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A gas-turbine power plant operates on the simple Brayton cycle with air as the working fluid and delivers 32 MW of power. The minimum and maximum temperatures in the cycle are 310 and 900 K, and the pressure of air at the compressor exit is 8 times the value at the compressor inlet. Assuming an isentropic efficiency of 80 percent for the compressor and 86 percent for the turbine, determine the mass flow rate of air through the cycle. Account for the variation of specific heats with temperature.

The main objective of this question is to calculate the air cycle’s mass flow rate.

This question uses the concept of mass flow rate. The mass of such a liquid passing in one unit of time is known as the mass flow rate. In other terms, the rate at which liquid passes across a unit area is defined as the mass flow rate. The mass flow is a direct function of the liquid’s density, speed, and cross-sectional area

Expert Answer

We know that:

\[ \space h_1 \space = \space 310.24 \space \frac {kj}{kg} \]

\[ \space P_{r1} \space = \space 1.5546 \]

The relative pressure is:

\[ \space P_{r2} \space = \space \frac{P_2}{P_1} P_{rl} \]

By putting values, we get:

\[ \space = \space 8 \space \times \space 1.5546  \]

\[ \space = \space 12.44 \]

Now:

\[ h_{2s} \space = \space 526.58 \frac{kj}{kg} \]

Now:

\[ \space h_3 \space = \space 932.93 \frac{kj}{kg} \]

\[ \space P_{r3} \space = \space \frac{P_4}{P_3} P_{r3} \]

By putting values, we get:

\[ \space = \space \frac{1}{8} 75.29 \]

\[ \space = \space 9.41 \]

Now:

\[ \space h_{4s} \space = \space 519.3 \frac{kj}{kg} \]

Now the mass flow rate can be calculated as:

\[ \space W \space = \space Wtask, \space outPSK  \space – \space W_c in \]

\[ \space Q \space = \space mn_T(h_3 \space – \space h_{4s}) \space – \space \frac{m}{n_C} (h_2 \space – \space h_1) \]

By putting the values and simplifying results in:

\[ \space = \space \frac{32000}{0.86(932.93 \space – \space 519.3) \space – \space \frac{1}{0.8}(562.58 \space – \space 310.24)} \]

\[ \space = \space 794 \frac{kg}{s} \]

Numerical Answer

The air cycle’s mass flow rate is:

\[ \space = \space 794 \frac{kg}{s} \]

Example

In the above question, if the power is $ 31.5MW $, determine the air cycle’s mass flow rate.

We know that:

\[ \space h_1 \space = \space 310.24 \space \frac {kj}{kg} \]

\[ \space P_{r1} \space = \space 1.5546 \]

The relative pressure is:

\[ \space P_{r2} \space = \space \frac{P_2}{P_1} P_{rl} \]

By putting values, we get:

\[ \space = \space 8 \space \times \space 1.5546  \]

\[ \space = \space 12.44 \]

Now:

\[ h_{2s} \space = \space 526.58 \frac{kj}{kg} \]

Now:

\[ \space h_3 \space = \space 932.93 \frac{kj}{kg} \]

\[ \space P_{r3} \space = \space \frac{P_4}{P_3} P_{r3} \]

By putting values, we get:

\[ \space = \space \frac{1}{8} 75.29 \]

\[ \space = \space 9.41 \]

Now:

\[ \space h_{4s} \space = \space 519.3 \frac{kj}{kg} \]

Now the mass flow rate can be calculated as:

\[ \space W \space = \space Wtask, \space outPSK  \space – \space W_c in \]

\[ \space Q \space = \space mn_T(h_3 \space – \space h_{4s}) \space – \space \frac{m}{n_C} (h_2 \space – \space h_1) \]

By putting the values and simplifying results in:

\[ \space = \space \frac{3 1 5 0 0}{0 . 8 6(9 3 2 . 9 3 \space – \space 5 1 9 .  3) \space – \space \frac{1}{0 . 8}(5 6 2 . 5 8 \space – \space 3 1 0 . 2 4 )} \]

\[ \space = \space 7 8 1 . 6 \frac{kg}{s} \]

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