

m) m/s)(0.4 (0.5Re23

avg =×

==−ν

hDV

which is less than 2300. Therefore, the flow is laminar, and the thermal entry length is roughly m 367,44)m 4.0)(750,28)(16.77(05.0PrRe05.0 === DLt which is much longer than the total length of the pipe. Therefore, we assume thermally developing flow, and determine the Nusselt number from

[ ]

73.13

)750,28)(16.77(m 1500

m 4.004.01

)750,28)(16.77(m 1500

m 4.0065.066.3

PrRe)/(04.01PrRe)/(065.066.3

3/23/2=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+

⎟⎠⎞

⎜⎝⎛

+=+

+==LDLD

khDNu

and C. W/m011.5)73.13(m 4.0

C W/m.1460.0 2 °=°

== NuDkh

Next we determine the exit temperature of oil

kg/s 56.15=m/s) (0.5

4m) (0.4)kg/m 6.893(

4

m 1885=m) m)(1500 4.0(2

3avg

2

avg

2

ππρρρ

ππ

=⎟⎟⎠

⎞⎜⎜⎝

⎛===

==

VDVAm

DLA

c

s

V&&

C 9.13 °=−−=−−=−− )1839)(15.56(

)1885)(011.5( )/( )100(0)( eeTTTT ps cmhA

isse&

(b) The logarithmic mean temperature difference and the rate of heat loss from the oil are

kW 90.3==°°=Δ=

°=⎟⎠⎞

⎜⎝⎛

−−−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=Δ

W300,90)C56.9)(m 1885)(C. W/m011.5(

C56.9

10013.90ln

1013.9

ln

22ln

ln

ThAQ

TTTT

TTT

s

is

es

ie

&

The friction factor is

8294.016.77

64Re64

===f

Then the pressure drop in the pipe and the required pumping power become

kW 21.8=⎟⎠⎞

⎜⎝⎛

⋅=Δ=Δ=

=⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅==Δ

/smkPa 1kW 1)kPa 4.347(m/s) (0.5

4m) (0.4

kPa 4.347kN/m 1

kPa 1m/skg 1000

kN 12

)m/s 5.0)(kg/m 6.893(m 4.0m 15008294.0

2

3

2

upump,

22

232avg

π

ρ

PVAPW

VDLfP

avgcV&&

Discussion The power input determined is the mechanical power that needs to be imparted to the fluid. The shaft power will be much more than this due to pump inefficiency; the electrical power input will be even more due to motor inefficiency.


13-24

13-49 Laminar flow of a fluid through an isothermal square channel is considered. The change in the pressure drop and the rate of heat transfer are to be determined when the mean velocity is doubled. Assumptions 1 The flow is fully developed. 2 The effect of the change in ΔTln on the rate of heat transfer is not considered. Analysis The pressure drop of the fluid for laminar flow is expressed as

2avg

2avg

avg

2avg

2avg

1 322

642Re

642 D

LVV

DL

DVV

DLV

DLfP ρνρνρρ

====Δ

When the free-stream velocity of the fluid is doubled, the pressure drop becomes

2avg

2avg

avg

2avg

2avg

2 642

42

642

4Re64

2)2(

DLV

VDL

DVV

DLV

DLfP ρνρνρρ

====Δ

Their ratio is

Laminar flow Vavg

L

2==ΔΔ

3264

1

2

PP

The rate of heat transfer between the fluid and the walls of the channel is expressed as

lnlnln1 98.2 TADkTNuA

DkThAQ sss Δ=Δ=Δ=&

When the effect of the change in ΔTln on the rate of heat transfer is disregarded, the rate of heat transfer remains the same. Therefore,

1=1

2

QQ&

&

Therefore, doubling the velocity will double the pressure drop but it will not affect the heat transfer rate.


13-25

13-50 Turbulent flow of a fluid through an isothermal square channel is considered. The change in the pressure drop and the rate of heat transfer are to be determined when the free-stream velocity is doubled. Assumptions 1 The flow is fully developed. 2 The effect of the change in ΔTln on the rate of heat transfer is not considered. Analysis The pressure drop of the fluid for turbulent flow is expressed as

DLDV

VDLDVV

DLV

DLfP ρ

ν

ρ

ν

ρρ 2.01.8avg

2avg

2.0

2.00.2avg

2avg2.0

2avg

1 092.02

184.02

Re184.02

−

−

−−− ⎟

⎠⎞

⎜⎝⎛====Δ

When the free-stream velocity of the fluid is doubled, the pressure drop becomes

DLDV

VDLDVV

DLV

DLfP

ρν

ρ

ν

ρρ

2.01.8avg

2.0

2avg

2.0

2.02.0avg

2avg2.0

2avg

2

)2(368.0

24)

184.02

4Re184.0

2)2(

−−

−

−−−

⎟⎠⎞

⎜⎝⎛=

===Δ(2

Turbulent flowVavg

L

Their ratio is

3.48===ΔΔ −

−2.0

1.8avg

1.8avg

2.0

1

2 )2(4092.0

)2(368.0

V

VPP

The rate of heat transfer between the fluid and the walls of the channel is expressed as

ln3/1

8.00.8

avg

ln3/18.0

lnln1

Pr023.0

PrRe023.0

TADkDV

TADkTNuA

DkThAQ

Δ⎟⎠⎞

⎜⎝⎛=

Δ=Δ=Δ=

ν

&

When the free-stream velocity of the fluid is doubled, the heat transfer rate becomes

ln3/1

8.08.0

avg2 Pr)2(023.0 TADkDVQ Δ⎟

⎠⎞

⎜⎝⎛=ν

&

Their ratio is

1.74=== 8.00.8

avg

8.0avg

1

2 2)2(

V

V

QQ&

&

Therefore, doubling the velocity will increase the pressure drop 3.8 times but it will increase the heat transfer rate by only 74%.


13-26

13-51E Water is heated in a parabolic solar collector. The required length of parabolic collector and the surface temperature of the collector tube are to be determined. Assumptions 1 Steady operating conditions exist. 2 The thermal resistance of the tube is negligible. 3 The inner surfaces of the tube are smooth. Properties The properties of water at the average temperature of (55+200)/2 = 127.5°F are (Table A-15E)

Water 55°F

4 lbm/s 200°F

(Inside glass tube)

Solar absorption, 350 Btu/h.ft

L

D = 1.25 in

368.3Pr

FBtu/lbm.999.0/sft 105681.0/

FBtu/ft. 374.0lbm/ft 59.61

25-

3

=

°=×==

°==

pcv

k

ρμ

ρ

Analysis The total rate of heat transfer is

Btu/h 102.086=Btu/s 4.579F)55200)(FBtu/lbm. 999.0)(lbm/s 4()( 6×=°−°=−= iep TTcmQ &&

The length of the tube required is

ft 5960=×

==Btu/h.ft 350

Btu/h 10086.2 6

QQ

L total&

&

The velocity of water and the Reynolds number are

ft/s 621.7

4)ft 12/25.1(

)lbm/m 59.61(

lbm/s 42

3avg ===

πρ cAmV&

525

avg 10397.1/sft 105681.0ft) 12m/s)(1.25/ (7.621Re ×=

×==

−νhDV

which is greater than 10,000. Therefore, we can assume fully developed turbulent flow in the entire tube, and determine the Nusselt number from

5.488)368.3()10397.1(023.0PrRe023.0 4.08.054.08.0 =×===k

hDNu h

The heat transfer coefficient is

F.Btu/h.ft 1754)5.488(ft 12/25.1

FBtu/h.ft. 374.0 2 °=°

== NuDkh

h

The heat flux on the tube is

26

Btu/h.ft 1070)ft 5960)(ft 12/25.1(

Btu/h 10086.2=

×==πsA

Qq&

&

Then the surface temperature of the tube at the exit becomes

F200.6°=°

°=+=⎯→⎯−=F.Btu/h.ft 1754

Btu/h.ft 1070+F200)(2

2

hqTTTThq eses&

&


13-27

13-52 A circuit board is cooled by passing cool air through a channel drilled into the board. The maximum total power of the electronic components is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat flux at the top surface of the channel is uniform, and heat transfer through other surfaces is negligible. 3 The inner surfaces of the channel are smooth. 4 Air is an ideal gas with constant properties. 5 The pressure of air in the channel is 1 atm. Properties The properties of air at 1 atm and estimated average temperature of 25°C are (Table A-22)

7296.0Pr

CJ/kg. 1007/sm 10562.1

C W/m.02551.0kg/m 184.1

25-

3

=

°=×=

°==

pc

k

ν

ρ

Air 15°C 4 m/s

Electronic components, 50°C

L = 20 cm

Air channel 0.2 cm × 14 cm

Analysis The cross-sectional and heat transfer surface areas are

2

2

m 028.0)m 2.0)(m 14.0(

m 00028.0)m 14.0)(m 002.0(

==

==

s

c

A

A

To determine heat transfer coefficient, we first need to find the Reynolds number,

m 003944.0m) 0.14+m 002.0(2)m 00028.0(44 2

===PA

D ch

1010/sm 10562.1m) 944m/s)(0.003 (4Re

25avg =

×==

−νhDV

which is less than 2300. Therefore, the flow is laminar and the thermal entry length is m 0.20< m 0.1453=m) 003944.0)(7296.0)(1010(05.0PrRe05.0 == ht DL

Therefore, we have developing flow through most of the channel. However, we take the conservative approach and assume fully developed flow, and from Table 13-1 we read Nu = 8.24. Then the heat transfer coefficient becomes

C. W/m30.53)24.8(m 003944.0

C W/m.02551.0 2 °=°

== NuDkh

h

Also,

kg/s 001326.0)m 00028.0)(m/s 4)(kg/m 184.1( 23 === cVAm ρ&

Heat flux at the exit can be written as )( es TThq −=& where C50°=sT at the exit. Then the heat transfer

rate can be expressed as , and the exit temperature of the air can be determined from

