Goal Solve for the efficiency of a heat engine using a fives

Goal Solve for the efficiency of a heat engine using a five-step process the includes: Making a state table. Making a process table. Calculating the totals for Work, Heat, and Internal -Energy -Change. Identifying the heat input (hot reservoir) and output (cold reservoir). Calculating the efficiency of the engine. Problem Shown in the figure to the right is a cyclic process undergone by a heat engine. Your heat engine shall use 4.0 moles of nitrogen gas (diatomic). During the process a rightarrow b, the pressure rises by a factor of 5.0. Fill in the State Table (all pressures in Pascals, all volumes in cubic meters, all temperatures in K). Fill in the Process Table (all entries in Joules). Find the Totals: Find the heat input (from \"hot reservoir\") and the heat output (to \"cold reservoir\"): Find the efficiency of the engine:

Solution

Number of moles n = 4 mol

Pressure at b is P \' = 5 ( Pressure at a )

                             = 5(10 ⁵ Pa)

Pressure at point a is P = 10 ⁵ Pa

Temprature at point a is T = 300 K

From the relation PV = nRT

Volume at point a is V = nRT / P

Where R = Gas constant = 8.314 J / mol K

Substitute values you get , V = 4(8.314)(300)/(10 ⁵)

= 99.768x10 ^-3 m ³

Pressure at point b is P \' = 5 x10 ⁵ Pa

point a to point b is constant volume process.

At constant volume P \' / P = T \' / T

From this temprature at point b is T \' = T ( P \' /P)

                                                     = 300 ( 5 )

                                                    = 1500 K

Volume at point b is V \' = V = 99.768 x10 ^-3 m ³

Point b to c is isothermal process.

In isothermal process, PV = constant

So, P \' V \' = P \" V \"
Where Pressure at point c is P \" = 10 ⁵ Pa

            P \' = 5 x10 ⁵ Pa

            V \' = 99.768 x10 ^-3 m ³

From above Volume at point c is V \" = P \' V \' / P \"

                                                     = (P \' / P \" ) V \'

                                                     = 5 (99.768 x10 ^-3 m ³ )

                                                    = 498.84x10 ^-3 m ³

(2). a --->b process :

Work done W = 0            Since change in volume in this process is zero.

Change in internal energy,U = n Cv (T \' - T )

Where Cv = Specific heat at constant volume

= 2.5 R                                

So, U = 4(2.5R)(1500-300)

         = 10 R (1200)

        = 10 x1200x 8.314

        = 99768 J

heat Q = U+W

           = 99768 J

Process b ---> c :

Work done W \' = nRT \' ln ( V \" / V \')

                       = 4(8.314)(1500) ln (5)

                       = 80285.2 J

Change in internal energy U \' = 0

Heat Q \' = U \' + W \'

             = 80285.2 J

Process c ---> a :

Work done W \" = P ( V -V \")

                       = 10 ⁵ [ ( 99.768 x10 ^-3 ) - (498.84 x10 ^-3 )]

                       = -39907.2 J

Change in internal energy U \" = nCV (T - T \")

                                           = 4 x 2.5 R ( 300 -1500)

                                          = - 99768 J

Heat Q \" = U \" + W \"

             = -139675.2 J

(C). Total work = W + W \'+W \"

Total change in internal energy = U + U \' + U \"

Total heat = Q + Q \'+Q \"