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Unformatted text preview: PROBLEM 4.37 Nomenclature: 1. ̇ = heat transfer rate ̇ = rate of working (or power) 2. 3. = pressure 4. = temperature 5. = velocity 6. = altitude 7. ℎ = specific enthalpy 8. = gravity constant Model: 1. Control volume encloses the diffuser 2. Steady state 3. ̇ = 0, ̇ = 0 and no potential energy effect 4. Ideal gas Solution: Mass conservation for a control volume in steady state ̇=̇=̇ (1) First law for a control volume in steady state ̇ ̇ + ̇ [(ℎ − ℎ ) + ]=0 +( − )] = 0 (2) (3) ̇ [(ℎ − ℎ ) + ℎ = 215.97 kJ/kg, ℎ = 250.05 kJ/kg from Table A22 = + 2(ℎ − ℎ ) (4) = (265 ) + 2(215.97 − 250.05) = 45 (5) PROBLEM 4.47 Nomenclature: 1. ̇ = heat transfer rate ̇ = rate of working (or power) 2. 3. = pressure 4. = temperature 5. ̇ = mass flow rate 6. = velocity 7. = altitude 8. ℎ = specific enthalpy 9. = gravity constant Model: 1. Control volume encloses the turbine 2. Steady state 3. Ideal gas Solution: Mass conservation for a control volume in steady state ̇=̇=̇ (1) First Law for a control volume in steady state ̇
̇ ̇ ̇ + ̇ [(ℎ − ℎ ) + +( − +( )] = −
̇ ̇ )] = 0 (2) (3) + [(ℎ − ℎ ) + Where: ̇ ̇ = −1.1 = 800 ℎ − ℎ = (3100 − 2300)
[ ] = ( − = −0.56 (3 ) = 0.03 ) = 9.81 Therefore:
̇ ̇ = [−1.1 + 800 − 0.56 + 0.03] = 10 798.3 = 133.1 = 798.3 ̇ PROBLEM 4.68 Nomenclature: ̇ = heat transfer rate 1. ̇ = rate of working (power) 2. 3. = pressure 4. = temperature 5. = velocity 6. = altitude 7. ℎ = specific enthalpy 8. = specific volume 9. = area of cross section 10. = gravity constant Model: 1. 2. 3. 4. Control volume encloses the power washer, including the inlet and delivery hoses Steady state Water in liquid phase Temperature change is ignored Solution: First law for a control volume in steady state ̇ Where ̇ + ̇ [(ℎ − ℎ ) + +( − )] = 0 (1) ̇ = 0.1 ̇ ̇= =
( ) = 0.0010018 = from Table A2 , = ) ℎ = ℎ (because Therefore
̇ ( ̇ ( ) ) = = . [ +( −
( )]
) (2) + 9.8 (−5 ) = 0.000077 (3) (.) . (. ) Electricity per liter = 0.000077 = 0.00062 < 0.05 (4) The cost of the water is significantly higher than the cost of electricity to delivery it. PROBLEM 4.82 Nomenclature: 1. ̇ = heat transfer rate ̇ = power 2. 3. = pressure 4. = temperature 5. = velocity 6. = altitude 7. ℎ = specific enthalpy 8. = specific volume 9. = area of cross section 10. = gravity constant Model: 1. Steady state 2. ̇ = 0, ̇ = 0 and no kinetic and potential energy change ( ) = 0.002339 3. = 0.3 > , thus liquid water at the inlet is compressed liquid. The standard approximation for compressed water at , is ( , ) = ( ). From this one then obtains ℎ( , ) = ℎ ( ) + ( )( − ( )). (Table A5 only has properties for compressed water at levels of 4.0 MPa or higher) Solution: Mass conservation for a control volume in steady state ̇+̇=̇ First law for a control volume in steady state ̇ ̇ +̇ ℎ+ + +̇ ℎ+ + −̇ ℎ+ + =0 (2) (1) ̇ ℎ + ̇ ℎ − ̇ ℎ =0 ̇ = ̇[ Where ℎ( , ℎ 20
° (3) ) = ℎ (20 ° )+ , (20 20
° ° )(0.3 − 0.002339 ) = 83.96 ) = 83.96 / / = 0.0010018 (300 from Table A2 − 2.339 ) = 84.26 ℎ( , + 0.0010018 from Table A4 from Table A3 ℎ = 2865 .5 ℎ = 2725 .3 Therefore ̇ = 6.37 . . . . = 120 .00 (4) PROBLEM 4.94 Nomenclature: 1. ̇ = heat transfer rate ̇ = power 2. 3. = pressure 4. = temperature 5. = velocity 6. = altitude 7. ℎ = specific enthalpy 8. = specific volume 9. = area of cross section 10. = gravity constant Model: 1. 2. 3. 4. Control volume encloses the turbine Steady state Ideal gas No kinetic and potential energy change Solution: (a) Mass conservation for a control volume in steady state ̇=̇ First for a control volume in steady state
. (1) ̇ =( )=̇ = 5.4 . ( ) = 5.89 (2) (b) ̇ ̇ Where ̇ = 900 + 1400 / = 2300 , ℎ = 821.95 / from Table A22 ̇ ̇ + ̇ [(ℎ − ℎ ) + + ̇ (ℎ − ℎ )= 0 +( − )] = 0 (1) (2) ℎ = 350.77 Therefore ̇ =̇  ̇ (ℎ − ℎ ) (3) = 2300 kW – 5.4 kg/s (821.95350.77)kJ/kg = 82 kW ...
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This note was uploaded on 10/24/2009 for the course ENGRD 2020 at Cornell University (Engineering School).
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