Homework Cover page Student Last Name Student First Name UIN Section Select 204 OR 504 HW Date Submitted Solution 6 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 Homework Assignment 6 Due Mar 07 2017 At the beginning of the class Problem 1 Refrigerant 134a enters a well insulated compressor as a saturated vapor at 32 C with a volumetric flow rate of 0 5 m3 s The R134a exits the compressor at a pressure of 9 0 bar and a temperature of 75 C a b c Determine the mass flow rate through the compressor in kg s Determine the volumetric flow rate at the compressor exit in m3 s Calculate the compressor power in kW Solution W Find m 2 CV Assume steady state uniform flow at inlet and exit Q CV 0 PE KE 0 dmCV 0 m 1 m 2 m m 1 m 2 dt V2 dECV 0 Q CV W CV m i hi i g zi m e Basic Equations dt 2 i e AV m AV v v Solution At state 1 T1 32 C V2 he e g ze 2 x1 1 0 m3 v1 v g 32 C 0 2451 kg m3 s 2 04 kg m m1 m1 1 m3 v1 s 0 2451 kg 0 5 h1 32 C 227 90 kJ kg 1 points for accuracy At state 2 T2 75 C p2 10 0 bar Tsat p 9 0 bar 39 39 C state 2 is SHV From Table A 13 m3 75 70 v 2 0 02423 0 02538 0 02423 0 02481 kg 80 70 m2 2 kg 2 04 v2 s 2 0 0506 kg m3 2 m2 v 2 2 04 0 02481 s kg m3 s 1 points for accuracy kJ kJ 75 70 h2 302 34 307 77 313 20 302 34 kg kg 80 70 From the first law dECV V12 0 QCV WCV m1 h1 g z1 dt 2 V22 g z1 m2 h2 2 kg kJ WCV m h1 h2 2 04 227 9 307 77 s kg WCV 162 93 kJ 162 93 kW s 1 points for accuracy Problem 2 Steam enters an adiabatic turbine at a mass flow rate of 12 kg s and at 4 MPa and 500 C The exit pressure is 60 kPa and the turbine produces 7 MW of mechanical power on the output shaft Neglecting kinetic and potential energy of the steam determine a the temperature C at the outlet b the volumetric flow rate at the outlet m3 s repeat a and b assuming the mechanical power of the turbine is 5 MW For each case draw the process on a P v diagram Inlet H2O 12 kg s 500 C 4 MPa Wout 7MW or 5MW Outlet 60 kPa Find T2 2 P n diagram Assume Steady State uniform flow at inlet and exit Qcv 0 DKE 0 DPE 0 dmCV 0 m1 m2 dt dECV V2 V2 0 QCV WCV mi hi i gzi me he e gze Basic Equations dt 2 2 i e V mn Inlet look up properties of water in saturated table Table A 3 At P1 4MPa Tsat T1 h1 3445kJ kg From conservation of mass m1 m2 12kg s dECV V2 V2 0 QCV WCV mi hi i gzi me he e gze dt 2 2 i e W m h1 h2 12kg s 3445kJ kg h2 7 MW h2 2861 67kJ kg From saturated water table Table A 3 At 60kPa hg 2653 5 h2 So State 2 is a superheated water vapor From Table A 4 we do not have a pressure table for P 60kPa so we need to interpolating between p 35kPa and p 70kPa to obtain state corresponding to h2 2861 67kJ kg and P2 60kPa Here we need a double interpolation First find the temperature range at P2 60kPa where the given enthpy falls within 60 35 h 160 2800 6 2798 2 2800 6 2798 9kJ kg 70 35 60 35 h 200 2878 4 2876 7 2878 4 2877 2kJ kg 70 35 Interpolating using h2 2861 67kJ kg for temperature T2 using above values 2861 67 2798 9 0 T2 160 200 160 192 06 C 2877 2 2798 9 Specific volume n 2 can also be interpolated similarly from Table A 4 for data between p 35kPa and p 70kPa to obtain data for P 60kPa 60 35 3 n 160 5 696 2 841 5 696 3 6567m kg 70 35 60 35 3 6 228 3 108 3 9994m kg 70 35 Interpolating using T2 192 060 C new specific volume n 200 6 228 192 160 3 3 9994 3 6567 3 9309m kg 200 160 Volume flow rate is given by n 2 T2 3 6567 V2 mn 2 12kg s 3 9309m3 kg 47 17m3 s 1665 793 ft 3 s Assume If the output power is 5MW repeat the a and b a At inlet look up properties of water in saturated table At P 4MPa Tsat T1 so it is superheated vapor P1 4MPa Tsat T1 h1 3445kJ kg m1 m2 12kg s dECV V2 V2 0 QCV WCV mi hi i gzi me he e gze dt 2 2 i e W m h1 h2 12kg s 3445kJ kg h2 5 MW h2 3028 3kJ kg At 60kPa h2 hg it is a superheated water vapor From Table A 4 interpolating data between p 35kPa and p 70kPa to obtain data for P 60kPa 60 35 h 240 2956 8 2955 5 2956 8 2955 9kJ kg 70 35 60 35 h 240 3036 3035 3036 3035 3kJ kg 70 35 3028 3 2955 9 0 T2 240 280 240 276 47 C 3035 3 2955 9 Specific volume n 2 can also be interpolated similarly from Table A 4 for data between p 35kPa and p 70kPa to obtain data for P 60kPa 60 35 3 3 374 6 758 4 3409m kg 70 35 60 35 3 n 280 7 287 3 640 7 287 4 6820m kg 70 35 Interpolating using T2 276 470 C new specific volume n 240 6 758 276 47 240 3 4 6820 4 3409 4 6519m kg 280 240 Volume flow rate is given by n 2 T2 4 3409 V2 mn 2 12kg s 4 6519m3 kg 55 82m3 s 1971 27 ft 3 s Problem 3 A gas turbine accepts gas air at a mass flow rate of 10 lbm s and at 200 psia and 2800 R The combustion products leave the turbine at 14 psia and 1500 R Due to insufficient insulation of the turbine heat is lost from the turbine at a rate of 420 Btu s The power produced by the gas turbine is used …