8 Chapter Internal Forced Convection
Review Problems 8-61 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 = 950.6 kg/m3, = 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 P1 V12 P2 V2 z1 hpump,u z 2 hturbine h L hpump,u g 2g g 2g 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 1 Water V V 1.5 m 3 /s Vm 5.305 m/s Ac D2 / 4 (0.60 m) 2 / 4 1.5 m3/s
hL
2
D = 60 cm
Re
Vm D
(950.6 kg/m 3 )(5.305 m/s)(0.60 m) 0.255 10
3
kg/m s
1.186 10 7
L = 12 km
which is greater than 10,000. Therefore, the flow is turbulent. The relative roughness of the pipe is
/D 2 10 6 m 0.60 m 3.33 10
6
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),
1 f 2.0 log /D 3.7 2.51 Re f 1 f 2.0 log 3.33 10 3.7
6
2.51 1.187 10 7 f
It gives f = 0.00829. Then the pressure drop, the head loss, and the required power input become
P
f
2 L Vm D 2
0.00829
Wpump,u
pump- motor
12,000 m (950.6 kg/m 3 )(5.305 m/s) 2 1 kN 0.60 m 2 1000 kg m/s
V P
pump- motor
1 kPa 1 kN/m 2
5118 kW
2218 kPa
Welect
(1.5 m 3 /s)(2218 kPa ) 1 kW 0.65 1 kPa m 3 /s
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,
Amount Welect, in t (5118 kW)(24 h/day) 122,832 kWh/day
8-42
Chapter 8 Internal Forced Convection
Cost Amount Unit cost (122,832 kWh/day)($0.06/kWh) $7370/day
(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
Welect VC p T T
pump- motorW elect, in
0.65 (5118 kJ/s) (950.6 kg/m 3 )(1.5 m 3 /s)(4.229 kJ/kg C)
VC p
0.55 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.
8-43
Chapter 8 Internal Forced Convection 8-62 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 = 950.6 kg/m3, = 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 P1 V12 P2 V2 z1 hpump,u z 2 hturbine h L hpump,u g 2g g 2g 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 1 V V 1.5 m 3 /s Water Vm 5.305 m/s Ac D2 / 4 (0.60 m) 2 / 4
hL
2
D = 60 cm
Re
Vm D
(950.6 kg/m )(5.305 m/s)(0.60 m) 0.255 10
0.00026 m 0.60 m
3
3
1.5 m3/s
1.187 10 7
L = 12 km
kg/m s
which is greater than 10,000. Therefore, the flow is turbulent. The relative roughness of the pipe is
/D 4.33 10
4
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),
1 f 2.0 log /D 3.7 2.51 Re f 1 f 2.0 log 4.33 10 3.7
4
2.51 1.187 10 7 f
It gives f = 0.01623 Then the pressure drop, the head loss, and the required power input become
P f
2 L Vm D 2
0.01623
12,000 m (950.6 kg/m 3 )(5.305 m/s)2 1 kN 0.60 m 2 1000 kg m/s
1 kPa 1 kN/m 2
4341 kPa
Welect
Wpump,u
pump-motor
V P
pump-motor
(1.5 m 3 /s)(4341 kPa ) 1 kW 0.65 1 kPa m 3 /s
10,017 kW
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,
Amount Welect,in t (10,017 kW)(24 h/day) 240,429 kWh/day
Cost
Amount Unit cost
(240,429 kWh/day)($0.06/kWh)
$14,426/da y
(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
8-44
Chapter 8 Internal Forced Convection 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
Welect VC p T T
pump-motorWelect,in
0.65 (10,017 kJ/s) (950.6 kg/m 3 )(1.5 m 3 /s)(4.229 kJ/kg C)
VC p
1.08 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.
