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?
[tex]Q= m_{air}C_p(T_2-T_1)[/tex]
5.087 = [tex]m_{air}[/tex] (1.004)(20)
[tex]m_{air}[/tex] = [tex]\frac{5.087}{1.004*20}[/tex]
[tex]m_{air}[/tex] = 0.253 kg/sec
∴ The mass flow rate of air to achieve a temperature of 10°C = 0.253 kg/sec
The mass flow rate of the R‐134a and the low-temperature heat transfer rate and the the mass flow rate of air to achieve a temperature of 10°C are;
m'_R-134a = 0.0403 kg/s
Q'_L = 5.089 kW
m_air = 0.2534 kg/s
From the Standard Refrigeration Cycle Table B.5 attached;
At 200 kPa, we have;
Enthalpy of saturation vapour; h₁ = h_g = 392.28 kJ/kg
Entropy of saturation vapour; s_1 = s_g = 1.7391 kJ/kg.k
At 1200 kPa and by interpolation, we have;
Enthalpy of saturation liquid; h₃ = h₄ = h_f = 266 kJ/kg
Enthalpy of saturation vapour; h₂ = h_g = 429.5 kJ/kg
A) Now, if we assume compressor to be ideal, then;
specific compression work from h₁ to h₂ is;
w_c = h₂ - h₁
w_c = 429.5 - 392.28
w_c = 37.2 kJ/kg
Now, formula for the mass flow rate is;
m' = W'_c/w_c
we are given power input; W'_c = 1.5 kW
Thus;
m' = 1.5/37.2
m' = 0.0403 kg/s
B) The formula for the rate of heat transfer at low temperature for the Evaporator is;
Q'_L = m'(h₁ - h₄)
Q'L = 0.0403(392.28 - 266)
Q'L = 5.089 kW
For the mass flow rate of the air cooler, we will use the formula;
Q'L = m _air * C_p * ∆T
where c_p is a constant = 1.004 kJ/kg
Thus;
m_air = Q'L/(C_p * ∆T)
m_air = 5.089/(1.004 * (30 - 10))
m_air = 0.2534 kg/s
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