MEEN 315 PRINCIPLES OF THERMODYNAMICS SPRING 2017 Homework Cover page Student Last Name Student First Name UIN Section circle one HW Date Submitted Solution 204 504 4 Grading Completion All problems attempted Quality Neatness clarity legible Accuracy Accuracy of selected problem s Cover page this page attached on the top and all pages stapled together Total 10 points Max Score 4 1 4 1 Your Score 4 1 4 1 10 10 This cover page should be completed and stapled to the top of your homework solution MEEN 315 PRINCIPLES OF THERMODYNAMICS SPRING 2017 Homework Assignment 2 Due Feb 14 2017 At the beginning of the class Topics covered Properties and processes for pure substances Problem 1 Determine the missing properties for water H2O for each state given Show your three step decision process i e 1 list two known properties 2 determine the phase 3 look up properties from the correct table and show all calculations made for each case Fill in the table with your final answers Phase quality T Phase name X F X 1 0 230 450 212 400 P psia v ft3 lbm u Btu lbm h Btu lbm 200 2 0 1000 Problem 2 Determine the missing properties for refrigerant R134a CF3CH2F for each state given Show your three step decision process i e 1 list two known properties 2 determine the phase 3 look up interpolate properties from the correct table and show all calculations made for each case Fill in the table with your final answers Phase quality T Phase name X C Sat vapor P kPa 425 180 800 2000 v m3 kg u kJ kg h kJ kg 0 1224 305 70 Problem 3 A rigid sealed cylinder is filled completely with 100 lbm of pure water at 70 F and atmospheric pressure Determine a the volume of the tank ft3 Later a pump is used to extract 15 lbm of water from the cylinder The water remaining in the cylinder eventually reaches thermal equilibrium with the surroundings at 70 F Determine b the final specific volume and c the final pressure psia Initial State 100 lbm H2O 70 F 1 atm Final State m2 m1 15 lbm 70 F Problem 4 A piston cylinder apparatus contains 5 kg of refrigerant R 134a at 200 kPa and 20 C The piston is slowly pressed downward by a force applied to the piston The MEEN 315 PRINCIPLES OF THERMODYNAMICS SPRING 2017 compression process is slow such that the temperature of the system remains constant At some point the system becomes nearly incompressible regardless of the amount of applied force For the process a determine the initial and final volume m3 c determine the final pressure kPa minimum c draw the process on a P v diagram labeling the initial and final states and including the vapor dome on the diagram Initial State Final State piston R134a Applied Force F 200 kPa 20 C piston 20 C Problem 5 A rigid tank contains a partition that divides the tank into two sections each having a volume of 2 ft3 The left side contains 1 5 lbm of refrigerant R 134a at a pressure of 50 psia and the right side contains 2 lbm of refrigerant R 134a at 30 psia The partition is then ruptured and the contents are allowed to mix Heat transfer occurs with the surroundings until the entire tank contents reach a new equilibrium temperature of 60 F For this process determine a the initial temperature of the left and right sections F b the final pressure in the tank psia Initial State 2 ft3 1 5 lbm R134a 50 psia Partition 2 ft3 2 lbm R134a 30 psia Final State 60 F 4 ft3 3 5 lbm MEEN 315 PRINCIPLES OF THERMODYNAMICS SPRING 2017 Problem 1 Always follow the three step decision process 1 List two known properties 2 Determine the phase 3 Look up for properties from correct table Case 1 Known properties x 1 0 T 2300 F Assume the water is at saturated vapor phase x 1 0 then look up in saturated water table English Unit A 2E p 20 78 psia n 19 39 ft 3 lbm Therefore u 1082 6 Btu lbm h 1157 1Btu lbm Case 2 Known properties T 4500 F p 200 psia Since the saturated pressure for T 4500 F is psat 422 1 psia We have psat p So case 2 is superheated vapor Look up in superheated vapor table A 4E n 2 548 ft 3 lbm u 1146 4 Btu lbm Therefore h 1240 Btu lbm Case 3 Known properties T 2120 F n 2 0 ft 3 lbm Look up saturated water table A 3E first find out if n f n n g at T 2120 F Then it is a mixture of liquid and vapor So x n g 1 x n f x 26 80 1 x 0 01672 2 x 0 074 u x ug 1 x u f 246 6Btu lbm h x hg 1 x h f 252 2Btu lbm Case 4 Known properties T 4000 F p 1000 psia Look up in saturated water table find out if psat p So it is compressed water after looking up in compressed liquid water n 0 01864 ft 3 lbm u 374 3 lbm h 375 1Btu lbm Phase quality Phase name X sat vap X 1 0 Sup heat vapor Mix 0 074 Cpmp liquid T F 230 450 212 400 P psia 20 78 200 14 709 1000 1 points for accuracy 1 points for accuracy v u 3 ft lbm Btu lbm 19 39 1082 6 2 548 1146 4 2 0 246 6 0 01864 374 3 h Btu lbm 1157 1 1240 252 375 1 MEEN 315 PRINCIPLES OF THERMODYNAMICS SPRING 2017 Problem 2 Similar to the first problem three step decision process also hold for this problem Case 1 Known properties x 1 0 P 425kPa Look up in saturated refrigerant table A 11 After interpolation 8 93 3 15 74 T 10 630 C 4 0 0509 3 0 0409 n 0 0484m3 kg 4 231 97 3 235 64 u 232 88kJ kg 4 252 32 3 256 07 h 253 26kJ kg 4 Case 2 Known properties P 180kPa n 0 1224m3 kg Look up in saturated refrigerant table first Since n n sat so it is a superheated refrigerant Then look up in superheated refrigerant table T 100 C u 237kJ kg h 259 41kJ kg Case 3 Known properties P 800kPa u 305kJ kg Look up in saturated refrigerant table A 10 Since u usat so it is a superheated refrigerant Then look up in superheated refrigerant table Interpolation is conducted 305 299 37 T 100 90 90 95 890 C 308 93 299 37 305 299 37 n 0 0352 0 034 0 034 0 03467 m3 kg 308 93 299 37 305 299 37 h 337 08 326 52 326 52 332 74 kJ kg 308 93 299 37 Case 4 Known properties P 2000kPa h 70kJ kg Look up in saturated refrigerant table A 11 …