USA Elementary Steady Flow Chapter 05
October 14, 2022 | Author: Anonymous | Category: N/A
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1. A propulsion turbine receives steam at the throttle at 875 psia and
A TH
940F at the rate of 100,000 lb/hr. After an irreversible expansion process, the steam exhaust s from the turbine at a pressure of 0.60 psia with a moisture content of 10%. Assume the difference between the entrance and exit kinetic k inetic energies is negligible and find:
TH =
(A) The work done, Btu/lb (B) The power develop, hp
m2 0.10
T1 940F
From Steam Table:
(A )
wk12 J
Wk12 '
2545J
550g c
125,000(2864.8)(32.2) 3600(550)(32.2)
= 180.9 hp
W HP 180.9 = = 301.5 hp ep 0.60
10,000(481.5) 2545(1)
wk 12
= h1 - h2 218.59 222.30 = (-) 3.17Btu/lb J wk12(actual) wk12(isentropic) (-) 3.17 (E) = = = (-) 6.18 Btu/lb J 0.60 J epump
= 481.5Btu/lb
M' wk12
M TH TH g
s1 = s2 0.36772 and P2 = 1200 psia h2 = 22 222. 2.30 30
(B) M' = 100,000 lb/hr ; 1 hp = 2545 Btu/hr
= 2864.8 ft
'
W HP
= h1 - h2 = 1475.6 - 994.1
= 18,920 hp
wk12
(F) h 2 = h1 -
= 218.59 218.59 + 6.18 = 224.77 224.77 Btu/lb J Entering Table 4 and interpolating between 1000 and 500 psi gives:
2. A boiler receives feed water at 1200 psia and 250F and delivers
t
h
steam from the superheater at 900 psia and 950F. Find the heat added, Btu/lb. Refer to Fig. 5-3 5-3
250C
22 221.0 1.03 3
300C
27 271.8 1.83 3
224.77 - 221.03
t =
P1 = 12 1200 00 ps psia ia
t1 = 250 250F
P2 = 90 900 0 ps psiia
t 2 = 95 950 0F
x 50 50 3.68 271.83 - 221.03 t 2 = 250 250 + 3.7 = 253.7 253.7F
q12 = h2 - h1 From Table 4: h1 = 220.61 + 2 (1.04) = 221.0 5 From From Table 3: h2 = 14 1480.5 80.5 Btu/lb q12 = 148 1480.5 0.5 - 221 = 125 1259.5 9.5 Btu/ Btu/lb lb
5. A water cooled reciprocating air compressor takes in air at 15 psia and 60 F and discharges it at 60 psia and 200F. Heat is removed in the amount of 21.4 Btu/lb. Assume steady flow conditions and the work done, Btu/lb.
3. If, in the preceding example, the feed water entering has a velocity of 3 m/s and the steam leaving the super heater has a velocity of 50 m/s. Find: (A) The additional heat required to accommodate the change in kinetic energy across the boiler, J/kg J/kg
P1 = 15 psi psia a
T1 = 40+ 460 = 520R
P2 = 60 ps psia ia
T2 = 2 200 00 + 46 460 0 = 66 660 0R
q12 =
- 21.4 Btu/lb
(B) The percentage error introduced by neglecting the kinetic energy change.
wk12 = q12 + h1 - h2 J where : h 1 - h2 C p T 0.24(520 660) 33.6
wk12 = -21.4 - 33.6 33.6 = - 55 Btu/l Btu/lb b (on) (on) J (A A)) q12 = h2 - h1 -
(D) Interpolating Interpolating in Table 4 of the steam tables with:
h1 1475.6 Bt Btu/lb h2 1098.6 (0.10)(1045.4) 994.1
58.82
(C) BHP =
P2 = 0.60 psia
;
144120 1200-2 0-29.8 9.8
B
P1 = 875 psia
1 g 32.2 where: = wh = = 58.82 lb f /ft 3 g 0 . 0 1 7 ( 3 2 . 2 ) 1 c
P2 P 1
V2 2 -V12 (50)2 (3)3 1245 J / kg 2gc 2(1)
J / kg 3 J//kg 2326 = 2930 x 10 J lb Btu Btu / lb 1245.5 x 100 Error = 0.043% 3
(B)q12 = 1259.2
6. A sample of steam is removed from a steam line where the pressure is 215 psia and passed through an Ellison throttling calorimeter. The calorimeter thermometer reads 250 F and the barometer is standard. Find for the line steam:
Btu
2930 x 10
A. B. C. D.
The enthalpy The quality The entropy The temperature to the nearest whole degree
4. Saturated water at 250F enters a centrifugal main feed pump and is discharges at 1200 psia. The pump efficiency is 60% and the delivery rate is 125,000 lb/hr. Find: (A) (B) (C) (D) (E) (F)
The total head developed by the pump, ft ft The water horsepower, WHP The brakepower, BHP The ideal (isentropic) pump work, Btu/lb The actual pump work, Btu/lb The estimated temperature of the water at discharge.
