6 Chapter Fundamentals of Convection 6-39 The oil in a journal bearing is considered. The velocity and temperature distributions, the maximum temperature, the rate of heat transfer, and the mechanical power wasted in oil are to be determined. Assumptions 1 Steady operating conditions exist. 2 Oil is an incompressible substance with constant properties. 3 Body forces such as gravity are negligible. Properties The properties of oil at 50 C are given to be k = 0.17 W/m-K and = 0.05 N-s/m2
Analysis (a) Oil flow in journal bearing can be approximated as parallel flow between two large plates with one plate moving and the other stationary. We take the x-axis to be the flow direction, and y to be the normal direction. This is parallel flow between two plates, and thus v = 0. Then the continuity equation reduces to u u v 0 0 Continuity: u = u(y) 3000 rpm x x y Therefore, the x-component of velocity does not change in the flow direction (i.e., the velocity profile remains unchanged). Noting that u = u(y), v = 0, and P / x 0 (flow is maintained by the motion of the upper plate rather than the pressure gradient), the xmomentum equation reduces to x-momentum: 12 m/s 6 cm
20 cm
0 y2 dy 2 This is a second-order ordinary differential equation, and integrating it twice gives
u( y) C1 y C 2
u u x
u v y
2
u
P x
d 2u
The fluid velocities at the plate surfaces must be equal to the velocities of the plates because of the no-slip condition. Taking x = 0 at the surface of the bearing, the boundary conditions are u(0) = 0 and u(L) = V , and applying them gives the velocity distribution to be
u( y) y V L
The plates are isothermal and there is no change in the flow direction, and thus the temperature depends on y only, T = T(y). Also, u = u(y) and v = 0. Then the energy equation with viscous dissipation reduce to
2
Energy: since u / y
0
dy 2 V / L . Dividing both sides by k and integrating twice give
k
T
y2
u y
2
k
d 2T
V L
2
T ( y)
y V 2k L
2
C3 y C 4
Applying the boundary conditions T(0) = T0 and T(L) = T0 gives the temperature distribution to be
T ( y ) T0 V2 2k y L y2 L2
The temperature gradient is determined by differentiating T(y) with respect to y,
dT dy y V2 1 2 2kL L
The location of maximum temperature is determined by setting dT/dy = 0 and solving for y,
6-11
Chapter 6 Fundamentals of Convection
dT dy y V2 1 2 2kL L 0
y
L 2
Therefore, maximum temperature will occur at mid plane in the oil. The velocity and the surface area are 1 min V Dn (0.06 m)(3000 rev/min ) 9.425 m/s 60 s
A DLbearing (0.06 m)(0.20 m) 0.0377 m 2
The maximum temperature is
Tmax T ( L / 2) T0 T0 V2 2k L/2 L ( L / 2) 2 L2 53.3 C
(0.05 N s/m 2 )(9.425 m/s) 2 V2 1W 50 C 8k 8(0.17 W/m C) 1 N m/s (b) The rates of heat transfer are
Q0 kA dT dy kA
y 0
V2 1 0 2kL
A
V2 2L 419 W
(0.0377 m 2 )
QL kA dT dy
(0.05 N s/m 2 )(9.425 m/s) 2 1W 2(0.0002 m) 1 N m/s
kA V2 1 2 2kL A V2 2L Q0 419 W
y L
Therefore, rates of heat transfer at the two plates are equal in magnitude but opposite in sign. The mechanical power wasted is equal to the rate of heat transfer. Wmech Q 2 419 838 W
6-12
Chapter 6 Fundamentals of Convection 6-40 The oil in a journal bearing is considered. The velocity and temperature distributions, the maximum temperature, the rate of heat transfer, and the mechanical power wasted in oil are to be determined. Assumptions 1 Steady operating conditions exist. 2 Oil is an incompressible substance with constant properties. 3 Body forces such as gravity are negligible. Properties The properties of oil at 50 C are given to be k = 0.17 W/m-K and = 0.05 N-s/m2