)( esss TThAAqQ −== &&

C5.33

)C15)(CJ/kg. 1007)(kg/s 001326.0()C50)(m 028.0)(C. W/m30.53(

)()(22

°=°−°=−°°

−=−

e

ee

iepess

TTT

TTcmTThA &

Then the maximum total power of the electronic components that can safely be mounted on this circuit board becomes

W24.7=°−°=−= )C155.33)(CJ/kg. 1007)(kg/s 001326.0()(max iep TTcmQ &&


13-28

13-53 A circuit board is cooled by passing cool helium gas through a channel drilled into the board. The maximum total power of the electronic components is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat flux at the top surface of the channel is uniform, and heat transfer through other surfaces is negligible. 3 The inner surfaces of the channel are smooth. 4 Helium is an ideal gas. 5 The pressure of helium in the channel is 1 atm. Properties The properties of helium at the estimated average temperature of 25°C are (from EES)

6636.0Pr

CJ/kg. 5193/sm 10214.1

C W/m.1553.0kg/m 1635.0

24-

3

=

°=×=

°==

pc

k

ν

ρ

Helium 15°C 4 m/s

Electronic components, 50°C

L = 20 cm


Analysis The cross-sectional and heat transfer surface areas are

2

2

m 028.0)m 2.0)(m 14.0(

m 00028.0)m 14.0)(m 002.0(

==

==

s

c

A

A

To determine heat transfer coefficient, we need to first find the Reynolds number

m 003944.0m) 0.14+m 002.0(2)m 00028.0(44 2

===pA

D ch

0.130/sm 10214.1m) 944m/s)(0.003 (4Re

24avg =

×==

−νhDV

which is less than 2300. Therefore, the flow is laminar and the thermal entry length is m 0.20 << m 0.0170=m) 003944.0)(6636.0)(0.130(05.0PrRe05.0 == ht DL

Therefore, the flow is fully developed flow, and from Table 13-3 we read Nu = 8.24. Then the heat transfer coefficient becomes

C. W/m5.324)24.8(m 003944.0

C W/m.1553.0 2 °=°

== NuDkh

h

Also,

kg/s 0001831.0)m 00028.0)(m/s 4)(kg/m 1635.0( 23 === cVAm ρ&

Heat flux at the exit can be written as )( es TThq −=& where C50°=sT at the exit. Then the heat transfer

rate can be expressed as , and the exit temperature of the air can be determined from

)( esss TThAAqQ −== &&

C68.46

)C50)(m 028.0)(C. W/m5.324()C15)(CJ/kg. 5193)(kg/s 0001831.0(

)()(22

°=−°°=°−°

−=−

e

ee

essiep

TTT

TThATTcm&

Then the maximum total power of the electronic components that can safely be mounted on this circuit board becomes

W30.1=°−°=−= )C1568.46)(CJ/kg. 5193)(kg/s 0001831.0()(max iep TTcmQ &&


13-29

13-54 EES Prob. 13-52 is reconsidered. The effects of air velocity at the inlet of the channel and the maximum surface temperature on the maximum total power dissipation of electronic components are to be investigated.

Analysis The problem is solved using EES, and the solution is given below. "GIVEN" L=0.20 [m] width=0.14 [m] height=0.002 [m] T_i=15 [C] Vel=4 [m/s] T_s=50 [C] "PROPERTIES" Fluid$='air' c_p=CP(Fluid$, T=T_ave)*Convert(kJ/kg-C, J/kg-C) k=Conductivity(Fluid$, T=T_ave) Pr=Prandtl(Fluid$, T=T_ave) rho=Density(Fluid$, T=T_ave, P=101.3) mu=Viscosity(Fluid$, T=T_ave) nu=mu/rho T_ave=1/2*(T_i+T_e) "ANALYSIS" A_c=width*height A=width*L p=2*(width+height) D_h=(4*A_c)/p Re=(Vel*D_h)/nu "The flow is laminar" L_t=0.05*Re*Pr*D_h "Taking conservative approach and assuming fully developed laminar flow, from Table 13-1 we read" Nusselt=8.24 h=k/D_h*Nusselt m_dot=rho*Vel*A_c Q_dot=h*A*(T_s-T_e) Q_dot=m_dot*c_p*(T_e-T_i)

Vel [m/s] Q [W] 1 9.453 2 16.09 3 20.96 4 24.67 5 27.57 6 29.91 7 31.82 8 33.41 9 34.76

10 35.92


13-30

1 2 3 4 5 6 7 8 9 105

10

15

20

25

30

35

40

Vel [m/s]

Q [

W]

Ts [C] Q [W] 30 10.59 35 14.12 40 17.64 45 21.15 50 24.67 55 28.18 60 31.68 65 35.18 70 38.68 75 42.17 80 45.65 85 49.13 90 52.6

30 40 50 60 70 80 9010

15

20

25

30

35

40

45

50

55

Ts [C]

Q [

W]


13-31

13-55 Air enters a rectangular duct. The exit temperature of the air, the rate of heat transfer, and the fan power are to be determined. Assumptions 1 Steady operating conditions exist. 2 The inner surfaces of the duct are smooth. 3 Air is an ideal gas with constant properties. 4 The pressure of air in the duct is 1 atm. Properties We assume the bulk mean temperature for air to be 40°C since the mean temperature of air at the inlet will drop somewhat as a result of heat loss through the duct whose surface is at a lower temperature. The properties of air at this temperature and 1 atm are (Table A-22)

/sm 10702.1

C W/m.02662.0kg/m 127.1

25-

3

×=

°==

ν

ρk

7255.0Pr

CJ/kg. 1007

=

°=pc

Air 50°C 7 m/s

L = 7 m

Air duct 15 cm × 20 cm

Ts = 10°C

Analysis (a) The hydraulic diameter, the mean velocity of air, and the Reynolds number are

[ ] m 1714.0m) (0.20+m) 15.0(2

m) m)(0.20 15.0(44===

pA

D ch

500,70/sm 10702.1

m) 4m/s)(0.171 (7Re

25avg =

×==

−νhDV



9.157)7255.0()500,70(023.0PrRe023.0 3.08.03.08.0 ====k

hDNu h


C. W/m52.24)9.157(m 1714.0

C W/m.02662.0 2 °=°

== NuDkh

h

Next we determine the exit temperature of air

kg/s 0.2367=)m m/s)(0.03 )(7kg/m 127.1(

m 0.03=m) m)(0.20 15.0(

m 4.9=m)] (0.20+m) 15.0[(72

23

2

2

==

=

×=

c

c

s

VAm

A

A

ρ&

C34.16°=−−=−−=−− )1007)(2367.0(

)9.4)(52.24()/( )5010(10)( eeTTTT ps cmhA

isse&

(b) The logarithmic mean temperature difference and the rate of heat loss from the air are

W3775=°°=Δ=

°=⎟⎠⎞

⎜⎝⎛

−−−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=Δ

)C42.31)(m 9.4)(C. W/m52.24(

C42.31

501016.3410ln

5016.34

ln

22ln

ln

ThAQ

TTTT

TTT

s

is

es

ie

&

(c) The friction factor, the pressure drop, and then the fan power can be determined for the case of fully developed turbulent flow to be 01973.0)500,70(184.0Re184.0 2.02.0 === −−f

W4.67==Δ

=

===Δ

3

2

2232

avg

kg/m 127.1)N/m 25.22)(kg/s 2367.0(

N/m 25.222

)m/s 7)(kg/m 127.1(m) 1714.0(

m) 7(01973.02

ρ

ρ

PmW

VDLfP

pump&&


13-32

13-56 EES Prob. 13-55 is reconsidered. The effect of air velocity on the exit temperature of air, the rate of heat transfer, and the fan power is to be investigated.

Analysis The problem is solved using EES, and the solution is given below. "GIVEN" L=7 [m] height=0.15 [m] width=0.20 [m] T_i=50 [C] Vel=7 [m/s] T_s=10 [C] "PROPERTIES" Fluid$='air' c_p=CP(Fluid$, T=T_ave)*Convert(kJ/kg-C, J/kg-C) k=Conductivity(Fluid$, T=T_ave) Pr=Prandtl(Fluid$, T=T_ave) rho=Density(Fluid$, T=T_ave, P=101.3) mu=Viscosity(Fluid$, T=T_ave) nu=mu/rho T_ave=1/2*(T_i+T_e) "ANALYSIS" "(a)" A_c=width*height p=2*(width+height) D_h=(4*A_c)/p Re=(Vel*D_h)/nu "The flow is turbulent" L_t=10*D_h "The entry length is much shorter than the total length of the duct." Nusselt=0.023*Re^0.8*Pr^0.3 h=k/D_h*Nusselt A=2*L*(width+height) m_dot=rho*Vel*A_c T_e=T_s-(T_s-T_i)*exp((-h*A)/(m_dot*c_p)) "(b)" DELTAT_ln=(T_e-T_i)/ln((T_s-T_e)/(T_s-T_i)) Q_dot=h*A*DELTAT_ln "(c)" f=0.184*Re^(-0.2) DELTAP=f*L/D_h*(rho*Vel^2)/2 W_dot_pump=(m_dot*DELTAP)/rho

Vel [m/s] Te [C] Q [W] Wpump [W] 1 29.01 715.6 0.02012

1.5 30.14 1014 0.06255 2 30.92 1297 0.1399

2.5 31.51 1570 0.2611 3 31.99 1833 0.4348

3.5 32.39 2090 0.6692 4 32.73 2341 0.9722

4.5 33.03 2587 1.352 5 33.29 2829 1.815

5.5 33.53 3066 2.369


13-33

6 33.75 3300 3.022 6.5 33.94 3531 3.781 7 34.12 3759 4.652

7.5 34.29 3984 5.642 8 34.44 4207 6.759

8.5 34.59 4427 8.008 9 34.72 4646 9.397

9.5 34.85 4862 10.93 10 34.97 5076 12.62

1 2 3 4 5 6 7 8 9 1029

30

31

32

33

34

35

0

1000

2000

3000

4000

5000

6000

Vel [m/s]

