The air conditioner in an automobile uses R-134a as the refrigerant. The compressor power input is 1.5 kW which brings the R-134a from 200 kPa to 1200 kPa by compression. The cold-space heat exchanger cools atmospheric air from an outside temperature of 30°C down to 10°C and blows it into the car. What is the mass flow rate of the R‐134a and what is the low-temperature heat transfer rate? What is the mass flow rate of air to achieve a temperature of 10°C? Assume air is an ideal gas and there is 100% heat transfer in the evaporator.

Mass flow rate of R-134a is m = 0.0403 kg/s

Hence,  the low-temperature heat transfer rate (Q) = 5.087 kW

The mass flow rate of air to achieve a temperature of 10°C = 0.253 kg/sec

Explanation:

Given that:

The compressor power input is 1.5 kW,

Which brings the R-134a from 200 kPa to 1200 kPa by compression.

i.e The compressor inlet pressure P₁ = 200 kPa and the compressor outlet pressure P₂ = 1200 kPa

The cold-space heat exchanger cools atmospheric air from an outside temperature of 30°C down to 10°C and blows it into the car.

This is telling us more information about the Maximum Temperature and Minimum Temperature.

Maximum Temperature = 30 °C

Minimum Temperature = 10 °C

Change in temperature (ΔT) = 20 °C

At P₁ = 200 kPa from saturated P-134a tables.

h₁ = hg = 392.28 kJ/kg

s₁ =sg = 1.7319 kJ/kg. K

At P₂ = 1200 kPa from super-heated tables

s₂ = sg = 1.7319 kJ/kg.K

      h₂ = 429.526 kJ/kg

To determine the mass flow rate of the R-134a,we need to first find the Specific compressor work which is given  by the formula:

w = h₁ - h₂

   where h₁ = 392.28 kJ/kg

               h₂ = 429.526 kJ/kg

w = (392.28 - 429.526) kJ/kg

w = -37.246 kJ/kg

W = mw

-1.5 = m × (-37.246)

m = 0.0403 kg/s

Mass flow rate of R-134a is m = 0.0403 kg/s

If P₂ = 1200kPa, then hf= h₃= h₄= 266.055 kj/kg

Therefore; the low-temperature heat transfer rate(Q) can  be calculated by the formula;

Q= m(h₁ - h₄)

Q= 0.0403 ( 392.28 - 266.055)

Q= 5.087 kW

Hence,  the low-temperature heat transfer rate (Q) = 5.087 kW

What is the mass flow rate of air to achieve a temperature of 10°C?

5.087 = (1.004)(20)

=

= 0.253 kg/sec

∴ The mass flow rate of air to achieve a temperature of 10°C = 0.253 kg/sec

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explanation: