An ideal vapor-compression refrigerant cycle operates at steady state with H,0 a

Question

An ideal vapor-compression refrigerant cycle operates at steady state with H,0 as the working fluid. S aturated vapor enters the compressor at 100°C, and saturated liquid leaves the condenser at working fluid is 60 kg/min. Complete the table below and determine the Qc (MW), Qn The fluid leaves the valve (state 1) as a two-phase mixture. Note that you are not required to compute S for state 1. 1 MPa. The mass flow rate of the . Hint: ae Refrigeration Cycle Valve s Evapoetor LA State- T (°C) P (MI Pal h (kl/kg) | S (kl/Kg K) - Refrigerated unit at low T 100 reservoer 4 1 Compressor

Explanation / Answer

For state 2:

Phase: saturated vapor at 100 C

From the saturated steam table at 100 C,

Psat = 0.101418 MPa

Hv = 2675.57 kJ/kg

Hl = 419.099 kJ/kg

Sv = 7.354 kJ/kg-K

Sl = 1.307 kJ/kg-K

P2 = Psat = 0.101418 MPa

H2 = Hv = 2675.57 kJ/kg.............since saturated vapor

S2 = Sv = 7.354 kJ/kg-K

State 4:

Phase: saturated water at P = 1 MPa

From saturated steam table at 1 MPa,

Tsat = 179.885 C

Hv = 2777.119 kJ/kg

Hl = 762.68 kJ/kg

Sv = 6.584979 kJ/kg-K

Sl = 2.138 kJ/kg-K

T4 = Tsat = 179.885 C

H4 = Hl = 762.68 kJ/kg

S4 = Sl = 2.138 kJ/kg-K

Now,

from state 4 to 1, the fluid passes through the valve, which is an isenthalpic process.

Therefore, H1 = H4 = 762.68 kJ/kg

From state 1 to 2 , pressure remains constant as both the streams are saturated at 100 C.

.: P1 = P2 = 0.101418 MPa

Similarly,

For state 3:

T3 = T4 = 179.885 C

For ideal vapor compression cycle, compression is isentropic.

.: S3 = S2 = 7.354 kJ/kg-K ( this is greater than the Sv at Tsat = 179.885 C, therefore, steam is superheated)

From superheated steam table, look for S = 7.354 kJ/kg-K and T = 179.885 C, we get

P3 = 0.23 MPa

H3 = 2828.448 kJ/kg

Now,

m = 60 kg/min = 1 kg/sec

Qc = m(H2- H1) = 1*(2675.57 - 762.68) = 1912.89 kW = 1.91289 MW

Qh = m(H3 - H4) = 1*(2828.448 - 762.68) = 2065.768 kW = 2.065768 MW

W = m(H3 - H2) = 152.878 kW = 0.152878 MW

COP = Qc / W = 1.91289 / 0.152878 = 12.51

State T C P MPa h kJ/kg S kJ/kg-K 1 100 0.101418 762.68 2 100 0.101418 2675.57 7.354 3 179.885 0.23 2828.448 7.354 4 179.885 1 762.68 2.138

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