PROBLEM 6.100

Refrigerant 22 in a refrigeration system enters one side of a counter-flow heat exchanger at 12

bar, 28oC. The refrigerant exits at 22 bar, 20

oC. A separate stream of R-22 enters the other side

of the heat exchanger as saturated vapor at 2 bar and exits as superheated vapor at 2 bar. The

mass flow rates of the two streams are equal. Stray heat transfer from the heat exchanger to its

surroundings and kinetic and potential energy effects are negligible. Determine the entropy

production in the heat exchanger, in kJ/K per kg of refrigerant flowing. What gives rise to the

entropy production in this application?

KNOWN: Two streams of R-22 pass though opposite sides of a counter-flow heat exchanger

operating at steady state with equal mass flow rates. Data are known for each stream.

FIND: Determine the entropy production for the heat exchanger per unit mass of refrigerant

flowing.

SCHEMATIC AND GIVEN DATA:

ENGINEERING MODEL: (1) The control volume is at

steady state. (2) . (3) Kinetic and potential energy

effects are negligible.

ANALYSIS: To fix state 4, we write mass and energy rate balances. The mass balances reduce

at steady state to and . Further,

0 = + [(h1 – h2) + (h3 – h4)] → h4 = h1 – h2 + h3

From Table A-7: h1 ≈ hf(28oC) = 79.05 kJ/kg and s1 ≈ sf(28

oC) = 0.2936 kJ/kg∙K

h2 ≈ hf(20oC) = 69.09 kJ/kg and s2 ≈ sf(20

oC) = 0.2607 kJ/kg∙K

From Table A-8: h3 = hg(2 bar) = 239.88 kJ/kg and s3 = sg(2 bar) = 0.9691 kJ/kg∙K

h4 = h1 – h2 + h3 = 79.05 – 69.09 + 239.88 = 249.84 kJ/kg

Interpolating in Table A-9: T4 ≈ - 9.815oC and s4 ≈ 1.0081 kJ/kg∙K

The entropy rate balance reduces as follows: 0 =

+ [(s1 – s2) + (s3 – s4)] +

Thus

= (0.2607 – 0.2936) + (1.0081 – 0.9691) = 0.0061 kJ/ kg∙K

The entropy production is due to irreversible heat transfer between the two streams. There

would be a small effect of frictional pressure drop, but pressure drops have been ignored.

12 bar (1)

28oC

(2) 12 bar

20oC

(3)

2 bar

sat. vapor

(4)

2 bar

sup. vapor

s

T

(1) 28

oC

-25.8oC

T4 2 bar

12 bar

(2)

(3)

(4)

20oC

. .

. .

PROBLEM 6.107

The figure

PROBLEM 6.107 (CONTINUED)

PROBLEM 6.119

PROBLEM 6.121

PROBLEM 6.138