7 Chapter External Forced Convection Special Topic: Thermal Insulation 7-73C Thermal insulation is a material that is used primarily to provide resistance to heat flow. It differs from other kinds of insulators in that the purpose of an electrical insulator is to halt the flow of electric current, and the purpose of a sound insulator is to slow down the propagation of sound waves. 7-74C In cold surfaces such as chilled water lines, refrigerated trucks, and air conditioning ducts, insulation saves energy since the source of "coldness" is refrigeration that requires energy input. In this case heat is transferred from the surroundings to the cold surfaces, and the refrigeration unit must now work harder and longer to make up for this heat gain and thus it must consume more electrical energy. 7-75C The R-value of insulation is the thermal resistance of the insulating material per unit surface area. For flat insulation the R-value is obtained by simply dividing the thickness of the insulation by its thermal conductivity. That is, R-value = L/k. Doubling the thickness L doubles the R-value of flat insulation. 7-76C The R-value of an insulation represents the thermal resistance of insulation per unit surface area (or per unit length in the case of pipe insulation). 7-77C Superinsulations are obtained by using layers of highly reflective sheets separated by glass fibers in an evacuated space. Radiation between two surfaces is inversely proportional to the number of sheets used and thus heat loss by radiation will be very low by using this highly reflective sheets. Evacuating the space between the layers forms a vacuum which minimize conduction or convection through the air space. 7-78C Yes, hair or any other cover reduces heat loss from the head, and thus serves as insulation for the head. The insulating ability of hair or feathers is most visible in birds and hairy animals.
7-79C The primary reasons for insulating are energy conservation, personnel protection and comfort, maintaining process temperature, reducing temperature variation and fluctuations, condensation and corrosion prevention, fire protection, freezing protection, and reducing noise and vibration. 7-80C The optimum thickness of insulation is the thickness that corresponds to a minimum combined cost of insulation and heat lost. The cost of insulation increases roughly linearly with thickness while the cost of heat lost decreases exponentially. The total cost, which is the sum of the two, decreases first, reaches a minimum, and then increases. The thickness that corresponds to the minimum total cost is the optimum thickness of insulation, and this is the recommended thickness of insulation to be installed.
7-63
Chapter 7 External Forced Convection 7-81 The thickness of flat R-8 insulation in SI units is to be determined when the thermal conductivity of the material is known. Assumptions Thermal properties are constant. Properties The thermal conductivity of the insulating material is given to be k = 0.04 W/m C. Analysis The thickness of flat R-8 insulation (in m2. C/W) is determined from the definition of R-value to be L Rvalue L Rvalue k (8 m2 . C / W)(0.04 W / m. C) 0.32 m k
R-8
L
7-82E The thickness of flat R-20 insulation in English units is to be determined when the thermal conductivity of the material is known. Assumptions Thermal properties are constant. Properties The thermal conductivity of the insulating material is given to be k = 0.02 Btu/h ft F. Analysis The thickness of flat R-20 insulation (in h ft2 F/Btu) is determined from the definition of R-value to be L R value L R value k (20 h.ft 2 . F/Btu)(0.02 Btu/h.ft. F) 0.4 ft k
R-20
L
7-64
Chapter 7 External Forced Convection 7-83 A steam pipe is to be covered with enough insulation to reduce the exposed surface temperature to 30 C . The thickness of insulation that needs to be installed is to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal properties are constant. 4 The thermal contact resistance at the interface is negligible. Properties The thermal conductivities are given to be k = 52 W/m C for cast iron pipe and k = 0.038 W/m C for fiberglass insulation. Analysis The thermal resistance network for this problem involves 4 resistances in series. The inner radius of the pipe is r1 = 2.0 cm and the outer radius of the pipe and thus the inner radius of insulation is r2 = 2.3 cm. Letting r3 represent the outer radius of insulation, the areas of the surfaces exposed to convection for a L = 1 m long section of the pipe become
A1 A3 2 r1 L 2 r3 L 2 (0.02 m)(1 m) = 0.1257 m2 2 r3 (1 m) = 2 r3 m 2 (r3 in m)
Ri Ti T1
Rpipe
Rinsulation T2 T3
Ro To
Then the individual thermal resistances are determined to be
Ri R1
R2 Rinsulation Ro
Rconv,1 Rpipe
1 hi A1 ln( r2 / r1 ) 2 k1 L
1 (80 W/m 2 . C)(0.1257 m 2 ) ln( 0.023 / 0.02) 2 (52 W/m. C)(1 m)
0.9944 C/W
0.00043 C/W
ln(r3 / r2 ) 2 k2 L Rconv,2 1 ho A3
ln(r3 / 0.023) 2 (0.038 W / m. C)(1 m) 1 (22 W / m . C)(2 r3 m )
2 2
4.188 ln(r3 / 0.023) C / W 1 138.2r3 C/W
Noting that all resistances are in series, the total resistance is
Rtotal Ri R1 R2 Ro 0.09944 0.00043 4.188 ln( r3 / 0.023) 1/(138.2r3 ) C/W
Then the steady rate of heat loss from the steam becomes
Q Ti To R total (110 22) C [0.09944 0.00043 4.188 ln( r3 / 0.023) 1 /(138.2r3 )] C/W
Noting that the outer surface temperature of insulation is specified to be 30 C, the rate of heat loss can also be expressed as
Q T3 To Ro (30 22) C 1/ (138.2r3 ) C / W 1106r3
Setting the two relations above equal to each other and solving for r3 gives r3 = 0.0362 m. Then the minimum thickness of fiberglass insulation required is t = r3 - r2 = 0.0362 - 0.0230 = 0.0132 m = 1.32 cm Therefore, insulating the pipe with at least 1.32 cm thick fiberglass insulation will ensure that the outer surface temperature of the pipe will be at 30 C or below.
