Question

A rigid tank with a total volume of 0.8m3 initially contains water in the form of a saturated liquid-vapour mixture at P1=500kPa. The liquid portion of the mixture initially occupies a volume of only 0.5L (the remainder is vapour). Steam from a supply line at Pi=1000kPa and Ti=500°C is then injected into the tank until the pressure in the tank reaches P2=800kPa. At this point (final state), the mass in the tank is m2=2.8kg.

a) Find the initial quality x1 and the mass added to the tank mi during the process. (7 marks)

b) Determine the final temperature T2 in the tank. (3 marks)

c) Calculate the heat transfer during the process. (8 marks)

Tank Supply Line

Answer 1

solution? Given Data? V=0.8m3. ca) at state 1: P = 500 kPa f = 0.5L 20.0005m3. So; Vg = V-ve. =0.8-0.0005 vg D.7995m3 from the steam table at p-500kpar. Ve=0.001093 m3lkg vg = 0.37466 m3/kg So, density of fluid; Pe= Kre 0.001093 -914.913 kgim density of steam; Pag - Kg 0.37466 22-669 kg (m3 Moys of Liquid; me =R x ve =914.913X 0.0005 = 0.4574565 kg

2 Mots of steam; mg spy xVg = 2.669x0.7995 -2.1338655 kg Initial Quality; x = mg mgtme - 2.1338655 2.133865570.4574565 x=0.82B466 Now; v = vetx, (vg - Vf) -0.001 093 +0.823466(0-373567). U=0.308713 m3/kg. ei - You 1 0.308 713 = 3239257 kg (m 3. mis eixu = 3.239257xo.g. = 2.5914 kg so; by mass balance; mitmi - ma 2.5914 etmi -2.8. moto-2086 kg

13 (b) from steam table at P=1000kpa and T=500°c hi=3478.3kJ/kg from steam Table at a=500kpa hf=640.1 kJ/kg hfg=2107. URJkg hehe+Gheg - 640.1 +0.& 23466(2107,4). hi=2375-47225 kJ/kg so, by Energy balance; m, hitmihi maha. 2-5914 (237547225) to 2086 (3478. 3)=2.8h2 ha - 2457.633 kJ/kg from steam Table at p=800kpa. hf=697 1 Kilkg hg = 2767.4 kJikg. ha=hf+xalag hf). 2457.683 = 6971+x2 (2767.4-697 D. 1x2=0.85 since ; g <1. so it is in the vapour come. so, the temperature to will be the satura- tom temperature at p=sookpa. 2 T2 = 170.4°C

min. u (c) Acdeling to the ist law of Thermodyna Q=sutu. since there is no wok output; so wro. 80; Q=AU = maugomu. from steam table at p=500kpa and 2,-0.823466 J,= 2221.57187 RJ1Kg From Steam table at p-800kpa and, (2=0.85 U2=22976533 kJlig Q- 2-862297.6533 )-2-5914 (22215787) Q=676-4545 o please like. It helps a lot to me.