Analysis (a) Oil flow in journal bearing can be approximated as parallel flow between two large plates with one plate moving and the other stationary. We take the x-axis to be the flow direction, and y to be the normal direction. This is parallel flow between two plates, and thus v = 0. Then the continuity equation reduces to u u v 0 0 Continuity: u = u(y) 3000 rpm x x y Therefore, the x-component of velocity does not change in the flow direction (i.e., the velocity profile remains unchanged). Noting that u = u(y), v = 0, and P / x 0 (flow is maintained by the motion of the upper plate rather than the pressure gradient), the xmomentum equation reduces to x-momentum: 12 m/s 6 cm
20 cm
0 y2 dy 2 This is a second-order ordinary differential equation, and integrating it twice gives
u( y) C1 y C 2
u u x
u v y
2
u
P x
d 2u
The fluid velocities at the plate surfaces must be equal to the velocities of the plates because of the no-slip condition. Taking x = 0 at the surface of the bearing, the boundary conditions are u(0) = 0 and u(L) = V , and applying them gives the velocity distribution to be
u( y) y V L
Frictional heating due to viscous dissipation in this case is significant because of the high viscosity of oil and the large plate velocity. The plates are isothermal and there is no change in the flow direction, and thus the temperature depends on y only, T = T(y). Also, u = u(y) and v = 0. Then the energy equation with dissipation reduce to
2
Energy: since u / y
0
dy 2 V / L . Dividing both sides by k and integrating twice give
V k L
2
k
T
y2
u y
2
k
d 2T
V L
2
dT dy T ( y)
y C3
2
y V 2k L
C3 y C 4
Applying the two boundary conditions give B.C. 1: y=0 B.C. 2: y=L
T ( 0) T1 C4 T1
k
dT dy
0
y L
C3
V2 kL
Substituting the constants give the temperature distribution to be
6-13
Chapter 6 Fundamentals of Convection
T ( y ) T1 V2 kL y y2 2L
The temperature gradient is determined by differentiating T(y) with respect to y, y V2 dT 1 dy kL L The location of maximum temperature is determined by setting dT/dy = 0 and solving for y, y V2 dT 1 0 y L dy kL L This result is also known from the second boundary condition. Therefore, maximum temperature will occur at the shaft surface, for y = L. The velocity and the surface area are
V DN (0.06 m)(3000 rev/min ) 1 min 60 s 9.425 m/s
A
DLbearing
(0.06 m)(0.20 m)
V2 L2 L kL 2L
0.0377 m 2
V2 1 1 k 2 V2 2k
The maximum temperature is
Tmax T ( L) T1 50 C T1 T1
(0.05 N s/m 2 )(9.425 m/s) 2 1W 2(0.17 W/m C) 1 N m/s (b) The rate of heat transfer to the bearing is V2 V2 dT Q0 kA kA 1 0 A dy y 0 kL L (0.0377 m 2 )
63.1 C
(0.05 N s/m 2 )(9.425 m/s) 2 1W 837 W 0.0002 m 1 N m/s The rate of heat transfer to the shaft is zero. The mechanical power wasted is equal to the rate of heat transfer, Wmech Q 837 W
6-14
Chapter 6 Fundamentals of Convection 6-41 "!PROBLEM 6-41" "GIVEN" D=0.06 "[m]" "N_dot=3000 rpm, parameter to be varied" L_bearing=0.20 "[m]" L=0.0002 "[m]" T_0=50 "[C]" "PROPERTIES" k=0.17 "[W/m-K]" mu=0.05 "[N-s/m^2]" "ANALYSIS" Vel=pi*D*N_dot*Convert(1/min, 1/s) A=pi*D*L_bearing T_max=T_0+(mu*Vel^2)/(8*k) Q_dot=A*(mu*Vel^2)/(2*L) W_dot_mech=Q_dot