7-65
Chapter 7 External Forced Convection 7-84 "!PROBLEM 7-84" "GIVEN" T_i=110 "[C]" T_o=22 "[C]" k_pipe=52 "[W/m-C]" r_1=0.02 "[m]" t_pipe=0.003 "[m]" "T_s_max=30 [C], parameter to be varied" h_i=80 "[W/m^2-C]" h_o=22 "[W/m^2-C]" k_ins=0.038 "[W/m-C]" "ANALYSIS" L=1 "[m], 1 m long section of the pipe is considered" A_i=2*pi*r_1*L A_o=2*pi*r_3*L r_3=r_2+t_ins*Convert(cm, m) "t_ins is in cm" r_2=r_1+t_pipe R_conv_i=1/(h_i*A_i) R_pipe=ln(r_2/r_1)/(2*pi*k_pipe*L) R_ins=ln(r_3/r_2)/(2*pi*k_ins*L) R_conv_o=1/(h_o*A_o) R_total=R_conv_i+R_pipe+R_ins+R_conv_o Q_dot=(T_i-T_o)/R_total Q_dot=(T_s_max-T_o)/R_conv_o
Ts, max [C] 24 26 28 30 32 34 36 38 40 42 44 46 48
tins [cm] 4.45 2.489 1.733 1.319 1.055 0.871 0.7342 0.6285 0.5441 0.4751 0.4176 0.3688 0.327
7-66
Chapter 7 External Forced Convection
7-67
Chapter 7 External Forced Convection 7-85 A cylindrical oven is to be insulated to reduce heat losses. The optimum thickness of insulation and the amount of money saved per year are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the insulation is onedimensional. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is negligible. 5 The surface temperature of the furnace and the heat transfer coefficient remain constant. 6 The surfaces of the cylindrical oven can be treated as plain surfaces since its diameter is greater than 1 m. Properties The thermal conductivity of insulation is given to be k = 0.038 W/m C. Analysis We treat the surfaces of this cylindrical furnace as plain surfaces since its diameter is greater than 1 m, and disregard the curvature effects. The exposed surface area of the furnace is
Ao 2 Abase Aside 2 r2 2 rL 2 (1.5 m) 2 2 (1.5 m)(6 m) 70.69 m 2
The rate of heat loss from the furnace before the insulation is installed is
Q ho Ao (Ts T ) (30 W/m 2 . C)(70.69 m 2 )(90 27) C = 133,600 W
Noting that the plant operates 52 80 = 4160 h/yr, the annual heat lost from the furnace is
Q Q t (133.6 kJ / s)(4160 3600 s / yr) = 2.001 109 kJ / yr
Rinsulation Ts
Ro T
The efficiency of the furnace is given to be 78 percent. Therefore, to generate this much heat, the furnace must consume energy (in the form of natural gas) at a rate of
Qin Q/
oven
(2.001 109 kJ / yr) / 0.78 = 2.565 109 kJ / yr = 24,314 therms / yr
since 1 therm = 105,500 kJ. Then the annual fuel cost of this furnace before insulation becomes
Annual Cost Q in Unit cost (24,314 therm/ yr)($0.50/ therm) = $12,157 / yr
We expect the surface temperature of the furnace to increase, and the heat transfer coefficient to decrease somewhat when insulation is installed. We assume these two effects to counteract each other. Then the rate of heat loss for 1-cm thick insulation becomes
Qins Ts T R total Ts Rins T Rconv Ao (Ts T ) t ins 1 k ins ho (70.69 m 2 )(90 27) C 0.01 m 1 0.038 W/m. C 30 W/m 2 . C 15,021 W
Also, the total amount of heat loss from the furnace per year and the amount and cost of energy consumption of the furnace become
Qins Qin,ins Annual Cost Qins t Qins / Q in,ins (15.021 kJ / s)(4160 3600 s / yr) = 2.249 108 kJ / yr
oven
(2.249 108 kJ / yr) / 0.78 = 2.884 108 kJ / yr = 2734 therms (2734 therm / yr)($0.50 / therm) = $1367 / yr
Unit cost
Cost savings = Energy cost w/o insulation - Energy cost w/insulation = 12,157 - 1367 = $10,790/yr The unit cost of insulation is given to be $10/m 2 per cm thickness, plus $30/m2 for labor. Then the total cost of insulation becomes
Insulation Cost
(Unit cost)(Surface area) = [($10 / cm)(1 cm) +$30 / m2 ](70.69 m2 ) = $2828
7-68
Chapter 7 External Forced Convection
To determine the thickness of insulation whose cost is equal to annual energy savings, we repeat the calculations above for 2, 3, . . . . 15 cm thick insulations, and list the results in the table below. Insulation thickness 0 cm 1 cm 5 cm 10 cm 11 cm 12 cm 13 cm 14 cm 15 cm Rate of heat loss W 133,600 15,021 3301 1671 1521 1396 1289 1198 1119 Cost of heat lost $/yr 12,157 1367 300 152 138 127 117 109 102 Cost savings $/yr 0 10,790 11,850 12,005 12,019 12,030 12,040 12,048 12,055 Insulation cost $ 0 2828 3535 9189 9897 10,604 11,310 12,017 12,724
Therefore, the thickest insulation that will pay for itself in one year is the one whose thickness is 14 cm.
