Solution manual heat and mass transfer a practical approach 3rd edition cengel CH08 1

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

Special Topic: Transitional Flow

8-65E A liquid mixture flowing in a tube with a bell-mouth inlet is subjected to uniform wall heat flux

The friction coefficient is to be determined

Assumptions Steady operating conditions exist

Properties The properties of the ethylene glycol-distilled water mixture are given to be Pr = 13.8, ν =

18.4×10-6 ft2/s and μb /μ s = 1.12

Analysis: For the calculation of the non-isothermal fully developed friction coefficient, it is necessary to

determine the flow regime before making any decision regarding which friction coefficient relation to use The Reynolds number at the specified location is

6425gal/min

8.448

/sft1/s

ft104.18

ft12/622.0)]

ft1011.2/(

)gal/min16.2[(

)/(

Re

3 6

2 3

0

6425

0791.0Re

0791

s

b f,

C

μμ

Repeating the calculations when the volume flow rate is increased by 50%, we obtain

9639gal/min

8.448

/sft1/s

ft104.18

ft12/622.0)]

ft1011.2/(

)gal/min16.2(5.1[)/(

Re

3 6

2 3

0

9639

0791.0Re

0791

s

b f,

C

μμ

8-66 A liquid mixture flowing in a tube with a bell-mouth inlet is subjected to uniform wall heat flux The

friction coefficient is to be determined

Assumptions Steady operating conditions exist

Properties The properties of the ethylene glycol-distilled water mixture are given to be Pr = 14.85, ν =

1.93×10-6 m2/s and μb /μ s = 1.07

Analysis: For the calculation of the non-isothermal fully developed friction coefficient, it is necessary to

determine the flow regime before making any decision regarding which friction coefficient relation to use The Reynolds number at the specified location is

/sm1093.1

m0158.0)]

m10961.1/(

)/sm1043.1[(

)/(

Re

2

6

2 4 3

From Table 8-6, we see that for a bell-mouth inlet and a heat flux of 3 kW/m2 the flow is in the transition

region Therefore, Eq 8-81 applies Reading the constants A, B, C and m1, m2, m3, and m4 from Table 8-5, the friction coefficient is determined to be

trans

,

07.15340

59731

Re1

m

s b

C B f

A C

μμ

8-67 A liquid mixture flowing in a tube with a bell-mouth inlet is subjected to uniform wall heat flux The

friction coefficient is to be determined

Assumptions Steady operating conditions exist

Properties The properties of the ethylene glycol-distilled water mixture are given to be Pr = 14.85, ν =

1.93×10-6 m2/s and μb /μ s = 1.07

Analysis: For the calculation of the non-isothermal fully developed friction coefficient, it is necessary to

determine the flow regime before making any decision regarding which friction coefficient relation to use

If the volume flow rate is increased by 50%, the Reynolds number becomes

/sm1093.1

m0158.0)]

m10961.1/(

)/sm1043.15.1[(

)/(

Re

2

6

2 4 3

0 turb

8960

0791.0Re

0791

s

b f

C

μμ

8-68 A liquid mixture flowing in a tube is subjected to uniform wall heat flux The Nusselt number at a

specified location is to be determined for two different tube inlet configurations

Assumptions Steady operating conditions exist

Properties The properties of the ethylene glycol-distilled water mixture are given to be Pr = 33.46, ν =

3.45×10-6 m2/s and μb /μ s = 2.0

Analysis For a tube with a known diameter and volume flow rate, the type of flow regime is determined

before making any decision regarding which Nusselt number correlation to use The Reynolds number at the specified location is

/sm1045.3

m0158.0)]

m10961.1/(

)/sm1005.2[(

)/(

Re

2

6

2 4 3

Therefore, the flow regime is in the transition region for all three inlet configurations (thus use the

information given in Table 8-8 with x/D = 10) and therefore Eq 8-83 should be used with the constants a,

b, c found in Table 8-7 However, Nulam and Nuturb are the inputs to Eq 8-83 and they need to be evaluated first from Eqs 8-84 and 8-85, respectively It should be mentioned that the correlations for Nulam and Nuturb

have no inlet dependency

From Eq 8-84:

(60,000)(33.46)] (2.0) 35.4[

025.010

)46.33)(

4790(24.1

GrPr025.0PrRe24.1Nu

14 0 3 / 1 75 0

14 0 3 / 1 75 0 lam

D

μμ

From Eq 8-85:

( )10 ( )2.0 85.1)

