ES-7A Thermodynamics HW 4: 6-63, 70, 95, 98, 100, 103, 108, 152, 157 Spring 2003 Page 1 of 9

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1 ES-7A ermodynamics HW 4: 6-63, 70, 95, 98, 00, 03, 08, 5, 57 Spring 003 Page of Entropy of Air Given: Insulated piston/cylinder device wit 300 L of air at 0 kpa and 7 C. e air is eated by a 00 W resistance eater for 5 minutes at constant pressure. Find: Entropy cange during tis process using a) constant specific eat, and b) variable specific eat. a) Using constant specific eat: e properties of air are: R kj/kgk, c p.005 kj/kgk, c v 0.78 kj/kgk. e mass of te air is found using ideal gas law: m P /R (0 0.3)/( ) kg. e temperature at state will be found using te st Law: Q m (u u ) + W m ( ) for constant pressure m c p ( ). Q 00W 5 min 60 sec/min W 80 kw. Q/mc p + 80/( ) K. e entropy cange will be found by: P 704. S m c p ln R ln ln kj/k. P 90 b) Using variable specific eat: e mass of te air is still found using te ideal gas law, so m kg. At state (90 K), we look up 90.6 kj/kg and s.6680 kj/kgk. From te st Law, Q m( ), so Q/m + 80/ kj/kg. From te table, we see tat tis is between 690 K ( 70.5 kj/kg, s.5573 kj/kgk) and 700 K ( 73.7 kj/kg, s.5777 kj/kgk). We interpolate to find s : s )/( ) + ( )( )/( ) kj/kgk. S m(s s ) 0.435( ) kj/k.

2 ES-7A ermodynamics HW 4: 6-63, 70, 95, 98, 00, 03, 08, 5, 57 Spring 003 Page of Reversible adiabatic process w/ air Given: Air is compressed in a piston-cylinder device from 00 kpa and 7 C to 800 kpa in a reversible, adiabatic process. Find: Final temperature and te work done using a) constant specific eat, and b) variable specific eat. a) Using constant specific eat: e properties of air are: R kj/kgk, c p.005 kj/kgk, c v 0.78 kj/kgk, k.4. Since tis is a reversible and adiabatic process, we can use te isentropic ratio: k k k k 800 P K. P 00 o find te work, we ll use te st Law: Q 0 m c v ( ) + W w c v ( ) 0.78( ) kj/kg (work consumed). b) Using variable specific eat: At state, u 06.9 kj/kg and P r.3. Since tis is an isentropic process, we ll use te relative pressure ratio: P /P P r /P r P r P r (P /P ).3(800/00) Interpolate between 50 K (P r 9.684, u kj/kg) and 530 K (P r 0.37, u kj/kg): (P r P r@50 )(530 50)/(P r@530 P r@50 ) + 50 ( )(0)/( ) K. u ( )/(530 50) + (5.4 50)( )/ kj/kg. From te st Law, m(u u ) + W 0 w (u u ) kj/kg.

3 ES-7A ermodynamics HW 4: 6-63, 70, 95, 98, 00, 03, 08, 5, 57 Spring 003 Page 3 of Isentropic efficienfy of steam turbine Given: Steam enters an adiabatic turbine at 8 MPa and 500 C wit a mass flow rate of 3 kg/s, and leaves at 30 kpa. e turbine as an isentropic efficiency of 90 %. Find: a) emperature of te steam at te exit, and b) power output of te turbine. a) Inlet conditions of te steam: kj/kg and s kj/kgk. e ideal turbine as s s s kj/kgk and P 30 kpa. is is a saturated mixture since at P, s s falls between s f and s g. s f kj/kgk, s fg kj/kgk, f 89.3 kj/kg, fg 336. kj/kg. x s (s s s f )/s fg ( )/ s f + x s fg (336.) kj/kg. For a turbine, te efficiency is defined as: η ( )/( s ) η( s ) + 0.9( ) kj/kg. is is a saturated mixture at 30 kpa, so sat 69. C. b) e power output is found from te First Law: Q & 0 m& + W& W& m& 3( ) kw. ( ) ( ) 6-98 Isentropic efficienfy of gas turbine Given: Argon gas enters an adiabatic turbine at 800 C and.5 MPa at a rate of 80 kg/min and exits at 00 kpa. Power output of te turbine is 370 kw. Find: Isentropic efficiency of te turbine. Using constant specific eat: Properties of argon are: R 0.08 kj/kgk, c p kj/kgk, c v 0.3 kj/kgk, k.667. For te ideal turbine, use te isentropic ratio to find s : s k k k k 00 P s K. P 500 e power output of te ideal turbine is: ideal min ( ) 0.503( ) W & & 4.97 kw. mc p s 80 60sec e isentropic efficiency is: W& η W& actual ideal , or 89.8 percent. 4.97