7-69
Chapter 7 External Forced Convection 7-86 A cylindrical oven is to be insulated to reduce heat losses. The optimum thickness of insulation and the amount of money saved per year are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the insulation is onedimensional. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is negligible. 5 The surface temperature of the furnace and the heat transfer coefficient remain constant. 6 The surfaces of the cylindrical oven can be treated as plain surfaces since its diameter is greater than 1 m. Properties The thermal conductivity of insulation is given to be k = 0.038 W/m C. Analysis We treat the surfaces of this cylindrical furnace as plain surfaces since its diameter is greater than 1 m, and disregard the curvature effects. The exposed surface area of the furnace is
Ao 2 Abase Aside 2 r2 2 rL 2 (1.5 m) 2 2 (1.5 m)(6 m) 70.69 m 2
The rate of heat loss from the furnace before the insulation is installed is
Q ho Ao (Ts T ) (30 W/m 2 . C)(70.69 m 2 )(75 27) C = 101,794 W
Noting that the plant operates 52 80 = 4160 h/yr, the annual heat lost from the furnace is
Q Q t (101794 kJ / s)(4160 3600 s / yr) = 1.524 109 kJ / yr .
Rinsulation Ts
Ro T
The efficiency of the furnace is given to be 78 percent. Therefore, to generate this much heat, the furnace must consume energy (in the form of natural gas) at a rate of
Qin Q/
oven
(1524 109 kJ / yr) / 0.78 = 1.954 109 kJ / yr = 18,526 therms / yr .
since 1 therm = 105,500 kJ. Then the annual fuel cost of this furnace before insulation becomes
Annual Cost Q in Unit cost (18,526 therm/ yr)($0.50/ therm) = $9,263/ yr
We expect the surface temperature of the furnace to increase, and the heat transfer coefficient to decrease somewhat when insulation is installed. We assume these two effects to counteract each other. Then the rate of heat loss for 1-cm thick insulation becomes
Qins Ts T R total Ts Rins T Rconv Ao (Ts T ) t ins 1 k ins ho (70.69 m 2 )(75 27) C 0.01 m 1 0.038 W/m. C 30 W/m 2 . C 11,445 W
Also, the total amount of heat loss from the furnace per year and the amount and cost of energy consumption of the furnace become
Qins Qin,ins Annual Cost Qins t Qins / Qin,ins (11445 kJ / s)(4160 3600 s / yr) = 1.714 108 kJ / yr .
oven
(1714 108 kJ / yr) / 0.78 = 2.197 108 kJ / yr = 2082 therms . (2082 therm / yr)($0.50 / therm) = $1041/ yr
Unit cost
Cost savings = Energy cost w/o insulation - Energy cost w/insulation = 9263 - 1041 = $8222/yr The unit cost of insulation is given to be $10/m 2 per cm thickness, plus $30/m2 for labor. Then the total cost of insulation becomes
Insulation Cost
(Unit cost)(Surface area) = [($10 / cm)(1 cm) +$30 / m2 ](70.69 m2 ) = $2828
7-70
Chapter 7 External Forced Convection
To determine the thickness of insulation whose cost is equal to annual energy savings, we repeat the calculations above for 2, 3, . . . . 15 cm thick insulations, and list the results in the table below. Insulation Rate of heat loss Cost of heat lost Cost savings Insulation cost Thickness W $/yr $/yr $ 0 cm 101,794 9263 0 0 1 cm 11,445 1041 8222 2828 5 cm 2515 228 9035 3535 9 cm 1413 129 9134 8483 10 cm 1273 116 9147 9189 11 cm 1159 105 9158 9897 12 cm 1064 97 9166 10,604 Therefore, the thickest insulation that will pay for itself in one year is the one whose thickness is 9 cm. The 10-cm thick insulation will come very close to paying for itself in one year.
