Basic Mechanical Engineering (3110006)

Winter-2019

Question-3a-OR

BE | Semester-1 Winter-2019 | 03-01-2020

Q3) (a)

One kg gas is compressed adiabatically by following the law pV = C from initial temperature of 290 K. The initial pressure of gas is 1 bar. The initial and final volumes of gas are 0.50 $m^{3}$ and 0.05 $m^{3}$ respectively. Find the final temperature and pressure of gas. Assume γ =1.4.

Given data
$m = 1 kg$
$T_{1} = 290 K$
$P_{1} = 1 bar = 100 kPa$
$V_{1} = 0.50 m^{3}$
$V_{2} = 0.05 m^{3}$

To Find $T_{2} = ?, P_{2} = ?$

Now,

$\frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1}$

$\Rightarrow T_{2} = T_{1} \times {(\frac{V_{1}}{V_{2}})}^{γ - 1}$

$\Rightarrow T_{2} = 290 \times {(\frac{0.50}{0.05})}^{1.4 - 1}$

$\Rightarrow T_{2} = 728.447 K$

Now,

$P_{1} {V_{1}}^{γ} = P_{2} {V_{2}}^{γ}$

$\Rightarrow P_{2} = P_{1} \times {(\frac{V_{1}}{V_{2}})}^{γ}$

$\Rightarrow P_{2} = 100 \times {(\frac{0.50}{0.05})}^{1.4}$

$\Rightarrow P_{2} = 2511.886 kPa$

Basic Mechanical Engineering (3110006)

Q3) (a)

One kg gas is compressed adiabatically by following the law pV = C from initial temperature of 290 K. The initial pressure of gas is 1 bar. The initial and final volumes of gas are 0.50 $m^{3}$ and 0.05 $m^{3}$ respectively. Find the final temperature and pressure of gas. Assume γ =1.4.

Questions

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Basic Mechanical Engineering (3110006)

Q3) (a)

One kg gas is compressed adiabatically by following the law pV = C from initial temperature of 290 K. The initial pressure of gas is 1 bar. The initial and final volumes of gas are 0.50 m3 and 0.05 m3 respectively. Find the final temperature and pressure of gas. Assume γ =1.4.

Questions

Go to Question Paper

One kg gas is compressed adiabatically by following the law pV = C from initial temperature of 290 K. The initial pressure of gas is 1 bar. The initial and final volumes of gas are 0.50 $m^{3}$ and 0.05 $m^{3}$ respectively. Find the final temperature and pressure of gas. Assume γ =1.4.