Question

A 10 kW cooling load is to be served by operating an ideal vapor-compression refrigeration cycle with its evaporator at 400 kPa and its condenser at 800 kPa. Calculate the refrigerant mass flow rate and the compressor power requirement when refrigerant-134a is used. (Take the required values from the saturated refrigerant-134a tables.)

Answer 1

Answer:

$\dot m = 0.062456 kg/sec$

b) w = 0.927211 kW

Explanation:

from table R-134 a

state 1  

P-1 = 400 kPa

h_1 = [email protected] kpa = 255.55 kJ?kg

s1 = s2 = 0.92961 kJ/kg K

state 2

P-2 = 800 kPa

s2 =s1 = 0.92691

s2> [email protected] 800 kPa = 0.92691 kJ/ kg K

from superheated table for R-134 a at pressure of 800 kPa

using interpolation for obtaining desire value of temperature and enthalapy

$\frac{T_2 - 31.31}{40 - 31.31} = \frac{0.9306 - 09183}{0.9480 - 0.9183}$

T_2 = 34.75 degree C

FOR  ENTHALAPY

$\frac{h_2 - 267.29}{276.45 - 267.29} = \frac{0.92691 - 0.9183}{0.9480 - 0.9183}$

h_2 = 270.3928 kJ/kg

state 3

$P_3 = 800 kPA$

$H_3 = Hf@800 kPa = 95.47 kJ/kg$

state 4  ( h remain constant)

$h_4 =h_3 = 95.47 kJ/kg$

a) mass flow rate $\dot m = \frac{cooling head}{\phi _{in}}$

$\dot m = \frac{10 kW}{255.55 - 95.47}$

$\dot m = 0.062456 kg/sec$

b) compressor power $w = \dot m [h_2 -h_1]$

w = 0.062456(270.3928 - 255.55)

w = 0.927211 kW