7-71
Chapter 7 External Forced Convection 7-87E Steam is flowing through an insulated steel pipe, and it is proposed to add another 1-in thick layer of fiberglass insulation on top of the existing one to reduce the heat losses further and to save energy and money. It is to be determined if the new insulation will pay for itself within 2 years. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal properties are constant. 4 The heat transfer coefficients remain constant. 5 The thermal contact resistance at the interface is negligible. Properties The thermal conductivities are given to be k = 8.7 Btu/h ft F for steel pipe and k = 0.020 Btu/h ft F for fiberglass insulation. Analysis The inner radius of the pipe is r1 = 1.75 in, the outer radius of the pipe is r2 = 2 in, and the outer radii of the existing and proposed insulation layers are r3 = 3 in and 4 in, respectively. Considering a unit pipe length of L = 1 ft, the individual thermal resistances are determined to be
Ri
R1
Rconv,1
Rpipe
1 hi A1
1 hi (2 r1 L)
1 (30 Btu / h.ft . F)[2 (1.75 / 12 ft)(1 ft)]
0.00244 h. F / Btu
2
0.0364 h. F / Btu
ln(r2 / r1 ) 2 k1 L
ln(2 / 175) . 2 (8.7 Btu / h.ft. F)(1 ft )
Ri Ti Current Case:
Rinsulation ln(r3 / r2 ) 2 k ins L ln(3 / 2) 2 (0.020 Btu / h.ft. F)(1 ft )
Rpipe T1
Rinsulation T2 T3
Ro To
3.227 h. F / Btu
Ro
Rconv,2
1 ho A3
1 ho (2 r3 )
1 (5 Btu / h.ft . F)[2 (3 / 12 ft)(1 ft )]
2
01273 h. F / Btu .
Then the steady rate of heat loss from the steam becomes
Qcurrent T Rtotal Ri Ti Rpipe To Rins Ro (400 60) F (0.0364 0.00244 3.227 01273) h. F / Btu . 100.2 Btu / h
Proposed Case:
Rinsulation ln(r3 / r2 ) 2 kins L ln(4 / 2) 2 (0.020 Btu / h.ft. F)(1 ft ) 5516 h. F / Btu .
Ro
Rconv,2
1 ho A3
1 ho (2 r3 )
1 (5 Btu / h.ft . F)[2 (4 / 12 ft)(1 ft )]
2
0.0955 h. F / Btu
Then the steady rate of heat loss from the steam becomes
Qprop T Rtotal Ri Ti Rpipe To Rins Ro (400 60) F (0.0364 0.00244 5516 0.0955) h. F / Btu . 60.2 Btu / h
Therefore, the amount of energy and money saved by the additional insulation per year are
Qsaved Qsaved Qprop Qcurrent 100.2 60.2 40.0 Btu/h 350,400 Btu/yr $3.504 / yr
Qsaved t
(40.0 Btu/h )(8760 h/yr)
Money saved = Qsaved ( Unit cost)
(350,400 Btu/yr )($0.01 / 1000 Btu )
or $7.01 per 2 years, which is barely more than the $7.0 minimum required. But the criteria is satisfied, and the proposed additional insulation is justified.
7-72
Chapter 7 External Forced Convection 7-88 The plumbing system of a plant involves some section of a plastic pipe exposed to the ambient air. The pipe is to be insulated with adequate fiber glass insulation to prevent freezing of water in the pipe. The thickness of insulation that will protect the water from freezing under worst conditions is to be determined. Assumptions 1 Heat transfer is transient, but can be treated as steady at average conditions. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal properties are constant. 4 The water in the pipe is stationary, and its initial temperature is 15 C. 5 The thermal contact resistance at the interface is negligible. 6 The convection resistance inside the pipe is negligible. Properties The thermal conductivities are given to be k = 0.16 W/m C for plastic pipe and k = 0.035 W/m C for fiberglass insulation. The density and specific heat of water are to be = 1000 kg/m3 and Cp = 4.18 kJ/kg. C (Table A-15). Analysis The inner radius of the pipe is r1 = 3.0 cm and the outer radius of the pipe and thus the inner radius of insulation is r2 = 3.3 cm. We let r3 represent the outer radius of insulation. Considering a 1-m section of the pipe, the amount of heat that be must transferred from the water as it cools from 15 to 0 C is determined to be
m Qtotal
V ( r12 L) (1000 kg/m 3 )[ (0.03 m) 2 (1 m)] 2.827 kg mC p T (2.827 kg)(4.18 kJ/kg. C)(15 - 0) C = 177.3 kJ
Ri
0.821 W
Then the average rate of heat transfer during 60 h becomes
Qave Qtotal t 177,300 J (60 3600 s)
0 T1
Rpipe
Rinsulation T2 T3
Ro To
Ti
The individual thermal resistances are
R1 Rpipe
ln(r3 / r2 ) 2 k2 L Ro Rconv
ln(r2 / r1 ) 2 k pipe L
ln(0.033 / 0.03) 2 (0.16 W / m. C)(1 m)
0.0948 C / W
Rinsulation
ln(r3 / 0.033) 2 (0.035 W / m. C)(1 m) 1 ho A3 1
2
4.55 ln(r3 / 0.033) C / W 1 188.5r3 C/W
(30 W / m . C)(2 r3 m )
2
Then the rate of average heat transfer from the water can be expressed as
Q Ti ,ave Rtotal To 0.821 W [7.5 ( 10)] C [0.0948 4.55 ln(r3 / 0.033) 1 / (188.5r3 )] C / W r3 350 m .
Therefore, the minimum thickness of fiberglass needed to protect the pipe from freezing is t = r3 - r2 = 3.50 - 0.033 = 3.467 m which is too large. Installing such a thick insulation is not practical, however, and thus other freeze protection methods should be considered.