T e [

C]

Q [

W]

Te

Q

1 2 3 4 5 6 7 8 9 100

2

4

6

8

10

12

14

Vel [m/s]

Wpu

mp

[W]


13-34

13-57 Hot air enters a sheet metal duct located in a basement. The exit temperature of hot air and the rate of heat loss are to be determined. Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the duct are smooth. 3 The thermal resistance of the duct is negligible. 4 Air is an ideal gas with constant properties. 5 The pressure of air is 1 atm. Properties We expect the air temperature to drop somewhat, and evaluate the air properties at 1 atm and the estimated bulk mean temperature of 50°C (Table A-22),

7228.0Pr

CJ/kg. 1007 /s;m 10798.1

C W/m.02735.0 ;kg/m 092.125-

3

=

°=×=

°==

pc

k

ν

ρ

Air 60°C 4 m/s

L = 12 mε = 0.3


T∞ = 10°C

Analysis The surface area and the Reynolds number are m 9.6m) m)(12 2.0(44 =×== aLAs

m 2.04

44 2==== a

aa

pA

D ch

494,44/sm 10798.1

m) m/s)(0.20 (4Re

25avg =

×==

−νhDV


which is much shorter than the total length of the duct. Therefore, we can assume fully developed turbulent flow for the entire duct, and determine the Nusselt number from

2.109)7228.0()494,44(023.0PrRe023.0 3.08.03.08.0 ====k

hDNu h

and

C. W/m93.14)2.109(m 2.0

C W/m.02735.0 2 °=°

== NuDkh

h

The mass flow rate of air is

kg/s 0.1747m/s) )(4m 0.2)(0.2kg/m 092.1( 23 =×== VAm cρ&

In steady operation, heat transfer from hot air to the duct must be equal to the heat transfer from the duct to the surrounding (by convection and radiation), which must be equal to the energy loss of the hot air in the duct. That is, airhot outrad,+convinconv, EQQQ &&&& Δ===

Assuming the duct to be at an average temperature of Ts , the quantities above can be expressed as

& :Qconv,in

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−°=→

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=Δ=

60ln

60)m 6.9)(C. W/m93.14(

ln

22ln

s

es

e

is

es

iesisi

TTT

TQ

TTTT

TTAhTAhQ &&

& :Qconv+rad,out ( )

[ ] 4444282

2244

K)27310()273().K W/m1067.5)(m 0.3(9.6+

C)10)(m C)(9.6. W/m10( )(

+−+×

°−°=→−+−=−

s

sossosso

T

TQTTATTAhQ && σε

Δ & :Ehot air C)C)(60J/kg. kg/s)(1007 1747.0( )( °−°=→−= eiep TQTTcmQ &&&

This is a system of three equations with three unknowns whose solution is

C3.33and , , °=°== se TTQ C45.1 W2622&

Therefore, the hot air will lose heat at a rate of 2622 W and exit the duct at 45.1°C.


13-35

13-58 EES Prob. 13-57 is reconsidered. The effects of air velocity and the surface emissivity on the exit temperature of air and the rate of heat loss are to be investigated.

Analysis The problem is solved using EES, and the solution is given below.

"GIVEN" T_i=60 [C] L=12 [m] side=0.20 [m] Vel=4 [m/s] epsilon=0.3 T_o=10 [C] h_o=10 [W/m^2-C] T_surr=10 [C] "PROPERTIES" Fluid$='air' c_p=CP(Fluid$, T=T_ave)*Convert(kJ/kg-C, J/kg-C) k=Conductivity(Fluid$, T=T_ave) Pr=Prandtl(Fluid$, T=T_ave) rho=Density(Fluid$, T=T_ave, P=101.3) mu=Viscosity(Fluid$, T=T_ave) nu=mu/rho T_ave=T_i-10 "assumed average bulk mean temperature" "ANALYSIS" A=4*side*L A_c=side^2 p=4*side D_h=(4*A_c)/p Re=(Vel*D_h)/nu "The flow is turbulent" L_t=10*D_h "The entry length is much shorter than the total length of the duct." Nusselt=0.023*Re^0.8*Pr^0.3 h_i=k/D_h*Nusselt m_dot=rho*Vel*A_c Q_dot=Q_dot_conv_in Q_dot_conv_in=Q_dot_conv_out+Q_dot_rad_out Q_dot_conv_in=h_i*A*DELTAT_ln DELTAT_ln=(T_e-T_i)/ln((T_s-T_e)/(T_s-T_i)) Q_dot_conv_out=h_o*A*(T_s-T_o) Q_dot_rad_out=epsilon*A*sigma*((T_s+273)^4-(T_surr+273)^4) sigma=5.67E-8 "[W/m^2-K^4], Stefan-Boltzmann constant" Q_dot=m_dot*c_p*(T_i-T_e)

Vel [m/s] Te [C] Q [W] 1 33.85 1150 2 39.43 1810 3 42.78 2273 4 45.1 2622 5 46.83 2898 6 48.17 3122 7 49.25 3310 8 50.14 3469 9 50.89 3606

10 51.53 3726


13-36

ε Te [C] Q [W]

0.1 45.82 2495 0.2 45.45 2560 0.3 45.1 2622 0.4 44.77 2680 0.5 44.46 2735 0.6 44.16 2787 0.7 43.88 2836 0.8 43.61 2883 0.9 43.36 2928 1 43.12 2970

1 2 3 4 5 6 7 8 9 1032.5

36.5

40.5

44.5

48.5

52.5

1000

1500

2000

2500

3000

3500

4000

Vel [m/s]

T e [

C]

Q [

W]

Te

Q

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 143

43.5

44

44.5

45

45.5

46

2400

2500

2600

2700

2800

2900

3000

ε

T e [

C]

Q [

W]

Te

Q


13-37

13-59 The components of an electronic system located in a rectangular horizontal duct are cooled by forced air. The exit temperature of the air and the highest component surface temperature are to be determined. Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the duct are smooth. 3 The thermal resistance of the duct is negligible. 4 Air is an ideal gas with constant properties. 5 The pressure of air is 1 atm. Properties We assume the bulk mean temperature for air to be 35°C since the mean temperature of air at the inlet will rise somewhat as a result of heat gain through the duct whose surface is exposed to a constant heat flux. The properties of air at 1 atm and this temperature are (Table A-22)

7268.0Pr

CJ/kg. 1007/sm 10655.1

C W/m.02625.0kg/m 145.1

25-

3

=

°=×=

°==

pc

k

ν

ρ

Air 27°C

0.65 m3/min

L = 1 m


180 W

Analysis (a) The mass flow rate of air and the exit temperature are determined from

kg/s 0.0124=kg/min 0.7443=/min)m )(0.65kg/m 145.1( 33== V&& ρm

C39.3°=°

°=+=→−=C)J/kg. kg/s)(1007 0124.0(

W)(0.85)(180+C27)(

pieiep cm

QTTTTcmQ

&

&&&

(b) The mean fluid velocity and hydraulic diameter are

m 16.0

m) 16.0(4m) m)(0.16 16.0(44

m/s 0.4232=m/min 4.25m) m)(0.16 (0.16

m/min 65.0

===

===

pA

D

AV

ch

cm

V&

Then

4091/sm 10655.1m) m/s)(0.16 (0.4232

Re25

avg =×

==−ν

hDV

which is not greater than 10,000 but the components will cause turbulence and thus we can assume fully developed turbulent flow in the entire duct, and determine the Nusselt number from

69.15)7268.0()4091(023.0PrRe023.0 4.08.04.08.0 ====k

hDNu h

and

C. W/m574.2)69.15(m 16.0

C W/m.02625.0 2 °=°

== NuDkh

h

The highest component surface temperature will occur at the exit of the duct. Assuming uniform surface heat flux, its value is determined from

[ ]C132°=

°°=+=

−=

C. W/m2.574m) m)(1 4(0.16 W)/(0.85)(180

+C2.39/

)(/

2,

,

hAQ

TT

TThAQ

sehighests

ehighestss

&

&


13-38

13-60 The components of an electronic system located in a circular horizontal duct are cooled by forced air. The exit temperature of the air and the highest component surface temperature are to be determined. Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the duct are smooth. 3 The thermal resistance of the duct is negligible. 4 Air is an ideal gas with constant properties. 5 The pressure of air is 1 atm. Properties We assume the bulk mean temperature for air to be 35°C since the mean temperature of air at the inlet will rise somewhat as a result of heat gain through the duct whose surface is exposed to a constant heat flux. The properties of air at 1 atm and this temperature are (Table A-22)

7268.0Pr

CJ/kg. 1007/sm 10655.1

C W/m.02625.0kg/m 145.1

25-

3

=

°=×=

°==

pc

k

ν

ρ

Air 27°C

0.65 m3/min

Electronics, 180 W

L = 1 m

D = 15 cm


kg/s 0.0124=kg/min 0.7443=/min)m )(0.65kg/m 145.1( 33== V&& ρm

C 39.3 °=°

°=+=→−=C)J/kg. kg/s)(1007 0124.0(

W)(0.85)(180+C 27)(

pieiep cm

QTTTTcmQ

&

&&&

(b) The mean fluid velocity is

m/s 0.613=m/min 8.36/4m) (0.15

m/min 65.02avg ===

πcAV V&

Then,

5556/sm 10655.1m) m/s)(0.15 (0.613

Re25

avg =×

==−ν

hDV

which is not greater than 10,000 but the components will cause turbulence and thus we can assume fully developed turbulent flow in the entire duct, and determine the Nusselt number from

05.20)7268.0()5556(023.0PrRe023.0 4.08.04.08.0 ====k

hDNu h

and

C. W/m51.3)05.20(m 15.0

C W/m.02625.0 2 °=°

== NuDkh

h

The highest component surface temperature will occur at the exit of the duct. Assuming uniform heat flux, its value is determined from

