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:
\[ \space = \space \frac{3 1 5 0 0}{0 . 8 6(9 3 2 . 9 3 \space – \space 5 1 9 . 3) \space – \space \frac{1}{0 . 8}(5 6 2 . 5 8 \space – \space 3 1 0 . 2 4 )} \]
\[ \space = \space 7 8 1 . 6 \frac{kg}{s} \]