- 2074 Bhadra · 8 marks
Air compressor: heat transfer rate
Short answer
Mass flow rate from ideal gas: 0.02904 kg/s. Using steady-flow energy equation, q = c_p(T2-T1) + w = 100.5 - 250 = -149.5 kJ/kg. Heat transfer rate Q̇ = ṁ q = -4.34 kW (heat loss).
Air enters a compressor operating at steady state at 100 kPa, 300 K and leaves at 1000 kPa, 400 K, with a volumetric flow rate of 1.5 m³/min. The work consumed by the compressor is 250 kJ per kg of air. Neglecting the effects of potential and kinetic energy, determine the heat transfer rate in kW. [Take R = 287 J/kgK and Cp = 1005 J/kgK]
Given: steady flow compressor, air: P₁ = 100 kPa, T₁ = 300 K; P₂ = 1000 kPa, T₂ = 400 K; inlet volume flow = 1.5 m³/min; work consumed w = 250 kJ/kg (so w_out = −250 kJ/kg). ΔKE = ΔPE = 0.
Step 1: Mass flow rate (ideal gas at inlet)
V̇₁ = 1.5/60 = 0.025 m³/s
ṁ = P₁ V̇₁/(R T₁) = (100 × 0.025)/(0.287 × 300) = 0.02904 kg/s
Step 2: Energy equation (per unit mass)
q − w = h₂ − h₁ = c_p (T₂ − T₁)
h₂ − h₁ = 1.005 × (400 − 300) = 100.5 kJ/kg
q = c_p ΔT + w = 100.5 + (−250) = −149.5 kJ/kg
Step 3: Heat transfer rate
Q̇ = ṁ q = 0.02904 × (−149.5) = −4.34 kW
Answer: Q̇ ≈ −4.34 kW; heat is lost from the compressor at a rate of about 4.34 kW.
✓ Answer checked by Claude Opus 5.5 (high effort)
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