THERMOFLOW MTV410 MEMO FOR APRIL 2011 TEST

QUESTION 1 [15]

a. Nusselt number is the dimensionless convection heat transfer coefficient, and it represents the enhancement of heat transfer through a fluid layer as a result of convection relative to conduction across the same fluid layer. [2]

b. The Prandtl number is the relative thickness of the velocity and thermal boundary layers. [2] c. If the Prandtl number is high then heat diffuses very slow as the thermal boundary layer is

very thin in comparison to the velocity boundary layer. [2]

d. (2.0E-3)(100)

Pr 20.1

Cp

k

, and

1/3 1/3

10.794 mm [2]

Pr 2t

e. The properties of air at 20°C and 1 atm are (Table A-15):

ρ = 1.204 kg/m3, Cp = 1.007 kJ/kg.K, Pr = 0.7309

For flat plates, the drag force is equivalent to friction force. The average friction coefficient Cf can be determined from

2

2

2

(1.204)102.4 = (4x4)

2

0.002492

f f s

f

f

VF C A

C

C

[2]

The average heat transfer coefficient can then be obtained from the modified Reynolds analogy, which is

2/3

2/3

2

Pr2

0.0024920.7309

2 (1.204)(1.007 3)(10)

18.62 W/m .C

f

p

C h

C V

h

E

h

[2] Then the rate of heat transfer becomes

( )

(18.62)(16)(80 20) 17.9 kW

s sQ hA T T

[3] QUESTION 2 [15]

Use the properties of air at 30°C (Table A-15): k = 0.02588 W/m°C, ν = 1.608E-5 m2/s, Pr = 0.72 The Reynolds number is

1000(70 )(8)

3600Re 9.674E61.608 5

VL

E

[3]

as this Reynolds number is greater than the critical Reynolds number, we have combined laminar and turbulent flow.

0.8 1/3

0.8 1/3

2

(0.037 Re 871) Pr [2]

(8)(0.027(9.674 6) 871)(0.7282)

0.02588

39.21 W/m . C [3]

L

hLNu

k

hE

h

The equilibrium temperature of the top surface is then determined by taking convection and radiation heat fluxes to be equal to each other

( ) [3]

300 39.21( 30)

35.1 C [4]

rad conv s

s

s

q q h T T

T

T

QUESTION 3 [20]

The oil properties were given as: ρ = 888.1 kg/m3, ν = 9.429E-4 m2/s, k = 0.145 W/m.°C, Cp = 1 880 J/kg.°C, and Pr = 10 863

2(0.3)

Re 636, thus the flow is laminar9.429 4

VD

E

The thermal entry length for laminar flow can be estimated from

0.05RePr (0.5)(636)(10 863)(0.3) = 103 600 m = 103.4 kmtL D

This is much greater than the total length of the pipe which is 200 m. This is typical of fluids with high Prandtl numbers. Therefore, we assume thermally developing flow and determine the Nusselt number from

2/3

2/3

0.065( / ) Re Pr3.66

1 0.04[( / ) Re Pr]

0.065(0.3 / 200)(636)(10863)3.66

1 0.04[(0.3 / 200)(636)(10863)]

37.31

hD D LNu

k D L

Therefore the average heat transfer coefficient over a distance of 200 m can be determined as

2

(0.3)37.31

0.145

18.04 W/m . C

hDNu

k

h

h

Also,

2

2

(0.3)(200) 188.6 m

(888.1) 0.3 (2) 125.6 kg/s4

s

c

A DL

m AV

(a) The exit temperature of the oil can be estimated as

(18.03)(188.6)

125.6(1880)20 (0 20)

19.71 C [2]

shA

mCp

e i s iT T T T e

e

It is not necessary to iterate the properties as the bulk temperature of 19.9°C is very close to the original selection of 20°C. The logarithmic mean temperature can be determined as

(20 0) (19.71 0)

(20 0)ln

(19.71 0)

19.86

LMTDT

(b) The heat transfer rate can be determined as

1

2

( ) (125.6)(1880)(20 19.86) 67.5 kW [2]

(18.04)(188.5)(19.86) 67.5 kW [2]

i e

s LMTD

Q mCp T T

Q hA T

Thus the heat transfer rates of the two methods are the same.

(c) The laminar flow of oil is hydrodynamically developed. Therefore, the friction factor can be determined from

64 640.1006

Re 636f

Then the pressure drop in the pipe is

2 2200 (888.1)2

(0.1006) 119 kW [2]2 0.3 2

L VP f

D

(d) and the required pumping power become

pump

(125.6)(119158)16.9 kW [2]

888.1

m PW

(e) If the pipe length is 207.2 km it would mean the first half is developing flow and for the

second half the flow will be fully developed (thus Nu = 3.66). Consider the first 103.6 km, and repeat all the above calculations:

Nu = 4.664, h = 2.254 W/m2.°C, Te = 7.87°C, Q = 2.87 MW, ∆P = 61.7 MPa, pumpW = 8.7 MW

For the second part of the pipe it is assumed that the water inlet temperature is 7.87°C and Nu = 3.66, then:

h = 1.7692.254 W/m2.°C, Te = 3.78°C, Q = 964 kW, ∆P = 61.7 MPa, pumpW = 8.7 MW

Thus for the full 207.2 kW, Te = 3.78°C, Q = 3.83 MW, ∆P = 61.7 MPa, pumpW = 17.4 MW

Iterate with better properties at the bulk temperature of 14°C for the first part of the pipe and 5.8°C for the second part. Source: Cengel 2006