Basic Mechanical Engineering (3110006)

Question-3c-OR

BE | Semester-1 Summer-2019 | 04-06-2023

Q3) (c)

Given data:

p_{1} = 1.013 bar = 101.3 kPa

T_{1} = 15 ° C = 288 k

p_{2} = 7 bar

V_{1} = 0.3 \frac{m^{3}}{\min}

n = 1.25

Calculate:

I . P . = ?

Solution:

Case 1: I.P. When the compression process is Isentropic.

I . P . = \frac{γ}{γ - 1} P_{1} V_{1} [{(\frac{P_{2}}{P_{1}})}^{\frac{γ - 1}{γ}} - 1]

I . P . = \frac{1.4}{1.4 - 1} \times 101.3 \times \frac{0.3}{60} [{(\frac{7}{1.013})}^{\frac{1.4 - 1}{1.4}} - 1]

I . P . = 1.306 kW

Case 2: I.P. when the compression is reversible isothermal

I . P . = P_{1} V_{1} \ln (\frac{P_{2}}{P_{1}})

I . P . = 101.3 \times \frac{0.3}{60} \times \ln (\frac{7}{1.013})

I . P . = 0.977 kW

Case 3: I.P. when the compression process is Polytropic

I . P . = \frac{n}{n - 1} P_{1} V_{1} [{(\frac{P_{2}}{P_{1}})}^{\frac{n - 1}{n}} - 1]

I . P . = \frac{1.25}{1.25 - 1} \times 101.3 \times \frac{0.3}{60} [{(\frac{7}{1.013})}^{\frac{1.25 - 1}{1.25}} - 1]

I . P . = 1.195 kW