--- title: "Air compressor: heat transfer rate" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 4 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:compressor-air-heat-transfer-rate" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/compressor-air-heat-transfer-rate site: BCT Engineering --- # Air compressor: heat transfer rate Fundamental of Thermodynamics & Heat Transfer · Chapter 4: First Law of Thermodynamics - Asked in: 2074 Bhadra, 8 marks - Video: [4.58 Air enters a compressor operating at steady state with a pressure of 14.7 lbf/in2 a temperature](https://www.youtube.com/watch?v=G824IV-F6z8) — GR Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/compressor-air-heat-transfer-rate - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-4 ## Question 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] ## 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). ## Answer **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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Mass flow rate comes from: A. ṁ = P1 V̇1/(R T1) = 0.02904 kg/s B. ṁ = V̇1 R T1/P1 C. ṁ = V̇1 at 1000 kPa D. ṁ = 1.5 kg/s Answer: A. Convert 1.5 m³/min to 0.025 m³/s and use ideal gas at inlet. (Key point: Mass flow) 2. The compressor work consumed of 250 kJ/kg is entered as: A. w = +250 kJ/kg B. w = -250 kJ/kg C. w = +100.5 kJ/kg D. w = 0 Answer: B. Work done on the system is negative in q - w. (Key point: Work sign) 3. q is: A. +349.5 kJ/kg B. -100.5 kJ/kg C. q = c_p ΔT + w = 100.5 - 250 = -149.5 kJ/kg D. +149.5 kJ/kg Answer: C. ΔKE = ΔPE = 0, so q - w = Δh. (Key point: Heat per kg) 4. The negative heat rate means: A. Heat is added to the air B. No heat transfer C. The compressor is adiabatic D. Heat is lost from the compressor, about 4.34 kW Answer: D. Q̇ = 0.02904 × (-149.5) = -4.34 kW. 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