8-45
Chapter 8 Internal Forced Convection 8-63 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
V (r ) Vmax 1 r2 R2
V(r)=Vmax(1-r2/R2)
The velocity profile in this case is given by
V( r ) 6(1 100r 2 )
R r
Comparing the two relations above gives the pipe radius, the maximum velocity, and the mean velocity to be 1 R2 R 0.10 m 100 Vmax = 6 m/s
Vm Vmax 2 6 m/s 2 3 m/s
0
Vmax
8-46
Chapter 8 Internal Forced Convection 8-64E 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 = 1.039 10-3 lbm/ft s, respectively (Table A-9E).
r2 R2
V(r)=Vmax(1-r2/R2)
= 62.42 lbm/ft3 and
= 3.74 lbm/ft h
Analysis The velocity profile in fully developed laminar flow in a circular pipe is
V (r ) Vmax 1
The velocity profile in this case is given by
V (r ) 0.8(1 625r 2 )
Comparing the two relations above gives the pipe radius, the maximum velocity, and the mean velocity to be 1 R2 R 0.04 ft 625 Vmax = 0.8 ft/s
Vm Vmax 2 0.8 ft/s 2 0.4 ft/s
R r 0
Vmax
Then the volume flow rate and the pressure drop become
V
V horiz
Vm Ac
Vm ( R 2 )
(0.4 ft/s)[ (0.04 ft) 2 ] 0.00201 ft 3 /s
( P) (0.08 ft) 4 128(1.039 10
3
P D4 128 L
0.00201 ft 3 /s
32.2 lbm ft/s 2 1 lbf lbm/ft s)(80 ft)
It gives
P 5.16 lbf/ft 2 0.0358 psi
Then the useful pumping power requirement becomes
Wpump,u
V P
(0.00201 ft 3 /s)(5.16 lbf/ft 2 )
1W 0.737 lbf ft/s
0.014 W
Checking The flow was assumed to be laminar. To verify this assumption, we determine the Reynolds number:
Re Vm D (62.42 lbm/ft 3 )(0.4 ft/s)(0.08ft) 1.039 10
3
1922
lbm/ft s
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.
8-47
Chapter 8 Internal Forced Convection 8-65 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)
1.225 kg/m 3 k 0.02476 W/m. C
Cp Pr
1007 J/kg. C 0.7323
Indoors 20 C Air 10 C, 95 kPa 3 2 0.27 m /s
1.568 10 -5 m 2 /s The density and kinematic viscosity at 95 kPa are 95 kPa P 0.938 atm 101.325 kPa (1.225 kg/m 3 )(0.938) 1.149 kg/m 3
D = 20 cm
(1.568 10 -5 m 2 /s)/(0.938) = 1.673 10 -5 m /s Analysis The mean velocity of air is L = 11 m 3 V 0.27 m /s Vm 8.594 m/s Ac (0.2 m) 2 /4 Vm Dh (8.594 m/s)(0.2 m) Then Re 1.0275 10 5 1.673 10 5 m 2 /s which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly Lh Lt 10D 10(0.2 m) 2 m 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
f (0.790 ln Re 1.64) 2 0.790 ln(1.0275 10 5 ) 1.64 0.01789 The pressure drop and the compressor power required to overcome this pressure drop are m V (1.149 kg/m 3 )(0.27 m 3 /s) 0.3101 kg/s
0.2
P W pump
f
L Vm 2 D 2
(0.01789)
(11 m) (1.149 kg/m 3 )(8.594 m/s) 2 (0.2 m) 2
41.74 N/m 2
11.3 W 1.149 kg/m 3 (b) For the fully developed turbulent flow, the Nusselt number is hD Nu 0.023 Re 0.8 Pr 0.4 0.023(1.0275 105 )0.8 (0.7323)0.4 207.5 k k 0.02476 W/m. C h Nu (207.5) 25.69 W/m 2 . C and Dh 0.2 m Disregarding the thermal resistance of the duct, the rate of heat transfer to the air in the duct becomes As DL (0.2 m)(11m) 6.912 m 2 T 1 T 2 20 10 Q 497.5 W 1 1 1 1 h1 As h2 As (25.69)(6.912) (10)(6.912)
m P
(0.3101 kg/s)(41.74 N/m 2 )
