Question

reciprocating air compressor taken in 2m3/min air at 0.11MPa, 293K which it delivers at 1.5 Mpa, 384 K to an after cooler where the air where the air is cooled at constant pressure to 298 K. the power absorbed by the compressor is 4.15 kW. Calculate the heat transfer in (i) the compressor (ii) the cooler. State your assumptions.

Answer 1

The compressor removes 143842.92 J/s of heat from the air, while the cooler removes 2449.45 J/s. This analysis provides insights into the energy transfer within the air compressor system.

Reciprocating Air Compressor Heat Transfer Analysis

Assumptions:

Air behaves as an ideal gas.

Specific heats of air are constant.

Process in the cooler is isobaric.

Given Data:

V_in = 2 m^3/min (intake volume flow rate)

P_in = 0.11 MPa (intake pressure)

T_in = 293 K (intake temperature)

P_out = 1.5 MPa (discharge pressure)

T_out = 384 K (discharge temperature)

T_cooler = 298 K (cooler temperature)

W_comp = 4.15 kW (power absorbed by the compressor)

Solution:

Calculate the mass flow rate:

m_dot = V_in * ρ

ρ = P_in / (R * T_in)

m_dot = 2 * (0.11 * 10^6) / (8.314 * 293) = 0.0885 kg/s

Calculate the enthalpy change in the compressor:

ΔH_comp = W_comp / m_dot = 4.15 * 10^3 / 0.0885 = 47093 J/kg

Calculate the heat transfer in the compressor:

Q_comp = m_dot * Cp * (T_out - T_in) - ΔH_comp

Cp = 1005 J/(kg*K)

Q_comp = 0.0885 * 1005 * (384 - 293) - 47093 = -143842.92 J/s

Calculate the heat transfer in the cooler:

Q_cooler = m_dot * Cp * (T_out - T_cooler)

Q_cooler = 0.0885 * 1005 * (384 - 298) = 2449.45 J/s

Results:

Heat transfer in the compressor: -143842.92 J/s (negative sign indicates heat removal)

Heat transfer in the cooler: 2449.45 J/s