Question

A gas mixture at a pressure of \(800\) kPa and a density of \(5\) kg/m\(^3\) enters a turbine stage. The temperature of the gas at the nozzle exit and the stage exit are \(790\) K and \(750\) K, respectively. Assume the specific heats are constant for the gas mixture in the range of temperatures considered. The specific heat at constant pressure is \(0.72\) kJ/kg-K and the ratio of specific heats is \(1.33\). The value of the degree of reaction of the turbine stage is _______ (rounded off to 2 decimal places).

Hint:

Degree of reaction is the fraction of the stage's static enthalpy drop that happens across the rotor: \(\Lambda = (T_2-T_3)/(T_1-T_3)\). You need the stage inlet temperature \(T_1\) from the ideal gas law, using \(R = c_p(\gamma-1)/\gamma\).

Answer 1

Correct Answer: 0.27

Step 1: Get cv first, then R.
Instead of using the combined formula for $R$, get $c_v$ from $\gamma$ first:
\[ c_v = \frac{c_p}{\gamma} = \frac{0.72}{1.33} = 0.5414 \text{ kJ/kg-K} \]
Then use $R = c_p - c_v$:
\[ R = 0.72 - 0.5414 = 0.1786 \text{ kJ/kg-K} \]
(same value as before, reached from the other direction).

Step 2: Work with specific volume instead of density.
Specific volume is the reciprocal of density, $v_1 = 1/\rho_1 = 1/5 = 0.2$ m$^3$/kg. Write the ideal gas law as $p_1 v_1 = R T_1$:
\[ T_1 = \frac{p_1 v_1}{R} = \frac{(800)(0.2)}{0.1786} = \frac{160}{0.1786} = 895.9 \text{ K} \]
(matches the density-based calculation, small rounding difference only).

Step 3: Use the enthalpy form of degree of reaction directly.
\[ \Lambda = \frac{h_2-h_3}{h_1-h_3} = \frac{c_p(T_2-T_3)}{c_p(T_1-T_3)} = \frac{T_2-T_3}{T_1-T_3} = \frac{790-750}{895.9-750} = \frac{40}{145.9} \]

Final Answer:
\[ \boxed{\Lambda \approx 0.27} \]