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