Question

A compressor receives 0.1 kg / s of R-134a at 150 kPa, − 0 ° C and delivers it at 1200 kPa, 50°C. The power input is measured to be 3 kW. The compressor has heat transfer to air at 100 kPa coming in at 20°C and leaving at 30°C. What is the mass-flow rate of air?

Answer 1

Answer:

The mass flow rate of air through the compressor is = 1.044 $\frac{kg}{s}$

Explanation:

Mass flow rate refrigerant $m_{ref}$ = 0.1 $\frac{kg}{s}$

$T_{1}$ = - 0 °c = 273 K

$T_{2}$ = 50 °c = 323 K

$W_{in}$ = 3 KW

$T_{3}$ = 20 °c = 293 K

$T_{4}$ = 30 °c = 303 K

From the continuity equation

$m_{1} h_{1} + W_{in} + m_{a} h_{3} = m_{2} h_{2} + m_{a} h_{4}$

$m_{1} = m_{2}$ = 0.1 $\frac{kg}{s}$

$m_{1} h_{1} - m_{2} h_{2} + W_{in} = m_{a} h_{4} - m_{a} h_{3}$

$m C_{p} (T_{1} - T_{2} ) + W_{in} = m_{a} C_{a} (T_{4} - T_{3})$

Put all the values in above formula

0.1 × 1.5 × ( 273 - 323 )  + 3 = $m_{a}$ × 1.005 × ( 303 - 293 )

10.5 = $m_{a}$ × 10.05

$m_{a}$ = 1.044 $\frac{kg}{s}$

Therefore the mass flow rate of air through the compressor is = 1.044 $\frac{kg}{s}$