46.33()4790(023.0

PrRe023.0Nu

14 0 0054 0 385 0 8

0

14 0 0054 0 385 0 8 0 turb

x

μμ

Then the transition Nusselt number can be determined from Eq 8-83,

Discussion Comparing the two results, it can be seen that under the same conditions, the Nusselt number

for the re-entrant inlet is much higher than that for the bell-mouth inlet To verify this trend, refer to Fig

8-35

8-69 A liquid mixture flowing in a tube is subjected to uniform wall heat flux The Nusselt number at a

specified location is to be determined for two different tube inlet configurations

Assumptions Steady operating conditions exist

Properties The properties of the ethylene glycol-distilled water mixture are given to be Pr = 33.46, ν =

3.45×10-6 m2/s and μb /μ s = 2.0

Analysis For a tube with a known diameter and volume flow rate, the type of flow regime is determined

before making any decision regarding which Nusselt number correlation to use The Reynolds number at the specified location is

/sm1045.3

m0158.0)]

m10961.1/(

)/sm1005.2[(

)/(

Re

2

6

2 4 3

Therefore, the flow regime is in the transition region for all three inlet configurations (thus use the

information given in Table 8-8 with x/D = 90) and therefore Eq 8-83 should be used with the constants a,

b, c found in Table 8-7 However, Nulam and Nuturb are the inputs to Eq 8-83 and they need to be evaluated first from Eqs 8-84 and 8-85, respectively It should be mentioned that the correlations for Nulam and Nuturb

have no inlet dependency

From Eq 8-84:

(60,000)(33.46)] (2.0) 20.0[

025.090

)46.33)(

4790(24.1

GrPr025.0PrRe24.1Nu

14 0 3 / 1 75 0

14 0 3 / 1 75 0 lam

D

μμ

From Eq 8-85:

( )90 ( )2.0 84.1)

46.33()4790(023.0

PrRe023.0Nu

14 0 0054 0 385 0 8

0

14 0 0054 0 385 0 8 0 turb

x

μμ

Then the transition Nusselt number can be determined from Eq 8-83,

Discussion Comparing the two results, it can be seen that under the same conditions, the Nusselt number

for the re-entrant inlet is much higher than that for the bell-mouth inlet To verify this trend, refer to Fig

8-35

Review Problems

8-70 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

0

C W/m

607

0

kg/m997

2 3 -

CJ/kg

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

kg/s0001.050

kg/s005.0channel

)20050(4)(2

)(4

+

×

=+

×

=

W H

W H p

A

D h

898s)kg/m10891.0)(

m1020050(

)m10kg/s)(80001.0(Re

3 2

ρ

ρμ

ρ

c h c

h h

A

D m A

VD m

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

C W/m

607

mm0.021

=)2.005.0(2)01.005.0(22

A

K 297.8

0001.0(

)101.2)(

770,27(exp)290350(350)

i s s

e T T T e

Then the rate of heat transfer becomes

W82.21C)8.297350)(

CJ/kg

4180)(

kg/s0001.0()

=

=

=nchannelQonechannel 50(21.82 W)

W&e &

8-71 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-9)

4180

/sm10891

0

C W/m

607

0

kg/m997

2 3 - 3

10 kg/s

5 tubes

D = 5 cm 35ºC 60ºC

Analysis (a) The rate of heat transfer is

W 836,000

kg/s )5/10(

kg/m10891.0

m)m/s)(0.05)(1.02

kg/m(997Re

14.6()067,57(023.0PrRe023

C W/m

607

m6.4845

5

m484.6C

)2560()C W/m3684( W000,836)

(

2

2 2

avg ,

ππ

π

D

A L DL

A

A A

T T hA

8-72 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-9)

4180

/sm10891

0

C W/m

607

0

kg/m997

2 3 - 3

20 kg/s

5 tubes

D = 5 cm 35ºC 60ºC

Analysis (a) The rate of heat transfer is

W 1,672,000

kg/s )5/20(

kg/m10891.0

m)m/s)(0.05)(2.04

kg/m(997

14.6()320,114(023.0PrRe023

C W/m

607

m7.4385

5

m438.7C

)2560()C W/m6423( W000,672,1)(

2

2 2

avg ,

ππ

π

D

A L DL

A

A A

T T hA

8-73 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-9)

4180

/sm10891

0

C W/m

607

0

kg/m997

2 3 - 3

CJ/kg

4180)(

kg/s3600/1500()

kg/s )3600/1500(

kg/m10891.0

m)m/s)(0.01)(5.32

kg/m(997

14.6()542,59(023.0PrRe023

C W/m

607

A

A A

T T hA

m)01.0(m0.1142

m1142.0C

)2549()C W/m056,19( W250,52)

(

2

2 2

avg ,

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

C3.53

2

1049)1142.02)(

056,19()10)(

4180)(

3600/

1500

(

)(

b s s i e p

T

T T

T T hA T T c

m&

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

8-74 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-9) The roughness of stainless steel pipes is 2×10-6 m (Table 8-3)

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

2

2 2 2 u pump, 1

2 1

1

L

h h

z g

V g

P h

z g

V

g

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

7 3

3 avg

2

3 2

avg

10186.1s

kg/m10255.0

m)m/s)(0.60305

.5)(

kg/m6.950(Re

m/s305.54/m)(0.60

/sm1.54

21

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

1033.3m60.0

m102

−

f f

51.27

.3

1033.3log0.21 Re

51.27.3

/log0.2

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

kN/m1

kPa1m/skg1000

kN12

m/s)305.5)(

kg/m6.950(m0.60

m000,1200829.0

2 3

=

=

/smkPa1

kW10.65

)kPa2218)(

/sm(1.5

3 3

motor - pump motor

pump

-u pump,

P V W

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/day832

,122h/day)kW)(245118(

(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

→Δ

=

)CkJ/kg229.4(/s)m5.1)(

kg/m6.950(

kJ/s)(51180.65

3 3 in

elect, motor - pump elect

p

p

c

W T

T c

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

8-75 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-9) The roughness of cast iron pipes is 0.00026 m (Table 8-3)

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

2

2 2 2 u pump, 1

2 1

1

L

h h

z g

V g

P h

z g

V

g

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

7 3

3 avg

2

3 2

avg

10187.1s

kg/m10255.0

m)m/s)(0.60305

.5)(

kg/m6.950(Re

m/s305.54/m)(0.60

/sm1.54

21

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

1033.4m60.0

m00026.0

−

f f

51.27

.3

1033.4log0.21 Re

51.27.3

/log0.2

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

kPa4342kN/m

1

kPa1m/skg1000

kN12

m/s)305.5)(

kg/m6.950(m0.60

m000,1201623.0

2 3

=

=

/smkPa1

kW10.65

)kPa4342)(

/sm(1.5

3 3

motor - pump motor

pump

-u pump,

P W

&

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,

kWh/day480

,240h/day)kW)(24020,10(

(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

→Δ

=

)CkJ/kg229.4(/s)m5.1)(

kg/m6.950(

kJ/s)(10,0200.65

3 3

in elect, motor - pump elect

p

p

c

W T

T c

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

8-76 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

(

R

r V

)1001(6

)

Comparing the two relations above gives the pipe radius, the

maximum velocity, and the mean velocity to be

m 0.10

max avg

V

V

8-77E 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-9E)

Analysis The velocity profile in fully developed laminar flow in a circular pipe is

)6251(8.0

)

Comparing the two relations above gives the pipe radius,

the maximum velocity, and the mean velocity to be

ft04.0 625

ft/s0.82

lbf1

ft/slbm2.32ft)s)(140lbm/ft10

308.1(128

ft)(0.08)(/s

ft0.00201 128

2 3

4 3

W1)lbf/ft37.11)(

/sft00201.0

lbm/ft10

308.1

ft)ft/s)(0.084

.0)(

lbm/ft42.62(

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

8-78 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-15)

7323.0Pr C,J/kg

1007

/s,m10470

1

C W/m

02476.0 ,kg/m225

1

2 5 -

=)/s)/(0.938m

5

-3 3

Indoors 20°C

L = 11 m

D = 20 cm

Analysis The mean velocity of air is

m/s594.8/4m)(0.2

/sm27.02

m)m/s)(0.2(8.594

which is greater than 10,000 Therefore, the flow is turbulent and the entry lengths in this case are roughly

m2m)2.0(10

64.1Reln790

kg/m149.1

=

=Δ

2 2

3 2

avg

kg/m149.1

)N/m09.40)(

kg/s3102.0(

N/m09.402

)m/s594.8)(

kg/m149.1(m)2.0(

m)11()01718.0(2

ρ

ρ

P m W

V D

L f

7323.0()10247.1(023.0PrRe023

C W/m

02476

=+

−

=+

−

)912.6)(

10(

1)

912.6)(

00.30(1

10201

1

2 1

2 1

s

s h A A h

T T

Q&

(c) The temperature rise of air in the duct is

C 1.7°

=Δ

→Δ

°

=

→Δ

Q& & p 518.4 W (0.3102kg/s)(1007J/kg C)

8-79 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-15)

