8-9
8-24 Combustion gases passing through a tube are used to vaporize waste water. The tube length and the rate of evaporation
of water are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the pipe is constant. 3 The thermal resistance
of the pipe is negligible. 4 Air properties are to be used for exhaust gases.
Properties The properties of air at the average temperature of (250+150)/2=200°C are (Table A-15)
c p = 1023 J/kg.°C
Ts=110°C
R = 0.287 kJ/kg.K
Also, the heat of vaporization of water at 1 atm or 100°C is
h fg = 2257 kJ/kg (Table A-9).
Analysis The density of air at the inlet and the mass flow rate of
exhaust gases are
ρ=
150°C
Exh. gases
250°C
5 m/s
P
115 kPa
=
= 0.7662 kg/m 3
RT (0.287 kJ/kg.K)(250 + 273 K)
D = 5 cm
L
2
⎛ πD 2 ⎞
⎟Vavg = (0.7662 kg/m 3 ) π (0.05 m) (5 m/s) = 0.007522 kg/s
m& = ρAcVavg = ρ ⎜⎜
⎟
4
⎝ 4 ⎠
The rate of heat transfer is
Q& = m& c p (Ti − Te ) = (0.007522 kg/s)(1023 J/kg.°C)(250 − 150°C) = 769.5 W
The logarithmic mean temperature difference and the surface area are
∆Tlm =
Te − Ti
⎛ T − Te
ln⎜⎜ s
⎝ Ts − Ti
⎞
⎟
⎟
⎠
=
150 − 250
= 79.82°C
⎛ 110 − 150 ⎞
ln⎜
⎟
⎝ 110 − 250 ⎠
Q& = hAs ∆Tlm ⎯
⎯→ As =
Q&
769.5 W
=
= 0.08034 m 2
h∆Tlm (120 W/m 2 .°C)(79.82°C)
Then the tube length becomes
As = πDL ⎯
⎯→ L =
As
0.08034 m 2
=
= 0.5115 m = 51.2 cm
πD
π (0.05 m)
The rate of evaporation of water is determined from
0.7695 kW
Q&
⎯→ m& evap =
=
= 0.0003409 kg/s = 1.23 kg/h
Q& = m& evap h fg ⎯
h fg 2257 kJ/kg
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
8-19
8-40 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 80°C assuming that the air does not loose much heat to the attic, the
properties of air are (Table A-15)
ρ = 0.9994 kg/m 3
k = 0.02953 W/m.°C
ν = 2.097 × 10 -5 m 2 /s
c p = 1008 J/kg.°C
Pr = 0.7154
Analysis The mean velocity of air, the hydraulic diameter, and the Reynolds number are
V=
V&
=
A
Dh =
Re =
0.15 m 3 /s
(0.2 m) 2
= 3.75 m/s
Air
80ºC
0.15 m3/s
2
4 A 4a
=
= a = 0.2 m
P
4a
VDh
ν
=
(3.75 m/s)(0.2 m)
2.097 × 10
−5
a = 0.2 m
L=8m
= 35,765
which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly
Lh ≈ Lt ≈ 10 D h = 10(0.2 m) = 2 m
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
Nu =
hD h
= 0.023 Re 0.8 Pr 0.3 = 0.023(35,765) 0.8 (0.7154) 0.3 = 91.4
k
Heat transfer coefficient is
h=
k
0.02953 W/m.°C
Nu =
(91.4) = 13.5 W/m 2 .°C
D
0.2 m
Next we determine the exit temperature of air
A = 4aL = 4(0.2 m)(8 m) = 6.4 m 2
Te = Ts − (Ts − Ti )e
− hA /( m& c p )
= 60 − (60 − 80)e
−
(13.5)( 6.4)
( 0.9994)(0.15)(1008)
= 71.3°C
Then the rate of heat transfer becomes
Q& = m& c p (Te − Ti ) = (0.9994 kg/m 3 )(0.15 m 3 /s)(1008 J/kg.°C)(80 − 71.3)°C = 1315 W
From Moody chart:
Re = 35,765 and ε/D = 0.001 → f = 0.026
Then the pressure drop is determined to be
∆P = f
ρV 2
2D
L = (0.026)
(0.9994 kg/m 3 )(3.75 m/s) 2
(8 m) = 7.3 Pa
2(0.2 m)
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.