The net work output of the heat engine following path ABCDA is -2996 J. The negative sign signifies work done on the engine, characteristic of a heat engine cycle.
To calculate the net work output of the heat engine following the path ABCDA, you can use the formula for work done in a thermodynamic process:
Work = ∫(V₁ to V₂) P dV
The work done during each segment of the cycle (AB, BC, CD, and DA) will be the area under the corresponding segment on the P-V diagram.
Work done in segment AB (Isochoric process):
W_AB = P_A * (V_B - V_A)
Work done in segment BC (Isobaric process):
W_BC = P_B * (V_C - V_B)
Work done in segment CD (Isochoric process):
W_CD = P_C * (V_D - V_C)
Work done in segment DA (Isobaric process):
W_DA = P_D * (V_A - V_D)
The net work output is the sum of the work done in each segment:
W_net = W_AB + W_BC + W_CD + W_DA
Now, plug in the given values:
W_AB = (2.85 × 10^6 N/m²) * (3.3 × 10^(-3) m³ - 0.75 × 10^(-3) m³)
W_AB = 2.85 × 10^6 N/m² * 2.55 × 10^(-3) m³
W_AB = 7275 J
W_BC = (2.01 × 10^6 N/m²) * (0.45 × 10^(-3) m³ - 3.3 × 10^(-3) m³)
W_BC = 2.01 × 10^6 N/m² * (-2.85 × 10^(-3) m³)
W_BC = -5728.5 J
W_CD = (0.45 × 10^6 N/m²) * (0.75 × 10^(-3) m³ - 0.45 × 10^(-3) m³)
W_CD = 0.45 × 10^6 N/m² * 0.3 × 10^(-3) m³
W_CD = 135 J
W_DA = (1.05 × 10^6 N/m²) * (0.75 × 10^(-3) m³ - 3.3 × 10^(-3) m³)
W_DA = 1.05 × 10^6 N/m² * (-2.55 × 10^(-3) m³)
W_DA = -2677.5 J
Finally, calculate the net work:
W_net = W_AB + W_BC + W_CD + W_DA
W_net = 7275 J + (-5728.5 J) + 135 J + (-2677.5 J)
W_net = -2996 J
Therefore, the net work output of the heat engine following the path ABCDA is -2996 J. Note that the negative sign indicates that the net work is done on the engine, which is typical for a heat engine cycle.