[ ]

C131.7°=°

°=+=→−=C. W/m3.51

m) m)(1 (0.15 W)/(0.85)(180+C2.39)(2,,

πhqTTTThq ehighestsehighests&

&


13-39

13-61 Air enters a hollow-core printed circuit board. The exit temperature of the air and the highest temperature on the inner surface are to be determined. Assumptions 1 Steady flow conditions exist. 2 Heat generated is uniformly distributed over the two surfaces of the PCB. 3 Air is an ideal gas with constant properties. 4 The pressure of air is 1 atm. Properties We assume the bulk mean temperature for air to be 40°C since the mean temperature of air at the inlet will rise somewhat as a result of heat gain through the hollow core whose surface is exposed to a constant heat flux. The properties of air at 1 atm and this temperature are (Table A-22)

kg/m.s 10008.2

kg/m.s 10918.1

7255.0Pr

CJ/kg. 1007/sm 10702.1

C W/m.02662.0kg/m 127.1

5C60@,

5

25-

3

−°

−

×=

×=

=

°=×=

°==

s

b

pc

k

μ

μ

ν

ρ

Air 32°C

0.8 L/s

Electronic components,

20 W

L = 18 cm



kg/s 109.02=/s)m 10)(0.8kg/m 127.1( -43-33 ××== V&& ρm

C54.0°=°×

°=+=→−=− C)J/kg. kg/s)(1007 1002.9(

W20+C 32)(4

pieiep cm

QTTTTcmQ

&

&&&


m 0049.0

m)] (0.0025+m) 12.0[(2m) m)(0.0025 12.0(44

m/s 67.2m) m)(0.0025 (0.12

/sm 108.0 33

avg

===

=×

==−

PA

D

AV

ch

c

V&

Then,

769/sm 10702.1m) 9m/s)(0.004 (2.67

Re25

avg =×

==−ν

hDV

which is less than 2300. Therefore, the flow is laminar and the thermal entry length in this case is m 0.14=m) 0049.0)(7255.0)(769(05.0PrRe05.0 == ht DL

which is shorter than the total length of the duct. Therefore, we assume thermally developing flow , and determine the Nusselt number from

58.410008.210918.1

0.18)0049.0)(7255.0)(769(

86.1PrRe86.114.0

5

53/114.03/1

=⎟⎟⎠

⎞⎜⎜⎝

⎛

×

×⎥⎦

⎤⎢⎣

⎡=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛==

−

−

s

bh

LD

khD

Nuμμ

and,

C. W/m9.24)58.4(m 0049.0

C W/m.02662.0 2 °=°

== NuDkh

h

The highest component surface temperature will occur at the exit of the duct. Its value is determined from

[ ] C72.2°=××°

°=

+=→−=

22

,,

0.18)m0.0025+0.182(0.12C). W/m(24.9 W20+C0.54

)(s

ehighestsehighestss hAQ

TTTThAQ&

&


13-40

13-62 Air enters a hollow-core printed circuit board. The exit temperature of the air and the highest temperature on the inner surface are to be determined. Assumptions 1 Steady flow conditions exist. 2 Heat generated is uniformly distributed over the two surfaces of the PCB. 3 Air is an ideal gas with constant properties. 4 The pressure of air is 1 atm. Properties We assume the bulk mean temperature for air to be 40°C since the mean temperature of air at the inlet will rise somewhat as a result of heat gain through the hollow core whose surface is exposed to a constant heat flux. The properties of air at 1 atm and this temperature are (Table A-22)

kg/m.s 10096.2

kg/m.s 10918.1

7255.0Pr

CJ/kg. 1007/sm 10702.1

C W/m.02662.0kg/m 127.1

5C80@,

5

25-

3

−°

−

×=

×=

=

°=×=

°==

s

b

pc

k

μ

μ

ν

ρ

Air 32°C

0.8 L/s

Electronic components, 35 W

L = 18 cm

Air channel 0.25 cm × 12


kg/s 109.02=/s)m 10)(0.8kg/m 127.1( -43-33 ××== V&& ρm

C70.5°=°×

°=+=→−=− C)J/kg. kg/s)(1007 1002.9(

W35+C32)(4

pieiep Cm

QTTTTcmQ

&

&&&


m 0049.0

m)] (0.0025+m) 12.0[(2m) m)(0.0025 12.0(44

m/s 67.2m) m)(0.0025 (0.12

/sm 108.0 33

avg

===

=×

==−

PA

D

AV

ch

c

V&

Then,

769/sm 10702.1m) 9m/s)(0.004 (2.67Re

25avg =

×==

−νhDV

which is less than 2300. Therefore, the flow is laminar and the thermal entry length in this case is m 0.14=m) 0049.0)(71.0)(783(05.0PrRe05.0 == ht DL

which is shorter than the total length of the duct. Therefore, we assume thermally developing flow , and determine the Nusselt number from

55.410096.210918.1

0.18)0049.0)(7255.0)(769(

86.1PrRe86.114.0

5

53/114.03/1

=⎟⎟⎠

⎞⎜⎜⎝

⎛

×

×⎥⎦

⎤⎢⎣

⎡=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛==

−

−

s

bh

LD

khD

Nuμμ

and,

C. W/m7.24)55.4(m 0049.0

C W/m.02662.0 2 °=°

== NuDkh

h

The highest component surface temperature will occur at the exit of the duct. Its value is determined from

C102.6°=

××°°=

+=⎯→⎯−=

]0.18)m0.0025+0.18C)[2(0.12. W/m(24.7 W35+C5.70

)(

22

,,s

ehighestsehighestss hAQTTTThAQ&

&


13-41

13-63E Water is heated by passing it through thin-walled copper tubes. The length of the copper tube that needs to be used is to be determined. Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the tube are smooth. 3 The thermal resistance of the tube is negligible. 4 The temperature at the tube surface is constant. Properties The properties of water at the bulk mean fluid temperature of F1002/)14060(, °=+=avgbT are (Table A-15E)

54.4Pr

FBtu/lbm. 999.0/sft 10738.0/

FBtu/h.ft. 363.0lbm/ft 0.62

25-

3

=

°=×==

°==

pc

k

ρμν

ρ

Analysis (a) The mass flow rate and the Reynolds number are

Water 60°F

0.4 lbm/s

140°F

250°F

L

D = 0.75 in

ft/s 10.2/4]ft) (0.75/12)[lbm/ft (62

lbm/s 4.023avgavg ===→=

πρρ

cc A

mVVAm&

&

810,17/sft 10738.0ft) /12ft/s)(0.75 (2.10Re

25avg =

×==

−νhDV

which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly in 5.7in) 75.0(1010 ==≈≈ DLL th

which is probably shorter than the total length of the pipe we will determine. Therefore, we can assume fully developed turbulent flow in the entire duct, and determine the Nusselt number from

9.105)54.4()810,17(023.0PrRe023.0 4.08.04.08.0 ====k

hDNu h

and

F.Btu/h.ft 615)9.105(ft )12/75.0(

FBtu/h.ft. 363.0 2 °=°

== NuDkh

h

The logarithmic mean temperature difference and then the rate of heat transfer per ft length of the tube are

Btu/h675,17)F4.146(ft)] 1)(ft 12/75.0()[F.Btu/h.ft 615(

F4.146

60250140250ln

60140

ln

2ln

ln

=°°=Δ=

°=⎟⎠⎞

⎜⎝⎛

−−−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=Δ

πThAQ

TTTT

TTT

s

is

es

ie

&

The rate of heat transfer needed to raise the temperature of water from 60°F to 140°F is Btu/h 115,085=F60)F)(140Btu/lbm. 99lbm/h)(0.9 36004.0()( °−°×=−= iep TTcmQ &&

Then the length of the copper tube that needs to be used becomes

ft 6.51==Btu/h 675,17Btu/h 085,115Length

(b) The friction factor, the pressure drop, and then the pumping power required to overcome this pressure drop can be determined for the case of fully developed turbulent flow to be 02598.0)810,17(184.0Re184.0 2.02.0 === −−f

hp 0.00016=⎟⎠

⎞⎜⎝

⎛⋅

=Δ

=

=⎟⎠

⎞⎜⎝

⎛

⋅==Δ

ft/slbf 550hp 1

lbm/ft 62)lbf/ft 58.13)(lbm/s 4.0(

lbf/ft 58.13ft/slbm 32.174

lbf 12

)ft/s 10.2)(lbm/ft 62(ft) 12/75.0(

ft) 69.7(02598.0

2

3

2

22

232avg

ρ

ρ

PmW

VDLfP

pump&&


13-42

13-64 A computer is cooled by a fan blowing air through its case. The flow rate of the air, the fraction of the temperature rise of air that is due to heat generated by the fan, and the highest allowable inlet air temperature are to be determined. Assumptions 1 Steady flow conditions exist. 2 Heat flux is uniformly distributed. 3 Air is an ideal gas with constant properties. 4 The pressure of air is 1 atm. Properties We assume the bulk mean temperature for air to be 25°C. The properties of air at 1 atm and this temperature are (Table A-22)

0.7296Pr

CJ/kg. 1007/sm 10562.1

C W/m.02551.0kg/m 184.1

25-

3

=

°=×=

°==

pc

k

ν

ρ

Analysis (a) Noting that the electric energy consumed by the fan is converted to thermal energy, the mass flow rate of air is

Cooling air

kg/s 0.01043=°°

+×=

−

+=→−=

C)C)(10J/kg. (1007 W25) 10(8

)()( fan elect,

iepiep TTc

WQmTTcmQ

&&&&&

(b) The fraction of temperature rise of air that is due to the heat generated by the fan and its motor is

