A gas turbine with air as the working fluid has an isentropic efficiency of 0.70 when operating at a pressure ratio of 3. Now, the pressure ratio of the turbine is increased to 5, while maintaining the same inlet conditions. Assume air as a perfect gas with specific heat ratio γ = 1.4. If the specific work output remains the same for both the cases, the isentropic efficiency of the turbine at the pressure ratio of 5 is ______ (round off to two decimal places)

The correct answer is 0.51.

Step-by-Step Explanation:

For a gas turbine, the actual specific work output ($w_a$) is given in terms of the inlet temperature ($T_1$), pressure ratio ($r_p$), specific heat ratio ($\gamma$), and isentropic efficiency (${\eta }_t$) as follows:

$w_a={\eta }_t·c_p·T_1\left[1-{\left(\frac{1}{r_p}\right)}^{\frac{\gamma -1}{\gamma }}\right]$

Given that the inlet conditions (specifically $T_1$) and the specific work output ($w_a$) remain the same for both cases, we can write:

$w_{a1}=w_{a2}$

This simplifies to:

${\eta }_{t1}\left[1-{\left(\frac{1}{r_{p1}}\right)}^{\frac{\gamma -1}{\gamma }}\right]={\eta }_{t2}\left[1-{\left(\frac{1}{r_{p2}}\right)}^{\frac{\gamma -1}{\gamma }}\right]$

Given data:
- Case 1: $r_{p1}=3$, ${\eta }_{t1}=0.70$
- Case 2: $r_{p2}=5$
- Specific heat ratio: $\gamma =1.4$

Let us compute the exponent:
$\frac{\gamma -1}{\gamma }=\frac{1.4-1}{1.4}=\frac{0.4}{1.4}=\frac{2}{7}\approx 0.2857$

Now, compute the terms inside the brackets for both cases:

For Case 1 (with $r_{p1}=3$):
$1-{\left(\frac{1}{3}\right)}^{0.2857}\approx 1-0.7306=0.2694$

For Case 2 (with $r_{p2}=5$):
$1-{\left(\frac{1}{5}\right)}^{0.2857}\approx 1-0.6310=0.3690$

Substitute these values back into the relation to solve for ${\eta }_{t2}$:

$0.70·0.2694={\eta }_{t2}·0.3690$

${\eta }_{t2}=\frac{0.70·0.2694}{0.3690}\approx 0.511$

Rounding off to two decimal places, the isentropic efficiency of the turbine at a pressure ratio of 5 is 0.51.