N [rpm] 0 250 500 750 1000 1250 1500 1750 2000 2250 2500 2750 3000 3250 3500 3750 4000 4250 4500 4750 5000
Wmech [W] 0 2.907 11.63 26.16 46.51 72.67 104.7 142.4 186 235.5 290.7 351.7 418.6 491.3 569.8 654.1 744.2 840.1 941.9 1049 1163
6-15
Chapter 6 Fundamentals of Convection
6-16
Chapter 6 Fundamentals of Convection 6-42 A shaft rotating in a bearing is considered. The power required to rotate the shaft is to be determined for different fluids in the gap. Assumptions 1 Steady operating conditions exist. 2 The fluid has constant properties. 3 Body forces such as gravity are negligible. Properties The properties of air, water, and oil at 40 C are (Tables A-15, A-9, A-13) Air: Water: Oil: = 1.918 10-5 N-s/m2 = 0.653 10-3 N-s/m2 = 0.212 N-s/m2 2500 rpm 12 m/s 5 cm
Analysis A shaft rotating in a bearing can be approximated as parallel flow between two large plates with one plate moving and the other stationary. Therefore, we solve this problem considering such a flow with the plates separated by a L=0.5 mm thick fluid film similar to the problem given in Example 6-1. By simplifying and solving the continuity, momentum, and energy equations it is found in Example 6-1 that
W mech Q0 QL kA dT dy kA
y 0
10 cm
V2 1 0 2kL
A
V2 2L
A
V2 2L
First, the velocity and the surface area are
V DN (0.05 m)(2500 rev/min ) 1 min 60 s 6.545 m/s
A
DLbearing
V2 2L
(0.05 m)(0.10 m)
0.01571 m 2
N s/m 2 )(6.545 m/s) 2 1W 2(0.0005 m) 1 N m/s
3
5
(a) Air:
Wmech A (0.01571 m 2 ) (1.918 10 0.013 W
(b) Water:
Wmech Q0 A V2 2L
V2 2L
(0.01571 m 2 )
(0.653 10
N s/m 2 )(6.545 m/s) 2 1W 2(0.0005 m) 1 N m/s
0.44 W
(c) Oil:
Wmech Q0 A (0.01571 m 2 ) (0.212 N s/m 2 )(6.545 m/s) 2 1W 2(0.0005 m) 1 N m/s 142.7 W
6-17
Chapter 6 Fundamentals of Convection 6-43 The flow of fluid between two large parallel plates is considered. The relations for the maximum temperature of fluid, the location where it occurs, and heat flux at the upper plate are to be obtained. Assumptions 1 Steady operating conditions exist. 2 The fluid has constant properties. 3 Body forces such as gravity are negligible. Analysis We take the x-axis to be the flow direction, and y to be the normal direction. This is parallel flow between two plates, and thus v = 0. Then the continuity equation reduces to u u v 0 0 Continuity: u = u(y) V x x y Therefore, the x-component of velocity does not change T0 in the flow direction (i.e., the velocity profile remains unchanged). Noting that u = u(y), v = 0, and L Fluid P / x 0 (flow is maintained by the motion of the upper plate rather than the pressure gradient), the xmomentum equation reduces to x-momentum:
0 y2 dy 2 This is a second-order ordinary differential equation, and integrating it twice gives
u( y) C1 y C 2
u
u x
v
u y
2
u
P x
d 2u
The fluid velocities at the plate surfaces must be equal to the velocities of the plates because of the no-slip condition. Therefore, the boundary conditions are u(0) = 0 and u(L) = V , and applying them gives the velocity distribution to be
u( y) y V L
Frictional heating due to viscous dissipation in this case is significant because of the high viscosity of oil and the large plate velocity. The plates are isothermal and there is no change in the flow direction, and thus the temperature depends on y only, T = T(y). Also, u = u(y) and v = 0. Then the energy equation with dissipation (Eqs. 6-36 and 6-37) reduce to
2
Energy: since u / y
0
