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CHAPTER 3

1. A Freon 12 refrigeration system, with a 10-TR capacity and a coefficient of performance of 3.23, operates with a condenser pressure of 984.5 kPa and an evaporator pressure of 150.8 kPa. The temperature entering and leaving the compressor are –10 C and 80 C, respectively. The temperature entering the expansion valve is 34 C. The compressor is water jacketed and operated with unknown pressure drops through the valves. Compute (a) the mass flow rate per minute, (b) the indicated work, and (c) the heat removed by the jacket water.

Solution:

Important Properties State points 1 2 3 4 (a) Mass flow rate

p , kPa 150.8 984.5 984.5 150.8

t, C -10 80 34 -20

h , kJ/kg 348.5 396.5 232.5 232.5

CHAPTER 3

QA (10)(3.516) = 0.3031 kg s = h1 − h4 348.5 − 232.5 m& = (0.3031)(60 ) = 18.20 kg min m& =

Q (10)(3.516) = 10.89 kW (b) W& = A = COP 3.23

(c) Qloss = W& − m& ∆h ∆h = h2 − h1 = 396.5 − 348.5 = 48 Btu lb m& (h2 − h1 ) = (0.3031)(396.5 − 348.5) = 14.55 kW since W& < ∆H , there is an error in this problem. (80 C). 2. A test of a 10-TR ammonia vapor compression following results: Condenser pressure Evaporator pressure Temperature leaving evaporator coils Temperature entering compressor Temperature leaving compressor Temperature entering condenser Temperature leaving condenser Coefficient of performance (A) (B) (C) Solution:

refrigeration system gave the 1600 kPa 191 kPa -10 C 0C 100 C 80 C 35 C 3.1

Find heat lost or gained, (a) between evaporator coils and compressor, (b) between compressor and condenser, and (c) to condenser water. Find, (a) temperature in the evaporator coils in saturated state, and (b) quantity of the vapor in the evaporator coils following expansion through valve. Find, (a) the work, and (b) the heat absorbed by jacket water.

CHAPTER 3

Important Properties State points 1 2 3 4 5 6

p , kPa 191 1600 1600 1600 191 191

t, C 0 100 80 35 -20 -10

QA (10)(3.516) = 0.032 kg s = h1 − h4 1460 − 366.1 m& = (0.032)(60 ) = 1.92 kg min m& =

(A)

(a) Q1 (b/w evaporator coils and compressor)

Q1 = m& (h1 − h6 ) = (1.92 )(1481 − 1460 ) = 40.32 kJ min (b) Q2 (b/w compressor and condenser)

Q2 = m& (h2 − h3 ) = (1.92)(1661 − 1602) = 113.3 kJ min (c) QR (to condenser water)

QR = m& (h3 − h4 ) = (1.92)(1602 − 366.1) = 2377 kJ min (B)

(a) t5 = temperature in the evaporator coils = − 20 C (b) x = quantity of vapor in the evaporator coils.

h , kJ/kg 1481 1661 1602 366.1 366.1 1460

CHAPTER 3

x=

h5 − h f h fg

at 20 C, h f = 108.6 kJ kg h fg = 1328.6 kJ kg x=

(C)

366.1 − 108.6 (100% ) = 19.38% 1328.6

(10)(3.516) = 11.34 kW Q& (a) W& = A = COP 3.1 & & (b) Q j = W − m& (h2 − h1 ) = 11.34 − 0.032(1661 − 1481) = 5.58 kW Q& = (5.58)(60 ) = 334.8 kJ min j

3. A refrigerant 12 refrigerating system operates with a condensing temperature of 38 C and an evaporating temperature of –12 C. The refrigerant leaves the evaporator and enters the compressor at –2 C. The compressor is a 4-cylinder, Varrangement single-acting type, direct driven by an electric motor at 875 rpm. The clearance is 5 percent, the capacity is 40 tons, bore-to-stroke ratio is 1.25 and the compression is polytropic with n = 1.275 . Determine (a) the bore and stroke, (b) the piston speed, and (c) the indicated hp of the compressor. Solution:

