Question

A piston-cylinder device contains 1.5 kg of nitrogen initially at 100 kPa and 17°C. The nitrogen is now compressed slowly in a polytropic process during which PlA3 = constant until the volume is reduced by one-half. Determine the work done and the heat transfer for this process.

Answer 1

Answer:

W_in = 99.5022 KJ

Q_out = 24.7 KJ

Explanation:

Given:

- mass of N_2 = 1.5 kg

- Initial Pressure = 100 KPa

- Initial Temperature = 17 C = 290 K

- Final Volume V_2 = 0.5*V_1

- Gas constant N_2 : R = 0.2968 KPa m^3 / kg.K

- C_v = 0.744 KJ / kg.K  (@ T = 350 K anticipated)

Find:

Determine the work done (W_net) and the heat transfer (Q_net) for this process.

Assumptions:

- N_2 is an ideal gas with constant specific heats

- E_k and E_p are negligible.

- The compression or expansion is a quasi-equilibrium process

Solution:

-The final Pressure P_2 and temperature T_2 can be determined as follows:

                          P_2 = P_1 * (V_1 / V_2)^1.3

                          P_2 = 100 KPa * (V_1 / 0.5V_1)^1.3

                          P_2 = 246.23 KPa

                          T_2 = T_1 * (P_2 * V_2) / (P_1*V_1)

                          T_2 = 290 K * (246.23 * 0.5*V_1) / (100*V_1)

                          T_2 = 357.03 K

-The boundary work for this poly-tropic process can be determined as follows:

                          $W_in = \int\limits^2_1 {P} \, dV = \frac{P_2 * V_2 - P_1 * V_1}{1 - n} = \frac{- m*R*(T_2 - T_1)}{1-n} \\\\W_in = \frac{- 1.5*0.2968*(357.03 - 290)}{1-1.3}\\\\Win = \frac{29.8417}{0.3}\\\\W_in = 99.5022 KJ$

- We take the contents of the cylinder as our system which constitutes a closed system. The respective energy balance is:

                         W_in - Q_out = m*C_v*(T_2 - T_1)

                         Q_out =  W_in - m*C_v*(T_2 - T_1)

                         Q_out = 99.5022 - 1.5*0.744*(357.03 - 290 )

                         Q_out = 24.7 KJ