dy 2 V / L . Dividing both sides by k and integrating twice give
V k L
2
k
T
y2
u y
2
k
d 2T
V L
2
dT dy T ( y)
y C3
2
y V 2k L
C3 y C 4
Applying the two boundary conditions give dT k 0 C3 0 B.C. 1: y=0 dy y 0
V2 2k Substituting the constants give the temperature distribution to be
B.C. 2: y=L
T ( L) T0
C4
T0
T ( y ) T0
y2 V2 1 2k L2
The temperature gradient is determined by differentiating T(y) with respect to y,
6-18
Chapter 6 Fundamentals of Convection
V2 dT y dy kL2 The location of maximum temperature is determined by setting dT/dy = 0 and solving for y,
dT dy
V2 kL2
y
0
y
0
Therefore, maximum temperature will occur at the lower plate surface, and it s value is
Tmax T (0) T0 V2 2k
.The heat flux at the upper plate is
qL k dT dy k
y L
V2 kL2
L
V2 L
6-19
Chapter 6 Fundamentals of Convection 6-44 The flow of fluid between two large parallel plates is considered. Using the results of Problem 6-43, a relation for the volumetric heat generation rate is to be obtained using the conduction problem, and the result is to be verified. Assumptions 1 Steady operating conditions exist. 2 The fluid has constant properties. 3 Body forces such as gravity are negligible. V Analysis The energy equation in Prob. 6-44 was determined to be T0 2 d 2T V (1) k L dy 2 L Fluid The steady one-dimensional heat conduction equation with constant heat generation is d 2T g 0 (2) 0 2 k dy Comparing the two equation above, the volumetric heat generation rate is determined to be
V L Integrating Eq. (2) twice gives g0
dT dy T ( y) g0 y C3 k g0 2 y C3 y C4 2k
2
Applying the two boundary conditions give dT k 0 C3 0 B.C. 1: y=0 dy y 0
g0 2 L 2k Substituting, the distribution temperature becomes
B.C. 2: y=L
T ( L ) T0
C4
T0
T ( y ) T0
g 0 L2 y2 1 2 2k L
Maximum temperature occurs at y = 0, and it value is g 0 L2 Tmax T (0) T0 2k which is equivalent to the result Tmax
T (0) T0 V2 obtained in Prob. 6-43. 2k
6-20
Chapter 6 Fundamentals of Convection 6-45 The oil in a journal bearing is considered. The bearing is cooled externally by a liquid. The surface temperature of the shaft, the rate of heat transfer to the coolant, and the mechanical power wasted are to be determined. Assumptions 1 Steady operating conditions exist. 2 Oil is an incompressible substance with constant properties. 3 Body forces such as gravity are negligible. Properties The properties of oil are given to be k = 0.14 W/m-K and conductivity of bearing is given to be k = 70 W/m-K. = 0.03 N-s/m2. The thermal
Analysis (a) Oil flow in a journal bearing can be approximated as parallel flow between two large plates with one plate moving and the other stationary. We take the x-axis to be the flow direction, and y to be the normal direction. This is parallel flow between two plates, and thus v = 0. Then the continuity equation reduces to u u v 0 0 Continuity: u = u(y) 4500 rpm x x y Therefore, the x-component of velocity does not change 12 m/s 5 cm in the flow direction (i.e., the velocity profile remains unchanged). Noting that u = u(y), v = 0, and P / x 0 (flow is maintained by the motion of the upper plate rather than the pressure gradient), the xmomentum equation reduces to 15 cm x-momentum:
0 y2 dy 2 This is a second-order ordinary differential equation, and integrating it twice gives
u( y) C1 y C 2
u
u x
v
u y
2
u
P x
d 2u
