Mechanical Engineering - Heat Transfer, Refrigeration and Air Conditioning - Discussion

To find the network of compression for a refrigerating system operating on a reversed Brayton refrigeration cycle, you can use the following formula for the work of compression:

Work of Compression (Wc) = m * Cp * ΔT

Where:
- Wc is the work of compression.
- m is the mass flow rate of the working substance (air in this case).
- Cp is the specific heat capacity at constant pressure (given as 1 kJ/kg·K for air).
- ΔT is the change in temperature during compression.

In the Brayton refrigeration cycle, the air undergoes two isentropic processes: compression and expansion.

Given:
- Temperature at the end of constant pressure cooling (T2) = 300 K,
- Desired temperature inside the refrigerator (T1) = 250 K,
- Rise in temperature of air in the refrigerator (ΔTrise) = 50 K.

The change in temperature during compression (ΔT) can be calculated as:

ΔT = T2 - T1.
ΔT = 300 K - 250 K,
ΔT = 50 K

Now, plug the values into the formula for the work of compression:

Wc = m * C_p * ΔT
Wc = m * 1 kJ/kg * 50 K

Now, we need to find the mass flow rate (m). In a refrigeration cycle, the mass flow rate is related to the cooling load (Qc) and the change in enthalpy (ΔH) during the cooling process:

Qc = m * ΔH

Since the air is cooled from T2 to T1 at constant pressure, ΔH can be calculated as:
ΔH = Cp * ΔTrise
ΔH = 1 kJ/kg * 50 K
ΔH = 50 kJ/kg

Now, rearrange the formula for Qc to solve for m:
m = Qc/ΔH

Now, we need to calculate Qc. The cooling load is the heat transfer required to maintain the desired temperature inside the refrigerator:
Qc = m * Cp * ΔTrise
Qc = m * 1 kJ/kg * 50 K

Now, plug in the values for Qc and ΔH to calculate m:
m = (m * 1 kJ/kg * 50 K)/(50 kJ/kg)
m = m

The mass flow rate (m) cancels out, which means it does not affect the calculation of the work of compression. Therefore, we can use the initial value of ΔT to calculate the work of compression:
Wc = m * 1 kJ/kg * 50 K
Wc = 50 kJ

So, the network of compression is 50 kJ.

Thank you @Atul Palsara.

The question is tricky in its statement.

The rise in the temperature of the air is 50 K in the refrigerator (it could be in the evaporator or the compressor). The temperature rise of 50 K may be due to compression hence W compressor = m cp dt = 1* 50 = 50 Kj/Kg and option B is the right answer.

You are right @Lalit Aheer,

If the temperature rise mentioned here is in the evaporator then the right answer is option D.

25 KJ/Kg is the correct answer.

T2*T4 = T1*T3.
T2 = (250*300)/200 = 375 K.
Work of compression = (h2-h1)-(h3-h4).
= Cp(T2-T1-T3+T4).
= 25 KJ/Kg.

T2/t1=t3/t4.
T1=250.
T2?
T3=300.
t1-t4=50, t4= 200 and t2 =375.
Comp work =Cp(t2-t1) 1*(375-250)=125 kj/ kg is the right answer.

How the Ambient air is 350 k?

Here ambient air will be at 350K.

Air after cooling effect = 300K (air temp inc as it act as refrigerant).

Air before cooling effect = 250K.

Cooling effect = (300-250) = 50 KJ/KG.

Total heat rejected to atm= (350-250) = 100KJ/KG.

Hence work done by compressor= thr - ce = 100-50= 50KJ/KG.

Please explain.