Question

refrigerant 134a enters a diffuser steadily at 0.5 mpa, 50°c, and 120 m/s at a rate of 1.2 kg/s. the inlet area of the diffuser is

Answer 1

The inlet area of the diffuser in the refrigeration system is 0.01 square meters.

Step-by-step explanation:

The question is asking for the inlet area of the diffuser in a refrigeration system. To find the inlet area, we can use the flow rate equation:

Flow Rate = Inlet Area x Inlet Velocity

Given that the refrigerant enters the diffuser at a rate of 1.2 kg/s and a velocity of 120 m/s, and assuming incompressible flow, we can rearrange the equation to solve for the inlet area:

Inlet Area = Flow Rate / Inlet Velocity

Substituting the given values, we have:

Inlet Area = 1.2 kg/s / 120 m/s

Inlet Area = 0.01 m²

Therefore, the inlet area of the diffuser is 0.01 square meters.

Answer 2

The inlet area of the diffuser is approximately
`\(0.00122 \, \text{m²}\)`.

To determine the inlet area of the diffuser, we can use the mass flow rate equation for a compressible fluid in a steady-flow process. The mass flow rate
`(\( \dot{m} \))` can be expressed as:

`\[ \dot{m} = \rho A V \]`

where:

`\( \dot{m} \)`is the mass flow rate (1.2 kg/s),

`\( \rho \)`is the density of the fluid,

`\( A \)`is the cross-sectional area, and

V \is the velocity of the fluid.

To find the density
`(\( \rho \))`, you can use the ideal gas law for the given conditions. The ideal gas law is given by:

`\[ PV = mRT \]`

where:

P is the pressure (0.5 MPa),

V is the specific volume (inverse of density),

m is the mass (1.2 kg),

R is the specific gas constant for refrigerant 134a, and

T is the temperature (50°C).

Let's convert the pressure to Pascals and the temperature to Kelvin and then use the ideal gas law to find the density:

`\[ P = 0.5 \, \text{MPa} = 500,000 \, \text{Pa} \]`

`\[ T = 50 + 273.15 \, \text{K} = 323.15 \, \text{K} \]`

Now, rearrange the ideal gas law to solve for density
`(\( \rho \))`:

`\[ \rho = (m)/(V) = (P)/(RT) \]`

Once you have the density, you can rearrange the mass flow rate equation to solve for the cross-sectional area (\( A \)):

`\[ A = \frac{\dot{m}}{\rho V} \]`

Firstly, let's calculate the density
`(\( \rho \))` using the ideal gas law:

`\[ \rho = (P)/(RT) \]`

where:

`\( P = 500,000 \, \text{Pa} \)` (pressure in Pascals),

R is the specific gas constant for refrigerant 134a, and

`\( T = 323.15 \, \text{K} \)`(temperature in Kelvin).

The specific gas constant ( R ) for refrigerant 134a is approximately 188.92 J/(kg·K).

`\[ \rho = \frac{500,000 \, \text{Pa}}{188.92 \, \text{J/(kg·K)} * 323.15 \, \text{K}} \]`

Now, calculate
`\( \rho \)`. The result will be in kg/m³.

`\[ \rho \approx (500,000)/(61,088.238) \, \text{kg/m³} \approx 8.18 \, \text{kg/m³} \]`

Now, use the mass flow rate equation to find the cross-sectional area
`(\( A \))`:

`\[ A = \frac{\dot{m}}{\rho V} \]`

where:

`\( \dot{m} = 1.2 \, \text{kg/s} \)`(mass flow rate),

`\( \rho = 8.18 \, \text{kg/m³} \)` (density), and

`\( V = 120 \, \text{m/s} \)` (velocity).

`\[ A = (1.2)/(8.18 * 120) \]`

Now, calculate A .

`\[ A \approx (1.2)/(981.6) \, \text{m²} \approx 0.00122 \, \text{m²} \]`

Therefore, the inlet area of the diffuser is approximately
`\(0.00122 \, \text{m²}\)`.