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2006 Thermodynamics Homework Spring #9 Solutions 1. A rigid wall, insulated container is divided into two regions by a removable wall. One region contains 1 lbm of kerosene at 100F, while the other region contains 2 lbm of kerosene at 150F. A stirrer is inserted into the container and when the wall between the two regions is removed the stirrer provides a shaft work input of 70 Btu. Determine the final temperature of the kerosene. Solution: System Type: Closed System Working Fluid: kerosene (incompressible substance) Process: Constant Volume Initial State Final State State 1 State 2 State 3 T1 = 100F = 560 R T2 = 150F = 610 R T3 = 640 R = 182F m1 = 1 lbm m2 = 2 lbm m3 = 3 lbm Bold values are calculated. Q=0 Wsh = -70 Btu (into system) Wbnd = 0 Initial State: Fixed Final State: UNKNOWN Conservation of Mass: m1 + m2 = m3 1st Law: Ufinal - Uinitial = -Wsh Approach: To fix the final state we will use our process description along with the conservation of mass to determine the final mass. Then using the conservation of energy, we can determine the final temperature. This will then fix the final state. Using conservation of mass m 3 = m1 + m 2 = 1 + 2 = 3 lb m Applying our conservation of energy we can write m 3 u 3 - m1 u1 - m 2 u 2 = - Wsh For an incompressible substance we can write u = cPT 1 ME 201 ME 201 Thermodynamics Spring 2006 Using this in our conservation of energy equation and solving for T3 - Wsh + m1c P T1 + m 2 c P T2 T3 - = m 3c P - (-70) + (1)(0.478)(560) + (2)(0.478)(610) = (3)(0.478) = 642 R or 182 F 2. An ice cube tray containing 0.5 kg of ice is removed from the freezer at -20C and placed on the kitchen counter. Determine the heat transfer required to (a) raise the ice temperature to its melting point (b) melt the ice (c) raise the resulting water to the room temperature of 23C The kitchen pressure is 100 kPa. Solution: For this problem we will consider the kitchen as our system. System Type: Closed System Working Fluid: Water (compressible substance) 0.001091 _333.40 State 1 T1 = -20C P1 = 100 kPa v1 = 0.001087 m3/kg State 2 T2 = 0C P2 = 100 kPa v2 = 0.001091 Constant m3/kg Process: pressure State 3 T3 = 0C P3 = 100 kPa v3 = 0.001000 u1 = -374.06 kJ/kg u3 = 0.0 kJ/kg u4 = 96.46 kJ/kg u2 = -333.4 kJ/kg phase: sub.solid phase: sat.solid phase: sat.liq. phase: sub.liq.. Process 1-2 Process 2-3 Process 3-4 Q1-2 = UNKNOWN Q2-3 = UNKNOWN Q3-4 = UNKNOWN Wsh = 0 Wsh = 0 Wsh = 0 Wbnd,1-2 = UNKNOWN Wbnd,2-3 = UNKNOWN Wbnd,3-4 = UNKNOWN Initial State: Fixed Initial State: Fixed Initial State: Fixed Final State: Fixed Final State: Fixed Final State: Fixed Conservation of Mass m1 = m2 1st Law u2 - u1 = q1-2 - wbnd,1-2 Wbnd wbnd,1-2 = P1(v2-v1) m3/kg State 4 T4 = 23C P4 = 100 kPa v4 = 0.001003 m3/kg Italicized values are from steam tables. m2 = m3 u3-u2 = q2-3 - wbnd,2-3 wbnd,2-3 = P2(v3-v2) Approach: The three heat transfers will be calculated from the first law, but to do this we must first evaluate the boundary work using the appropriate expression for Pdv. Since all of our states are fixed it will simply be a matter of evaluating our needed 2 m3 = m4 u4-u3 = q3-4 - wbnd,3-4 wbnd,3-4 = P3(v4-v3) ME 201 Thermodynamics Spring 2006 properties. We begin by noting that all four states can be treated as incompressible substances, so that if we desire to use the steam tables we can write v1 = vi(at -20C) = 0.001087 m3/kg u1 = ui(at -20C) = -374.06 kJ/kg v2 = vi(at 0C) = 0.001091 m3/kg u2 = ui(at 0C) = -333.4 kJ/kg v3 = vf(at 0C) = 0.001000 m3/kg u3 = uf(at 0C) = 0.0 kJ/kg v4 = vf(at 23C) = 0.001003 m3/kg u4 = uf(at 23C) = 96.46 kJ/kg Now calculating our boundary work wbnd,1-2 = P1(v2-v1) = (100)(0.001091 - 0.001087) = 4.0 x 10-4 kJ/kg wbnd,2-3 = P2(v3-v2) = (100)(0.001000 - 0.001091) = -9.10 x 10-3 kJ/kg wbnd,3-4 = P3(v4-v3) = (100)(0.001003 - 0.001000) = 2.60 x 10-4 kJ/kg and our heat transfer is q1-2 = u2 - u1 + wbnd,1-2 = (-333.4) - (-374.03) + (4.0 x 10-4) = 40.66 kJ/kg q2-3 = u3 - u2 + wbnd,2-3 = 0.0 - (-333.4) + (-9.1 x 10-3) = 333.39 kJ/kg q3-4 = u4 - u3 + wbnd,3-4 = 96.46 - 0.0 + (2.6 x 10-4) = 96.46 kJ/kg or on a total basis Q1-2 = (0.5)(40.66) = 20.33 kJ/kg Q2-3 = (0.5)(333.39) = 166.70 kJ/kg Q3-4 = (0.5)(96.46) = 48.23 kJ/kg 3

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