Respuesta :
The mass flow rate of nitrogen through the compressor, assuming the compression process is as given in each of the options are;
A) m' = 0.047 kg/s
B) m' = 0.051 kg/s
C) m' = 0.062 kg/s
D) m' = 0.056 kg/s
We are given;
Initial Pressure; P₁ = 80 kPa
Final Pressure; P₂ = 480 kPa
Temperature; T₁ = 27 °C = 300 K
Work Input; W = 10 kW
- A) We are told that the compression process is isentropic. Thus; γ = 1.4
Formula for the mass flow rate is;
[tex]m' = \frac{W}{\frac{\gamma RT_{1}}{\gamma - 1}[(\frac{P_{2}}{P_{1}})^{\frac{\gamma - 1}{\gamma}} - 1]}[/tex]
where R is ideal gas constant for nitrogen = 0.2968 kJ/kg.K
Plugging in the relevant values gives;
[tex]m' = \frac{10}{\frac{1.4 * 0.2968 * 300}{1.4 - 1}[(\frac{480}{80})^{\frac{1.4 - 1}{1.4}} - 1]}[/tex]
m' = 0.047 kg/s
- B) We are told that the compression process is polytropic with γ = 1.3. Thus;
[tex]m' = \frac{10}{\frac{1.3 * 0.2968 * 300}{1.3 - 1}[(\frac{480}{80})^{\frac{1.3 - 1}{1.3}} - 1]}[/tex]
m' = 0.051 kg/s
- C) We are told that the compression process is isothermal and as such γ = 1.4 but the formula for the mass flow rate is;
[tex]m' = \frac{W}{RT_{1}In(\frac{P_{2}}{P_{1}}) }[/tex]
plugging in the relevant values gives;
[tex]m' = \frac{10}{0.2968 * 300In(\frac{480}{80}) }[/tex]
m' = 0.062 kg/s
- D) We are told that the compression process is ideal two-stage polytropic with γ = 1.3. The formula for the mass flow rate is;
[tex]m' = \frac{W}{\frac{2\gamma RT_{1}}{\gamma - 1}[(\sqrt{\frac{P_{2}}{P_{1}})}^{\frac{\gamma - 1}{\gamma}} - 1]}[/tex]
Plugging in the relevant values gives;
[tex]m' = \frac{10}{\frac{2 * 1.3 * 0.2968 * 300}{1.3 - 1}[(\sqrt{\frac{480}{80})}^{\frac{1.3 - 1}{1.3}} - 1]}[/tex]
m' = 0.056 kg/s
Read more about compressor mass flow rate at; https://brainly.com/question/16014998