7-73
Chapter 7 External Forced Convection 7-89 The plumbing system of a plant involves some section of a plastic pipe exposed to the ambient air. The pipe is to be insulated with adequate fiber glass insulation to prevent freezing of water in the pipe. The thickness of insulation that will protect the water from freezing more than 20% under worst conditions is to be determined. Assumptions 1 Heat transfer is transient, but can be treated as steady at average conditions. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal properties are constant. 4 The water in the pipe is stationary, and its initial temperature is 15 C. 5 The thermal contact resistance at the interface is negligible. 6 The convection resistance inside the pipe is negligible. Properties The thermal conductivities are given to be k = 0.16 W/m C for plastic pipe and k = 0.035 W/m C for fiberglass insulation. The density and specific heat of water are to be = 1000 kg/m3 and Cp = 4.18 kJ/kg. C (Table A-15). Analysis The inner radius of the pipe is r1 = 3.0 cm and the outer radius of the pipe and thus the inner radius of insulation is r2 = 3.3 cm. We let r3 represent the outer radius of insulation. The latent heat of freezing of water is 33.7 kJ/kg. Considering a 1-m section of the pipe, the amount of heat that must be transferred from the water as it cools from 15 to 0 C is determined to be
m Q total Qfreezing Q total V ( r1 2 L) (1000 kg/m 3 )[ (0.03 m) 2 (1 m)] 2.827 kg mC p T (2.827 kg)(4.18 kJ/kg. C)(15 - 0) C = 177.3 kJ 0.2 mhif Qcooling 0.2 (2.827 kg)(333.7 kJ/kg) 188.7 kJ Qfreezing 177.3 188.7 366.0 kJ
Then the average rate of heat transfer during 60 h becomes
Qave Qtotal t 366,000 J (60 3600 s)
Ri
1694 W .
0 T1
Rpipe
Rinsulation T2 T3
Ro To
Ti
The individual thermal resistances are
R1 Rpipe
ln(r3 / r2 ) 2 k2 L Ro Rconv
ln(r2 / r1 ) 2 k pipe L
ln(0.033 / 0.03) 2 (0.16 W / m. C)(1 m)
0.0948 C / W
Rinsulation
ln(r3 / 0.033) 2 (0.035 W / m. C)(1 m) 1 ho A3 1
2
4.55 ln(r3 / 0.033) C / W 1 188.5r3 C/W
(30 W / m . C)(2 r3 m )
2
Then the rate of average heat transfer from the water can be expressed as
Q Ti ,ave Rtotal To 1.694 W [7.5 ( 10)] C [0.0948 4.55 ln(r3 / 0.033) 1 / (188.5r3 )] C / W r3 0.312 m
Therefore, the minimum thickness of fiberglass needed to protect the pipe from freezing is t = r3 - r2 = 0.312 - 0.033 = 0.279 m which is too large. Installing such a thick insulation is not practical, however, and thus other freeze protection methods should be considered.
7-74
Chapter 7 External Forced Convection Review Problems 7-90 Wind is blowing parallel to the walls of a house. The rate of heat loss from the wall is to be determined. Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5 105. 3 Radiation effects are negligible. 4 Air is an ideal gas with constant properties. 5 The pressure of air is 1 atm. Properties Assuming a film temperature of Tf = 10 C for the outdoors, the properties of air are evaluated to be (Table A-15) k 0.02439 W/m. C
1.426 10 -5 m 2 /s Pr 0.7336
L=8m T Air V = 50 km/h T 2=4 C WALL
1
= 22 C
Analysis Air flows along 8-m side. The Reynolds number in this case is
Re L V L (50 1000 / 3600) m/s (8 m) 1.426 10
5
m 2 /s
7.792 10 6
which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow. Using the proper relation for Nusselt number, heat transfer coefficient is determined to be
Nu ho h0 L (0.037 Re L 0.8 871) Pr1/ 3 0.037(7.792 10 6 ) 0.8 k k 0.02439 W/m. C Nu (10,096) 30.78 W/m 2 . C L 8m
(3 m)(8 m) = 24 m 2
1 hi As 1 (8 W/m . C)(24 m 2 )
2
871 (0.7336)1/ 3
10,096
The thermal resistances are
As wL
Ri Rinsulation Ro
Ri T
1
Rinsulation
Ro T
2
0.0052 C/W
( R 3.38) value 3.38 m 2 . C/W 0.1408 C/W As 24 m 2 1 1 0.0014 C/W 2 ho As (30.78 W/m . C)(24 m 2 )
Rinsulation
1 T Rtotal 2
Then the total thermal resistance and the heat transfer rate through the wall are determined from
Rtotal Q Ri T Ro 0.0052 0.1408 0.0014 122.1 W 0.1474 C/W (22 4) C 0.1474 C/W
7-75
Chapter 7 External Forced Convection 7-91 A car travels at a velocity of 60 km/h. The rate of heat transfer from the bottom surface of the hot automotive engine block is to be determined for two cases. Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5 105. 3 Air is an ideal gas with constant properties. 4 The pressure of air is 1 atm. 5 The flow is turbulent over the entire surface because of the constant agitation of the engine block. 6 The bottom surface of the engine is a flat surface. Properties The properties of air at 1 atm and the film temperature of (Ts + T )/2 = (75+5)/2 = 40 C are (Table A-15)
k 0.02662 W/m. C 1.702 10 -5 m 2 /s Pr 0.7255
L = 0.7 m
Engine block
Air V = 60 km/h T =5 C
5 2
Analysis The Reynolds number is
Re L V L (60 1000 / 3600) m/s (0.7 m) 1.702 10 m /s 6.855 10 5
Ts = 75 C = 0.92
Ts = 10 C
which is less than the critical Reynolds number. But we will assume turbulent flow because of the constant agitation of the engine block.
