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7-1. A convector whose performance characteristics are shown in Fig. 7-4 is supplied with a flow rate of 0.04 kg/s of water at 90 C. The length of the convector is 4 m, and the room-air temperature is 18 C. What is the rate of heat transfer from the convector to the room air?
Solution: See Fig. 7- m = 0.04 kg/s t 1 = 90 C L = 4 m ti = 18 C
p
(^2 1) mc
q t =t −
cp = 4.19 kJ/kg.K
q t 2 = 90 −
t 2 = 90 − 0. 0059666 q Mean Water Temp.
t m = 21 ( t 1 +t 2 )
t m = 21 ( 90 + 90 - 0.0059666q)
t (^) m = 90 − 0.0029833q Equation for Fig. 7-4.
16t 560 W/m L
q = m −
q = ( 4 )(16t m − 560 )
q =64tm − 2240 Substituting: tm = 90 - 0.0029833 (64tm - 2240) tm = 81.182 C
q = 64tm - 2,240 Watts q = 64(81.182) - 2,240 Watts q = 2,956 Watts q = 2.956 kW ---- Ans.
7-2. Compute the pressure drop in pascals per meter length when a flowrate of 8 L/s of 60 C water flows through a Schedule 40 steel pipe of nominal diameter 75 mm (a) using Eq. (7-1) and (b) using Figs. 7-6 and 7-7.
Solution: (a) Eq. 7-1.
∆ = ρ 2
V
D
L
p f
2
From Table 7-3 at 60 C. ρ = 983.19 kg/m 3 μ = 0.476 mPa.s = 0.000476 Pa.s
75-mm Schedule 40 Steel Pipe, Table 7-1, ID = 77.92 mm
1.678m/s 0.07792m 4
0.008m/s V (^2)
3
π
Table 6-1, ε = 0.000046 commercial steel.
D
ε
DV 1.678 0.07792983.
Re = μ
π
Re = 270,
From the Moody Chart, Fig. 6-1. Re = 270,067,
D
ε
f = 0.
∆ = ρ 2
V
D
L
p f
2
L
p 2
∆∆∆∆p/L = 338 Pa/m ---- Ans.
7-3. In the piping system shown schematically in Fig. 7-14 the common pipe has a nominal 75 mm diameter, the lower branch 35 mm, and the upper branch 50 mm. The pressure of water at the entrance is 50 kPa above atmospheric pressure, and both branches discharged to atmospheric pressure. The water temperature is 20 C. What is the water flow rate in liters per second in each branch?
Solution: ∆p = 50 kPa - 0 - 50 kPa = 5000 Pa Use Fig. 7-6, water temperature of 20 C
Table 7-4. For 75-mm pipe Elbow = 4 x 3 m = 12 m Straight Pipe = 8 m + 4 m + 5 m + 7 m + 15 m = 39 m L 1 = 12 m + 39 m = 51 m
2 5 (^ )^32
2 2 3
3 Q
8Q
L
p
- 02 998. 21 = 306 , 352 , 668
π
L
p 51 L
p 2
2 1
1
(5,633,748Q 1 2 )( 51 ) +(40,535,176 Q 22 )(30.9 ) =50,
(5,633,748Q 1 2 )( 51 ) +(40,535,176 Q 22 )(30.9 ) =50,
287,321,148Q 12 + 1,253,093,138Q 22 = 50,
2
2 L
p
L
p 3
3
306,352,668Q 3 2 =1.036913 (40,535,176 Q 22 )
Q 2 = 2.7Q 3
Q 1 = Q 2 + Q 3
Q 1 = 2.7Q 3 + Q 3
Q 1 = 3.7Q 3
Then. 287,321,148Q 12 + 1,253,093,138Q 22 = 50, 287,321,148(3.7Q 3 ) 2
- 1,253,093,138(2.7Q 3 ) 2 = 50, Q 3 = 0.00196 m^3 /s Q 3 = 1.96 L/s - - - Ans. Q 2 = 2.7Q 3 Q 2 = 5.29 L/s - - - - Ans. Q 1 = 3.8Q 3 Q 1 = 7.25 L/s - - - - Ans.
7-4. A centrifugal pump with the characteristics shown in Fig. 7-9 serves a piping network and delivers 10 L/s. An identical pump is placed in parallel with the original one to increase the flow rate. What is (a) the new flow rate in liters per second and (b) the total power required by the two pumps?
Solution: Use Fig. 7-9. At 10 L/s. Pressure Rise, ∆p = 130 kPa Efficiency = η = 62 % P = (0.10)(130) / (0.62) = 2.097 kw
For the pipe network. ∆p α Q
Q 1 = 10 L/s ∆p 1 = 130 kPa
Use trial and error to find Q 2 and ∆p 2 that will lie along the pump curve in Fig. 7-9.
Trial 1, Q 2 = 15 L/s
2
1
2 (^2 1) Q
Q
p p
( ) 292.5kPa
p 130
2 2 =
Each pump = Q = 7.5 L/s
From Fig. 7-9. ∆p = 210 kPa < 292.5 kPa Trial 2, Q 2 = 13.3 L/s
2
1
2 (^2 1) Q
Q
p p
( ) 292.5kPa
p 130
2 2 =
Each pump = Q = 6.65 L/s
From Fig. 7-9. ∆p = 230 kPa ~ 230 kPa Efficiency = 80 % Then Q 2 = 13.3 L/s total - - - Ans. Power = 2(0.00665)(230) / 0. Power = 3.82 kW - - - - Ans.
7-5. An expansion tank is to be sized so that the change in air volume between the cold-water conditions (25 C) and the operating water temperature (85 C) is to be one fourth the tank volume. If pi = 101 kPa abs and pc = 180 kPa abs,., what will ph be?
Solution: Eq. 7-7.
B C
t
h
i b
i V V
V
p
p p
p
VB - VC = 0.25Vt
t
t
h
i b
i 0.25V
V
p
p p
p
p
p p
p h
i c
i (^) − =
p
h
ph = 325 kPa abs. - - - Ans.
