KNOWN: Separate streams of air and water flow through a compressor and heat exchanger arrangement.

Transcription

1 .00 Separate tream of air and water flow through the compreor and heat exchanger arrangement hown in Fig. P.00. Steady-tate operating data are provided on the figure. Heat tranfer with the urrounding can be neglected, a can all kinetic and potential energy effect. The air i modeled a an ideal ga. Determine (a) the total power required by both compreor, in kw. (b) the ma flow rate of the water, in kg/. KNOWN: Separate tream of air and water flow through a compreor and heat exchanger arrangement. FIND: (a) The total power required by both compreor, in kw, and (b) the ma flow rate of the water, in kg/. SCHEMATIC AND GIVEN DATA: Air p = bar T = 00 K m 0.6 kg/ W A p = 9 bar T = 800 K W B Compreor A Compreor B p = bar T = 600 K p = p T = 50 K 6 5 T 6 = 0 o C p 6 = p 5 Heat exchanger Water T 5 = 0 o C p 5 = bar ENGINEERING MODEL:. Control volume at teady tate encloe the compreor and heat exchanger.. For each control volume, heat tranfer with the urrounding i negligible and kinetic and potential effect can be ignored.. The air i modeled a an ideal ga. ANALYSIS:

2 (a) A ma balance for the air flowing through compreor A, the heat exchanger, and compreor B give m m m m = 0.6 kg/ The energy rate balance for compreor A 0 = Q W A A m [(h h ) + ½ (V V ) + g(z z )] implifie to W A m ( h h ) The pecific enthalpie for air at tate and tate are obtained from Table A-: h = 00.9 /kg and h = /kg. Subtituting value and olving yield kg kw W A = 8. kw kg kg Similarly for compreor B, W B m ( h h ) The pecific enthalpie for air at tate and tate are obtained from Table A-: h = 5.80 /kg and h = 8.95 /kg. Subtituting value and olving yield kg kw W B =. kw kg kg The total power required by both compreor i W total W A W B = ( 8. kw) + (. kw) = 06. kw The negative ign indicate power i to the compreor. (b) Since the air and water do not mix in the heat exchanger, the teady tate ma balance reduce to m m = 0.6 kg/ The teady tate form of the energy rate balance m 5 m 6

3 implifie to 0 Q W i V m i i h i gzi V m e e h e e 0 m h m 5h5 m h m 6h6 gze or ubtituting reult from the ma balance Solving for the ma flow rate of water give 0 m ( h h ) m 5( h5 h6 ) m( h h ) m 5 h6 h5 The pecific enthalpie of water at tate 5 and tate 6 are obtained from Table A-: h 5 h f5 = 8.96 /kg and h 6 h f6 = 5.79 /kg. Subtituting value and olving yield kg kg kg m kg kg m 5 =. kg/

4 .99 Figure P.99 how a turbine-driven pump that provide water to a mixing chamber located 5 m higher than the pump. Steady-tate operating data for the turbine and pump are labeled on the figure. Heat tranfer from the water to it urrounding occur at a rate of kw. For the turbine, heat tranfer with the urrounding and potential energy effect are negligible. Kinetic energy effect at all numbered tate can be ignored. Determine (a) The power required by the pump, in kw, to upply water to the inlet of the mixing chamber. (b) The ma flow rate of team, in kg/, that flow through the turbine. KNOWN: A team turbine drive a pump through which water flow to a mixing chamber located 5 m higher than the pump. FIND: (a) The power required by the pump, in kw, to upply water to the inlet of the mixing chamber and (b) the ma flow rate of team, in kg/, that flow through the turbine. SCHEMATIC AND GIVEN DATA: h = 7.69 /kg Mixing chamber Q kw Steam p = 0 bar T = 00 o C 5 m W p = 5 bar T = 80 o C Turbine Pump Saturated liquid water p = bar m 50 kg/ ENGINEERING MODEL:. All component operate at teady tate.. Define control volume A to encompa the pump and the line to the mixing chamber.. Define control volume B to encompa the turbine.. For both control volume A and B, kinetic energy effect can be ignored. 5. No tray heat tranfer occur between control volume B and it urrounding. 6. For control volume B, potential energy effect can be ignored.

5 ANALYSIS: (a) The energy rate balance for control volume A 0 = Q W m [(h h ) + ½ (V V ) + g(z z )] implifie to W Q m [(h h ) + g(z z )] ince m m. The pecific enthalpy at tate i obtained from Table A-: h = h f = 7.6 /kg. Since the elevation at tate i higher than the elevation at tate, (z z ) = 5 m. Subtituting value and olving yield W kg kw kg 7.69 W kg m N 9.8 ( 5 m) kg m = 5.8 kw kw 000 N m Since the value for power i negative, work i to the pump. (b) The energy rate balance for control volume B implifie to 0 = Q W m [(h h ) + ½ (V V ) + g(z z )] 0 W m (h h ) ince m m. Solving for ma flow rate of team yield m W h h Since both tate and are uperheated vapor, their pecific enthalpie are obtained from Table A-: h = 0.9 /kg and h = 8.0 /kg. Since the turbine produce the power required by the pump, W (turbine) W (pump) = ( 5.8 kw) = 5.8 kw. Subtituting value and olving yield 5.8 kw m kw kg kg m = kg/