(c) The temperature rise of air in the duct is Q mC p T 497.5 W (0.3101 kg/s)(1007 J/kg. C) T
T
1.6 C
8-48
Chapter 8 Internal Forced Convection 8-66 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)
1.204 kg/m 3 k Cp Pr 0.02514 W/m. C 1.516 10 -5 m 2 /s 1007 J/kg. C 0.7309
Air 25 C 3 m/s River water 15 C
D = 20 cm
Analysis The Reynolds number is
Re V m Dh (3 m/s)(0.2 m) 1.516 10
5
L = 15 m
m 2 /s
3.959 10
4
which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly
Lh Lt 10D 10(0.2 m) 2m
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
Nu hD h k 0.023 Re 0.8 Pr 0.3 0.023(3.959 10 4 ) 0.8 (0.7309) 0.3 99.75
and
h k Nu Dh 0.02514 W/m. C (99.75) 12.54 W/m 2 . C 0.2 m
Next we determine the exit temperature of air,
As m DL Vm Ac (0.2 m)(15 m) = 9.425 m 2 (1.204 kg/m3 )(3 m/s) (0.2 m)2 = 0.1135 kg/s 4
and
Te Ts (Ts Ti )e
hAs /(mC p )
15 (15 25)e
(12.54 )( 9.425) (0.1135)(1007)
18.6 C
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
f (0.790 ln Re 1.64)
2
0.790 ln( 3.959 10 4 ) 1.64
0.2
0.02212
P
f
2 L Vm D 2
0.02212
15 m (1.204 kg/m 3 )(3 m/s) 2 1N 0.2 m 2 1 kg m/s 2
1 Pa 1 N/m 2
8.992 Pa
Wfan
Wpump,u
pump-motor
V P
pump-motor
(0.1135 m 3 /s)(8.992 Pa ) 0.55
1W 1 Pa m 3 /s
1.54 W
8-49
Chapter 8 Internal Forced Convection 8-67 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-15)
1.204 kg/m 3 k Cp Pr 0.02514 W/m. C 1.516 10 -5 m 2 /s 1007 J/kg. C 0.7309
Mineral deposit 0.15 mm River water 15 C
Water 25 C 3 m/s
D = 20 cm
Analysis The Reynolds number is
Re V m Dh (3 m/s)(0.2 m) 1.516 10
5
L = 15 m
m 2 /s
3.959 10 4
which is greater than 4000. Therefore, the flow is turbulent and the entry lengths in this case are roughly
Lh
Lt
10D 10(0.2 m)
2m
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
Nu hD h k 0.023 Re 0.8 Pr 0.3 0.023(3.959 10 4 ) 0.8 (0.7309) 0.3 99.75
and
h k Nu Dh 0.02514 W/m. C (99.75) 12.54 W/m 2 . C 0.2 m
Next we determine the exit temperature of air,
As m DL Vm Ac (0.2 m)(15 m) = 9.425 m 2 (1.204 kg/m3 )(3 m/s) (0.2 m)2 = 0.1135 kg/s 4
The unit thermal resistance of the mineral deposit is
Rmineral L k
1 h
0.0015 m 3 W / m. C
1
0.0005 m2 . C / W
which is much less than (under 1%) the unit convection resistance,
Rconv 12.54 W/m 2 . C 0.0797 m 2 . C/W
Therefore, the effect of 0.15 mm thick mineral deposit on heat transfer is negligible. Next we determine the exit temperature of air,
Te Ts (Ts Ti )e
hA /(mC p )
15 (15 25)e
(12.54 )( 9.425) (0.1135)(1007)
18.6 C
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
f (0.790 ln Re 1.64)
2
0.790 ln( 3.959 10 4 ) 1.64
0.2
0.02212
8-50
Chapter 8 Internal Forced Convection
P f
2 L Vm D 2
0.02212
15 m (1.204 kg/m 3 )(3 m/s) 2 1N 0.2 m 2 1 kg m/s 2
1 Pa 1 N/m 2
8.992 Pa
Wfan
Wpump,u
pump-motor
V P
pump-motor
(0.1135 m 3 /s)(8.992 Pa ) 0.55
1W 1 Pa m 3 /s
1.54 W
8-51
Chapter 8 Internal Forced Convection 8-68E 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)
0.03422 lbm/ft 3 k 0.0280 Btu/h.ft. F 0.5902 10 -3 ft 2 /s
Cp Pr
0.2535 Btu/lbm. F 0.694
Exhaust 800 F 0.2 lbm/s
80 F
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,
L = 8 ft
(0.03422 lbm/ft 3 )(1.054) 0.03608 lbm/ft 3 (0.5902 10 -3 ft 2 /s)/(1.054) = 0.5598 10 -3 ft 2 /s
Analysis (a) The velocity of exhaust gases at the inlet of the exhaust pipe is m 0.2 lbm/s m Vm Ac Vm 82.97 ft/s Ac (0.03608 lbm/ft 3 ) (3.5/12 ft) 2 / 4 (b) The Reynolds number is Vm Dh (82.97 ft/s)(3.5/12 ft) Re 43,231 0.5598 10 3 ft 2 /s which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths this in case are roughly Lh Lt 10D 10(3.5 / 12 ft) 2.917 ft 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 hDh Nu 0.023 Re 0.8 Pr 0.3 0.023(43,231)0.8 (0.694)0.3 105.4 k k 0.03422 Btu/h.ft. F hi h Nu (105.4) 10.12 Btu/h.ft 2 . F and Dh (3.5 / 12) ft
As DL (3.5 / 12 ft)(8 ft) = 7.33 ft 2 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, Q Q Q E
internal external exhaust gases
Assuming the duct to be at an average temperature of Ts , the quantities above can be expressed as Te Ti T 800 F Q hi As Tln hi As Q (10.12 Btu/h.ft 2 . F)(7.33 ft 2 ) e Qinternal : T Te T Te ln s ln s Ts Ti Ts 800
Qexternal : E
exhaust gases :
Q Q
ho As (Ts To )
mC p (Te Ti )
Q (3 Btu/h.ft 2 . F)(7.33ft 2 )(Ts 80) F Q (0.2 3600 lbm/h)(0.2535 Btu/lbm. F)(800 Te ) F
This is a system of three equations with three unknowns whose solution is Q 11,635 Btu/h, Te 736.3 F, and Ts 609.1 F Therefore, the exhaust gases will leave the pipe at 865 F.
8-52
Chapter 8 Internal Forced Convection 8-69 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)
965.3 kg/m 3 ; / 0.326 10
-6 2
k m /s; C p
0.675 W/m. C 4206 J/kg. C
Water 90 C 0.8 m/s
10 C Di = 4 cm Do = 4.6 cm L = 15 m
Pr 1.96 Analysis (a) The mass flow rate of water is
m
AcV
(965.3 kg/m 3 )
(0.04 m) 2 (0.8 m/s) 0.9704 kg/s 4
The Reynolds number is Vm Dh (0.8 m/s)(0.04 m) Re 98,062 0.326 10 6 m 2 /s which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly Lh Lt 10D 10(0.04 m) 0.4 m 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 = 98,062 and /D = (0.026 cm)/(4 cm) = 0.0065 is determined from the Moody chart to be f = 0.034. Then the Nusselt number becomes hD h Nu 0.125 f Re Pr 1 / 3 0.125 0.034 98,062 1.961 / 3 521.6 k k 0.675 W/m. C hi h Nu (521.6) 8801 W/m 2 . C and Dh 0.04 m which is much greater than the convection heat transfer coefficient of 15 W/m 2. 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
Ao Q Qrad D0 L (0.046 m)(15 m) = 2.168 m2 (15 W/m 2 . C)(2.168 m 2 )(90 10) C = 2601 W
8
conv
ho Ao (Ts Tsurr )
A0 (Ts 4 Tsurr 4 ) (0.7)(2.168 m 2 )(5.67 10 Q Q Q 2601 + 942 = 3543 W
total conv rad
W/m 2 .K 4 ) (90 273 K) 4
(10 273 K) 4
942 W
(b) The temperature at which water leaves the basement is Q 3543 W Q mC p (Ti Te ) Te Ti 90 C mC p (0.9704 kg/s)(4206 J/kg. C)
89.1 C
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 ln( D2 / D1 ) ln( 4.6 / 4 ) R pipe 1.65 10 5 C/W 4 kL 4 (52 W/m. C)(15 m) T Q R (3543 W )(1.65 10 5 C/W ) 0.06 C
pipe total pipe
which justifies our assumption that the temperature drop across the pipe is negligible. 8-70 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.
8-53
Chapter 8 Internal Forced Convection 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)
965.3 kg/m 3 ; /
10 C
k
0.675 W/m. C 4206 J/kg. C
Water 90 C 0.8 m/s Di = 4 cm Do = 4.6 cm L = 15 m
0.326 10 -6 m 2 /s; C p
Pr 1.96 Analysis (a) The mass flow rate of water is
m
AcV
(965.3 kg/m 3 )
(0.04 m) 2 (0.8 m/s) 0.9704 kg/s 4
The Reynolds number is Vm Dh (0.8 m/s)(0.04 m) Re 98,062 0.326 10 6 m 2 /s which is greater than 4000. Therefore, the flow is turbulent and the entry lengths in this case are roughly Lh Lt 10D 10(0.04 m) 0.4 m which are much shorter than the total length of the pipe. Therefore, we can assume fully developed turbulent flow in the entire pipe. Assuming the copper pipe to be smooth, the Nusselt number is determined to be hD h Nu 0.023 Re 0.8 Pr 0.3 0.023 98,062 0.8 1.96 0.3 277.1 k k 0.675 W/m. C hi h Nu (277.1) 4676 W/m 2 . C and Dh 0.04 m which is much greater than the convection heat transfer coefficient of 15 W/m 2. 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
Ao D0 L (0.046 m)(15 m) = 2.168 m2
Qconv
Q rad
ho Ao (Ts Tsurr )
A0 (Ts 4 Tsurr 4 )
(15 W/m 2 . C)(2.168 m 2 )(90 10) C = 2601 W
(0.7)(2.168 m 2 )(5.67 10
8
W/m 2 .K 4 )[(90 273 K) 4
(10 273 K) 4 ] 942 W
Qtotal
Qconv
Qrad
2601 + 942 = 3543 W
(b) The temperature at which water leaves the basement is Q 3544 W Q mC p (Ti Te ) Te Ti 90 C Cp m (0.970 kg/s)(4206 J/kg. C)
89.1 C
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 ln( D2 / D1 ) ln( 4.6 / 4 ) R pipe 1.92 10 6 C/W 4 kL 4 (386 W/m. C)(15 m) T Q R (3543 W )(1.92 10 6 C/W ) 0.007 C
pipe total pipe
which justifies our assumption that the temperature drop across the pipe is negligible. 8-71 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.
8-54
Chapter 8 Internal Forced Convection 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 and the viscosity at the anticipated surface temperature of 25 C are (Table A-9)
998 kg/m 3 k Cp 0.598 W/m. C / 1.004 10 -6 m 2 /s 7.01
L = 1 cm
Micro-channel 0.3 mm 0.05 mm
Water 20 C
4182 J/kg. C; Pr
V
Analysis (a) The mass flow rate of water is
m (998 kg/m 3 )(0.01 10 -3 m 3 /s)
0.00998 kg/s
The temperature rise of water as it flows through the micro channels is Q 50 J/s Q mC p T T mC p (0.00998 kg/s)(4182 J/kg C) (b) The Reynolds number is
Vm Dh Re V Ac 4 Ac P Vm Dh 0.01 10 (0.05 10
3 3
1.2 C
m 3 /s
3
6.667 m/s m) 100
3 3
m)(0.3 10
3 3
4(0.05 10 2(0.05 10
m)(0.3 10 m + 0.3 10
5 6
m) m)
8.571 10 569.1
5
m
(6.667 m/s)(8.57 10 1.004 10
2
m)
m /s
5
which is less than 2300. Therefore, the flow is laminar, and the thermal entry length in this case is
Lt 0.05 Re Pr Dh 0.05(569.1)(7.01)(8.571 10 m) = 0.0171 m
which is longer than the total length of the channels. Therefore, we can assume thermally developing flow, and determine the Nusselt number from (actually, the relation below is for circular tubes)
0.065 Nu hD k 3.66 0.065( D / L) Re Pr 1 0.04 ( D / L) Re Pr
2/3
3.66 1 0.04
8.571 10 5 m (569.1)(7.01) 0.01 m 8.571 10 m (569.1)(7.01) 0.01 m
5 2/3
5.224
(5.224) 36,445 W/m 2 . C 8.571 10 5 m Then the average surface temperature of the base of the micro channels is determined to be As pL 2(0.3 0.05) 10 3 0.01 7 10 6 m 2 Q hA (T T )
and
h
k Nu Dh
0.598 W/m. C
s
s ,ave
m,ave
Ts ,ave
Tm,ave
Q hAs
20 21.2 2
C
(50 / 100) W (36,445 W/m 2 . C)(7 10
6
m2 )
22.6 C
8-55
Chapter 8 Internal Forced Convection 8-72 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)
k Cp 1.060 kg/m 3 0.02808 W/m. C 1.895 10 -5 m 2 /s 1007 J/kg. C
Micro-channel 0.3 mm 0.05 mm
Air 0.5 L/s
L = 1 cm
Pr 0.7202 Analysis (a) The mass flow rate of air is m V (1.060 kg/m 3 )(0.5 10 -3 m 3 /s)
5.298 10
4
kg/s
The temperature rise of air as it flows through the micro channels is Q 50 J/s Q mC p T T 4 mC p (5.298 10 kg/s)(1007 J/kg. C) (b) The Reynolds number is (0.5 10 3 /100) m 3 /s V Vm 333.3 m/s Ac (0.05 10 3 m)(0.3 10 3 m)
Dh Re 4 Ac P Vm Dh 4(0.05 10 2(0.05 10
3 3
93.7 C
m)(0.3 10 m + 0.3 10
5 5 2
3 3
m) m)
8.571 10
5
m
1508 1.895 10 m /s which is smaller than 2300. Therefore, the flow is laminar and the thermal entry length in this case is
(333.3 m/s)(8.57 10
m)
Lt
0.05 Re Pr Dh
0.05(1508)(0.7202)(8.571 10
5
m) 0.004653 m
which is 42% of the total length of the channels. Therefore, we can assume thermally developing flow, and determine the Nusselt number from (actually, the relation below is for circular tubes)
0.065 Nu hD k 3.66 0.065( D / L) Re Pr 1 0.04 ( D / L) Re Pr
2/3
3.66
8.571 10 5 m (1508)(0.7202) 0.01 m
2/3
4.174
8.571 10 5 m 1 0.04 (1508)(0.7202) 0.01 m
and
h
k Nu Dh
0.02808 W/m. C 8.571 10
5
(4.174) 1368 W/m 2 . C
m
3
Then the average surface temperature of the base of the micro channels becomes
As Q Ts ,ave pL 2(0.3 0.05) 10 Tm,ave ) (50 / 100) W 20 113.7 C 2 (1368 W/m 2 . C)(7 10
6
0.01 7 10
6
m2
hAs (Ts ,ave Tm,ave Q hAs
m2 )
119.1 C
8-56