Nu h hL 0.037 Re L 0.8 Pr 1 / 3 0.037(6.855 10 5 ) 0.8 (0.7255)1 / 3 k k 0.02662 W/m. C Nu (1551) 58.97 W/m 2 . C L 0.7 m 1551
Qconv
hAs (T
Ts )
(58.97 W/m 2 . C) (0.6 m)(0.7 m) (75 - 5) C = 1734 W
The heat loss by radiation is then determined from Stefan-Boltzman law to be Q A (T 4 T 4 )
rad s s surr
(0.92)(0.6 m)(0.7 m)(5.67 10 -8 W/m 2 .K 4 ) (75 + 273 K) 4
Qtotal Qconv Qrad
(10 + 273 K) 4
181 W
Then the total rate of heat loss from the bottom surface of the engine block becomes
1734 181 1915 W
The gunk will introduce an additional resistance to heat dissipation from the engine. The total heat transfer rate in this case can be calculated from
Q T 1 hAs Ts L kAs (75 - 5) C 1 (58.97 W/m 2 . C)[(0.6 m)(0.7 m)] (0.002 m ) (3 W/m. C)(0.6 m 0.7 m) = 1668 W
The decrease in the heat transfer rate is 1734-1668 = 66 W
7-76
Chapter 7 External Forced Convection 7-92E A minivan is traveling at 60 mph. The rate of heat transfer to the van is to be determined. Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5 105. 3 Radiation effects are negligible. 4 Air flow is turbulent because of the intense vibrations involved. 5 Air is an ideal gas with constant properties. 5 The pressure of air is 1 atm. Properties Assuming a film temperature of Tf = 80 F, the properties of air are evaluated to be (Table A-15E)
k Pr 0.01481 Btu/h.ft. F 0.1697 10 ft /s 0.7290
L = 11 ft
-3 2
Air V = 60 mph T = 50 F
Minivan
Analysis Air flows along 11 ft long side. The Reynolds number in this case is
Re L V L [(60 5280 / 3600) ft/s](11 ft) 0.1697 10
3
ft 2 /s
5.704 10 6
which is greater than the critical Reynolds number. The air flow is assumed to be entirely turbulent because of the intense vibrations involved. Then the Nusselt number and the heat transfer coefficient are determined to be
ho L 0.037 Re L 0.8 Pr1 / 3 0.037(5.704 10 6 ) 0.8 (0.7290)1 / 3 8461 k k 0.01481 Btu/h.ft. F ho Nu (8461) 11.39 Btu/h.ft 2 . F Ri L 11 ft T 1 The thermal resistances are Nu
As 2 (3.2 ft)(6 ft) + (3.2 ft)(11ft) + (6 ft)(11ft)
Ri Rinsulation Ro 1 hi As 1 (1.2 Btu/h.ft . F)(240.8 ft 2 )
2
Rinsulation
Ro T
2
240.8 ft 2
0.0035 h. F/Btu
( R 3) value 3 h.ft 2 . F/Btu 0.0125 h. F/Btu As (240.8 ft 2 ) 1 1 0.0004 h. F/Btu ho As (11.39 Btu/h.ft 2 . F)(240.8 ft 2 )
Rinsulation
1 T Rtotal 2
Then the total thermal resistance and the heat transfer rate into the minivan are determined to be
Rtotal Q Ri T Ro 0.0035 0.0125 0.0004 1220 Btu/h 0.0164 h. F/Btu (90 70) F 0.0164 h. F/Btu
7-77
Chapter 7 External Forced Convection 7-93 Wind is blowing parallel to the walls of a house with windows. The rate of heat loss through the window is to be determined. Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5 105. 3 Radiation effects are negligible. 4 Air flow is turbulent because of the intense vibrations involved. 5 The minivan is modeled as a rectangular box. 6 Air is an ideal gas with constant properties. 6 The pressure of air is 1 atm. Properties Assuming a film temperature of 5 C, the properties of air at 1 atm and this temperature are evaluated to be (Table A-15)
k Pr 0.02401 W/m. C 1.382 10 0.7350
-5
T Air V = 60 km/h T 2 = -2 C
1
= 22 C
WINDOW
m /s
L = 1.2 m
2
Analysis Air flows along 1.2 m side. The Reynolds number in this case is
Re L V L (60 1000 / 3600) m/s (1.2 m) 1.382 10
5
m 2 /s
1.447 10 6
which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow. Using the proper relation for Nusselt number, heat transfer coefficient is determined to be
