19–81
19–86 Water is flowing between two parallel 1-m wide plates with 12.5-mm spacing. Hydrogen gas flows width-wise
in parallel over the upper and lower surfaces of the two plates. The outlet mean temperature of the water, the surface
temperature of the plates, and the total rate of heat transfer are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 Isothermal parallel plates. 4 The thermal
resistance of the plates is negligible (thin plates). 5 The bulk mean fluid temperature of the water is 30°C (this will be
validated). 6 The film temperature of the H2 gas is 100°C (this will be validated).
(Table A-23)
Analysis The Reynolds number, hydrodynamic and thermal entry lengths can be determined to be
19–83
19–87 Reconsider Prob. 19–86. Water is flowing between two parallel 1-m wide plates with 12.5-mm spacing.
Hydrogen gas flows width-wise in parallel over the upper and lower surfaces of the two plates. The effect of water mass flow
rate on the free-stream velocity of the H2 gas and the surface temperature of the parallel plates, and the effect of the free–
stream velocity of the H2 gas on the total heat transfer rate are to be evaluated.
Analysis The problem is solved using EES, and the solution is given below.
c_p=cP(water, T=T_b, x=0)*Convert(kJ/kg-C, J/kg-C)
k=Conductivity(water, T=T_b, x=0)
rho=Density(water, T=T_b, x=0)
Pr=Prandtl(water, T=T_b, x=0)
mu=Viscosity(water, T=T_b, x=0)
p=2*(width+spacing) “Perimeter”
D_h=(4*A_c)/p “Hydraulic diameter”
A_s=p*L “Surface area”
Q_dot=m_dot*c_p*(T_e-T_i)
“Flow over plates”
Re_H2=V_infinity*width/nu_H2
Nusselt_H2=0.664*Re_H2^0.5*Pr_H2^(1/3)
1
5
19–84
ṁ [kg/s] V∞ [m/s] Ts [°C] Q
̇ [W]
0.01 0.001348 40 836.7
0.05 0.0337 40 4183
0.10 0.1348 40 8367
0.12 0.1942 40.01 10040
0.14 0.2644 40.03 11713
0.16 0.3456 40.08 13387
0.18 0.4379 40.14 15060
0.20 0.5415 40.23 16733
0.22 0.6566 40.35 18407
0.24 0.7834 40.50 20080
0.26 0.9222 40.67 21753
0.28 1.073 40.87 23427
0.30 1.237 41.08 25100
0.32 1.413 41.32 26773
0.35 1.702 41.70 29283
0.40 2.252 42.41 33467
0.45 2.891 43.20 37650
0.50 3.625 44.04 41833
0.55 4.459 44.93 46017
0.58 5.009 45.48 48527
105.
0 1 2 3 4 5
0
10000
20000
30000
40000
50000
V¥ [m/s]
Q [W]
0 0.1 0.2 0.3 0.4 0.5 0.6
40
41
42
43
44
45
46
m [kg/s]
Ts [C]
19–86
19–89 Liquid glycerin is flowing through a 25-mm diameter and 10-m long tube, the constant surface temperature of the tube
is to be determined.
23007.38
)skg/m 6582.0)(m 025.0(
)kg/s 5.0(44
Re
D
m
(laminar flow)
31.13
)]5631)(7.38)(10/025.0[(0401
)5631)(7.38)(10/025.0(065.0
66.3
Pr]Re)/[(0401
PrRe)/(065.0
66.3Nu
3/2
3/2
.
LD.
LD
k
The surface temperature of the tube is
s
hA
)]/(exp[
psie
cmhATT
09757.0
)KJ/kg 2447)(kg/s 5.0(
)m 10)(m 025.0()K W/m152( 2
p
s
cm
hA
19–87
19-90 Air at 20°C (1 atm) enters into a 5-mm diameter and 10-cm long circular tube, the convection heat transfer coefficient
Analysis The Reynolds number, hydrodynamic and thermal entry lengths are
)m 005.0)(m/s 5(
avg
DV
19–90
ṁ [kg/s] h [W/m2∙K] Te [°C] Q
̇ [W]
0.05 87.34 35 1224
0.1 87.34 35.01 2447
0.15 87.34 35.08 3671
0.2 87.34 35.28 4894
0.25 87.34 35.59 6118
0.3 87.34 35.99 7341
0.6 94.99 38.69 14682
0.7 96.15 39.73 17129
0.8 97.29 40.76 19576
0.9 98.40 41.80 22023
1.0 99.49 42.82 24470
1.2 101.6 44.85 29364
1.4 103.7 46.82 34258
1.6 105.7 48.75 39152
1.8 107.6 50.62 44046
2.0 109.5 52.44 48940
2.5 114.1 56.79 61175
3.0 118.4 60.89 73410
4.0 126.4 68.46 97880
5.0 133.7 75.37 122350
6.0 140.5 81.76 146820
0 1 2 3 4 5 6
0
20000
40000
60000
80000
100000
120000
140000
160000
Q [W]
0 1 2 3 4 5 6
30
40
50
60
70
80
90
m [kg/s]
Ts [°C]
19–97
19–99 Air flows in a square cross section pipe. The rate of heat loss and the pressure difference between the inlet and outlet
sections of the duct are 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.
Properties Taking a bulk mean fluid temperature of 80C based on the problem statement (this assumes that the air does not
loose much heat to the attic), the properties of air are (Table A-22)
7154.0Pr
CJ/kg. 1008
C W/m.02953.0
kg/m 9994.0
25–
3
p
c
k
Analysis The mean velocity of air, the hydraulic diameter, and the Reynolds number are
m/s 75.3
)m 2.0(
/sm 15.0
2
3
A
V
V
m 2.0
4
44 2
a
a
a
P
A
Dh
765,35
10097.2
)m 2.0)(m/s 75.3(
Re 5
h
VD
which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly
m 2m) 2.0(1010 hth DLL
which is much shorter than the total length of the pipe. Therefore, we can assume fully developed turbulent flow in the entire
duct, and determine the Nusselt number from
4.91)7154.0()765,35(023.0PrRe023.0 3.08.03.08.0 k
hD
Nu h
Air
80ºC
0.15 m3/s
L = 8 m
a = 0.2 m
19-100
19-102 Hot air enters a sheet metal duct located in a basement. The exit temperature of hot air and the rate of heat loss 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 evaluate the air properties at 1 atm and the estimated bulk mean temperature of 50C based on the problem
7228.0Pr
Analysis The surface area and the Reynolds number are
/sm 10798.1
which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly
m 0.2m) 2.0(1010 hth DLL
which is much shorter than the total length of the duct. Therefore, we can assume fully developed turbulent flow for the
entire duct, and determine the Nusselt number from
2.109)7228.0()494,44(023.0PrRe023.0 3.08.03.08.0 k
hD
Nu h
m 2.0
C W/m.02735.0 2
D
k
h
The mass flow rate of air is
kg/s 0.1747m/s) )(4m 0.2)(0.2kg/m 092.1( 23 VAm c
In steady operation, heat transfer from hot air to the duct must be equal to the heat transfer from the duct to the surrounding
(by convection and radiation), which must be equal to the energy loss of the hot air in the duct. That is,
50C. Therefore, there is no need to repeat calculations.
Air duct
20 cm 20 cm