Question

3) Adding heat at constant volume:

You have a 10 L (10

-2

m

3

) fixed volume container that initially contains 1 kg of

water. The water exists in both vapor and liquid phases, and is at an initial

temperature of 180 C. Heat is then added until the final temperature of the water

in the container is 250 C. This problem involves calculating the amount of heat

that is added to achieve this final state. I will roughly lead you through this

problem – please put the answers in the correct parts for ease of grading.

a. What is the pressure of the initial and final states?

b. Calculate the mass fraction of the water in the vapor state – both in the

initial state and the final state. Yes, there is enough information to answer

this before the energy balance.

c. Choose a system (should be easy), and write down and simplify the energy

balance for this problem.

d. Calculate the heat added to the contents of the container to achieve the

final state. You can consider that the walls of the container to have

negligible heat capacity.

Answer 1

ca V = COL= 10 xi m = 6 m m = 1kg T=180°c T2= 250°C Intial spectic volume, Vis the one to her - B-dimlkg At Te = 186 c, 16 = 0.001127 milla, 19 = 0.1938479/kg Of <VI <vg .. Satorated mušture. Pi = Psat @ 180c Pi = 1002.8 kPa] At Tz -250°C, 24-0.001252mp/kg, Ug = 0.050085milley 0, =V2 = 0.01 m Ikg Vf <V2 <Vg - saturated mixterne P2 = Psat @ 25000 1 P2 = 3976.2 kPa] (b) ui= uf+ 2 (vg_vf) @ 180° 0.01 = 0.001127 + 2160.19384 -0.001127) 1761 = 0.046] 21 = 4.67 U2= of +82 (vg-uf) @ 2500 0:01 = 0.001252 +2260.050085 -0.001252) [X2 0.179 [12 = 17.97.]

---- V=LOL maika Ti = 180°C Rús Em - Eout = AE system. Qui A E system (ain = (12-21) (d Qui = 42-2, 41= wftx, Ufg @ 180°c - 761.92 +0.046 (1820.9) = 845.76 KJ kg U2 = uf+ K2 Ufg @ 2500 = 1080.77 0.179 (1521-1) = 1353.19 kJ lig Qui = m (12-2) -(1kg) (1353.19-845.76) kJkg Qum = 507.43 kJ