Air enters a compressor operating at steady state with a pressure of 14.7 lbf/in2, a temperature of 80 degrees Fahrenheit, and a volumetric flow rate of 18 ft3/s.

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Air enters a compressor operating at steady state with a pressure of 14.7 lbf/in2, a temperature of 80 degrees Fahrenheit, and a volumetric flow rate of 18 ft3/s. The air exits the compressor at a pressure of 90 lbf/in2. Heat transfer from the compressor to its surroundings occurs at a rate of 9.7 Btu per lb of air flowing. The compressor power input is 90 hp. Neglecting kinetic and potential energy effects and modeling the air as an ideal gas, determine the exit temperature, in degrees Fahrenheit.

The Correct Answer and Explanation is:

To solve this problem, we apply the First Law of Thermodynamics for a steady-flow process: Q˙−W˙=m˙(h2−h1)\dot{Q} – \dot{W} = \dot{m}(h_2 – h_1)

Where:

  • Q˙\dot{Q}: heat transfer rate per unit mass (Btu/lb)
  • W˙\dot{W}: work input rate per unit mass (Btu/lb)
  • h1,h2h_1, h_2: specific enthalpy at inlet and outlet (Btu/lb)
  • m˙\dot{m}: mass flow rate (lb/s)

Step 1: Convert given values to consistent units

Given:

  • Inlet pressure: P1=14.7 lbf/in2P_1 = 14.7 \, \text{lbf/in}^2
  • Inlet temperature: T1=80∘F=540 °RT_1 = 80^\circ F = 540 \, \text{°R}
  • Volumetric flow rate: V˙=18 ft3/s\dot{V} = 18 \, \text{ft}^3/\text{s}
  • Exit pressure: P2=90 lbf/in2P_2 = 90 \, \text{lbf/in}^2
  • Heat transfer: Q˙=−9.7 Btu/lb\dot{Q} = -9.7 \, \text{Btu/lb}
  • Compressor power: W˙=90 hp=90×2545=229050 ft\cdotplbf/s\dot{W} = 90 \, \text{hp} = 90 \times 2545 = 229050 \, \text{ft·lbf/s}

Step 2: Find the mass flow rate

Using the ideal gas law: m˙=P1V˙RT1\dot{m} = \frac{P_1 \dot{V}}{R T_1}

Where R=53.35 ft\cdotplbf/(lbm\cdotp°R)R = 53.35 \, \text{ft·lbf}/(\text{lbm·°R})

Convert P1=14.7×144=2116.8 lbf/ft2P_1 = 14.7 \times 144 = 2116.8 \, \text{lbf/ft}^2 m˙=2116.8×1853.35×540=38102.428809≈1.323 lbm/s\dot{m} = \frac{2116.8 \times 18}{53.35 \times 540} = \frac{38102.4}{28809} \approx 1.323 \, \text{lbm/s}

Step 3: Apply the First Law

Rewriting the energy balance: Q˙total−W˙totalm˙=h2−h1=cp(T2−T1)\frac{\dot{Q}_{total} – \dot{W}_{total}}{\dot{m}} = h_2 – h_1 = c_p(T_2 – T_1)

  • Q˙total=m˙⋅(−9.7)\dot{Q}_{total} = \dot{m} \cdot (-9.7)
  • W˙total=229050 ft\cdotplbf/s=229050778.17≈294.4 Btu/s\dot{W}_{total} = 229050 \, \text{ft·lbf/s} = \frac{229050}{778.17} \approx 294.4 \, \text{Btu/s}

So: −9.7⋅1.323−294.41.323=cp(T2−T1)\frac{-9.7 \cdot 1.323 – 294.4}{1.323} = c_p (T_2 – T_1) −12.83−294.41.323=cp(T2−T1)\frac{-12.83 – 294.4}{1.323} = c_p (T_2 – T_1) −232.1=cp(T2−T1)-232.1 = c_p (T_2 – T_1)

Use cp=0.24 Btu/lbm\cdotp°Rc_p = 0.24 \, \text{Btu/lbm·°R} T2−T1=−232.10.24≈−967.1 °RT_2 – T_1 = \frac{-232.1}{0.24} \approx -967.1 \, \text{°R} T2=T1+ΔT=540−967.1=−427.1 °RT_2 = T_1 + \Delta T = 540 – 967.1 = -427.1 \, \text{°R}

This is not physically possible, so let’s recheck the signs.

Since work is input (positive into the system) and heat is lost, we must set: Q˙=−9.7⋅m˙(out)W˙=+294.4(in)\dot{Q} = -9.7 \cdot \dot{m} \quad (\text{out}) \\ \dot{W} = +294.4 \quad (\text{in})

So: h2−h1=−12.83+294.41.323=281.571.323=212.87 Btu/lbh_2 – h_1 = \frac{-12.83 + 294.4}{1.323} = \frac{281.57}{1.323} = 212.87 \, \text{Btu/lb} T2−T1=212.870.24≈886.98 °RT_2 – T_1 = \frac{212.87}{0.24} \approx 886.98 \, \text{°R} T2=540+886.98=1426.98 °R=1426.98−459.67=967.3∘FT_2 = 540 + 886.98 = 1426.98 \, \text{°R} = 1426.98 – 459.67 = \boxed{967.3^\circ F}

✅ Final Answer:

967.3∘F\boxed{967.3^\circ \text{F}?

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