An exhaustive analysis of a heat engine. A heat engine goes through a four-step

Question

An exhaustive analysis of a heat engine. A heat engine goes through a four-step cycle, with its representation in the P-V plane shown at right. The step from point A to point B is a straight line in the P-V diagram. The step from point B to point C is iso baric. The step from point C to point D is isochoric. The step from point D back to point A is isothermal. The working fluid is a monatomic ideal gas. At point A the pressure is 0.500 MPa, the volume is 1.000 x 10^3 m^3, and the temperature is 400 K. At points B and C the pressure is 1.000 MPa. At points C and D the volume is 1.500 x 10^-3 m^3 The temperature at point B is 960 K. How many moles of working fluid are there in the engine? What is the temperature at point C? (C) How much work is done on each step in the cycle? (For the A rightarrow B step, remember that the work is the area under the curve; in this step, that area is a rectangle of width (V_b - V_A) and height P_A, plus a triangle with base (V_B - V_A) and height (P_B - P_A). For the D rightarrow A step, the working fluid is a monatomic ideal gas.) (D) Find the heat flow that occurs in each step.

Explanation / Answer

A)
Consider the state at point A
Pa = 0.5 x 106 Pa
Va = 1 x 10-3 m3
T = 400 K
Using the formula PV = nRT,
R = 8.314 JK1mol1.
n = PV/RT
= [(0.5 x 106) x (10-3)] / [8.314 x 400]
= 0.15 mol

B)
Again using the formula PV = nRT at C,
Pressure at C = 1 x 106 Pa
Volume at C = 1.5 x 10-3 m3
n = 0.15 mol

T = PV/nR
= [(1 x 106) x (1.5 x 10-3)] / [0.15 x 8.314]
= 1200 K.

C)
Step A to B
Work done = Area under AB curve.
= Area of rectangle + Area of triangle
We need to find the volume at B using the formula PV = nRT
V = nRT/P
= [0.15 x 8.314 x 960] / [1 x 106]
= 1.2 x 10-3 m3.

Vb - Va = 1.2 x 10-3 m3 - 1 x 10-3 m3
= 0.2 x 10-3 m3
Total area = Pa x (Vb - Va) + 1/2 (Pb - Pa) x (Vb - Va)
= [(0.5 x 106) x (0.2 x 10-3)] + [0.5 x (0.5 x 106) x (0.2 x 10-3)]
= 150 J

Step B to C
Work done in isochoric process = Area under the curve BC
= Pc x (Vc - Vb)
= 1 x 106 x [(1.5 x 10-3 m3) - (1.2 x 10-3 m3)]
= 300 J

Step C to D
Work done = 0 since there is no volume change.

Step D to A
Work done in an isothermal process, W = nRT ln[Va/Vd]
= 0.15 x 8.314 x 400 ln(1 x 10-3/1.5 x 10-3)
= - 202.73 J

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