Question

Helium is expanded in a turbine from 1500 kPa and \(300^{\circ} \mathrm{C}\) to \(100 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). Determine the maximum work this turbine can produce, in \(\mathrm{kJ} / \mathrm{kg}\). Does the maximum work require an adiabatic turbine?

Short answer

Answer: The maximum work produced by the turbine per unit mass is 414 kJ/kg, and this maximum work does indeed require an adiabatic turbine.

Step-by-step solution

Convert temperatures to Kelvin

First, we need to convert the temperatures from Celsius to Kelvin. This is done using the following relation: $$T_K = T_C + 273.15$$ Initial temperature, \(T_1 = 300^{\circ} \mathrm{C} = 573.15 \mathrm{K}\) \ Final temperature, \(T_2 = 25^{\circ} \mathrm{C} = 298.15 \mathrm{K}\)

Use the isentropic relationship to find the specific volume ratios

Under isentropic (constant entropy) conditions, the relationship between initial and final pressures and specific volumes is given by the following equation: $$\frac{v_2}{v_1} = \left(\frac{P_1}{P_2}\right)^{\frac{1}{\gamma}}$$ Where \(\gamma\) is the heat capacity ratio for helium (which, as a monatomic gas, is equal to \(\frac{5}{3}\)). The initial and final pressures, \(P_1\) and \(P_2\), are given: $$P_1 = 1500\, \mathrm{kPa}\,, \,\, P_2 = 100\,\mathrm{kPa}$$ Now, we can find the ratio of specific volumes: $$\frac{v_2}{v_1} = \left(\frac{1500\,\mathrm{kPa}}{100\,\mathrm{kPa}}\right)^{\frac{1}{\frac{5}{3}}} = 15^{\frac{3}{5}}$$

Apply the ideal gas law to the initial state

In order to find the specific volume at the initial state, we can use the ideal gas equation of state: $$P_1 v_1 = R T_1$$ Where \(R\) is the specific gas constant for helium (\(\approx 2077\,\mathrm{J \cdot kg^{-1} \cdot K^{-1}}\)). Solving for \(v_1\): $$v_1 = \frac{RT_1}{P_1} = \frac{(2077\,\mathrm{J \cdot kg^{-1} \cdot K^{-1}})(573.15\,\mathrm{K})}{(1500\,\mathrm{kPa})(1000\,\mathrm{J \cdot m^{-3} \cdot kPa^{-1}})} = 0.0806\,\mathrm{m^3 \cdot kg^{-1}}$$

Find the specific volume at the final state

Using the volume ratio found in step 2: $$v_2=\left(\frac{v_2}{v_1}\right)v_1 = (15^{\frac{3}{5}})(0.0806\,\mathrm{m^3 \cdot kg^{-1}}) = 0.424\,\mathrm{m^3 \cdot kg^{-1}}$$

Calculate the maximum work using the isentropic process

The maximum work output can be calculated using the isentropic work equation for an ideal gas: $$W_{\max} = \frac{R(T_1 - T_2)}{1 - \gamma}$$ Substituting the values: $$W_{\max} = \frac{(2077\,\mathrm{J \cdot kg^{-1} \cdot K^{-1}})(573.15\,\mathrm{K} - 298.15\,\mathrm{K})}{1 - \frac{5}{3}} \approx -414\,\mathrm{kJ \cdot kg^{-1}}$$ Since the work output is negative, the maximum work produced by the turbine is 414 kJ/kg.

Determine if the maximum work requires an adiabatic turbine

An adiabatic turbine operates without heat transfer, resulting in an isentropic process. Since we have assumed an isentropic process, we can conclude that the maximum work does indeed require an adiabatic turbine. Therefore, the maximum work that the turbine can produce is 414 kJ/kg, and this maximum work is produced by an adiabatic turbine.