Important Properties State points 1 2 3

p , kPa 203.90 914.23 914.23

t, C -2

h , kJ/kg 352.5

38

236.5

v , m3/kg 0.088

CHAPTER 3 4

203.90

-12

236.5

Solving for m , QA (40)(3.516) = 1.2124 kg s m& = = h1 − h4 352.5 − 236.5 1

1

 p n  914.23  1.25 η v = 1 + c − c 2  = 1 + 0.05 − 0.05  = 0.7113  203.90   p1  (a) For bore and stroke, D = 1.25 L V&1′ = m& ν 1 = (1.2124 )(0.088) = (0.1067 m 3 s )(60 s min ) = 6.4 m3 min

ην V&D = V&1′ (0.7113) π (1.25L )2 (L )(875)(4) = 6.4 4 L = 0.1280 m L = 12.80 cm D = 1.25L = 16.0 cm (b) Piston Speed, V p V p = 2 Ln = 2(0.128)(875) = 224 m min n −1 0.275       np1V1  p2 n ( 1.275)(203.90)(6.4)  914.23  1.275     − 1 = (c) IP = − 1     203.90   n − 1  p1  0.275     IP = 2312 kJ min

IHP =

2312 = 51.6 hp (60)(0.746)

4. A food freezing requires 20 tons of refrigeration at an evaporator temperature of –30 C and a condenser temperature of 35 C. The refrigerant Freon 22 is subcooled 4 C before entering the expansion valve and the vapor is superheated 5 C before leaving the evaporator. The clearance is 4%, the suction and discharge valve pressure drops are 35 kPa and 50 kPa, respectively. Compression is polytropic with n = 1.08 . The mechanical efficiency is 75%. An 11.5 x 11.5 – cm, six cylinder single acting compressor running at 1000 rpm is used. Determine (a) clearance volumetric efficiency, (b) piston displacement, (c) total volumetric efficiency, (d) theoretical hp, (e) actual hp, (f) heat rejected to condenser cooling

CHAPTER 3 water, and (h) quantity of cooling water in the condenser if the temperature rise is 6 C. Solution:

p1= p6 − 35 kPa p 2 = p3 + 50 kPa

Important Properties State points 1 2 3 4 5 6 m& =

p , kPa 128.5 1404.8 1354.8 1354.8 163.5 163.5

t, C

h , kJ/kg 396.0

31 -30 -25

237.9 237.9 396.0

QA (20)(3.516) = 0.4448 kg s = h1 − h4 396 − 237.9

V&1′ = m& ν 1 = (0.4448)(0.175)(60 ) = 4.67 m3 min n −1 0.08     np1V1  p2  n ( 1.08)(128.5)(4.67 )  1404.8  1.08     W= −1 =   −1   128.5   n − 1  p1  0.08    

v , m3/kg 0.175

CHAPTER 3  1 min   = 26.2 kW W = (1570 kJ min )  60 s 

v2  p1  =  v1  p2 

1 n

1

v2  128.5 1.08 =  0.175  1404.8  v2 = 0.019 m 3 kg at p2 = 1404.8 kPa , v2 = 0.019 m 3 kg h2 = 440 kJ kg h3 = h2 1 n

1

p   1404.8 1.08 (a) η v = 1 + c − c 2  = 1 + 0.04 − 0.04  = 0.6742 = 67.42%  128.7   p1  V& 4.67 (b) V&D = 1 = = 6.93 m3 min ηv 0.6742 V& (c) η vT = 1 V&DT

π 2 V&DT = (0.115) (0.115)(1000 )(6 ) = 7.17 m3 min 4 4.67 η vT = (100% ) = 0.6513(100% ) = 65.13% 7.17 (d) TP = W& = 26.2 kW 26.2 THP = = 35.12 hp 0.746 (e) actual hp =

THP

ηm

=

35.12 = 46.83 hp 0.75

(f) Q& j (heat rejected during compression)

CHAPTER 3 Q& j = W& − m& (h2 − h1 ) = 26.2 − 0.4448(440 − 396) = 6.63 kW Q& j = (6.63)(60) = 398 kJ min

(g) Q& R = m& (h3 − h4 ) = 0.4448(440 − 237.9 ) = 89.9 kW Q& = (89.9 )(60 ) = 5394 kJ min R

(h) Q& R = m& wcw∆t 5394 = m& w (4.187 )(6 ) m& w = 214.7 kg min