7309

0

Pr

CJ/kg

1007

/sm10516

1

C W/m

02514

0

kg/m204

1

2 5 - 3

5

avg

10958.3/sm10516.1

m)m/s)(0.2(3

Air 25°C

3 m/s

River water

L = 15 m

D = 20 cm

which is greater than 10,000 Therefore, the flow is turbulent and the entry lengths in this case are roughly

m2m)2.0(10

C W/m

02514

m)(0.09425kg/m

204.1(

=4

m)(0.2m/s))(3kg/m204.1(

m9.425

=m)m)(152.0(

3 3

2 3

ππ

c

s

A V

54 12 ( )

64.1Reln790

Pa1m/skg1

N12

m/s)3)(

kg/m204.1(m0.2

m1502212.0

2 3

W 1.54

=

=

/smPa1

W155

.0

)Pa988.8)(

/sm09425.0(

3 3

motor - pump motor

pump

-u pump,

P W

&

8-80 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

0

Pr

CJ/kg

1007

/sm10516

1

C W/m

02514

0

kg/m204

1

2 5 - 3

3 m/s

River water15°C

L = 15 m

D = 20 cm

Mineral deposit0.15 mm

Analysis The Reynolds number is

4 2

5

avg

10958.3/sm10516.1

m)m/s)(0.2(3

which is greater than 10,000 Therefore, the flow is turbulent and the entry lengths in this case are roughly

m2m)2.0(10

≈

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

C W/m

02514

m)(0.09425kg/m

204.1(

=4

m)(0.2m/s))(3kg/m204.1(

m9.425

=m)m)(152.0(

3 3

2 3

ππ

c

s

A V

3

m0025

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

C/W.m0797.0C W/m54.12

Therefore, the effect of 0.25 mm thick mineral deposit on heat transfer is negligible

Next we determine the exit temperature of air

C 18.6°

54 12 ( )

64.1Reln790

Pa1m/skg1

N12

m/s)3)(

kg/m204.1(m0.2

m1502212.0

2 3

W 1.54

=

&

8-81E 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°C 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-15)

/sft10

225

6

FBtu/h.ft

2535.0

=

°

=

p c

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,

/sft105.906

=)/s)/(1.054ft

lbm/ft03421

0

(

2 4 - 2

4 -

3 3

lbm/ft03606.0(

lbm/s0.2

2 3

avg avg

πρ

ρ

c c

A

m V

A V

&

(b) The Reynolds number is

990,40/sft10906.5

ft)12ft/s)(3.5/

(83.01Re

2 4

which is greater than 10,000 Therefore, the flow is turbulent and the entry lengths in this case are roughly

ft917.2ft)12/5.3(10

≈

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

FBtu/h.ft

0280

h

h i

2

ft7.33

=ft)ft)(812/5.3(π

Q

Q&= & = & =Δ&

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

=

800ln

F800)

ft33.7)(

F.Btu/h.ft70.9( ln

2 2

ln

s

e s e

i s

e s

i e s i s

i

T

T T

T Q

T T

T T

T T A h T A h

& :

Qexternal Q&=h o A s(T s−T o )→ Q& =(3Btu/h.ft2.°F)(7.33ft2)(T s −80)°F

Δ &Eexhaust gases: Q&=m&c p(T e−T i)→Q& =(0.2×3600lbm/h)(0.2535Btu/lbm.°F)(800−T e)°F

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

Q&=11,528Btu/h, T e =736.8°F, and T s =604.2°F

8-82 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-9)

4206

/s;

m10326

0

/

C W/m

675.0

;kg/m

3

965

2 6 - 3

2 3

m)m/s)(0.04(1.2

which is greater than 10,000 Therefore, the flow is turbulent and the entry lengths in this case are roughly

m4.0m)04.0(10

.1240,147032.0125.0PrRe125

C W/m

675

h

h i

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

=C)10)(90mC)(2.168

W/m12()

m168.2)(

7.0(

)(

4 4

4 2 8 2

4 4 0

=+

−+

Q& ε σ

W 3023

=942+2081

=+

= conv rad

total Q Q

Q& & &

(b) The temperature at which water leaves the basement is

C 89.5°

4206)(

kg/s456.1(

W3023C

90)

(

p i e e

i p

c m

Q T T T

T c

85.2)(

W3023(

C/W10

85.2m)C)(15 W/m

52(4

)4/6.4ln(

2

)/ln(

5

5 1

pipe

R Q T

kL

D D R

&

ππ

8-83 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-15)

4206

/s;

m10326

0

/

C W/m

675.0

;kg/m

3

965

2 6 - 3

2 3

m)m/s)(0.04(1.2

which is greater than 10,000 Therefore, the flow is turbulent and the entry lengths in this case are roughly

m4.0m)04.0(10

.1240,147032.0125.0PrRe125

C W/m

675

h

h i

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

=C)10)(90mC)(2.168

W/m12()

m168.2)(

7.0(

)(

4 4

4 2 8 2

4 4 0

=+

−+

Q& ε σ

W 3023

=942+2081

=+

= conv rad

total Q Q

Q& & &

(b) The temperature at which water leaves the basement is

C 89.5°

4206)(

kg/s456.1(

W3023C

90)

