KNOWN: Air undergoes a polytropic process in a piston-cylinder assembly. The work is known.

Size: px

Start display at page:

Download "KNOWN: Air undergoes a polytropic process in a piston-cylinder assembly. The work is known."

Thomas Carpenter
6 years ago
Views:

1 PROBLEM.7 A hown in Fig. P.7, 0 ft of air at T = 00 o R, 00 lbf/in. undergoe a polytropic expanion to a final preure of 5.4 lbf/in. The proce follow pv. = contant. The work i W = 94.4 Btu. Auming ideal ga behavior for the air, and neglecting kinetic and potential energy effect, determine (a) the ma of air, in lb, and the final temperature, in o R. (b) the heat tranfer, in Btu. KNOWN: Air undergoe a polytropic proce in a piton-cylinder aembly. The work i known. FIND: Determine the ma of air, the final temperature, and the heat tranfer. SCHEMATIC AND GIVEN DATA: Q Air W = 94.4 Btu ENGINEERING MODEL:. The air i a cloed ytem.. Volume change i the only work mode.. The proce i polytropic, with pv. = contant and W = 94.4 Btu. 4. Kinetic and potential energy effect can be neglected. T = 00 o R p = 00 lbf/in. V = 0 ft P = 5.4 lbf/in. pv. = contant p T pv. = contant. T ANALYSIS: (a) The ma i determined uing the ideal ga equation of tate. v m = = = 9.00 lb To get the final temperature, we ue the polytropic proce, pv. = contant, to evaluate V a follow. Now V = = (0 ft ) = 4.8 ft

2 PROBLEM.7 (CONTINUED) T = = = 57 o R Alternative olution for T The work for the polytropic proce can be evaluated uing W =. For the proce pv. = contant, and incorporating the ideal ga equation of tate, we get W = = Solving for T and inerting value T = + T = + (00 o R) = 57 o R (b) Applying the energy balance; ΔKE + ΔPE + ΔU = Q W. With ΔU = m(u u ), we get Q = m(u u ) + W From Table A-E: u(00 o R) = 0.4 Btu/lb and u(57 o R) = 9.5 Btu/lb. Thu, Q = (9.00 lb)( )Btu/lb + (94.4 Btu) = Btu (in)

3 Corrected December, Refrigerant 4a enter an air conditioner compreor at 4 bar, 0 C, and i compreed at teady tate to bar, 80 C. The volumetric flow rate of the refrigerant entering i 4 m /min. The power input to the compreor i 0 per of refrigerant flowing. Neglecting kinetic and potential energy effect, determine the heat tranfer rate, in kw. KNOWN: Refrigerant 4a with known inlet and exit condition flow through a compreor operating at teady tate. FIND: Determine the heat tranfer rate, in kw. SCHEMATIC AND GIVEN DATA: Refrigerant 4a p = 4 bar T = 0 o C A V = 4 m /min Compreor W m = 0 / Q =? p = bar T = 80 o C ENGINEERING MODEL: () The control volume hown in the accompanying chematic operate at teady tate. () Potential and kinetic energy change from inlet to exit can be neglected. ANALYSIS: To determine the heat tranfer rate, begin with the teady tate ma and energy balance. m m m V V 0 Q W m h h g z z Simplify baed on aumption and olve for Q.

4 Q W m h m W h m h h () To obtain m, in /, fix tate by referencing Table A- at 4 bar. T > T at and therefore a uperheated condition exit at tate. From Table A- at T and p : v = m /. m AV v m 4 min m min 0.5 From Table A-: h =.9 / and h = 0.4 /. To obtainq ubtitute into Eq. (). kw # Q kW, in kw,. The negative ign indicate that there i energy rejected from the ytem by heat tranfer.

6 PROBLEM.45 Steam undergoe an adiabatic expanion in a piton-cylinder aembly from 00 bar, 0 o C to bar, 0 o C. What i work in per of team for the proce? Calculate the amount of entropy produced, in /K per of team. What i the maximum theoretical work that could be obtained from the given initial tate to the ame final preure. Show both procee on a properly-labeled ketch of the T- diagram. KNOWN: Steam undergoe an adiabatic expanion in a piton-cylinder aembly. Data are given at the initial and final tate. FIND: Determine the work and entropy produced, each per unit ma of team. Find the maximum theoretical work that could be obtained from the given initial tate to the ame final preure. Show both procee on a T- diagram. SCHEMATIC AND GIVEN DATA: Steam p = 00 bar T = 0 o C p bar T = 0 o C ENGINEERING MODEL: () The team i a cloed ytem. () Q = 0. () ΔKE +ΔPE = 0. ANALYSIS: The work i determined by reducing the energy balance, a follow. ΔKE +ΔPE + ΔU = Q W W/m = (u u ) From Table A-: p = 00 bar and T = 0 o C; u = 79. / and =.000 / K From Table A-: p = bar and T = 0 o C; u = / and = / K W/m = ( ) =. / (out, a expected) Now, the entropy production can be evaluated uing the entropy balance, a follow. ΔS = + σ σ/m = ( ) = ( ) / K =.57 / K To find the maximum theoretical work, we note that ince ( ) = σ/m 0;