Nu h hL (0.037 Re L 0.8 871) Pr 1 / 3 0.037(1.447 10 6 ) 0.8 871 (0.7350) 1 / 3 k k 0.02401 W/m. C Nu (2046 ) 40.93 W/m 2 . C L 1.2 m 2046
The thermal resistances are
As 3(1.2 m)(1.5 m) = 5.4 m 2
1 1 0.0231 C/W 2 hi As (8 W/m . C)(5.4 m 2 ) L 0.005 m 0.0012 C/W kAs (0.78 W/m. C)(5.4 m 2 ) 1 1 0.0045 C/W 2 ho As (40.93 W/m . C)(5.4 m 2 )
Rconv ,i T
1
Rconv ,i Rcond Rconv ,o
Then the total thermal resistance and the heat transfer rate through the 3 windows become
Rtotal Q Rcond
2
Rconv ,o
0.0231 0.0012 0.0045 833.3 W
0.0288 C/W
T
Rtotal
[22 ( 2)] C 0.0288 C/W
7-78
Chapter 7 External Forced Convection 7-94 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. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The pressure of air is 1 atm. 4 The average human body can be treated as a 30-cm-diameter cylinder with an exposed surface area of 1.7 m2. Properties We assume the film temperature to be 35 C. The properties of air at 1 atm and this temperature are (Table A-15) k 0.02625 W/m. C
1.655 10 -5 m 2 /s Pr 0.7268
V = 5 m/s T = 32 C
Person, Ts 90 W
= 0.9
Analysis The Reynolds number is
Re V D (5 m/s)(0.3m) 1.655 10
5
D = 0.3 m
m 2 /s
9.063 10 4
The proper relation for Nusselt number corresponding to this Reynolds number is
Nu hD k 0.3 1 0.3 0.62 Re 0.5 Pr 1 / 3 0.4 / Pr
2 / 3 1/ 4
1
Re 282,000
5/8
4/5
0.62(9.063 10 4 ) 0.5 (0.7268)1 / 3 1 0.4 / 0.7268
2 / 3 1/ 4
1
9.063 10 4 282,000
5/8
4/5
203.6
Then
h k Nu D 0.02655 W/m. C (203.6) 18.02 W/m 2 . C 0.3 m
Considering that there is heat generation in that person's body at a rate of 90 W and body gains heat by radiation from the surrounding surfaces, an energy balance can be written as Q Q Q
generated radiation convection
Substituting values with proper units and then application of trial & error method yields the average temperature of the outer surface of the person.
90 W 90 (0.9)(1.7)(5.67 10
8
As (Tsurr 4 Ts 4 ) )[(40 273)
4 4
hAs (Ts
T )
Ts ] (18.02)(1.7)[Ts (32 273)] Ts 309.2 K 36.2 C
7-79
Chapter 7 External Forced Convection 7-95 The heat generated by four transistors mounted on a thin vertical plate is dissipated by air blown over the plate on both surfaces. The temperature of the aluminum plate is to be determined. Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5 105. 3 Radiation effects are negligible. 4 The entire plate is nearly isothermal. 5 The exposed surface area of the transistor is taken to be equal to its base area. 6 Air is an ideal gas with constant properties. 7 The pressure of air is 1 atm. Properties Assuming a film temperature of 40 C, the properties of air are evaluated to be (Table A-15) k 0.02662 W/m. C
Pr 1.702 10 -5 m 2 /s 0.7255
L= 22 cm
V = 250 m/min T = 20 C
12 W
Ts
Analysis The Reynolds number in this case is
Re L V L (250 / 60) m/s (0.22 m) 1.702 10
5
m 2 /s
5.386 10 4
which is smaller than the critical Reynolds number. Thus we have laminar flow. Using the proper relation for Nusselt number, heat transfer coefficient is determined to be
Nu h hL 0.664 Re L 0.5 Pr1 / 3 0.664(5.386 10 4 ) 0.5 (0.7255)1 / 3 k k 0.02662 W/m. C Nu (138.5) 16.75 W/m 2 . C L 0.22 m 138.5
The temperature of aluminum plate then becomes Q (4 12) W Q hAs (Ts T ) Ts T 20 C hAs (16.75 W/m 2 . C)[2(0.22 m) 2 ]
50.0 C
Discussion In reality, the heat transfer coefficient will be higher since the transistors will cause turbulence in the air.
