In a reversible isothermal expansion process fluid expands from 10 bar and 2 m³ to 2 bar and 10 m³. During the process the heat supplied is at the rate of 100 kW. What is the rate of work done during the process

Question

Question

This question was previously asked in

BPSC AE Paper 5 (Mechanical) 2019 Official Paper

Answer 1

Explanation:

Heat transfer in the constant temperature process by first law of thermodynamics is given as:
Q = U + W
where, Q = heat supplied. U = 0 = Internal energy, W = work done
So, the heat transfer equation is, Q = W = 100 kW

Important Points

Process	Work Done
Constant Pressure (Isobaric / Isopiestic)	W1-2 = p(V2 – V1)
Constant Volume (Isochoric)	W1-2 = 0
Isothermal	\({W_{1 - 2}} = {p_1}{V_1}\ln \left( {\frac{{{p_1}}}{{{p_2}}}} \right) = {p_2}{V_2}\ln \left( {\frac{{{p_1}}}{{{p_2}}}} \right)\)
Polytropic For Adiabatic (n = γ = 1.4)	\({W_{1 - 2}} = \frac{{{p_1}{V_1} - {p_2}{V_2}}}{{n - 1}} = \frac{{{p_1}{V_1}}}{{n - 1}}\left[ {1 - {{\left( {\frac{{{p_2}}}{{{p_1}}}} \right)}^{\frac{{n - 1}}{n}}}} \right]\)