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:

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

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