4 ES-7A ermodynamics HW 4: 6-63, 70, 95, 98, 00, 03, 08, 5, 57 Spring 003 Page 4 of Compressor w/ R-34a Given: R-43a enters an adiabatic compressor as saturated vapor at 0 kpa and exits at MPa. e flow rate at te inlet is 0.3 m 3 /min, and te isentropic efficiency of te compressor is 80 %. Find: a) emperature of te refrigerant at te exit, and b) power input in kw. Sow te process on a - s diagram wit respect to te saturation lines. a) e properties at te inlet are: v 0.64 m 3 /kg, kj/kg, s kj/kgk. For te ideal case, s s s kj/kgk. At MPa, tis is a supereated vapor between 40 C (s kj/kgk, kj/kg) and 50 C (s kj/kgk, 80.9 kj/kg): s (s s )( ) ( )( )/( ) kj/kg. Using te definition for isentropic efficiency, we can find te actual : η s s η kj/kg. 0.8 At MPa, tis is a supereated vapor between 50 C ( 80.9 kj/kg) and 60 C ( 9.36 kj/kg). )(60 ) + 50 ( )(0)/( ) C. b) e power input is found from te st Law: Q& 0 m& + W& W& m& e mass flow rate is: ( ) ( ) min & sec m& kg/s. v 0.64 ( ) ( 88.83) W & m& kw (work consumed). e -s diagram is sown on te rigt: o state is saturated vapor. o ideal case is isentropic (vertical line). o actual case slants to te rigt (entropy increases). P ideal actual o P is same for actual and ideal cases. P s

5 ES-7A ermodynamics HW 4: 6-63, 70, 95, 98, 00, 03, 08, 5, 57 Spring 003 Page 5 of Air compressor Given: Adiabatic air compressor wit inlet conditions of 95 kpa and 7 C and outlet conditions of 600 kpa and 77 C. Find: a) Isentropic efficiency and b) exit temperature for te ideal case, using variable specific eat. Using variable specific eat, te conditions at inlet ( 300 K) are: kj/kg, P r For te ideal case, we use te relative pressure ratio to find P rs : P rs /P r P /P P rs P r (P /P ).3860(600/95) is value falls between 500 K (P r 8.4, kj/kg) and 50 K (P r 9.03, 53.5 kj/kg). s (P rs P r@500 )(50 500)/(P r@50 P r@500 ) ( )(0)/( ) K. s ( )/(50 500) ( )( )/ kj/kg. e actual is found from 550 K: kj/kg. e isentropic efficiency is: η s , or 8.6 percent

6 ES-7A ermodynamics HW 4: 6-63, 70, 95, 98, 00, 03, 08, 5, 57 Spring 003 Page 6 of Nozzle wit combustion gas using variable specific eat. Given: Hot combustion gas enters a nozzle at 60 kpa, 747 C, and 80 m/s, and exits at 85 kpa. e nozzle as an isentropic efficiency of 9 percent. Find: a) exit velocity, and b) exit temperature. a) At te entrance, 00 K, so kj/kg and P r 3.4. For te ideal case, we will use te relative pressure ratio: P rs P r (P /P ) 3.4(85/60) is is between 760 K (P r 39.7, kj/kg) and 780 K (P r 43.35, kj/kg). s (P rs P )/(P r@780 P r@760 ) ( )( )/( ) kj/kg. o find s, we use te st Law: q 0 s ( ) s s ( ) s ( ) (m/s)². e actual velocity is found from te isentropic efficiency: η s ( η ) ( ) s m/s. b) From te st Law: q 0 ( ) kj/kg Interpolate between 780 K ( kj/kg) and 800 K ( 8.95 kj/kg) to find : ) ( )(0)/( ) K.