Chapter 8 Internal Forced Convection 8-73 Hot exhaust gases flow through a pipe. For a specified exit temperature, the pipe length is to be determined. Assumptions 1 Steady operating conditions exist. 2 The inner surface of the pipe is smooth. 3 Air is an ideal gas with constant properties. 4 The pressure of air is 1 atm. Properties The properties of air at 1 atm and the bulk mean temperature of (450+250)/2 = 350 C are (Table A-15)
0.5664 kg/m 3 k Cp Pr 0.04721 W/m. C 5.475 10 -5 m 2 /s 1056 J/kg. C 0.6937
Exhaust gases 450 C 3.6 m/s L
5
Ts = 180 C
D = 15 cm
250 C
Analysis The Reynolds number is
Re Vm D (3.6 m/s)(0.15 m) 5.475 10 m 2 /s 9864
which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly
Lh Lt 10D 10(0.15 m) = 1.5 m
which is probably 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
Nu hD k k Nu D
Te
0.023 Re 0.8 Pr 0.3
0.023(9864) 0.8 (0.6937) 0.3
32.31
Heat transfer coefficient is
h 0.04721 W/m. C (32.31) 10.17 W/m 2 . C 0.15 m
Ti 250 450 180 250 ln 180 450
The logarithmic mean temperature difference is
Tln T Te ln s Ts Ti 148.2 C
The rate of heat loss from the exhaust gases can be expressed as
Q hAs Tln (10.17 W/m 2 . C) (0.15 m) L (148.2 C) 710.25L
where L is the length of the pipe. The rate of heat loss can also be determined from
m VAc (0.5664 kg/m 3 )(3.6 m/s) (0.15 m) 2 /4 = 0.03603 kg/s
Q
mC p T
(0.03603 kg/s)(1056 J/kg. C)(450 250) C
7612 W
Setting this equal to rate of heat transfer expression above, the pipe length is determined to be
Q 710.25 L 7612 W L 10.72 m
8-57
Chapter 8 Internal Forced Convection 8-74 Water is heated in a heat exchanger by the condensing geothermal steam. The exit temperature of water and the rate of condensation of geothermal steam are to be determined. Assumptions 1 Steady operating conditions exist. 2 The inner surfaces of the tube are smooth. 3 Air is an ideal gas with constant properties. 4 The surface temperature of the pipe is 165 C, which is the temperature at which the geothermal steam is condensing. Properties The properties of water at the anticipated mean temperature of 85 C are (Table A-9)
968.1 kg/m 3 k Cp Pr 0.673 W/m. C 4201 J/kg. C 2.08 0.333 10 h fg @ 165 C
3
Ts = 165 C
Water
kg/m.s
20 C 0.8 kg/s 4 cm Te
968.1 kg/m 3 2066.5 kJ/kg
3.44 10 -7 m 2 /s
14 m
Analysis The velocity of water and the Reynolds number are
m
Re
AVm
Vm D
0.8 kg/s
7
(968.1 kg/m 3 )
m 2 /s
(0.04 m) 2 Vm 4
Vm
0.5676 m/s
(0.5676 m/s)(0.04 m) 3.44 10
76,471
which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly
Lh
Lt
10D 10(0.04 m) = 0.4 m
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
Nu hD k 0.023 Re 0.8 Pr 0.4 0.023(76,471) 0.8 (2.08) 0.4 248.7
Heat transfer coefficient is
h k Nu D 0.673 W/m. C (248.7) 0.04 m 4185 W/m 2 . C
Next we determine the exit temperature of air,
As DL (0.04 m)(14 m) = 1.759 m 2
hAs /(mC p )
Te
Ts
(Ts Ti )e
Te Ti T Te ln s Ts Ti
165 (165 20)e
( 4185)(1.759) (0.5676)( 4201)
148.8 C
The logarithmic mean temperature difference is
Tln 148.8 20 165 148.8 ln 165 20 58.8 C
The rate of heat loss from the exhaust gases can be expressed as
Q
Q
hAs Tln
mh fg
(4185 W/m 2 . C)(1.759 m 2 )(58.8 C)
432,820 W
The rate of condensation of steam is determined from
432.820 kW m(2066.5 kJ/kg) m 0.204 kg/s
8-58
