Complete Analysis of Heat Engine Goal Solve for the efficiency of a heat engine

Question

Complete Analysis of Heat Engine Goal Solve for the efficiency of a heat engine using a five-step process the includes: 1. Making a state table. 2. Making a process table. 3. Calculating the totals for Work, Heat, and Internal-Energy-Change. 4. Identifying the heat input (hot reservoir) and output (cold reservoir) 5. Calculating the efficiency of the engine. isothermal Problem Shown in the figure to the right is a cyclic process undergone by a heat engine. Your heat engine shall use 8.0 moles of nitrogen gas (diatomic). During the process a->b, the pressure rises by a factor of 5.0. T 300K Pa-100,000 Pa Engine Cycle

Explanation / Answer

at point "A" :

Va = Volume = ?

Pa = 105 Pa

Ta= 300 K

n = 8

Using the equation

Pa Va = n R Ta

(105) Va = (8) (8.314) (300)

Va = 0.2 m3

at Point "B" :

Pb = 5 Pa = 5 (1 x 105) = 5 x 105 Pa

Vb = Va = 0.2

Tb = ?

Using the equation

Pb Vb = n R Tb

(5 x 105) (0.2) = 8 (8.314) Tb

Tb = 1503.5 K

at c:

Tc = Tb = 1503.5 K

Pc = Pa = 0.2 m3

Vc = ?

Using the equation

Pc Vc = n R Tc

(1 x 105) Vc = 8 (8.314) (1503.5)

Vc = 1

2)

During the process A to B :

W = Pa (Vb - Va) = (1 x 105) (0.2 - 0.2) = 0 J

Q = n Cv (Tb - Ta) = n (2.5 R) (Tb - Ta) = 8 (2.5 x 8.314) (1503.5 - 300) = 2 x 105 J

Using the equation

Q = W + U

2 x 105 = 0 + U

U = 2 x 105 J

for b to c (isothermal process) :

U = 0

W = n R T log(Vc/Vb) = 8 (8.314) (1503.5) log(1/0.2) = 1.6 x 105 J

Q = W = 1.6 x 105 J

For process c to a :

W = Pc (Va - Vc) = (1 x 105) (0.2 - 1) = - 0.8 x 105 J

Q = n Cp (Ta - Tc) = 6 (3.5 x 8.314) (300 - 1503.5) = - 2.1 x 105 J

U = Q - W = - 2.1 x 105 - (- 0.8 x 105) = - 1.3 x 105 J

3)

W = Wab + Wbc + Wca = 0 x 105 + (1.6 x 105 ) + (- 0.8 x 105) = 0.8 x 105J

U = Uab + Ubc + Uca =  2 x 105 + 0 + (- 1.3 x 105 ) = 0.7 x 105 J

Q = Qab + Qbc + Qca = 2 x 105 + ( 1.6 x 105 ) + (- 2.1 x 105 ) = 1.5 x 105 J

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