ES 211: Thermodynamics Tutorial 5 & 6

Q1. An 80-L vessel contains 4 kg of refrigerant-134a at a pressure of 160 kPa. Determine (a) the temperature, (b) the quality, (c) the enthalpy of the refrigerant, and (d) the volume occupied by the vapor phase.

a) Refrigerant compressed liquid or saturated mixture region a) Refrigerant compressed liquid or saturated mixture region?? Finding specific volume first: At 160 kPa, per kg, Saturated mixture region! T = Tsat @ 160 kPa = -15.60 oC b) Quality x = 0.157 c) At 160 kPa find hf and hfg

h= 64. 2 kJ/kg d) Mass of the vapor mg= xmf = (0. 157)(4 kg) = 0 h= 64.2 kJ/kg d) Mass of the vapor mg= xmf = (0.157)(4 kg) = 0.628 kg Volume ocupied Vg = mgvg = (0.628)(0.12348 m3/kg) = 0.775 m3 (or 77.5 L) Rest volume (2.5 L) occupied by liquid.

Q2. A piston-cylinder device contains 25 g of saturated water vapor that is maintained at a constant pressure of 300 kPa. A resistance heater within the cylinder is turned on and passes a current of 0.2 A for 5 min from a 120-V source. At the same time, a heat loss of 3.7 kJ occurs. (a) Show that for a closed system the boundary work Wb and the change in the internal energy ∆U in the first-law relation can be combined into one term, ∆H for a constant pressure process. (b) Determine the final temperature of the steam.

Energy balance For constant pressure process, the boundary work becomes Substitution gives: However, Therefore,

Determining the electrical work Initial state: P1=300 kPa, h1=hg at 300kPa = 2724.9 kJ/kg For h2, using equation derived in part (a) Final state: P2=300 kPa, h2=2864.9 kJ/kg, T2= 200 oC

Q3 A rigid tank is divided into two equal parts by a partition. Initially, one side of the tank contains 5 kg of water at 200 kPa and 25 oC and the other side is evacuated. The partition is then removed, and the water expands into the entire tank. The water is allowed to exchange heat with its surrounding until the temperature in the tank returns to the initial value of 25 oC. Determine (a) the volume of the tank, (b) the final pressure, and (c) the heat transfer for this process.

a) Initially the water in the tank exists as a compressed liquid since its pressure (200 kPa) is greater than the saturation pressure at 25 oC (3.1698 kPa). Approximating the compressed liquid as a saturated liquid at the given temperature, we find Then, initial volume of water is The total volume of the tank is twice this amount b) At the final state, specific volume of water is At 25 oC The water therefore is a saturated liquid-vapor mixture. Thus the pressure is saturation pressure at 25 oC

c) The energy balance on the system can be expressed as Even though the water is expanding during the process, the system involves fixed boundaries only therefore the moving boundary work is zero. Quality at the final state is determined from the specific volume information: X2=2.3 ×10-5 Then Substituing:

Q4 Determine the missing properties and the phase description in the following table for water: T, oC P, kPa U, kJ/kg x Phase description (a) 200 0.6 (b) 125 1600 (c) 1000 2950 (d) 75 500 (e) 850 0.0

a) Quality x=0.6 (given). Therefore we have saturated liquid-vapor mixture at a pressure of 200 kPa. Then the temperature must be the saturation temperature at the given pressure: Read uf and ufg at 200 kPa. Then the average internal energy: b) At 125 oC we read the uf=524.83 kJ/kg and ug=2534.3 kJ/kg Since given value of u=1600 we have a liquid-vapor mixture. Therefore, the pressure must be the saturation pressure at the given temperature Quality:

c) At 1 MPa we have uf=761. 39 kJ/kg and ug=2582. 8 kJ/kg c) At 1 MPa we have uf=761.39 kJ/kg and ug=2582.8 kJ/kg. The specified u value is 2950 kJ/kg. Therefore we have superheated vapor and the temperature at this state is determined by interpolation to be T=395.2 oC d) At given pressure 500 kPa, we have Tsat=151.83 oC. Comparing the given T value to this Tsat value tells us that we have compressed liquid. The given pressure is much lower than the lowest pressure value in the compressed liquid table therefore, we are justified to treat the compressed liquid as saturated liquid at the given temperature: e) Quality given is zero thus we have a saturated liquid at the given pressure of 850 kPa. Temperature thus must be the saturation temperature at the given pressure, and the internal energy must have the saturated liquid value:

Q5 The power output of an adiabatic steam turbine is 5 MW and the inlet and the exit conditions of the steam are as indicated in figure below: (a) Compare the magnitudes of ∆h, ∆ke and ∆pe (b) Determine the work done per unit mass of the steam flowing through the turbine (c) Calculate the mass flow rate of the steam.

a) At the inlet, steam is in a superheated vapor state, and its enthalpy is P1=2 Mpa, T1=400 oC giving h1=3248.4 kJ/kg At the turbine exit, we obviously have a saturated liquid-vapor mixture at 15 kPa pressure. The enthalpy at this state is Then b) The energy balance for this steady flow system can be expressed as

The work done per unit mass of the steam is determined to be c) The required mass flow rate for a 5-MW power output is

Problem 6 In a shower system where water at 600C is mixed with cold water at 100C. If it is desired that a steady stream of warm water at 450C be supplied, determine the ratio of the mass flow rates of the hot to cold water. Assume heat loss is negligible and mixing to take place at a pressure at 150 kPa.