Chapter 8 Internal Forced Convection 8-75 Cold-air flows through an isothermal pipe. The pipe temperature is to be estimated. Assumptions 1 Steady operating conditions exist. 2 The inner surface of the duct is smooth. 3 Air is an ideal gas with constant properties. 4 The pressure of air is 1 atm. Properties The properties of air at 1 atm and the bulk mean temperature of (5+19) / 2 = 12 C are (Table A-15)
Ts
1.238 kg/m 3 k Cp Pr 0.02454 W/m. C 1.444 10 -5 m 2 /s 1007 J/kg. C 0.7331
20 m
Air
5C 2.5 m/s 12 cm 19 C
Analysis The rate of heat transfer to the air is
m Ac Vm (1.238 kg/m 3 ) (0.12 m) 2 (2.5 m/s) 0.03499 m/s 4 (0.03499 kg/s)(1007 J/kg. C)(19 5) C 493.1 W
Q
mC p T
Reynolds number is
Re
V D
(2.5 m/s)(0.12 m) 1.444 10
5
m 2 /s
20,775
which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly
Lh
Lt
10D 10(0.12 m) = 1.2 m
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
Nu hD k k Nu D 0.023 Re 0.8 Pr 0.4 0.023(20,775) 0.8 (0.7331) 0.4 57.79
Heat transfer coefficient is
h 0.02454 W/m. C (57.79) 11.82 W/m 2 . C 0.12 m
(11.82 W/m 2 . C) (0.12 m)(20 m) Tln
The logarithmic mean temperature difference is determined from
Q hAs Tln 493.1 W Tln 5.535 C
Then the pipe temperature is determined from the definition of the logarithmic mean temperature difference
Tln Te T ln s Ts Ti Te Ti 5.535 C 19 5 T 19 ln s Ts 5 Ts 3.8 C
8-59
Chapter 8 Internal Forced Convection 8-76 Oil is heated by saturated steam in a double-pipe heat exchanger. The tube length is to be determined. Assumptions 1 Steady operating conditions exist. 2 The surfaces of the tube are smooth. 3 Air is an ideal gas with constant properties. Properties The properties of oil at the average temperature of (10+30)/2=20 C are (Table A-13)
888 kg/m 3 k Cp Pr 0.145 W/m. C 1880 J/kg. C 2.08
Ts = 100 C
Oil
10 C 0.8 m/s 30 C 3 cm L 5 cm
Analysis The mass flow rate and the rate of heat transfer are
(0.03 m) 2 (888 kg/m 3 ) (0.8 m/s) 0.5022 kg/s 4
m
Ac Vm
Q
mC p (Te Ti )
(0.5022 kg/s)(1880 J/kg. C)(30 10) C 18,881 W
The Nusselt number is determined from Table 8-4 at Di /Do =3/5=0.6 to be Nui = 5.564. Then the heat transfer coefficient, the hydraulic diameter of annulus, and the logarithmic mean temperature difference are
hi k Nu i Dh 0.145 W/m. C (5.564) 0.02 m 40.34 W/m 2 . C
Dh Tln
Do
Di Ti Te
0.05 m 0.03 m 10 30 100 30 ln 100 10
0.02 m 79.58 C
ln
Ts Te Ts Ti
The heat transfer surface area is determined from Q 18,881 W Q hAs Tln As h Tln (40.34 W/m 2 . C)(79.58 C) Then the tube length becomes
As DL L As Di 5.881 m 2 (0.03 m 2 ) 62.4 m
5.881 m 2
8-77 .... 8-79 Design and Essay Problems
8-60
Chapter 8 Internal Forced Convection 8-79 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)
Cp R
1007 J/kg. C 0.287 kPa.m 3 /kg.K
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 Q 80 J/s Q mC p (Tout Tin ) m 0.007944 kg/s C p (Tout Tin ) (1007 J/kg. C)(60 50) C 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
P 70.12 kPa 0.7337 kg/m 3 RT (0.287 kPa.m 3 /kg.K)(60 + 273)K m 0.007944 kg/s 0.01083 m 3 /s 0.6497 m 3 /min 0.7337 kg/m 3
V
For an average velocity of 120 m/min, the diameter of the duct in which the fan is installed can be determined from
V AcV D2 V 4 D 4V V 4(0.6497 m 3 /min ) (120 m/min ) 0.083 m
Cooling air
8.3 cm
8-61