P1 = 215 psia P2 = 14.6 14.696 96 ps psia ia
T2 = 25 250 0F
(A) from table 3: h2 = 1 116 168. 8.8 8 h1 = h2 = 116 1168. 8.8 8B Btu tu/m /min in th thro rott ttli ling ng(h (h=c =con onst stan ant) t) (B) (B) from from ttab able le 2 fo forr 21 215 5 ps psia ia:: hg 12 1200 00.3 .3;; hfg 838.1
h1 hg - m1hfg
t1 = 250F (sat.w (sat.water) ater)
; P1 = 29.8 psia
1 0.017001 ft / lb
; P2 1200 psia
m1
h1 218.59 Btu/lb
; epump 60%
3
M'=125,000 lb/hr V1 V2 ; Z1 Z 2
hg - h1 hfg
1200.3 1168.8 31.5 0.0376 838.1 838.1
x1 = 1 - m1 = 1 - 0.037 0.0376 6 = 0.962 0.9624 4 = 96.24 96.24% %
(C) s1 sg m1sfg
1.54 5403 03 0.03 0.0376 76(0 (0.9 .988 887) 7) 1.50 1.5031 31 Btu/l Btu/lb. b.R R 1.
(D) t1 387.97 388 F
7. In a lube oil cooler, oil enters at 140 F and leaves at 100 F, at the
rate of 400 lb/min. The cooling medium is sea water , which enters at 60F. The average specific heat of the oil is 0.50 Btu/lb.F and the salt water is 0.94 Btu/lb.F. If the flow of the sea water is at the rate of 500 lb/min, find the t he overload discharge temperature.
t 2 100F
t1 = 40°F
t 3 60F
M' A = 40 400 0 lb/mi lb/min; n; C A = 0.50 0.50 Btu Btu/l /lb. b.F F
M'B = 50 500 0 lb/m lb/miin; CB 0. 0.94 94 Btu/ Btu/llb. b.F F M' A h1 - h2 M'B h4 - h3 Btu / min M' AC A t1 - t 2 M'BCB t 4 - t 3 Btu / min
t4
400 0.50 140-100
- t3 =
M'o = 60, 60,000 000 lb/hr lb/hr
= 17
Co 0.50 Btu/lb.F 2
A = 258 ft A. Q' = M'oCo t1 t 2 (60,000)(0.50)(145 120)
50 500 0 0. 0.94 94 t 4 - 60 = 17 t 4 = 77°F
= 750,000 Btu/hr
8. Steam enters the condenser of propulsion plant at 0.50 psia and a
quality of 89 percent at the rate of 100,000 lb/hr and with a velocity of 1000 ft/s. It leaves the condenser hotwell as saturated liquid without any change in pressure but a velocity of 10 ft/s. The salt water inlet ( injection ) temperature is 70 F and the discharge (overboard ) temperature is 85 F. Sea water has a specific heat of 0.94 Btu/lb.F and a density of 64 lb/ft 3 . The injection and overboard velocities are substantially equal. Calculate the following:
m
1 2 55 45 49.8F 1 55 ln ln 45 2
(A) The rate at which energy energy is extracted extracted from the conde condensing nsing steam as heat, (B) The flow of sea water required, gallons gallons per minu minute te (gpm)
B. For parallel flow: Q' 750,000 Btu/hr U = 58.4 58.4B Btu tu//hrhr-ft 2 F
P1 = P2 0.50 psia
t 3 70F
m1 1 0.89 0.11
t 4 85F
V1 1 0 00 00 ft/s t/s
CB 0.94 Btu/l /lb b.F
V2 10 ft/s
((A) A) q12 = h2 - h1
V22 V 12 2g c J
h1 1096.2 0.11 (1048.6 ) 980.9 Btu/lb
h2 47.7
q
2
2
10 1000 = 953.2 2 32.2 778
= 47.7 980.9 +
12
m
A=
1 2 70 30 47.2F 70 ln 1 ln 30 2
Q' 750, 000 = 272 ft 2 U m 58.4 47.2
10. A steam superheater has a net heat transfer area of 1620 ft 2
B = 64 lb/ft lb/ft 3
M' A = 100 100,00 ,000 0 lb/hr lb/hr ; 1 ft 3 = 7.4 7.481 81 gal
Btu lb
and a design capacity of 221,000 lb of steam/hr when receiving saturated steam at 650 psia and discharging at 850F with a pressure drop if not of not more than 25 psi and through heater. The design heat transfer coefficient is 30 Btu/hr.ft 2.F. In operation of superheater receives 220,000 lb of saturated steam per hour at at a pressure of 650 psia and discharges against a pressure of 630 psia with a temperature s at entrance and exit are 2100F and 1430F, respectively. Calculate the operating heat transfer coefficient using counter flow log mean temperature difference.
M' A q12 =
100, 00 000 953.2
(b) (-)Q'12 Q '34 M'B
1,588 ,588,7 ,700 00 Bt Btu/ u/mi min n 60 M 'B xCB t 4 t3
Q '34 1,588,700 = 112,670 lb/min CB t 4 t 3 0.94(85 70)
flow, (gpm) = 7.481
gal M ' ft 3 B
7.481112,670 112,670 64
= 13,170 gpm
9. A counter-flow lubricating oil cooler with a net heat transfer area of 258 ft2 cools 60,000 lb of oil per hour from a temperature of 145F
at inlet to 120F at discharge. The temperatures of the cooling water are 75F and 90F respectively, and the specific heat of the oil is 0.50 Btu/lb.F. Calculate:
Q ' Ms hs 220, 000 1434.2 1203.1 51x10 6 Btu / hr A. B.
2 The value conditions, of the overall heat transfer coefficient under these operating Btu/hr.ft F, and the required area for a parallel flow d device evice having the same capacity under identical operating conditions.
m 1250 935 1085F 1250 ln 935
51x 106 Q' 29 Btu / hr ft 2F U A 1620 1085
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