2-62E A steam pipe is subjected to convection on the inner surface and to specified temperature on the outer surface. The mathematical formulation, the variation of temperature in the pipe, and the rate of heat loss are to be determined for steady one-dimensional heat transfer.

Assumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its thickness, and there is thermal symmetry about the center line. 2 Thermal conductivity is constant. 3 There is no heat generation in the pipe.

Properties The thermal conductivity is given to be k = 7.2 Btu/hft°F.

Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this problem can be expressed as

and kdT r

drh T T r

( )[ ( )]1

1

T r T( )2 2 160 F

(b) Integrating the differential equation once with respect to r gives

rdT

drC 1

Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating,

dT

dr

C

r 1

T r C r C( ) ln 1 2

where C1 and C2 are arbitrary constants. Applying the boundary conditions give

r = r1: kC

rh T C r C1

11 1 2[ ( ln )]

r = r2: T r C r C T( ) ln2 1 2 2 2

Solving for C C1 2 and simultaneously gives

CT Tr

r

k

hr

C T C r TT Tr

r

k

hr

r12

2

1 1

2 2 1 2 22

2

1 1

2

lnln

lnln and

Substituting C C1 2 and into the general solution, the variation of temperature is determined to be

(c) The rate of heat conduction through the pipe is

Steam250Fh=1.25

L = 15 ft

T =160F

3-19 A double-pane window consists of two 3-mm thick layers of glass separated by a 12-mm wide stagnant air space. For specified indoors and outdoors temperatures, the rate of heat loss through the window and the inner surface temperature of the window are to be determined.Assumptions 1 Heat transfer through the window is steady since the indoor and outdoor temperatures remain constant at the specified values. 2 Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivities of the glass and air are constant. 4 Heat transfer by radiation is negligible.Properties The thermal conductivity of the glass and air are given to be kglass = 0.78 W/m°C and kair = 0.026 W/m°C.Analysis The area of the window and the individual resistances are

A ( . ( .12 2 2 4 m) m) m2

The steady rate of heat transfer through window glass then becomes

The inner surface temperature of the window glass can be determined from

Air

R1 R2 R3 RoRi

T1 T2

4-36 Large brass plates are heated in an oven. The surface temperature of the plates leaving the oven is to be determined.

Assumptions 1 Heat conduction in the plate is one-dimensional since the plate is large relative to its thickness and there is thermal symmetry about the center plane. 3 The thermal properties of the plate are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The properties of brass at room temperature are given to be k = 110 W/m.C, = 33.910-6 m2/s

Analysis The Biot number for this process is

BihL

k

(80 )( . )

( ).

W / m . C m

W / m. C

2 0 015

1100 0109

The constants 1 1 and A corresponding to this Biot number are, from Table 4-1,

1 101039 10018 . . and A

The Fourier number is

t

L2

633 9 10 10

0 01590 4 0 2

( .

( .. .

m / s)( min 60 s / min)

m)

2

2

Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the temperature at the surface of the plates becomes

Discussion This problem can be solved easily using the lumped system analysis since Bi < 0.1, and thus the lumped system analysis is applicable. It gives

which is almost identical to the result obtained above.

Plates25C

Furnace, 700C

7-50E A fan is blowing air over the entire body of a person. The average temperature of the outer surface of the person is to be determined for two cases.

Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gas with constant properties. 4 The average human body can be treated as a 1-ft-diamter cylinder with an exposed surface area of 18 ft2. 5 The local atmospheric pressure is 1 atm.

Properties We assume the film temperature to be 100 F . The properties of air at this temperature are (Table A-15E)

Analysis The Reynolds number is

The proper relation for Nusselt number corresponding this Reynolds number is

The heat transfer coefficient is

Then the average temperature of the outer surface of the person becomes

If the air velocity were doubled, the Reynolds number would be

The proper relation for Nusselt number corresponding this Reynolds number is

Heat transfer coefficient is

Then the average temperature of the outer surface of the person becomes

V = 6 ft/sT = 85F

Person, Ts

300 Btu/h

D = 1 ft

8-59E 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 ofare (Table A-9E)

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

which is greater than 4000. Therefore, the flow is turbulent and the entry lengths in this case are roughly

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

and

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

The rate of heat transfer needed to raise the temperature of water from 54 F to 140 F is

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

Water54F

0.7 lbm/s 140F

250F

L

D = 0.75 in

(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

11-61 The absorber plate of a solar collector is exposed to solar and sky radiation. The net rate of solar energy absorbed by the absorber plate is to be determined.

Properties The solar absorptivity and emissivity of the surface are given to s = 0.87 and = 0.09.

Analysis The net rate of solar energy delivered by the absorber plate to the water circulating behind it can be determined from an energy balance to be

( ) ( )

q q q

q G T T h T T

net gain loss

net s solar s sky s air

4 4

Then,

Therefore, heat is gained by the plate and transferred to water at a rate of 36.5 W per m2 surface area.

12-33 Two parallel disks whose back sides are insulated are black, and are maintained at a uniform temperature. The net rate of radiation heat transfer from the disks to the environment is to be determined.Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered.Properties The emissivities of all surfaces are = 1 since they are black. Analysis Both disks possess same properties and they are black. Noting that environment can also be considered to be blackbody, we can treat this geometry as a three surface enclosure. We consider the two disks to be surfaces 1 and 2 and the environment to be surface 3. Then from Figure 12-7, we read

F F

F12 21

13

0 26

1 0 26 0 74

.

. . ( )

summation rule

The net rate of radiation heat transfer from the disks into the environment then becomes

Insulation

AirT = 25CTsky = 15C

Absorber plateTs = 70Cs = 0.87 = 0.09

600 W/m2

Disk 1, T1 = 700 K, 1 = 1

Disk 2, T2 = 700 K, 2 = 1

0.40 mEnvironment

T3 =300 K1 = 1

D = 0.6 m

13-100 Steam is condensed by cooling water in a shell-and-tube heat exchanger. The rate of heat transfer and the rate of condensation of steam are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. 5 The thickness of the tube is negligible.

Properties The specific heat of the water is given to be 4.18 kJ/kg. C. The heat of condensation of steam at 30C is given to be 2430 kJ/kg.

Analysis (a) The heat capacity rate of a fluid condensing in a heat exchanger is infinity. Therefore,

C C m Cc c pcmin . (0.5 kg / s)(4.18 kJ / kg. C) kW/ C2 09

and C = 0

Then the maximum heat transfer rate becomes ( ) .max min , ,Q C T Th in c in (2.09 kW/ C)(30 C 15 C) kW3135

and

The NTU of this heat exchanger

Then the effectiveness of this heat exchanger corresponding to C = 0 and NTU = 6.76 is determined using the proper relation in Table 13-5

1 1 6 76 1exp( ) exp( . )NTU

Then the actual heat transfer rate becomes

(b) Finally, the rate of condensation of the steam is determined from

Steam30C

15C

Water1800 kg/h

30C