W t. Strategy: We need to step through the equipments and figure out the inlets and outlets to allow us to calculate the work and heat terms.

Transcription

1 Problem et G 10.1 olution pring 00 YT ) A schematic of the Rankine cycle and representation on a T- diagram: 1 Boiler P 0 W p pump 0 P 0 Condenser turbine 0 4 C 475 o C Depends on your last name. The information given to us: P 00 kpa, P 50kPa; T 500 o C 55 o C The solution will be given in this form (all solutions). We are asked to find the thermal efficiency, η (net work output) / (heat input). trategy: We need to step through the equipments and figure out the inlets and outlets to allow us to calculate the work and heat terms. Turbine ( ) We want to find the quality of the outlet stream and the shaft work produced. Need to find state. tate is already defined by the temperature and pressure. We can get and : W t T kj/kg K and 45.8 kj/kg from steam table at 00 kpa and T o C 500 o C 55 o C 0. Then we can find state, which is at P 50kPa and has the given above. When we look at the steam table for 50 kpa, we find that lies in between kj/kg K and sat vap kj/kg K Therefore the outlet of the turbine must be a mixture of liquid and vapor. To find the quality: x sat vap kj/kg K ( ) kj/kg K ee problem 5 d) for a case where the > sat vap and how we deal with that. Plugging in ( ) from above, gives us: 0.96 x 0.97 (ans) for T o C 500 o C 55 o C x fraction of vapor (1-x) sat vap sat vap The quality (x) allows us to calculate : x sat vap + (1 x)

2 From steam table (50 kpa): sat vap 646.0, kj/kg kj/kg W t kj/kg (since the turbine is adiabatic, 0 and the shaft work, W t - ) Condenser ( 4) All we need to know from the condenser is that the output is saturated liquid at P 4 P 50 kpa (since for condenser P 0). This output goes into the pump. Pump (4 1) W of a reversible pump, W p V liq P. From lecture notes. We have assumed V liq is constant because liquid is fairly incompressible. Looking at the steam table at 50kPa for saturated liquid: kj/kg. V cm /g m 1000 g 1kJ Wp V4 (P1 P4 ) 1.00 cm /g (00-50) kpa.5 kj/kg 1cm 1kg 1kPa m Note that W Then W p 4.9 kj/kg. pump is negligible compared to W turbine. Boiler (1 ) We want to calculate the heat required, 1. But we already have these values: kj/kg 4.9 kj/kg kj/kg Putting it all together Finally we can calculate the thermal efficiency: W η net Wt W p (ans) d) We can see from the calculation above that increasing the turbine-inlet temperature will increase the quality of the outlet. We can also see this graphically. T const P increasing T Increasing turbine inlet-temperature increases the thermal efficiency. Generally, for a power cycle, larger separation between T high and T low higher thermal efficiency.

3 e) What if the turbine s efficiency is 80%? T η W t / W t,rev The process is represented on the T- diagram We see that the only point affected by the change is point. (Point is the reversible turbine case). The work of the pump is the same as before, and so is the heat required by the boiler. Then: 1 4 rev 80% η const P η W net W t W p η W t,rev W p 0.80 W t,rev W p W t,rev kj/kg. W p.5 kj/kg kj/kg. Plugging in these values, we get η In general, the coefficient of performance of the cycle has dropped to roughly 80% of the reversible case. This is because W pump << W turbine. o the performance of the cycle is mainly determined by the efficiency of the turbine. Equipment sizing: ince we now only produce 80% of the maximum work possible, we need to use more steam than before; to be precise, by a factor of 1.5 ( 1/0.8). The equipment, then, has to accommodate this higher flow rate of steam.

4 4) Two-stage cascade refrigeration system. Cycle : Tetrafluoroethane Between 0 o F and 90 o F. Compressors are 75% efficient. Cycle 1: R or ammonia Between 55 o F and 0 o F., Condenser Τ 90 ο F throttle compressor 0 0 Τ 10 ο F Interchanger Τ 0 ο F throttle compressor 0 0 Τ 55 ο F Evaporator Cycle W c, W c,1 Cycle 1 C,1 ince we are given temperatures for both cycles, we can analyze each cycle separately. trategy: We want to find the heat and shaft work terms, for which we need values of enthalpy for the points 1,,, and 4. Refrigeration cycle (with throttle valve) on a P- diagram Point and 4 are usually defined in the problem ( anchor points ) because they are saturated vapor and saturated liquid, respectively. We can start with point 4 and identify point 1. Then we can start with point and identify point. P 4 1 Cycle analysis: Note: This solution will make a lot more sense if you follow it using Table 9.1 and Figure G.. Throttle valve (4 1) Inlet (point 4): aturated liquid at 90 o F. (temperature specified in problem statement) Table 9.1 (or Figure G.) gives us for 90 o F: P 4 P sat 119 psia; the enthalpy Btu/lbm. Outlet (point 1): Mixture of vapor and liquid. For throttle valve 0. Then Btu/lbm. There is no heat or work produced. Evaporator (1 ) Outlet (point ): aturated vapor at 10 o F. (temperature specified in problem statement) Table 9.1 (or Figure G.): P P sat psia; the enthalpy Btu/lbm. ince there is no work done, C, 1 ( ) Btu/lbm Btu/lbm.

