Calculate the net work output of a heat engine following path ABCDA in Figure 14.30, where horizontal axis is V and each unit is 0.510^-3 m^3, V1 = 1.010^-3 m^3 and V2 = 4.0*10^-3 m^3. The vertical axis is P(N/m^2);

Question

Calculate the net work output of a heat engine following path ABCDA in Figure 14.30, where horizontal axis is V and each unit is 0.510^-3 m^3, V1 = 1.010^-3 m^3 and V2 = 4.0*10^-3 m^3. The vertical axis is P(N/m^2);

A is (1.0*10^-3,2.6*10^6),
B is (4.0*10^-3,2.0*10^6),
C is (4.0*10^-3,0.6*10^6),
D is (1.0*10^-3,1.0*10^6).

I can't copy the figure. I hope someone can understand and help me to solve the problem.

Answer 1

To calculate the net work output of a heat engine following path ABCDA in Figure 14.30, we need to calculate the work done during each segment of the cycle and then find the sum of these works.

The work done during each segment of the cycle can be calculated using the formula:

W = ∫PdV,

where W is the work done, P is the pressure, and dV is the change in volume.

Let's calculate the work done during each segment:

Segment AB:
The volume change (dV) is V2 - V1 = (4.0*10^-3 m^3) - (1.0*10^-3 m^3) = 3.0*10^-3 m^3.
The pressure (P) is constant at 2.6*10^6 N/m^2.
So, the work done (W_AB) during this segment is:
W_AB = P * dV = (2.6*10^6 N/m^2) * (3.0*10^-3 m^3) = 7.8 Nm or J.

Segment BC:
The volume change (dV) is 0, as the volume remains constant.
The pressure (P) changes from 2.6*10^6 N/m^2 to 0.6*10^6 N/m^2.
So, the work done (W_BC) during this segment is:
W_BC = ∫PdV = ∫(0.6*10^6 + V)dV = (0.6*10^6 N/m^2) * (0) = 0 Nm or J.

Segment CD:
The volume change (dV) is V2 - V1 = (1.0*10^-3 m^3) - (4.0*10^-3 m^3) = -3.0*10^-3 m^3.
The pressure (P) is constant at 0.6*10^6 N/m^2.
So, the work done (W_CD) during this segment is:
W_CD = P * dV = (0.6*10^6 N/m^2) * (-3.0*10^-3 m^3) = -1.8 Nm or J.

Segment DA:
The volume change (dV) is 0, as the volume remains constant.
The pressure (P) changes from 0.6*10^6 N/m^2 to 2.6*10^6 N/m^2.
So, the work done (W_DA) during this segment is:
W_DA = ∫PdV = ∫(2.6*10^6 - V)dV = (2.6*10^6 N/m^2) * (0) = 0 Nm or J.

The net work output of the heat engine is the sum of the work done during each segment:

Net work output = W_AB + W_BC + W_CD + W_DA
= 7.8 Nm + 0 Nm + (-1.8 Nm) + 0 Nm
= 6 Nm or J.

Therefore, the net work output of the heat engine following path ABCDA in Figure 14.30 is 6 Nm or J.

Calculate the net work output of a heat engine following path ABCDA in Figure 14.30, where horizontal axis is V and each unit is 0.5*10^-3 m^3, V1 = 1.0*10^-3 m^3 and V2 = 4.0*10^-3 m^3. The vertical axis is P(N/m^2);

To calculate the net work output of a heat engine following path ABCDA in Figure 14.30, we need to calculate the work done during each segment of the cycle and then find the sum of these works.

Calculate the net work output of a heat engine following path ABCDA in Figure 14.30, where horizontal axis is V and each unit is 0.510^-3 m^3, V1 = 1.010^-3 m^3 and V2 = 4.0*10^-3 m^3. The vertical axis is P(N/m^2);