Question

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 10.0 moles of nitrogen gas (diatomic). During the process a->b, the pressure rises by a factor of 2.0. T= 300 K Pa = 100,000 Pa Engine Cycle Solution Pressure (1) Fill in the State Table (all pressures in Pascals, all volumes in cubic meters, all temperatures in K). Volume Temperature Work Heat du a->b (2) Fill in the Process Table (all entries in Joules). b-> C-> (3) Find the Totals: Work = Heat = dU = (4) Find the heat input (from "hot reservoir") and the heat output (to "cold reservoir"): Q-hot = Q-cold = (5) Find the efficiency of the engine: efficiency =

Answer 1

, 15 otherm Ta= 300k Pa=100,000 pa va = nRTA P Pa 2048.314*300 100000 0.25 ms Temperature pressure - a. 100000 pa b. 200000 Pa volume 0.25 m3 . 0.25 m? 3ook Toe Porno = 601.4K nr c. 100000 Pa = 0.5 T-601.4K m3 work Head do 62.64 kg agb b 88: Enco- = 10x5 6014-300) 42.64 K1 34.66 k} btc с MRT т-т, 2 , = 1048.314 x 6014 enero = 34.66 KB coa 5217 pe (na-re) I neesat =-100000 (0.5-0.25)=related =-25 ks -10 366014-300) .:-97.7k1 Scanned with CamScanner

Totals work=0+34.66-25 -9.66 kJ Head = 62.64+34.66-87.7 = 9.6 kJ du = 62.64 +0 -52.7 59.94 kg Q- hot = 62.64 +34.66 = 97. 30kg or cold=8717 kJ Efficiency = 12 Qcold o not 87.7. .. .ale to 21-0.90 -0,10 - 2017 Scanned with ÇamScanner