(

p i e e

i p

c m

Q T T T

T c

84.3)(

W3023(

C/W10

84.3m)C)(15 W/m

386(2

)4/6.4ln(

4

)/ln(

6

6 1

pipe

R Q T

kL

D D R

&

ππ

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

8-84 Integrated circuits are cooled by water flowing through a series of microscopic channels The

temperature rise of water across the microchannels and the average surface temperature of the

microchannels are to be determined

Assumptions 1 Steady flow conditions exist 2 The inner surfaces of the microchannels are smooth 3 Entrance effects are disregarded 4 Any heat transfer from the side and cover surfaces are neglected

Properties We assume the bulk mean temperature of water to be the inlet temperature of 20°C since the mean temperature of water at the inlet will rise somewhat as a result of heat gain through the microscopic channels The properties of water at 20°C are (Table A-9)

01.7Pr C;

J/kg

4182

/sm10004

1

/

C W/m

Water 20°C

Analysis (a) The mass flow rate of water is

m& =ρV&=(998kg/m3)(0.01×10-3m3/s)=0.00998kg/s

The temperature rise of water as it flows through the micro channels is

C 1.2°

=

°

=

=Δ

⎯→

⎯Δ

=

)CJ/kg4182)(

kg/s00998.0(

J/s50

p p

c m

Q T T

m10004.1

m)1057.8m/s)(

(6.667Re

m10571.8)m103.0+m1005.0(2

)m103.0)(

m1005.0(44

m/s667.6100)m103.0)(

m1005.0(

/sm1001.0

2 6

5 avg

5 3

3

3 3

3 3

3 3 avg

c

D V

P

A D

A

V V&

which is less than 2300 Therefore, the flow is laminar, and the thermal entry length in this case is

m0.0171

=m)10571.8)(

01.7)(

1.569(05.0PrRe05

1.569(m01.0

m10571.804.01

)01.7)(

1.569(m01.0

m10571.8065.066.3PrRe)/(04.01

PrRe)/(065.066

L D k

hD

Nu

m10571.8

C W/m

598

=

×

°+

C W/m445,36(

W)100/50(C

2

2.2120

)(

m10701.010)05.03.0(2

2 6 2

, ,

, ,

2 6 3

avg m avg

s

avg m avg s s s

hA

Q T

T

T T hA Q

pL A

&

8-85 Integrated circuits are cooled by air flowing through a series of microscopic channels The

temperature rise of air across the microchannels and the average surface temperature of the microchannels are to be determined

Assumptions 1 Steady flow conditions exist 2 The inner surfaces of the microchannels are smooth 3 Entrance effects are disregarded 4 Any heat transfer from the side and cover surfaces are neglected 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 60°C since the mean temperature of air at the inlet will rise somewhat as a result of heat gain through the microscopic channels whose base areas are exposed to uniform heat flux The properties of air at 1 atm and 60°C are (Table A-15)

0.7202

Pr

CJ/kg

1007

/sm10896

1

C W/m

02808

0

kg/m059

1

2 5 - 3

Air 0.5 L/s

Analysis (a) The mass flow rate of air is

kg/s10295.5)/sm10)(0.5kg/m059.1

→Δ

)CJ/kg

1007)(

kg/s10295.5(

J/s50

p p

c m

Q T T c

m10896.1

m)1057.8m/s)(

(333.3Re

m10571.8)m103.0+m1005.0(2

)m103.0)(

m1005.0(44

m/s3.333)m103.0)(

m1005.0(

/sm/100)105.0(

2 5

5 avg

5 3

3

3 3

3 3

3 3

c

D V

P

A D

A

V V&

which is smaller than 2300 Therefore, the flow is laminar and the thermal entry length in this case is

m 0.004651m)

10571.8)(

7202.0)(

1507(05.0PrRe05

1507(m01.0

m10571.804.01

)7202.0)(

1507(m01.0

m10571.8065.066.3PrRe)/(04.01

PrRe)/(065.066

L D k

hD

Nu

m10571.8

C W/m

02808

=

×

°+

C W/m1367(

W)100/50(C

2

8.11320

)(

m10701.010)05.03.0(2

2 6 2

, ,

, ,

2 6 3

s avg m avg

s

avg m avg s s s

hA

Q T

T

T T hA Q

pL A

&

&