23.8%

C2.38

==°°

°=°

==Δ→Δ=

238.0C10C2.38=

C)J/kg. kg/s)(1007 (0.01043 W25

f

cmQ

TTcmQp

p &

&&&

(c) The mean velocity of air is

[ ]

m/s 06.3)m 12.0)(m 003.0()kg/m (1.184

kg/s )8/01043.0(3avgavg ===→=

cc A

mVVAmρ

ρ&

&

and m 00585.0m) 12.0m 003.0(2m) m)(0.12 003.0(44

=+

==PA

D ch

Therefore,

1146/sm 10562.1m) 85m/s)(0.005 (3.06Re

25avg =

×==

−νhDV

which is less than 2300. Therefore, the flow is laminar. Assuming fully developed flow, the Nusselt number is determined from Table 13-4 corresponding to a/b = 12/0.3 = 40 to be Nu = 8.24. Then,

C. W/m9.35)24.8(m 00585.0

C W/m.02551.0 2 °=°

== NuDkh

h

The highest component surface temperature will occur at the exit of the duct. Assuming uniform heat flux, the air temperature at the exit is determined from

C7.61C. W/m35.9

]m 0.18)0.003+0.182(0.12 W]/[8)2580[(C70 )(2

2

max,max, °=°

×××+−°=−=→−=

hqTTTThq sees&

&

The highest allowable inlet temperature then becomes C51.7°=°−°=°−=→°=− C10C7.61C10C10 eiie TTTT Discussion Although the Reynolds number is less than 2300, the flow in this case will most likely be turbulent because of the electronic components that that protrude into flow. Therefore, the heat transfer coefficient determined above is probably conservative.


13-43

Review Problems 13-65 A silicon chip is cooled by passing water through microchannels etched in the back of the chip. The outlet temperature of water and the chip power dissipation are to be determined. Assumptions 1 Steady operating conditions exist. 2 The flow of water is fully developed. 3 All the heat generated by the circuits on the top surface of the chip is transferred to the water.

H

W

Circuits generating power eW&

T,i

Te

Chip, Ts

Cap, Ts

10 mm

10 mm

Properties Assuming a bulk mean fluid temperature of 25°C, the properties of water are (Table A-15)

/sm 10891.0

C W/m.607.0kg/m 997

23-

3

×=

°==

μ

ρk

14.6Pr

CJ/kg. 4180

=

°=pc

Analysis (a) The mass flow rate for one channel, the hydraulic diameter, and the Reynolds number are

kg/s 0001.050

kg/s 005.0

channel

total ===nm

m&

&

m 108m 80)20050(2)20050(4

)(2)(44 5-×==

+×

=+×

== μWHWH

pADh

898s)kg/m 10891.0)(m 1020050(

)m 10kg/s)(8 0001.0(Re

3212

-5=

⋅×××

×====

−−μμρρ

μρ

c

h

c

hh

ADm

AVDmVD &&

which is smaller than 2300. Therefore, the flow is laminar. We take fully developed laminar flow in the entire duct. The Nusselt number in this case is 66.3=NuHeat transfer coefficient is

C. W/m770,27)66.3(m 108

C W/m.607.0 25

°=×

°==

−Nu

Dkh

Next we determine the exit temperature of water 2-62 m 102.1mm 0.021=)2.005.0(2)01.005.0(222 ×=×+×=+= HLWLA

K 297.8=⎥⎥⎦

⎤

⎢⎢⎣

⎡ ×−−−=−−=

−−

)4180)(0001.0()101.2)(770,27(exp)290350(350)(

6)/( pcmhA

isse eTTTT &

Then the rate of heat transfer becomes W82.21C)8.297350)(CJ/kg. 4180)(kg/s 0001.0()( =°−°=−= iep TTcmQ &&

(b) Noting that there are 50 such channels, the chip power dissipation becomes W1091=== W)82.21(50channel onechannelQnWe

&&


13-44

13-66 Water is heated by passing it through five identical tubes that are maintained at a specified temperature. The rate of heat transfer and the length of the tubes necessary are to be determined. Assumptions 1 Steady flow conditions exist. 2 The surface temperature is constant and uniform. 3 The inner surfaces of the tubes are smooth. 4 Heat transfer to the surroundings is negligible. Properties The properties of water at the bulk mean fluid temperature of (15+35)/2=25ºC are (Table A-15)

14.6Pr

CJ/kg. 4180/sm 10891.0

C W/m.607.0kg/m 997

23-

3

=

°=×=

°==

pc

k

μ

ρ

Water 15ºC

10 kg/s

5 tubes

D = 5 cm

35ºC

60ºC

Analysis (a) The rate of heat transfer is

W836,000=°−°=−= C)1535)(CJ/kg. 4180)(kg/s 10()( iep TTcmQ &&

(b) The water velocity is

m/s 02.14/m) (0.05)kg/m 997(

kg/s )5/10(23

===πρ cA

mV&

The Reynolds number is

067,57skg/m 10891.0

m) m/s)(0.05 )(1.02kg/m (997Re

3

3=

⋅×==

−μρVD

which is greater than 10,000. Therefore, we have turbulent flow. Assuming fully developed flow in the entire tube, the Nusselt number is determined from

5.303)14.6()067,57(023.0PrRe023.0 4.08.04.08.0 ====k

hDNu h


C. W/m3684)5.303(m 05.0

C W/m.607.0 2 °=°

== NuDkh

Using the average fluid temperature and considering that there are 5 tubes, the length of the tubes is determined as follows:

m 8.26===⎯→⎯=

=⎯→⎯°−°⋅=⎯→⎯−=

m) 05.0(5m 6.484

55

m 484.6C)2560()C W/m3684( W000,836)(2

22avg,

πππ

DALDLA

AATThAQ sbs&


13-45

13-67 Water is heated by passing it through five identical tubes that are maintained at a specified temperature. The rate of heat transfer and the length of the tubes necessary are to be determined. Assumptions 1 Steady flow conditions exist. 2 The surface temperature is constant and uniform. 3 The inner surfaces of the tubes are smooth. 4 Heat transfer to the surroundings is negligible. Properties The properties of water at the bulk mean fluid temperature of (15+35)/2=25ºC are (Table A-15)

14.6Pr

CJ/kg. 4180/sm 10891.0

C W/m.607.0kg/m 997

23-

3

=

°=×=

°==

pc

k

μ

ρ

Water 15ºC

20 kg/s

5 tubes

D = 5 cm

35ºC

60ºC


W1,672,000=°−°=−= C)1535)(CJ/kg. 4180)(kg/s 20()( iep TTcmQ &&

(b) The water velocity is

[ ] m/s 04.24/m) (0.05)kg/m 997(

kg/s )5/20(23

===πρ cA

mV&


320,114skg/m 10891.0

m) m/s)(0.05 )(2.04kg/m (997Re

3

3=

⋅×==

−μρVD


0.529)14.6()320,114(023.0PrRe023.0 4.08.04.08.0 ====k

hDNu h


C. W/m6423)0.529(m 05.0

C W/m.607.0 2 °=°

== NuDkh

Using the average fluid temperature and considering that there are 5 tubes, the length of the tubes is determined as follows:

m 9.47===⎯→⎯=

=⎯→⎯°−°⋅=⎯→⎯−=

m) 05.0(5m 7.438

55

m 438.7C)2560()C W/m6423( W000,672,1)(2

22avg,

πππ

DALDLA

AATThAQ sbs&


13-46

13-68 Water is heated as it flows in a smooth tube that is maintained at a specified temperature. The necessary tube length and the water outlet temperature if the tube length is doubled are to be determined. Assumptions 1 Steady flow conditions exist. 2 The surface temperature is constant and uniform. 3 The inner surfaces of the tube are smooth. 4 Heat transfer to the surroundings is negligible. Properties The properties of water at the bulk mean fluid temperature of (10+40)/2=25ºC are (Table A-15)

14.6Pr

CJ/kg. 4180/sm 10891.0

C W/m.607.0kg/m 997

23-

3

=

°=×=

°==

pc

k

μ

ρ

Water 1500 kg/h

L

D = 1 cm


W250,52C)1040)(CJ/kg. 4180)(kg/s 3600/1500()( =°−°=−= iep TTcmQ &&

The water velocity is

[ ] m/s 32.54/m) (0.01)kg/m 997(

kg/s )3600/1500(23

===πρ cA

mV&


542,59skg/m 10891.0

m) m/s)(0.01 )(5.32kg/m (997Re

3

3=

⋅×==

−μρVD


9.313)14.6()542,59(023.0PrRe023.0 4.08.04.08.0 ====k

hDNu h


C. W/m056,19)9.313(m 01.0

C W/m.607.0 2 °=°

== NuDkh

Using the average fluid temperature, the length of the tubes is determined as follows:

m 3.6=⎯→⎯=⎯→⎯=

=⎯→⎯°−°⋅=⎯→⎯−=

LLDLA

AATThAQ sbs

m) 01.0(m 0.1142

m 1142.0C)2549()C W/m056,19( W250,52)(2

22avg,

ππ

&

(b) If the tube length is doubled, the surface area doubles, and the outlet water temperature may be obtained from an energy balance to be

C3.532

1049)1142.02)(056,19()10)(4180)(3600/1500(

)()( avg,

°=

⎟⎟⎠

⎞⎜⎜⎝

⎛ +−×=−

−=−

e

ee

bssiep

T

TT

TThATTcm&

which is greater than the surface temperature of the wall. This is impossible. It shows that the water reaches the surface temperature before the entire length of tube is covered and in reality the water will leave the tube at the surface temperature of 49ºC. This example demonstrates that the use of unnecessarily long tubes should be avoided.