7 ES-7A ermodynamics HW 4: 6-63, 70, 95, 98, 00, 03, 08, 5, 57 Spring 003 Page 7 of Nozzle wit combustion gas using constant specific eat. Given: Hot combustion gas enters a nozzle at 60 kpa, 747 C, and 80 m/s, and exits at 85 kpa. e nozzle as an isentropic efficiency of 9 percent. Find: a) exit velocity, and b) exit temperature. For constant specific eat, we ll use c p.005 kj/kgk and R kj/kgk. a) We can use te isentropic ratio to find s : s k k.4 k k 85.4 P s K P 60 o find s, we use te st Law: q 0 s ( ) s 000c s p ( ) s ( ) (m/s)². e actual velocity is found from te isentropic efficiency: η s b) From te st Law: ( η ) ( ) s m/s q 0 c p ( ) K c p

8 ES-7A ermodynamics HW 4: 6-63, 70, 95, 98, 00, 03, 08, 5, 57 Spring 003 Page 8 of Air compressor Given: Air is compressed by a compressor from 00 kpa and 7 C to 700 kpa at a rate of 5 kg/min. Find: Minimum power required if te process is a) adiabatic, and b) isotermal. Use variable specific eats. Using variable specific eat: a) e initial conditions are: 90.6 kj/kg, P r.3. For an adiabatic process, minimum power is acieved wen te process is isentropic. Using relative pressure ratios, find P r : P r P r (P /P ).3(700/00) Interpolate between 500 K (P r 8.4, kj/kg) and 50 K (P r 9.03, 53.3 kj/kg): (P r P )/(P r@50 P r@500 ) ( )( )/( ) kj/kg. From te st Law: Q & m& + W& W& m& min ( ) ( ) 5 ( ) 0 60 sec -8.0 kw (consumed). b) For isotermal process, tere is no cange in entropy ( 0), so te st Law becomes Q & W&. e work is minimized wen te process is reversible. For a reversible, isotermal process: P ( ) 700 Q& m & s ln 5 min s m & o o s s R ln kw 60 sec P 00 Q & W&, so te work is kw (consumed).

9 ES-7A ermodynamics HW 4: 6-63, 70, 95, 98, 00, 03, 08, 5, 57 Spring 003 Page 9 of Compressor wit R-34a Given: A 0.5 kw adiabatic compressor wit R-34a. Inlet conditions are 40 kpa and -0 C, exit conditions are 700 kpa and 60 C. Find: a) isentropic efficiency of te compressor, b) volumetric flow rate at te inlet in L/min, c) maximum volumetric flow rate at te inlet tat tis compressor can ave witout violating te Second Law. a) At te inlet, we ave supereated vapor wit kj/kg, v m 3 /kg, and s kj/kgk. At te exit, we ave supereated vapor wit kj/kg. For an ideal compressor, s s s kj/kgk, and at 700 kpa tis is supereated vapor between 40 C (s kj/kgk, kj/kg) and 50 C (s kj/kgk, kj/kg). Interpolate to find s : s (s s )( ) ( )( )/( ) kj/kg. e isentropic efficiency is: η s b) e flow rate is round using te st Law: Q& 0 m& ( ) m 3 /s m 3 sec 60 sec min + W& 000L 3 m m& & v 0.650, or 65 percent. W& 8.9 L/min. & v W & ( 0.5) c) e maximum volumetric flow rate is acieved wen te process is reversible and adiabatic. We repeat te calculation in part (b) using s instead of : ( 0.5) v W & & m 3 /s s m 3 sec 60 sec min 000L 3 m 5.9 L/min.