The fluid velocities at the plate surfaces must be equal to the velocities of the plates because of the no-slip condition. Therefore, the boundary conditions are u(0) = 0 and u(L) = V , and applying them gives the velocity distribution to be
u( y) y V L
where
V Dn (0.05 m)(4500 rev/min ) 1 min 60 s 11.78 m/s
The plates are isothermal and there is no change in the flow direction, and thus the temperature depends on y only, T = T(y). Also, u = u(y) and v = 0. Then the energy equation with viscous dissipation reduces to
2
Energy:
dy 2 since u / y V / L . Dividing both sides by k and integrating twice give
dT dy T ( y) V k L
2
0
k
T
y2
u y
2
k
d 2T
V L
2
y C3
2
y V 2k L
C3 y C 4
Applying the two boundary conditions give dT k 0 C3 0 B.C. 1: y=0 dy y 0
6-21
Chapter 6 Fundamentals of Convection
V2 2k Substituting the constants give the temperature distribution to be
B.C. 2: y=L
T ( L) T0
C4
T0
T ( y ) T0
y2 V2 1 2k L2
The temperature gradient is determined by differentiating T(y) with respect to y, V2 dT y dy kL2 .The heat flux at the upper surface is
qL k dT dy k
y L
V2 kL2
L
V2 L
Noting that heat transfer along the shaft is negligible, all the heat generated in the oil is transferred to the shaft, and the rate of heat transfer is
Q As q L ( DW ) V2 L (0.05 m)(0.15 m) (0.03 N s/m 2 )(11.78 m/s) 2 0.0006 m 163.5 W
(b) This is equivalent to the rate of heat transfer through the cylindrical sleeve by conduction, which is expressed as
Q k 2 W (T0 Ts ) ln( D0 / D) (70 W/m C) 2 (0.15 m)(T0 - 40 C) ln( 8 / 5) 163.5 W
which gives the surface temperature of the shaft to be To = 41.2 C (c) The mechanical power wasted by the viscous dissipation in oil is equivalent to the rate of heat generation, Wlost Q 163.5 W
6-22
Chapter 6 Fundamentals of Convection 6-46 The oil in a journal bearing is considered. The bearing is cooled externally by a liquid. The surface temperature of the shaft, the rate of heat transfer to the coolant, and the mechanical power wasted are to be determined. Assumptions 1 Steady operating conditions exist. 2 Oil is an incompressible substance with constant properties. 3 Body forces such as gravity are negligible. Properties The properties of oil are given to be k = 0.14 W/m-K and conductivity of bearing is given to be k = 70 W/m-K. = 0.03 N-s/m2. The thermal
Analysis (a) Oil flow in a journal bearing can be approximated as parallel flow between two large plates with one plate moving and the other stationary. We take the x-axis to be the flow direction, and y to be the normal direction. This is parallel flow between two plates, and thus v = 0. Then the continuity equation reduces to u u v 0 0 Continuity: u = u(y) 4500 rpm x x y Therefore, the x-component of velocity does not change 12 m/s 5 cm in the flow direction (i.e., the velocity profile remains unchanged). Noting that u = u(y), v = 0, and P / x 0 (flow is maintained by the motion of the upper plate rather than the pressure gradient), the xmomentum equation reduces to 15 cm x-momentum:
0 y2 dy 2 This is a second-order ordinary differential equation, and integrating it twice gives
u( y) C1 y C 2
u
u x
v
u y
2
u
P x
d 2u
The fluid velocities at the plate surfaces must be equal to the velocities of the plates because of the no-slip condition. Therefore, the boundary conditions are u(0) = 0 and u(L) = V , and applying them gives the velocity distribution to be
u( y) y V L
where
V Dn (0.05 m)(4500 rev/min ) 1 min 60 s 11.78 m/s