7-80
Chapter 7 External Forced Convection 7-96 A spherical tank used to store iced water is subjected to winds. The rate of heat transfer to the iced water and the amount of ice that melts during a 24-h period are to be determined. Assumptions 1 Steady operating conditions exist. 2 Thermal resistance of the tank is negligible. 3 Radiation effects are negligible. 4 Air is an ideal gas with constant properties. 5 The pressure of air is 1 atm. Properties The properties of air at 1 atm pressure and the free stream temperature of 30 C are (Table A15) Ts = 0 C k 0.02588 W/m. C V = 25 km/h
1.608 10 -5 m 2 /s 1.872 10
s, @ 0 C 5 5
T = 30 C
kg/m.s kg/m.s
Iced water Di = 3 m 0 C 1 cm
1.729 10 0.7282
Pr
Analysis (a) The Reynolds number is
Re V D (25 1000/3600) m/s (3.02 m) 1.608 10
5
Q
m 2 /s 1.304 10 6
The Nusselt number corresponding to this Reynolds number is determined from
Nu hD k
1/ 4
2
0.4 Re 0.5 0.06 Re 2 / 3 Pr 0.4
s
2
0.4(1.304 10 6 ) 0.5
0.06(1.304 10 6 ) 2 / 3 (0.7282) 0.4
1.872 10 1.729 10
5 5
1/ 4
1056
and
h
k Nu D
0.02588 W/m. C (1056) 3.02 m
h( D 2 )(Ts T )
9.05 W/m 2 . C
The rate of heat transfer to the iced water is
Q hAs (Ts T ) Q t (9.05 W/m 2 . C)[ (3.02 m) 2 ](30 0) C 7779 W
(b) The amount of heat transfer during a 24-hour period is
Q (7.779 kJ/s)(24 3600 s) 672,079 kJ
Then the amount of ice that melts during this period becomes
Q mhif m Q hif 672,079 kJ 333.7 kJ/kg 2014 kg
7-81
Chapter 7 External Forced Convection 7-97 A spherical tank used to store iced water is subjected to winds. The rate of heat transfer to the iced water and the amount of ice that melts during a 24-h period are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 7 The pressure of air is 1 atm. Properties The properties of air at 1 atm pressure and the free stream temperature of 30 C are (Table A15)
k 0.02588 W/m. C 1.608 10 1.872 10
-5 5
m /s kg/m.s
2
s, @ 0 C
1.729 10 0.7282
5
kg/m.s
Pr
V = 25 km/h T = 30 C
Ts, out 0C
Analysis (a) The Reynolds number is
Re V D (25 1000/3600) m/s (3.02 m) 1.608 10
5
Iced water Di = 3 m 0 C
1 cm
m 2 /s
1.304 10 6
The Nusselt number corresponding to this Reynolds number is determined from
Nu hD k
1/ 4
2
0.4 Re
0.5
0.06 Re
6 0.5
2/3
Pr
0.4 s
2
0.4(1.304 10 )
0.06(1.304 10 )
6 2/3
(0.7282)
0.4
1.872 10 1.729 10
5 5
1/ 4
1056
and
h
k Nu D
0.02588 W/m. C (1056) 3.02 m
9.05 W/m 2 . C
In steady operation, heat transfer through the tank by conduction is equal to the heat transfer from the outer surface of the tank by convection and radiation. Therefore,
Q Q Qthrough tank Ts,out Ts,in Rsphere Qfrom tank, conv+rad ho Ao (Tsurr Ts,out ) Ao (Tsurr 4 Ts,out 4 )
where R sphere
Ao
r2 r1 4 kr1r2
(1.51 1.50) m 4 (15 W/m. C)(1.51 m)(1.50 m)
2.342 10
5
C/W
D2
(3.02 m) 2
28.65 m 2
Substituting,
Q Ts ,out 2.34 10 0C
5
C/W
(9.05 W/m 2 . C)(28.65 m 2 )(30 Ts ,out ) C (0.9)(28.65 m 2 )(5.67 10
8
W/m 2 .K 4 )[(15 273 K) 4
(Ts ,out
273 K) 4 ]
whose solution is
Ts Q 0.23 C and Q Q t 9630 W 9.63 kW
(b) The amount of heat transfer during a 24-hour period is
(9.63 kJ/s)(24 3600 s) 832,032 kJ
Then the amount of ice that melts during this period becomes
Q mhif m Q hif 832,032 kJ 333.7 kJ/kg 2493 kg
7-82
Chapter 7 External Forced Convection 7-98E A cylindrical transistor mounted on a circuit board is cooled by air flowing over it. The maximum power rating of the transistor is to be determined. Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 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 film temperature of T f (180 120) / 2 150 F are (Table A-15)
k Pr 0.01646 Btu/h.ft. F 0.210 10 -3 ft 2 /s 0.7188
Air 500 ft/min 120 F
Power transistor D = 0.22 in
Analysis The Reynolds number is
727.5 0.210 10 3 ft 2 /s The Nusselt number corresponding to this Reynolds number is
Nu hD k 0.3 0.62 Re 0.5 Pr1 / 3 1 0.3
k Nu D
Re
V D
(500/60 ft/s)(0.22/12 ft)
0.4 / Pr
2 / 3 1/ 4
1
Re 282,000 1
5/8
4/5
0.62(727.5) 0.5 (0.7188)1 / 3 1 0.4 / 0.7188
2 / 3 1/ 4
727.5 282,000
5/8
4/5
13.72
and
h
0.01646 Btu/h.ft. F (13.72) 12.32 Btu/h.ft 2 . F (0.22 / 12 ft)
Then the amount of power this transistor can dissipate safely becomes Q hA (T T ) h DL (T T )
s s s
(12.32 Btu/h.ft 2 . F) (0.22/12 ft)(0.25/12 ft) (180 120) C 0.887 Btu/h = 0.26 W (1 W = 3.412 Btu/h)
7-83