13-47

13-69 Geothermal water is supplied to a city through stainless steel pipes at a specified rate. The electric power consumption and its daily cost are to be determined, and it is to be assessed if the frictional heating during flow can make up for the temperature drop caused by heat loss. Assumptions 1 The flow is steady and incompressible. 2 The entrance effects are negligible, and thus the flow is fully developed. 3 The minor losses are negligible because of the large length-to-diameter ratio and the relatively small number of components that cause minor losses. 4 The geothermal well and the city are at about the same elevation. 5 The properties of geothermal water are the same as fresh water. 6 The fluid pressures at the wellhead and the arrival point in the city are the same. Properties The properties of water at 110°C are ρ = 950.6 kg/m3, μ = 0.255×10-3 kg/m⋅s, and cp = 4.229 kJ/kg⋅°C (Table A-15). The roughness of stainless steel pipes is 2×10-6 m (Table 13-3). Analysis (a) We take point 1 at the well-head of geothermal resource and point 2 at the final point of delivery at the city, and the entire piping system as the control volume. Both points are at the same elevation (z2 = z2) and the same velocity (V1 = V2) since the pipe diameter is constant, and the same pressure (P1 = P2). Then the energy equation for this control volume simplifies to

22 upump,turbine2

222

upump,1

211

LL hhhhzg

Vg

Phz

gV

gP

=→++++=+++ρρ

That is, the pumping power is to be used to overcome the head losses due to friction in flow. The mean velocity and the Reynolds number are

73

3avg

2

3

2avg

10186.1skg/m 10255.0

m) m/s)(0.60 305.5)(kg/m 6.950(Re

m/s 305.54/m) (0.60

/sm 1.54/

×=⋅×

==

====

−μ

ρππ

DVDA

Vc

VV &&

L = 12 km

D = 60 cm

Water 1.5 m3/s

21


1033.3m 60.0

m 102/ 66

−−

×=×

=Dε


⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

×+

×−=→

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+−=

−

fffD

f 7

6

10187.151.2

7.31033.3log0.21

Re51.2

7.3/log0.21 ε

It gives f = 0.00829. Then the pressure drop, the head loss, and the required power input become

kPa2218kN/m 1

kPa 1m/skg 1000

kN 12

m/s) 305.5)(kg/m 6.950(m 0.60m 000,1200829.0

2 22

232avg =⎟

⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅==Δ

VDLfPρ

kW 5118=⎟⎠

⎞⎜⎝

⎛

⋅=

Δ==

/smkPa 1kW 1

0.65)kPa 2218)(/sm (1.5

3

3

motor-pumpmotor-pump

upump,elect ηη

PVWW

&&&

Therefore, the pumps will consume 5118 kW of electric power to overcome friction and maintain flow. (b) The daily cost of electric power consumption is determined by multiplying the amount of power used per day by the unit cost of electricity,

kWh/day 832,122h/day) kW)(24 5118(Amount inelect, ==Δ= tW&

$7370/day==×= 0.06/kWh)kWh/day)($ 832,122(costUnit AmountCost


13-48

(c) The energy consumed by the pump (except the heat dissipated by the motor to the air) is eventually dissipated as heat due to the frictional effects. Therefore, this problem is equivalent to heating the water by a 5118 kW of resistance heater (again except the heat dissipated by the motor). To be conservative, we consider only the useful mechanical energy supplied to the water by the pump. The temperature rise of water due to this addition of energy is

C0.55°=°⋅

×==Δ→Δ=

)CkJ/kg 229.4(/s)m 5.1)(kg/m 6.950(kJ/s) (51180.65

33

inelect,motor-pumpelect

pp c

WTTcW

VV

&

&&&

ρ

ηρ

Therefore, the temperature of water will rise at least 0.55°C, which is more than the 0.5°C drop in temperature (in reality, the temperature rise will be more since the energy dissipation due to pump inefficiency will also appear as temperature rise of water). Thus we conclude that the frictional heating during flow can more than make up for the temperature drop caused by heat loss. Discussion The pumping power requirement and the associated cost can be reduced by using a larger diameter pipe. But the cost savings should be compared to the increased cost of larger diameter pipe.


13-49

13-70 Geothermal water is supplied to a city through cast iron pipes at a specified rate. The electric power consumption and its daily cost are to be determined, and it is to be assessed if the frictional heating during flow can make up for the temperature drop caused by heat loss. Assumptions 1 The flow is steady and incompressible. 2 The entrance effects are negligible, and thus the flow is fully developed. 3 The minor losses are negligible because of the large length-to-diameter ratio and the relatively small number of components that cause minor losses. 4 The geothermal well and the city are at about the same elevation. 5 The properties of geothermal water are the same as fresh water. 6 The fluid pressures at the wellhead and the arrival point in the city are the same. Properties The properties of water at 110°C are ρ = 950.6 kg/m3, μ = 0.255×10-3 kg/m⋅s, and cp = 4.229 kJ/kg⋅°C (Table A-15). The roughness of cast iron pipes is 0.00026 m (Table 13-3). Analysis (a) We take point 1 at the well-head of geothermal resource and point 2 at the final point of delivery at the city, and the entire piping system as the control volume. Both points are at the same elevation (z2 = z2) and the same velocity (V1 = V2) since the pipe diameter is constant, and the same pressure (P1 = P2). Then the energy equation for this control volume simplifies to

22 upump,turbine2

222

upump,1

211

LL hhhhzg

Vg

Phz

gV

gP

=→++++=+++ρρ

That is, the pumping power is to be used to overcome the head losses due to friction in flow. The mean velocity and the Reynolds number are

73

3avg

2

3

2avg

10187.1skg/m 10255.0

m) m/s)(0.60 305.5)(kg/m 6.950(Re

m/s 305.54/m) (0.60

/sm 1.54/

×=⋅×

==

====

−μ

ρππ

DVDA

Vc

VV &&

L = 12 km

D = 60 cm

Water 1.5 m3/s

21


1033.4m 60.0

m 00026.0/ 4−×==Dε


⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

×+

×−=→

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+−=

−

fffD

f 7

4

10187.151.2

7.31033.4log0.21

Re51.2

7.3/log0.21 ε

It gives f = 0.01623. Then the pressure drop, the head loss, and the required power input become

kPa 4342kN/m 1

kPa 1m/skg 1000

kN 12

m/s) 305.5)(kg/m 6.950(m 0.60m 000,1201623.0

2 2

232avg =⎟

⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

==ΔV

DLfPρ

kW 10,020=⎟⎠

⎞⎜⎝

⎛

⋅=

Δ==

/smkPa 1kW 1

0.65)kPa 4342)(/sm (1.5

3

3


upump,elect ηη

PWW V&&&

Therefore, the pumps will consume 10,017 W of electric power to overcome friction and maintain flow. (b) The daily cost of electric power consumption is determined by multiplying the amount of power used per day by the unit cost of electricity,


13-50

kWh/day 480,240h/day) kW)(24 020,10(Amount inelect, ==Δ= tW&

y$14,430/da==×= 0.06/kWh)kWh/day)($ 480,240(costUnit AmountCost

(c) The energy consumed by the pump (except the heat dissipated by the motor to the air) is eventually dissipated as heat due to the frictional effects. Therefore, this problem is equivalent to heating the water by a 10,020 kW of resistance heater (again except the heat dissipated by the motor). To be conservative, we consider only the useful mechanical energy supplied to the water by the pump. The temperature rise of water due to this addition of energy is

C1.08°=°⋅

×==Δ→Δ=

)CkJ/kg 229.4(/s)m 5.1)(kg/m 6.950(kJ/s) (10,0200.65

33

inelect,motor-pumpelect

pp c

WTTcW

VV

&

&&&

ρ

ηρ

Therefore, the temperature of water will rise at least 1.08°C, which is more than the 0.5°C drop in temperature (in reality, the temperature rise will be more since the energy dissipation due to pump inefficiency will also appear as temperature rise of water). Thus we conclude that the frictional heating during flow can more than make up for the temperature drop caused by heat loss. Discussion The pumping power requirement and the associated cost can be reduced by using a larger diameter pipe. But the cost savings should be compared to the increased cost of larger diameter pipe.


13-51

13-71 The velocity profile in fully developed laminar flow in a circular pipe is given. The radius of the pipe, the mean velocity, and the maximum velocity are to be determined. Assumptions The flow is steady, laminar, and fully developed. Analysis The velocity profile in fully developed laminar flow in a circular pipe is

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

2

2

max 1)(RrVru

R r

0

Vmax

u(r)=Vmax(1-r2/R2)


)1001(6)( 2rru −=

Comparing the two relations above gives the pipe radius, the maximum velocity, and the mean velocity to be

m 0.10=→= RR 100

12

Vmax = 6 m/s

m/s 3===2m/s 6

2max

avgV

V


13-52

13-72E The velocity profile in fully developed laminar flow in a circular pipe is given. The volume flow rate, the pressure drop, and the useful pumping power required to overcome this pressure drop are to be determined. Assumptions 1 The flow is steady, laminar, and fully developed. 2 The pipe is horizontal. Properties The density and dynamic viscosity of water at 40°F are ρ = 62.42 lbm/ft3 and μ = 1.308×10-3 lbm/ft⋅s, respectively (Table A-15E). Analysis The velocity profile in fully developed laminar flow in a circular pipe is

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

2

2

max 1)(RrVru

R r

0

Vmax

u(r)=Vmax(1-r2/R2) The velocity profile in this case is given by

)6251(8.0)( 2rru −=

Comparing the two relations above gives the pipe radius, the maximum velocity, and the mean velocity to be

ft 04.0 62512 =→= RR

Vmax = 0.8 ft/s

ft/s 4.02ft/s 0.8

2max

avg ===V

V

Then the volume flow rate and the pressure drop become

/sft 0.00201 3==== ]ft) (0.04ft/s)[ 4.0()( 22 ππRVAV avgcavgV&

⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅

⋅×

Δ=→

Δ=

− lbf 1ft/slbm 2.32

ft) s)(140lbm/ft 10308.1(128ft) (0.08)(

/sft 0.00201 128

2

3

43

4

horizπ

μπ P

LDP

V&

It gives

psi 0.0790==Δ 2lbf/ft 37.11P

Then the useful pumping power requirement becomes

ft/slbf 0.737

W1)lbf/ft 37.11)(/sft 00201.0( 23upump, W0.031=⎟

⎠⎞

⎜⎝⎛

⋅=Δ= PW V&&

Checking The flow was assumed to be laminar. To verify this assumption, we determine the Reynolds number:

1527slbm/ft 10308.1

ft) ft/s)(0.08 4.0)(lbm/ft 42.62(Re3

3=

⋅×==

−μ

ρ DVavg

which is less than 2300. Therefore, the flow is laminar. Discussion Note that the pressure drop across the water pipe and the required power input to maintain flow is negligible. This is due to the very low flow velocity. Such water flows are the exception in practice rather than the rule.


