Chapter 05

CHAPTER 5 1. A two-stag two-stagee cascade refrige refrigerati ration on system system uses ammonia ammonia as the working working substanc substance. e. The evaporator is at –35 C and the high-pressure condenser is at 1514.2 kPa. The cascade cascade condenser condenser is a direct direct contact type. type. The refriger refrigeratio ation n load is 90 tons. Determine (a) the mass flow rate in the lower pressure loop, (b) the mass flow rate in the high-p h igh-pressur ressuree loop, (c) the power required, required, (d) the COP, and (e) the quantity of the fluid entering the evaporator. Solution:

For cascade condenser pressure  p5

=

 p 6 p1

 p6

=

1514.2 kPa

 p1

=

 p sat  at  − 35 C  = 93.5 kPa

 p5

=

(1514.2) ( 93.5) = 376.3 kPa

CHAPTER 5

Important properties p , kPa

State points 1 2 3 4 5 6 7 8

h , kJ/kg 1415.2 1610 183.6 183.6 1457.8 1670 385.7 385.7

93.5 376.3 376.3 93.5 376.3 1514.2 1514.2 93.5

 a = mass flow rate in the lower pressure loop. (a) m a m

Q A

=

h1 − h4

=

( 90)( 3.516) 1415.2 − 183.6

=

0.257 kg   s

 b = mass flow rate in the high pressure loop. (b) m  b ( h5 m

−

h8 )

=

 a ( h2 m

 b (1457.8 − 385) m a m

=

=

h3 )

=

0.257(1610 − 183.6 )

=

 a ( h2 − h1 ) + mb ( h6 m

0.342 kg   s

 (c) W  T   W  T 

−

=

1 + W  2 W 

−

h5 )

0.257(1610 − 1415.2 ) + 0.342(1670 − 1457.8)

Q  A (d) COP  =  W T 

=

( 90)( 3.516 ) 122.6

=

=

122.6 kW 

2.58

(e) at 4,  p = 93.5 kPa , h = 183.6 kJ  kg  t  = −35 C  h f   = 41.2 kJ  kg , h fg  = 1374.0 kJ  kg  h − h f   183.6 − 41.2 (100% ) = 10.36%  x = = h fg  1374.0 2. A two-stage cascade refrigeration system uses ammonia as the refrigerant. The mass flow rate in the high-pressure loop is 0.10 kg/s. The condenser saturated

CHAPTER 5 temperature is 38 C and the evaporator temperature is – 40 C. The cascade condenser is a direct contact type. Determine (a) the refrigerating effect in tons, (b) the power required, and (c) the COP. Solution: At 38 C,  p = 1472.4 kPa At – 40 C,  p = 72.0 kPa Therefore, cascade condenser pressure = t  sat  = −7.25 C 

(1472.4 )( 72.0 )

=

325.6 kPa

CHAPTER 5

Important properties t  ,

State points 1 2 3 4 5 6 7 8

b m

=

C -40

-7.25 -40 -7.25 38 -7.25

h,

kJ/kg 1407.3 1621 166.6 166.6 1453.5 1680 380.7 380.7

0.10 kg   s

a, For  m  a ( h2 m

−

h3 )

=

 b ( h5 m

 a (1621 − 166.6 ) m a m

=

=

−

h8 )

0.10(1453.5 − 380.7 )

0.074 kg   s

(a) Q  A = refrigerating effect Q  A

=

 a ( h1 − h4 ) m

=

0.074(1407.3 − 166.6 ) = 91.81 kW 

In tons, 91.81 Q  A = = 26.11 tons 3.516

 = m ( h2 (b) W  a

−

 b ( h6 h1 ) + m

−

h5 )

=

0.074(1621 − 1407.3) + 0.10(1680 − 1453.5 ) = 38.5 kW 

CHAPTER 5

(c) COP  =

Q A

 W 

=

91.81 38.5

=

2.38