Question

Problem 1

A piston-cylinder system contains a paddle stirrer (a paddle with blades that can stir the contents of the cylinder). The cylinder contains a gas that is initially at 150 kPa and occupies a volume of 0.196 m3 . The piston is unrestrained and has a cross-sectional area of 0.196 m2 . 20 kJ of electrical energy is supplied to the paddle stirrer, resulting in the piston moving up by 0.2 m. During this process, 5.27 kJ of heat are lost to the surroundings. The system boundary is the interior surface of the piston-cylinder system.

a. What is the mass of the piston, in kg?

b. What are Qin, Qout, Win, and Wout, in kJ?

c. What is the change in internal energy (ΔU), in kJ?

Problem 2 A piston-cylinder system contains 1 kg of a gas. The system is initially at a pressure of 100 kPa and has a volume of 10 m3 . For each of the following cases, if possible, calculate the boundary work (and its sign), in kJ, and accurately sketch the path between State 1 and State 2 on a p-v diagram. If such a calculation is not possible, indicate what information is missing.

a. Constant pressure compression to half the initial volume.

b. Constant volume heat addition until the pressure rises to 300 kPa.

c. Polytropic expansion with n = 5/3 to a final pressure of 50 kPa.

d. Heat removal until the pressure drops to 75 kPa and volume decreases to 8 m3 .

Answer 1

1 Ap = 0.196 m2 finial Pi - 150 kPa Vi = 0.196 me initial it do 20kJ mg = P; x Ap 101325 x 0.196 Piston displacement x = 0,2 m Heat lost is 5.27 kJ ie. Q . - 5.27 kJ Work Input is. We – 20 kJ Patm (0) Id ng d. Patmx Ap 19 r a ↑ m= 150x103 x 0. 196 9.81 m=972.5 kg. Dout = 5.27 KJ Win = 20 kJ Wout = pdv - As the gas is expanding. AV. Apa = 0.196x 0.2 m3 Also V₂ = V + DV V = 0.196+ 0.19680.2 V2 = 0.2352 Wout 150X10°x (0.196X0:2) Wout = 5880 J Wout = 5.88 kJ Qir=0 as no heat is supplied. (C) Q-W-AU from 1st Law of Thermodynamics. (-5.27)-(-20+5.88) - DV AU = 8.05 kJ.

n = 1 kg 2. Pi= 100 kPa Vi z 10 m3 To calculate work in each case - ie. Closed system Work bec. it is Piston - Cylindes. system P (V-Vi) W = 100x10 X10( 5-10) Wa – 5 x 10 J Prefssure Work = 100 kPa Cive means work is absorbed. ce. Work is done on the system. 0/2 = 5m3 Volume = 10m P 300 upg W = Pav W = 0 dreo 100x Po © P 100 kPa 513 PVC 50 HP C. 10m3 We P; V; – Pq Vt n-1 Vf can be found as :- PV 53 5/3 100 x 100 = 50x VS/ - 15.157 m² ...W= 100x10x10 - 50x10x15.157 V 5-1 W z 363 225 J Wz 3.63225x105 J & Not possible as pressure and volume cannot decrease simultaneously. P XX WOH possible. V