7 PROBLEM.45 (CONTINUED) Graphically T Not acceible adiabatically 00 bar () 0 o C... () () bar 0 o C Note that ince, tate i the limiting cae when = (reverible expanion with no entropy production). It i not poible to acce tate to the left of tate adiabatically from tate. Expanion to tate give the bigget difference in u, and hence the maximum theoretical work. From Table A-: p = bar, =.000 / K give and x = = = u = u f@ bar + x (u g@ bar u f@ bar ) = (0.775)( ) = 09. / Finally, the maximum theoretical work i (W/m) max = (u u ) = ( ) = 89.8 /

8 PROBLEM.40 m/ and

9 8.4 Compare the reult of Problem 8.40 with thoe for the ame cycle whoe procee of the working fluid are not internally reverible in the turbine and pump. Aume that both turbine tage and both pump have an ientropic efficiency of 8%. KNOWN: A regenerative vapor power cycle with one open feedwater heater operate with team a the working fluid. Operational data are provided. FIND: Determine (a) the cycle thermal efficiency, (b) the ma flow rate into the firt turbine tage, in /, and (c) the rate of entropy production in the open feedwater heater, in kw/k. Compare reult with thoe of Problem SCHEMATIC AND GIVEN DATA: Q in ( y) Steam Generator () p = MPa T = 50 o C Turbine h t = 8% W t W cycle 0 MW (y) p = MPa ( y) p = kpa () p 7 = p = MPa 7 Open Feedwater Heater 5 ( y) Q out Condener Pump p = p 5 = p = MPa x = 0 (aturated liquid) h p = 8% h p = 8% Pump 4 p 4 = p = kpa x 4 = 0 (aturated liquid) W p W p T p = MPa p = MPa p = kpa

10 ENGINEERING MODEL:. Each component of the cycle i analyzed a a control volume at teady tate. The control volume are hown on the accompanying ketch by dahed line.. All procee of the working fluid are internally reverible except for procee in the turbine and pump and mixing in the open feedwater heater.. The turbine, pump, and open feedwater heater operate adiabatically. 4. Kinetic and potential energy effect are negligible. 5. Saturated liquid exit the open feedwater heater, and aturated liquid exit the condener. ANALYSIS: Firt fix each principal tate. State : p = MPa (0 bar), T = 50 o C h = 50. /, =.840 / K State : p = p = MPa (0 bar), = =.840 / K h = 8. / State : p = MPa (0 bar), h = 99.4 / (ee below) =.974 / K h h h t h h h h h t( h h ) 50. (0.8)(50. 8.) = 99.4 / State : p = p = kpa (0.0 bar), = =.974 / K x = 0.89, h = 0.4 / State : p = kpa (0.0 bar), h = 7.9 / (ee below) x = 0.870, = 7.0 / K h h t h h h h h h t( h h ) 99.4 (0.8)( ) = 7.9 / State 4: p 4 = kpa (0.0 bar), aturated liquid h 4 = 5.5 /, v 4 = m /, 4 = 0.50 / K State 5: p 5 = p = MPa (0 bar), h 5 = 5.74 / (ee below) / K (auming the aturated liquid tate correponding to h 5 = h f in Table A- and interpolating for 5 = f ) v4( p5 p4) v4( p5 p4) hp h5 h4 h h h 5 4 p h m N ( )(000 ) kpa 000 m 0.8 kpa = 5.74 / 000 N m State : p = MPa (0 bar), aturated liquid h = 7.8 /, =.87 / K,

11 v = m / State 7: p 7 = p = MPa (0 bar), h 7 = / (ee below) h p v ( p7 p h h 7 ) h 7 h v ( p7 p h p ) h m N (0.007 )( ) kpa 000 m 0.8 kpa 000 N m = / (a) Applying energy and ma balance to the control volume encloing the open feedwater heater, the fraction of flow, y, extracted at location i h h5 ( ) / y = 0.89 h h ( ) / For the control volume urrounding the turbine tage 5 W m t ( h h ) ( y)( h h ) W m t ( ) ( 0.89)( ) = 09. / For the pump W p m ( h7 h ) ( y)( h5 h4 ) W p m ( ) ( 0.89)( ) = 5.89 / For the working fluid paing through the team generator Q m in h h 7 ( ) = 78.5 / Thu, the thermal efficiency i

12 W t / m W p / m ( ) / h = 0.94 (9.4%) Q in / m 78.5 / (b) The net power developed i W cycle m ( W t / m W p / m ) Thu, W cycle m ( W t / m W p / m ) MW m = 0.9 / ( ) MW (c) The rate of entropy production in the open feedwater heater i determined uing the teadytate form of the entropy rate balance: Q j 0 m ii m ee T j j i e Since the feedwater heater i adiabatic, the heat tranfer term drop. Thu, m ee e i m ii m m m 55 m [ y ( y) 5] 0.9 [.87 (0.89)(.974) ( 0.89)(0.549)] K kw / = 5.8 kw/k Compared to the ideal cycle in problem 8.40, the preence of internal irreveribilitie in the turbine tage and the pump reult in lower cycle thermal efficiency, higher required ma flow rate of team entering the firt-tage turbine, and greater rate of entropy production in the open feedwater heater. 4

KNOWN: Data are provided for a closed system undergoing a process involving work, heat transfer, change in elevation, and change in velocity.

KNOWN: Data are provided for a closed system undergoing a process involving work, heat transfer, change in elevation, and change in velocity. Problem 44 A closed system of mass of 10 kg undergoes a process during which there is energy transfer by work from the system of 0147 kj per kg, an elevation decrease of 50 m, and an increase in velocity