3-75
a) To find the mass of the refrigerant, you need to divide the volume of the tank by the specific volume at the initial state. Since the quality = 0.4, the v= vf + xvfg = 0.0007532 + [0.4 * (0.0993-0.0007532)] = 0.04017 m3/kg Since the V = 0.5 m3, the mass = 0.5 m3/0.04017 m3/kg = 12.45 kg. b) The energy balance simplifies to Q = DU since there's no work. Therefore, Q = m(u2 - u1). Then u1 = uf + xufg = 36.69 + [0.4*9221.43-36.69)] = 110.59 kJ/kg. For u2, you need two intensive, independent properties that set the system. You have pressure, and since this is a rigid tank, specific volume stays constant. So v2 = v1 = 0.04017 m3/kg. Looking at a pressure of 800 kPa, we find that this specific volume is greater than vg, so it's a superheated vapor. Looking in the superheated tables for 800 kPa, we interpolate the u2 to be 363.09. Putting the numbers in the energy balance gives us a Q = 3144 kJ.

3-79
Use the energy balance for this one. We need Q which equals DU + W. First find work, which is P(V2 - V1) = Pm(v2 - v1). The pressure is 800 kPa. The mass is 5 kg. At state 1, the refrigerant exists as a superheated vapor. The v = 0.02992 m3/kg and the u = 271.04 kJ/kg (you'll need that later). At state 2, it's a compressed liquid, so you look up the vf and uf at 20 C which are 0.0008157 m3/kg and 76.80 kJ/kg respectively. Putting everything together, Q = W + DU = Pm(v2 - v1) + m(u2 - u1) = m[P(v2 - v1) + u2 - u1)] = -1089 kJ. Note the negative sign since heat is being lost by the system.

3-84
This process takes place in three different states. There is the initial state which is a saturated vapor at 200 kPa. The next state the pressure increases to 300 kPa, but the volume stays constant. Finally, the pressure stays constant, but the volume doubles. a) To find the final temperature, we have to determine the state. We have pressure, and we can figure out the specific volume since the mass stays the same and the volume doubles from the initial state. At the initial state, the v1 = vg @ 200 kPa = 0.8857 m3/kg. The final state, the specific volume is twice that or 1.7715 m3/kg. This sets the state which is a superheated vapor, and the temperature is around 878.9 C. b) All the work done is in the final stage since there is no volume change from the initial to the second state. So W = Pm(v3 - v2). But v2 = v1 which equals 0.8857 m3/kg. To find the mass, divide the initial volume by the initial specific volume = 0.5645 kg. You should get W = 150 kJ. c) Heat transfer is determined from the energy balance Q = DU + W. The DU = m(u3 - u1). Interpolating u3 to be 3813.8 kJ/kg and looking up u1 as ug @ 200 kPa which equals 2529.5 kJ/kg, Q ends up being 875 kJ.

3-154
This is a similar problem
a) The initial temperature is Tsat @ 100 kPa = 99.63 C. The final temperature is found from P = 200 kPa and the specific volume. Since the volume increases by 20 %, so does the specific volume. Find v1 from the quality, v1 = vf + xvfg = 1.004 m3/kg. So v3 = 1.205 m3/kg. Looking at 200 kPa, we see that it's a superheated vapor, and looking at the superheated tables lets us estimate the T to be 252.6 C. b) Since v1 = v2 = 1.004 m3/kg, if we look that up on the saturated water table under pressure, we see that it is greater than the vg. So we have a superheated vapor even at state two, and there is no liquid left. c) The work done is P2(V3 - V2) = 200 kPa(6.025 - 5.02 m3) = 201 kJ.

3-166
This is exactly like the demo I did on Friday except instead of heat transfer between the metal and the water, we have heat transfer between the two gases. First find the masses of each gas using the ideal gas law. The mnitrogen = 4.77 kg. The mhelium = 0.808 kg. Then the energy balance becomes that 0=DUnitrogen + DUhelium = mCv(Tf - TI)nitrogen + mCv(Tf - TI)helium. Watch your signs. Putting in all the numbers gives a final temperature of 57.2 C. If the piston is not free to move, the answer would be the same since it would effect only the pressure and not the specific heats.