Question

Refrigerant-22 absorbs heat from a cooled space at 50°F as it flows through an evaporator of a refrigeration system. R-22 enters the evaporator at 10°F at a rate of 0.08 lbm/s with a quality of 0.3 and leaves as a saturated vapor at the same pressure. Determine:

a. The rate of cooling provided, in Btu/h.

b. The rate of exergy destruction in the evaporator.

c. The second-law efficiency of the evaporator.


Take T0 = 77°F. The properties of R-22 at the inlet and exit of the evaporator are: h1 = 107.5 Btu/lbm, s1 = 0.2851 Btu/lbm·R, h2 = 172.1 Btu/ lbm, s^2 = 0.4225 Btu/lbm·R.

Answer 1

a) the rate of cooling provided is 18604.8 Btu/h

b) the rate of exergy destruction in the evaporator is 0.46 Btu/Ibm

c) the second-law efficiency of the evaporator is 37.39%

Step-by-step explanation:

Given that;

Temperature of sink TL = 50°F = 510 R

Temperature at evaporator inlet TI = 10°F = 470 R

mass flow rate m" = 0.08 lbm/s

quality of refrigerant at evaporator inlet x1 = 0.3

quality of refrigerant at evaporator exit x2 = 1.0

T₀ = 77°F = 537 R

h1 = 107.5 Btu/lbm

s1 = 0.2851 Btu/lbm·R,

h2 = 172.1 Btu/ lbm,

s2 = 0.4225 Btu/lbm·R.

a) rate of cooling provided, in Btu/h.

QL = m"( h2 - h1)

we substitute

QL = 0.08( 172.1 - 107.5

= 0.08 × 64.6

= 5.168 Btu/s

we convert to Btu/h

5.168 × 60 × 60

QL = 18604.8 Btu/h

Therefore the rate of cooling provided is 18604.8 Btu/h

b) The rate of exergy destruction in the evaporator

Entropy generation can be expressed as;

S_gen = m"(s2 - s1) - QL/TL

so we substitute

S_gen = 0.08( 0.4225 - 0.2851 ) - 5.168 / 510

= 0.010992 - 0.01013

S_gen = 0.00086 Btu/ibm.R

now the energy destroyed expressed as;

X_dest = T₀ × S_gen

so

X_dest = 537 × 0.00086

X_dest = 0.46 Btu/Ibm

Therefore the rate of exergy destruction in the evaporator is 0.46 Btu/Ibm

c) The second-law efficiency of the evaporator.

Energy expended is expressed as;

X_exp = m"(h1 - h2) - m"T₀(s1 - s2)

we substitute

= 0.08( 107.5 - 172.1 ) - [0.08 × 537 ( 0.2851 - 0.4225 )

= -5.168 - [ - 5.9027)

= -5.168 + 5.9027

= 0.7347 Btu/s

Now second law efficiency is expressed as;

nH = 1 - (X_dest / X_esp)

= 1 - ( 0.46 / 0.7347 )

= 1 - 0.6261

= 0.3739

nH = 37.39%

Therefore the second-law efficiency of the evaporator is 37.39%