A two-stage compression refrigeration system operates with refrigerant-134a between the pressure limits of 1.4 and 0.10 MPa. The refrigerant leaves the condenser as a saturated liquid and is throttled to a flash chamber operating at 0.4 MPa. The refrigerant leaving the low-pressure compressor at 0.4 MPa is also routed to the flash chamber. The vapor in the flash chamber is then compressed to the condenser pressure by the high-pressure compressor, and the liquid is throttled to the evaporator pressure. Assuming the refrigerant leaves the evaporator as saturated vapor and both compressors are isentropic, determine (a) the fraction of the refrigerant that evaporates as it is throttled to the flash chamber, (b) the rate of heat removed from the refrigerated space for a mass flow rate of 0.25 kg/s through the condenser, and (c) the coefficient of performance.

A) q = 0.33

B) Q'_L = 28.565 Kw

C) C.O.P = 2.5

Explanation:

From the table first table i attached, in state 1,at P1 = 0.1MPa, we have;

Specific enthalpy, h1 = 234.46 KJ/Kg

Specific Entropy, s1 = 0.9519 KJ/Kg.k

Now, in state 2, from the second table i attached and by interpolation under the condition s1=s2 at P2 = 0.4MPa,we have;

Specific enthalpy, h2 = 262.74 KJ/Kg

Now, in state 3, from the first table i attached, at P3 = 0.4MPa,we have;

Specific enthalpy, h3 = 255.61 KJ/Kg

Now, in state 5 and 6, from the first table i attached, at P5 = 1.4MPa,we have;

h_f = h5 = h6 = 127.25 KJ/Kg

Now, in state 7 and 8, from the first table i attached, at P7 = 0.4MPa,we have;

h_f = h7 = h8 = 63.92 KJ/Kg

A) The amount of evaporated refrigerant is simply the quality at state 6.

Thus;

q = (h6 - h7)/(h_fg,0.4MPa)

Now,from the third table i attached,

(h_evap,0.4MPa) = h_f,g = 191.62 KJ/Kg

Thus, q = (127.25 - 63.92)/(191.62)

q = 0.33

B) The enthalpy at state 9 is determined from the energy balance of the heat exchanger as;

q = (h9 - h2)/(h3 - h2)

Making h9 the subject, we have;

h9 = qh3 + (1 - q)h2

Thus, plugging in the relevant values to get ;

h9 = 0.33(255.61) + (1 - 0.33)262.74

h9 = 84.35 + 176.04

h9 = 260.39 KJ/Kg

The entropy at this state and value of enthalpy is done by interpolation from the third table i attached at P9 = 0.4 MPa

This gives; s9 = 0.9483 KJ/Kg.k

The enthalpy at state 4 and is done by interpolation from the third table i attached under the condition s4 = s9 and P4 = 1.4MPa.This gives;

h4 = 287.13 KJ/Kg

The rate of heat removal is given by the formula;

Q'_L = m'_b(h1 - h8)

This can be simplified into;

Q'_L = m'_a(1 - q)•(h1 - h8)

Plugging in the relevant values to obtain;

Q'_L = 0.25(1 - 0.33)•(234.46 - 63.92)

Q'_L = (0.25 x 0.67) x 170.54

Q'_L = 28.565 Kw

C) The formula for COP is;

COP = Q'_L/W

This gives;

C.O.P = Q'_L/[m'_a(h4 - h9) + m_b(h2 - h1)]

C.O.P = Q'_L/[m'_a((h4 - h9) + (1 - q) (h2 - h1)]

Plugging in the relevant values to obtain;

C.O.P = 28.565/[0.25((287.13 - 260.39) + (1 - 0.33)(262.74 - 234.46)]

C.O.P = 28.565/(0.25(26.74 + 18.9476)

C.O.P = 28.565/11.4219 = 2.5