Basic Mechanical Engineering (3110006)

Winter-2019

Question-3c

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

Q3) (c)

One kg of gas at a pressure of 100 kPa and temperature of 17°C is compressed isothermally in a piston-cylinder arrangement to final pressure of 2500 kPa. The characteristic gas equation is given by the relation pV = 260 T per kg where T is in Kelvin. Find out (1) Final Volume (2) Compression Ratio (3) Change in enthalpy (4) Work done on the gas.

Given data
$m = 1 kg$
$p_{1} = 100 kN / m^{2}$
$T_{1} = 17 ℃ = 290 K$
$p_{2} = 2500 kN / m^{2}$
$pv = 260 T per kg$

Now, $p_{1} V_{1} = 260 T_{1}$

$\Rightarrow V_{1} = \frac{260 T_{1}}{p_{1}}$

$\Rightarrow V_{1} = \frac{260 \times 290}{100 \times 10^{3}}$

$\Rightarrow V_{1} = 0.754 \frac{m^{3}}{kg}$

Now, $p_{1} V_{1} = p_{2} V_{2}$

$\Rightarrow V_{2} = \frac{p_{1} V_{1}}{p_{2}}$

$\Rightarrow V_{2} = \frac{100 \times 0.754}{2500}$

$\Rightarrow V_{2} = 0.03016 \frac{m^{3}}{kg}$

Now, $∆ H = {mC}_{p} (T_{2} - T_{1}) = 0 kJ$

Now, $W = P_{1} V_{1} \ln (\frac{P_{1}}{P_{2}}) = 100 \times 0.754 \ln (\frac{100}{2500})$

$\Rightarrow W = - 242.7 kJ$

Basic Mechanical Engineering (3110006)

Q3) (c)

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