Answer:
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