The plates are isothermal and there is no change in the flow direction, and thus the temperature depends on y only, T = T(y). Also, u = u(y) and v = 0. Then the energy equation with viscous dissipation reduces to
2
Energy:
dy 2 since u / y V / L . Dividing both sides by k and integrating twice give
dT dy T ( y) V k L
2
0
k
T
y2
u y
2
k
d 2T
V L
2
y C3
2
y V 2k L
C3 y C 4
Applying the two boundary conditions give dT k 0 C3 0 B.C. 1: y=0 dy y 0
6-23
Chapter 6 Fundamentals of Convection
V2 2k Substituting the constants give the temperature distribution to be
B.C. 2: y=L
T ( L) T0
C4
T0
T ( y ) T0
y2 V2 1 2k L2
The temperature gradient is determined by differentiating T(y) with respect to y, V2 dT y dy kL2 .The heat flux at the upper surface is
qL k dT dy k
y L
V2 kL2
L
V2 L
Noting that heat transfer along the shaft is negligible, all the heat generated in the oil is transferred to the shaft, and the rate of heat transfer is
Q As q L ( DW ) V2 L (0.05 m)(0.15 m) (0.03 N s/m 2 )(11.78 m/s) 2 0.001 m 98.1 W
(b) This is equivalent to the rate of heat transfer through the cylindrical sleeve by conduction, which is expressed as
Q k 2 W (T0 Ts ) ln( D0 / D) (70 W/m C) 2 (0.15 m)(T0 - 40 C) ln( 8 / 5) 98.1 W
which gives the surface temperature of the shaft to be To = 40.7 C (c) The mechanical power wasted by the viscous dissipation in oil is equivalent to the rate of heat generation, Wlost Q 98.1 W
6-24
Chapter 6 Fundamentals of Convection
Momentum and Heat Transfer Analogies
6-47C Reynolds analogy is expressed as C f , x
Re L 2
Nu x . It allows us to calculate the heat transfer
coefficient from a knowledge of friction coefficient. It is limited to flow of fluids with a Prandtl number of near unity (such as gases), and negligible pressure gradient in the flow direction (such as flow over a flat plate).
6-48C
Modified
Reynolds
analogy
is
expressed
as
C f ,x
Re L 2
Nu x Pr
1/ 3
or
C f ,x 2
hx Pr 2/3 C pV
j H . It allows us to calculate the heat transfer coefficient from a knowledge of
friction coefficient. It is valid for a Prandtl number range of 0.6 < Pr < 60. This relation is developed using relations for laminar flow over a flat plate, but it is also applicable approximately for turbulent flow over a surface, even in the presence of pressure gradients. 6-49 A flat plate is subjected to air flow, and the drag force acting on it is measured. The average convection heat transfer coefficient and the rate of heat transfer are to be determined. Assumptions 1 Steady operating conditions exist. 2 The edge effects are negligible. Properties The properties of air at 20 C and 1 atm are (Table A-15) = 1.204 kg/m3, Cp =1.007 kJ/kg-K, Pr = 0.7309 Air 20 C 10 m/s
Analysis The flow is along the 4-m side of the plate, and thus the characteristic length is L = 4 m. Both sides of the plate is exposed to air flow, and thus the total surface area is
As 2WL 2(4 m)(4 m) 32 m
2
L=4 m
For flat plates, the drag force is equivalent to friction force. The average friction coefficient Cf can be determined from
Ff C f As V2 2 Cf Ff As V 2 / 2 1 kg m/s 2 1N (1.204 kg/m 3 )(32 m 2 )(10 m/s) 2 / 2 2.4 N 0.006229
Then the average heat transfer coefficient can be determined from the modified Reynolds analogy to be
2 Pr Them the rate of heat transfer becomes Q hAs (Ts T ) (46.54 W/m 2 C)(32 m 2 )(80 20) C 89,356 W
h
Cf
VCp
2/3
