A piston-cylinder device contains 0.1 [kg] of air at a pressure of 100 [kPa] and a temperature of 400 [K] that undergoes an expansion process. The volume of the piston-cylinder device expands from 1 [m3 ] to 3 [m3 ] at a constant pressure of 2,000 [kPa]. Then, as the piston-cylinder device expands from 3 [m3 ] to 5 [m3 ], the pressure linearly decreases from 2,000 to 1,000 [kPa]. (a) Determine the heat input.(b) Determine the work for the overall process

a) Q_in = 30 KJ

b)  W_net, out = 7,000 KJ

Explanation:

Given:

- mass of air m = 0.1 kg

State 1:                               State 2:                         State 3:

- P_1 = 2000 KPa             P_2 = 2000 KPa            P_3 = 1,000 KPa

- T_1 = 400 K                    V_2 =  3 m^3                  V_3 = 5 m^3

- V_1 = 1 m^3

Find:

(a) Determine the heat input.

(b) Determine the work for the overall process

Solution:

- Process 1: Constant pressure, Expansion Heat addition.

 Using Ideal Gas Law for Air.

                       P_1*V_1 / T_1 = P_2*V_2 / T_2

Where, P_1 = P_2,

                      V_1 / T_1 = V_2 / T_2

                      T_2 = (V_2 / V_1)*T_1

                      T_2 = (3 / 1 )*400 = 700 K

 The amount of Heat input Q_in:

                      Q_in = m*c_p*( T_2 - T_1)

                      Q_in = 0.1*1*( 700 - 400) = 30 KJ

The amount of work done by the system in the process:

                     W_out = P_1 * ( V_2 - V_1)

                     W_out = 2000 * ( 3 - 1) = 4,000 KJ

- Process 2: Linear Expansion.

                     W_out = Area under the P-V graph

                     W_out = Area of Trapezium

                     W_out = 0.5*(P_ 1 + P_2)*( V_3 - V_2)

                     W_out = 0.5*(3000)*(2)

                     W_out = 3,000 KJ

- The total work for overall process is:

                      W_net, out = 4,000 + 3,000 = 7,000 KJ

answer:

the formula for amps is watts divided by volts. to use the chart, cover up the a with your finger and use the remaining chart calculation of w divided by v. using our sample panel data, 60 watts divided by 12 volts equals 5 amps.

explanation: