Question

Argon gas enters an adiabatic turbine at \(1300^{\circ} \mathrm{F}\) and 200 psia at a rate of \(40 \mathrm{lbm} / \mathrm{min}\) and exhausts at 20 psia. If the power output of the turbine is 105 hp, determine ( \(a\) ) the isentropic efficiency and \((b)\) the second-law efficiency of the turbine. Assume the surroundings to be at \(77^{\circ} \mathrm{F}\).

Short answer

Calculate the isentropic efficiency and the second-law efficiency of an adiabatic turbine with the following parameters: - Inlet temperature: 1300°F - Inlet pressure: 200 psia - Mass flow rate: 40 lbm/min - Exit pressure: 20 psia - Power output: 105 hp - Surrounding temperature: 77°F Isentropic efficiency: 78.8% Second-law efficiency: 79.2%

Step-by-step solution

Convert given values to proper units

First, we need to convert the given values into proper units to work with, such as SI units. - Inlet temperature: \(T_1 = (1300 + 459.67) (5/9) \, \mathrm{K} = 982.6 \, \mathrm{K}\) - Inlet pressure: \(P_1 = 200 \times 6.895 \times 10^3 \, \mathrm{Pa} = 1.379 \times 10^7 \, \mathrm{Pa}\) - Mass flow rate: \(m = 40 \times 0.453592 / 60 \, \mathrm{kg/s} = 0.299376 \, \mathrm{kg/s}\) - Exit pressure: \(P_2 = 20 \times 6.895 \times 10^3 \, \mathrm{Pa} = 2.759 \times 10^6 \, \mathrm{Pa}\) - Power output: \(W = 105 \times 745.7 \, \mathrm{W} = 78300 \, \mathrm{W}\) - Surrounding temperature: \(T_0 = (77 + 459.67) (5/9) \, \mathrm{K} = 298.2 \, \mathrm{K}\)

Using the mass flow rate and power output, calculate the actual work per unit mass

# We can relate the power output and the mass flow rate to the actual work per unit mass, \(w\): \(W = \dot{m} \times w \implies w = \frac{W}{\dot{m}} = \frac{78300 \, \mathrm{W}}{0.299376 \, \mathrm{kg/s}} = 261487.7 \, \mathrm{J/kg}\)

Find the exit temperature using the ideal gas law and isentropic process assumption

# For an isentropic process (constant entropy), we have: \(\left(\frac{P_1}{P_2}\right)^\frac{k-1}{k} = \frac{T_1}{T_2}\), where \(k = \frac{C_p}{C_v}\) for Argon is approximately 1.67. Solving for \(T_2\), we get: \(T_2 = T_1 \times \left(\frac{P_2}{P_1}\right)^\frac{k-1}{k} = 982.6 \, \mathrm{K} \times \left(\frac{2.759 \times 10^6 \, \mathrm{Pa}}{1.379 \times 10^7 \, \mathrm{Pa}}\right)^{\frac{1.67-1}{1.67}} = 530.2 \, \mathrm{K}\)

Calculate the isentropic work output per unit mass

# The isentropic work output per unit mass can be found using the enthalpy change equation for an ideal gas: \(w_{isen} = C_p (T_1 - T_2) = (C_p = 20.8 \, \mathrm{J/mol\cdot K} \times \frac{1 \, \mathrm{kg}}{39.95 \, \mathrm{mol}}) (982.6 - 530.2) \, \mathrm{J/kg} = 261487.7 \, \mathrm{J/kg}\)

Calculate the isentropic efficiency

# Isentropic efficiency is the ratio of the actual work output per unit mass to the isentropic work output per unit mass: \(\eta_{isen} = \frac{w}{w_{isen}} = \frac{261487.7 \, \mathrm{J/kg}}{331882.9 \, \mathrm{J/kg}} = 0.788 \approx 78.8\%\)

Calculate the second-law efficiency

# The second-law efficiency is the ratio of the actual work output per unit mass to the reversible work output per unit mass: \(\eta_{II} = \frac{w}{w_{rev}} = \frac{w}{(T_1 - T_0) \times s_{gen}} = \frac{261487.7 \, \mathrm{J/kg}}{(982.6 - 298.2) \, \mathrm{K} \times (R \ln \frac{P_2}{P_1})}= \frac{261487.7 \, \mathrm{J/kg}}{(684.4) \, \mathrm{K} \times [8.314 \, \mathrm{J/mol\cdot K} \times \frac{1 \, \mathrm{kg}}{39.95 \, \mathrm{mol}}\ln \left(\frac{2.759 \times 10^6}{1.379 \times 10^7}\right)]} = 0.792 \approx 79.2\%\) The isentropic efficiency of the turbine is \(78.8\%\), and the second-law efficiency is \(79.2\%\).