0.006229 (1.204 kg/m 3 )(10 m/s)(1007 J/kg C) 2 (0.7309) 2 / 3
46.54 W/m 2 C
6-25
Chapter 6 Fundamentals of Convection 6-50 A metallic airfoil is subjected to air flow. The average friction coefficient is to be determined. Assumptions 1 Steady operating conditions exist. 2 The edge effects are negligible. Properties The properties of air at 25 C and 1 atm are (Table A-15) = 1.184 kg/m3, Cp =1.007 kJ/kg-K, Pr = 0.7296 Air 25 C 8 m/s
Analysis First, we determine the rate of heat transfer from
Q mC p,airfoil (T2 T1 ) t (50 kg)(500 J/kg C) (160 150) C (2 60 s) 2083 W
L=3 m
Then the average heat transfer coefficient is
Q hAs (Ts T ) h Q As (Ts T ) 2083 W (12 m )(155 25) C
2
1.335 W/m 2
C
where the surface temperature of airfoil is taken as its average temperature, which is (150+160)/2=155 C. The average friction coefficient of the airfoil is determined from the modified Reynolds analogy to be
Cf
2hPr 2/3 VCp
2(1.335 W/m 2 C)(0.7296) 2 / 3 (1.184 kg/m 3 )(8 m/s)(1007 J/kg C)
0.000227
6-51 A metallic airfoil is subjected to air flow. The average friction coefficient is to be determined. Assumptions 1 Steady operating conditions exist. 2 The edge effects are negligible. Properties The properties of air at 25 C and 1 atm are (Table A-15) = 1.184 kg/m3, Cp =1.007 kJ/kg-K, Pr = 0.7296 Air 25 C 12 m/s
Analysis First, we determine the rate of heat transfer from
Q mC p,airfoil (T2 T1 ) t (50 kg)(500 J/kg C) (160 150) C (2 60 s) 2083 W
L=3 m
Then the average heat transfer coefficient is
Q hAs (Ts T ) h Q As (Ts T ) 2083 W (12 m )(155 25) C
2
1.335 W/m 2
C
where the surface temperature of airfoil is taken as its average temperature, which is (150+160)/2=155 C. The average friction coefficient of the airfoil is determined from the modified Reynolds analogy to be
Cf
2hPr 2/3 VCp
2(1.335 W/m 2 C) (0.7296) 2 / 3 (1.184 kg/m 3 )(12 m/s)(1007 J/kg C)
0.0001512
6-26
Chapter 6 Fundamentals of Convection 6-52 The windshield of a car is subjected to parallel winds. The drag force the wind exerts on the windshield is to be determined. Assumptions 1 Steady operating conditions exist. 2 The edge effects are negligible. Properties The properties of air at 0 C and 1 atm are (Table A-15) = 1.292 kg/m3, Cp =1.006 kJ/kg-K, Pr = 0.7362 Air 0 C 80 km/h Windshield Ts=4 C
11.57 W/m
2
Analysis The average heat transfer coefficient is
Q h hAs (Ts T ) Q As (Ts T ) 50 W (0.6 1.8 m 2 )(4 0) C C
0.6 m
The average friction coefficient is determined from the modified Reynolds analogy to be
1.8 m
(1.292 kg/m 3 )(80 / 3.6 m/s)(1006 J/kg C) The drag force is determined from
Ff C f As V2 2 0.0006534(0.6 1.8 m 2 )
Cf
2hPr 2/3 VCp
2(11.57 W/m 2 C) (0.7362) 2 / 3
0.0006534
(1.292 kg/m 3 )(80 / 3.6 m/s) 2 1N 2 1 kg.m/s 2
0.225 N
6-53 An airplane cruising is considered. The average heat transfer coefficient is to be determined. Assumptions 1 Steady operating conditions exist. 2 The edge effects are negligible. Properties The properties of air at -50 C and 1 atm are (Table A-15) Cp =0.999 kJ/kg-K Pr = 0.7440 Air -50 C 800 km/h Wing Ts=4 C 3m
The density of air at -50 C and 26.5 kPa is
P RT 26.5 kPa (0.287 kJ/kg.K)(-50 273)K 0.4141 kg/m 3
Analysis The average heat transfer coefficient can be determined from the modified Reynolds analogy to be
h Cf 2 VCp Pr 2/3 0.0016 (0.4141 kg/m 3 )(800 / 3.6 m/s)(999 J/kg C) 2 (0.7440) 2 / 3
25 m
89.6 W/m 2 C
6-54, 6-55 Design and Essay Problems
6-27