13-53

13-73 A compressor is connected to the outside through a circular duct. The power used by compressor to overcome the pressure drop, the rate of heat transfer, and the temperature rise of air are to be determined. Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the duct are smooth. 3 The thermal resistance of the duct is negligible. 4 Air is an ideal gas with constant properties. Properties We take the bulk mean temperature for air to be 15°C since the mean temperature of air at the inlet will rise somewhat as a result of heat gain through the duct whose surface is exposed to a higher temperature. The properties of air at this temperature and 1 atm pressure are (Table A-22)

7323.0Pr C,J/kg. 1007 /s,m 10470.1

C W/m.02476.0 ,kg/m 225.125-

3

=°=×=

°==

pc

k

ν

ρ

The density and kinematic viscosity at 95 kPa are

atm 938.0kPa 101.325

kPa 95==P

/sm 101.378=)/s)/(0.938m 10470.1(

kg/m 149.1)938.0)(kg/m 225.1(25-25-

33

××=

==

ν

ρ

Air 10°C, 95 kPa

0.27 m3/s

Indoors 20°C

L = 11 m

D = 20 cm

Analysis The mean velocity of air is

m/s 594.8/4m) (0.2

/sm 27.02

3

avg ===πcA

V V&

Then 525

avg 10247.1/sm 10378.1m) m/s)(0.2 (8.594Re ×=

×==

−νhDV

which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly m 2m) 2.0(1010 ==≈≈ DLL th

which is shorter than the total length of the duct. Therefore, we assume fully developed flow in a smooth pipe, and determine friction factor from

[ ] 01718.064.1)10247.1ln(790.0)64.1Reln790.0(252 =−×=−=−−f

The pressure drop and the compressor power required to overcome this pressure drop are kg/s 3102.0)/sm 27.0)(kg/m 149.1( 33 === V&& ρm

W10.8==Δ

=

===Δ

3

2

2232

avg

kg/m 149.1)N/m 09.40)(kg/s 3102.0(

N/m 09.402

)m/s 594.8)(kg/m 149.1(m) 2.0(m) 11()01718.0(

2

ρ

ρ

PmW

VDLfP

pump&&

(b) For the fully developed turbulent flow, the Nusselt number is

3.242)7323.0()10247.1(023.0PrRe023.0 4.08.054.08.0 =×===k

hDNu

and C. W/m00.30)3.242(m 2.0

C W/m.02476.0 2 °=°

== NuDkh

h

Disregarding the thermal resistance of the duct, the rate of heat transfer to the air in the duct becomes 2m 912.6m) m)(11 (0.2 === ππDLAs

W518.4=+

−=

+

−= ∞∞

)912.6)(10(1

)912.6)(00.30(1

102011

21

21

ss AhAh

TTQ&

(c) The temperature rise of air in the duct is C1.7°=Δ→Δ°=→Δ= TTTcmQ p C)J/kg. kg/s)(1007 (0.3102 W4.518 &&


13-54

13-74 Air enters the underwater section of a duct. The outlet temperature of the air and the fan power needed to overcome the flow resistance are to be determined. Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the duct are smooth. 3 The thermal resistance of the duct is negligible. 4 The surface of the duct is at the temperature of the water. 5 Air is an ideal gas with constant properties. 6 The pressure of air is 1 atm. Properties We assume the bulk mean temperature for air to be 20°C since the mean temperature of air at the inlet will drop somewhat as a result of heat loss through the duct whose surface is at a lower temperature. The properties of air at 1 atm and this temperature are (Table A-22)

7309.0Pr

CJ/kg. 1007/sm 10516.1

C W/m.02514.0kg/m 204.1

25-

3

=

°=×=

°==

pc

k

ν

ρ


425

avg 10958.3/sm 10516.1

m) m/s)(0.2 (3Re ×=×

==−ν

hDV

Air 25°C 3 m/s

River water

L = 15 m

D = 20 cm



76.99)7309.0()10958.3(023.0PrRe023.0 3.08.043.08.0 =×===k

hDNu h

and

C. W/m54.12)76.99(m 2.0

C W/m.02514.0 2 °=°

== NuDkh

h


kg/s 0.1135/s)m )(0.09425kg/m 204.1(=

4m) (0.2m/s) )(3kg/m 204.1(

m 9.425=m) m)(15 2.0(

332

3avg

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛==

==

πρ

ππ

c

s

AVm

DLA

&

and

C18.6°=−−=−−=−− )1007)(1135.0(

)425.9)(54.12()/( )2515(15)( eeTTTT ps cmhA

isse&

The friction factor, pressure drop, and the fan power required to overcome this pressure drop can be determined for the case of fully developed turbulent flow in smooth pipes to be

[ ] 02212.064.1)10958.3ln(790.0)64.1Reln790.0(242 =−×=−=−−f

Pa 988.8N/m 1

Pa 1m/skg 1N 1

2m/s) 3)(kg/m 204.1(

m 0.2m 1502212.0

2 22

232avg =⎟

⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅==Δ

VDLfPρ

W1.54=⎟⎠

⎞⎜⎝

⎛

⋅==

Δ==

/smPa 1 W1

55.0)Pa 988.8)(/sm 09425.0(

3

3


upump,fan ηη

PWW V&&&


13-55

13-75 Air enters the underwater section of a duct. The outlet temperature of the air and the fan power needed to overcome the flow resistance are to be determined. Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the duct are smooth. 3 The thermal resistance of the duct is negligible. 4 Air is an ideal gas with constant properties. 5 The pressure of air is 1 atm. Properties We assume the bulk mean temperature for air to be 20°C since the mean temperature of air at the inlet will drop somewhat as a result of heat loss through the duct whose surface is at a lower temperature. The properties of air at 1 atm and this temperature are (Table A-22)

7309.0Pr

CJ/kg. 1007/sm 10516.1

C W/m.02514.0kg/m 204.1

25-

3

=

°=×=

°==

pc

k

ν

ρ

Water 25°C 3 m/s

River water15°C

L = 15 m

D = 20 cm

Mineral deposit0.15 mm


425

avg 10958.3/sm 10516.1

m) m/s)(0.2 (3Re ×=

×==

−νhDV


which is much shorter than the total length of the duct. Therefore, we can assume fully developed turbulent flow in the entire duct, and determine the Nusselt number and h from

76.99)7309.0()10959.3(023.0PrRe023.0 3.08.043.08.0 =×===k

hDNu h

and C. W/m54.12)76.99(m 2.0

C W/m.02514.0 2 °=°

== NuDkh

h


kg/s 0.1135/s)m )(0.09425kg/m 204.1(=

4m) (0.2m/s) )(3kg/m 204.1(

m 9.425=m) m)(15 2.0(

332

3avg

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛==

==

πρ

ππ

c

s

AVm

DLA

&

The unit thermal resistance of the mineral deposit is

C/W.m 00083.0C W/m.3m 0025.0 2

mineral °=°

==kLR

which is much less than (about 1%) the unit convection resistance,

C/W.m 0797.0C. W/m54.12

11 22conv °=°

==h

R

Therefore, the effect of 0.25 mm thick mineral deposit on heat transfer is negligible. Next we determine the exit temperature of air

C18.6°=−−=−−=−− )1007)(1135.0(

)425.9)(54.12()/( )2515(15)( eeTTTT pcmhA

isse&

The friction factor, pressure drop, and the fan power required to overcome this pressure drop can be determined for the case of fully developed turbulent flow in smooth pipes to be

[ ] 02212.064.1)10958.3ln(790.0)64.1Reln790.0(242 =−×=−=−−f

Pa 988.8N/m 1

Pa 1m/skg 1N 1

2m/s) 3)(kg/m 204.1(

m 0.2m 1502212.0

2 22

232avg =⎟

⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅==Δ

VDLfPρ

W1.54=⎟⎠

⎞⎜⎝

⎛

⋅==

Δ==

/smPa 1 W1

55.0)Pa 988.8)(/sm 09425.0(

3

3


upump,fan ηη

PWW V&&&


13-56

13-76E The exhaust gases of an automotive engine enter a steel exhaust pipe. The velocity of exhaust gases at the inlet and the temperature of exhaust gases at the exit are to be determined. Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the pipe are smooth. 3 The thermal resistance of the pipe is negligible. 4 Exhaust gases have the properties of air, which is an ideal gas with constant properties. Properties We take the bulk mean temperature for exhaust gases to be 700°F since the mean temperature of gases at the inlet will drop somewhat as a result of heat loss through the exhaust pipe whose surface is at a lower temperature. The properties of air at this temperature and 1 atm pressure are (Table A-22E)

/sft 10225.6

FBtu/h.ft. 0280.0lbm/ft 03421.0

24-

3

×=

°==

ν

ρk 694.0Pr

FBtu/lbm. 2535.0

=

°=pc

Exhaust 800°F

0.2 lbm/s

80°F

L = 8 ft

D = 3.5 in Noting that 1 atm = 14.7 psia, the pressure in atm is

P = (15.5 psia)/(14.7 psia) = 1.054 atm. Then,

/sft 105.906=)/s)/(1.054ft 10225.6(

lbm/ft 03606.0)054.1)(lbm/ft 03421.0(24-24-

33

××=

==

v

ρ

Analysis (a) The velocity of exhaust gases at the inlet of the exhaust pipe is

( ) ft/s 83.01===⎯→⎯=4/ft) (3.5/12)lbm/ft 03606.0(

lbm/s 0.223avgavg

πρρ

cc A

mVAVm&

&

(b) The Reynolds number is

990,40/sft 10906.5ft) 12ft/s)(3.5/ (83.01Re

24avg =

×==

−νhDV

which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly ft 917.2ft) 12/5.3(1010 ==≈≈ DLL th

which are shorter than the total length of the duct. Therefore, we can assume fully developed turbulent flow in the entire duct, and determine the Nusselt number from

0.101)6940.0()990,40(023.0PrRe023.0 3.08.03.08.0 ====k

hDNu h

and F.Btu/h.ft 70.9)0.101(ft )12/5.3(

FBtu/h.ft. 0280.0 2 °=°

=== NuDkhh

hi

2ft 7.33=ft) ft)(8 12/5.3(ππ == DLAs

In steady operation, heat transfer from exhaust gases to the duct must be equal to the heat transfer from the duct to the surroundings, which must be equal to the energy loss of the exhaust gases in the pipe. That is, gasesexhaust externalinternal EQQQ &&&& Δ===

Assuming the duct to be at an average temperature of Ts , the quantities above can be expressed as

& :Qinternal

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

°−°=→

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=Δ=

800ln

F800)ft 33.7)(F.Btu/h.ft 70.9(

ln

22ln

s

es

e

is

es

iesisi

TTT

TQ

TTTT

TTAhTAhQ &&

& :Qexternal F)80)(ft F)(7.33.Btu/h.ft 3( )( 22 °−°=→−= sosso TQTTAhQ &&

Δ & :Eexhaust gases F)F)(800Btu/lbm. 535lbm/h)(0.2 36002.0( )( °−°×=→−= eiep TQTTcmQ &&&

This is a system of three equations with three unknowns whose solution is

F2.604and , Btu/h, 11,528 °=°== se TTQ F736.8&


13-57

13-77 Hot water enters a cast iron pipe whose outer surface is exposed to cold air with a specified heat transfer coefficient. The rate of heat loss from the water and the exit temperature of the water are to be determined. Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the pipe are smooth. Properties We assume the water temperature not to drop significantly since the pipe is not very long. We will check this assumption later. The properties of water at 90°C are (Table A-15)

96.1Pr

CJ/kg. 4206 /s;m 10326.0/

C W/m.675.0 ;kg/m 3.96526-

3

=

°=×==

°==

pc

k

ρμν

ρ

Analysis (a) The mass flow rate of water is

kg/s 456.1m/s)2.1(4

m) (0.04)kg/m 3.965(

23

avg === πρ VAm c&

Water 90°C

1.2 m/s

10°C

L = 15 m

Di = 4 cm Do = 4.6 cm


240,147/sm 10326.0

m) m/s)(0.04 (1.2Re

26avg =

×==

−νhDV


which are much shorter than the total length of the pipe. Therefore, we can assume fully developed turbulent flow in the entire pipe. The friction factor corresponding to Re = 147,240 and ε/D = (0.026 cm)/(4 cm) = 0.0065 is determined from the Moody chart to be f = 0.032. Then the Nusselt number becomes

1.73796.1240,147032.0125.0PrRe125.0 3/13/1 =×××=== fk

hDNu h

and C. W/m440,12)1.737(m 04.0

C W/m.675.0 2 °=°

=== NuDkhh

hi

which is much greater than the convection heat transfer coefficient of 12 W/m2.°C. Therefore, the convection thermal resistance inside the pipe is negligible, and thus the inner surface temperature of the pipe can be taken to be equal to the water temperature. Also, we expect the pipe to be nearly isothermal since it is made of thin metal (we check this later). Then the rate of heat loss from the pipe will be the sum of the convection and radiation from the outer surface at a temperature of 90°C, and is determined to be 2

0 m 2.168=m) m)(15 046.0(ππ == LDAo

W2081=C)10)(90m C)(2.168. W/m12()( 22 °−°=−= surrsooconv TTAhQ&

[ ] W942K) 27310(K) 27390().K W/m1067.5)(m 168.2)(7.0(

)(444282

440

=+−+×=

−=−

surrsrad TTAQ σε&

W3023=942+2081=+= radconvtotal QQQ &&& (b) The temperature at which water leaves the basement is

C89.5°=°

−°=−=⎯→⎯−=)CJ/kg. 4206)(kg/s 456.1(

W3023C90)(p

ieeip cmQ

TTTTcmQ&

&&&

The result justifies our assumption that the temperature drop of water is negligible. Also, the thermal resistance of the pipe and temperature drop across it are

C09.0)C/W 1085.2)( W3023(

C/W 1085.2m) C)(15 W/m.52(4

)4/6.4ln(2

)/ln(

5

512

°=°×==Δ

°×=°

==

−

−

pipetotalpipe

pipe

RQT

kLDD

R

&

ππ

which justifies our assumption that the temperature drop across the pipe is negligible.


13-58

13-78 Hot water enters a copper pipe whose outer surface is exposed to cold air with a specified heat transfer coefficient. The rate of heat loss from the water and the exit temperature of the water are to be determined. Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the pipe are smooth. Properties We assume the water temperature not to drop significantly since the pipe is not very long. We will check this assumption later. The properties of water at 90°C are (Table A-22)

96.1Pr

CJ/kg. 4206 /s;m 10326.0/

C W/m.675.0 ;kg/m 3.96526-

3

=

°=×==

°==

pc

k

ρμν

ρ

Analysis (a) The mass flow rate of water is

kg/s 456.1m/s)2.1(4

m) (0.04)kg/m 3.965(

23

avg === πρ VAm c&

Water 90°C

1.2 m/s

10°C

L = 15 m

Di = 4 cm Do = 4.6 cm


240,147/sm 10326.0

m) m/s)(0.04 (1.2Re

26avg =

×==

−νhDV


which are much shorter than the total length of the pipe. Therefore, we can assume fully developed turbulent flow in the entire pipe. The friction factor corresponding to Re = 147,240 and ε/D = (0.026 cm)/(4 cm) = 0.0065 is determined from the Moody chart to be f = 0.032. Then the Nusselt number becomes

1.73796.1240,147032.0125.0PrRe125.0 3/13/1 =×××=== fk

hDNu h

and C. W/m440,12)1.737(m 04.0

C W/m.675.0 2 °=°

=== NuDkhh

hi

which is much greater than the convection heat transfer coefficient of 12 W/m2.°C. Therefore, the convection thermal resistance inside the pipe is negligible, and thus the inner surface temperature of the pipe can be taken to be equal to the water temperature. Also, we expect the pipe to be nearly isothermal since it is made of thin metal (we check this later). Then the rate of heat loss from the pipe will be the sum of the convection and radiation from the outer surface at a temperature of 90°C, and is determined to be 2

0 m 2.168=m) m)(15 046.0(ππ == LDAo

W2081=C)10)(90m C)(2.168. W/m12()( 22 °−°=−= surrsooconv TTAhQ&

[ ] W942K) 27310(K) 27390().K W/m1067.5)(m 168.2)(7.0(

)(444282

440

=+−+×=

−=−

surrsrad TTAQ σε&

W3023=942+2081=+= radconvtotal QQQ &&& (b) The temperature at which water leaves the basement is

C89.5°=°

−°=−=⎯→⎯−=)CJ/kg. 4206)(kg/s 456.1(

W3023C90)(p

ieeip cmQ

TTTTcmQ&

&&&

The result justifies our assumption that the temperature drop of water is negligible. Also, the thermal resistance of the pipe and temperature drop across it are

C012.0)C/W 1084.3)( W3023(

C/W 1084.3m) C)(15 W/m.386(2

)4/6.4ln(4

)/ln(

6

612

°=°×==Δ

°×=°

==

−

−

pipetotalpipe

pipe

RQT

kLDD

R

&

ππ

which justifies our assumption that the temperature drop across the pipe is negligible.


13-59

13-79 …. 13-81 Design and Essay Problems 13-81 A computer is cooled by a fan blowing air through the case of the computer. The flow rate of the fan and the diameter of the casing of the fan are to be specified. Assumptions 1 Steady flow conditions exist. 2 Heat flux is uniformly distributed. 3 Air is an ideal gas with constant properties. Properties The relevant properties of air are (Tables A-1 and A-15)

Cooling air

/kg.KkPa.m 287.0

CJ/kg. 10073=

°=

R

c p

Analysis We need to determine the flow rate of air for the worst case scenario. Therefore, we assume the inlet temperature of air to be 50°C, the atmospheric pressure to be 70.12 kPa, and disregard any heat transfer from the outer surfaces of the computer case. The mass flow rate of air required to absorb heat at a rate of 80 W can be determined from

kg/s 007944.0C)5060)(CJ/kg. 1007(

J/s 80)(

)( =°−°

=−

=⎯→⎯−=inoutp

inoutp TTcQ

mTTcmQ&

&&&

In the worst case the exhaust fan will handle air at 60°C. Then the density of air entering the fan and the volume flow rate becomes

/minm 0.6497 3====

===

/sm 01083.0kg/m 0.7337

kg/s 007944.0

kg/m 7337.0273)K+/kg.K)(60kPa.m 287.0(

kPa 70.12

33

33

ρ

ρ

m

RTP

&&V

For an average velocity of 120 m/min, the diameter of the duct in which the fan is installed can be determined from

cm 8.3====⎯→⎯== m 083.0)m/min 120(

)/minm 6497.0(444

32

πππ

VDVDVAc

VV

&&


Documents

Chapter 13 INTERNAL FORCED CONVECTION - …mbahrami/ENSC 388/Solution manual/IntroTHT_2e...13-2 General Flow Analysis 13-1C Liquids are usually transported in circular pipes because