5 Compressor ( ) We need to calculate the work of a reversible compressor first (to use the 75% efficiency information). For a reversible compressor, 0, or *. We want to compress from psia to 119 psia. On Figure G.: We follow a constant entropy line from point (sat. vap. at psia) to 119 psia. This gives us point * where * 10 Btu/lbm (and T * 116 o F). For this reversible compressor, W c,,rev * ( ) Btu/lbm 18.5 Btu/lbm. For our real compressor, η 75%. Then W c, W c,rev / Btu/lbm. Putting it all together For completion sake, we can calculate, (heat released in the condenser). ince for a cycle must be 0, W c + + C 0. Then, (W c, + C, ) Btu/lbm We are ready to answer the questions for cycle. The pressure limits, as we have seen, are the P sat at the specified temperatures: 16.6 psia at 10 o F and 119 psia at 90 o F (ans) W required by real compressor always larger than W required by reversible compressor. Thus we divide, not multiply, by Coefficient of performance COP c / W net C / W c (since the valve produces no work) For cycle, COP (60.1 Btu/lbm) / (4.6 Btu/lbm).44 (ans) Cycle 1 analysis: (this solution will show only the method for ammonia; method for R is same) Throttle valve (4 1) ammonia Inlet (point 4): aturated liquid at 0 o F. (temperature specified in problem statement) NIT webbook gives us for 0 o F: P 4 P sat 0.97 psia; the enthalpy Btu/lbm. Outlet (point 1): Mixture of vapor and liquid. For throttle valve 0. Then Btu/lbm. There is no heat or work produced. Evaporator (1 ) ammonia Outlet (point ): aturated vapor at 55 o F. (temperature specified in problem statement) NIT webbook: P P sat 6.51 psia; the enthalpy Btu/lbm. ince there is no work done, C,1 1 ( ) Btu/lbm Btu/lbm.

6 Compressor ( ) ammonia We need to calculate the work of a reversible compressor first (to use the 75% efficiency information). For a reversible compressor, 0, or *. We want to compress from 6.51 psia to 0.97 psia. On NIT webbook, we find Btu/lbm R (entropy of saturated vapor at 6.51). Then we search at P 0.97 psia for the point where *. This is the point where T * 11 o F and * Btu/lbm. For this reversible compressor, W c,1,rev * ( ) Btu/lbm Btu/lbm. For our real compressor, η 75%. Then W c,1 W c,rev / Btu/lbm. Putting it all together (ammonia) For completion sake, we can calculate,1 (heat released in the condenser). ince for a cycle must be 0, W c + + C 0. Then,1 (W c,1 + C,1 ) Btu/lbm We are ready to answer the questions for cycle. The pressure limits, as we have seen, are the P sat at the specified temperatures: 6.51 psia at 55 o F and 0.97 at 0 o F (ans) Coefficient of performance COP c / W net C / W c (since the valve produces no work) For cycle 1 (ammonia), COP 1 (548.7 Btu/lbm) / (114.9 Btu/lbm) (ans) For R We can follow exactly the same method as above with the data for R. The results are below. ummarizing the result for the three cycles: Cycle P low P high W c C COP (psia) (psia) (Btu/lbm) (Btu/lbm) (Btu/lbm) 1 ammonia R tetrafluoroethane

7 c) If cycle 1 uses ammonia: Cooling rate of 00 Btu/s at 55 o F. C, as calculated above, has units of Btu/lbm. Flow rate of ammonia m 1 (cooling per second) / (cooling per pound) (00 Btu/s) / ( C ) 00 / lbm ammonia/s (ans) ow are the two cycles connected? By the heat exchanger between the two. The heat released from the condenser in cycle 1 must equal the heat used by the evaporator in cycle. Or m 1,1 m C, (this way, the units are correct: this-many Btu/s equals this-many Btu/s) Flow rate of tetrafluoroethane m m 1 (,1 / C, ) 4.04 lbm tetrafluoroethane/s (ans) Power requirement of each compressor: Cycle 1: Power m 1 W c,1 (0.64 lbm ammonia /s)(114.9 Btu/lbm ammonia) Btu/s (ans) Cycle : Power m W c, (4.04 lbm tetrafluoro./s)(4.6 Btu/lbm tetrafluoro.) 99.0 Btu/s (ans) eat output from 1 m 1,1 (0.64 lbm ammonia/s)(66.6 Btu/lbm ammonia) Btu/s (ans) Overall COP (m 1 C,1 ) / (m 1 W c,1 + m W c, ) (00 Btu/s) / (41.89 Btu/s Btu/s) 1.4 (ans) If cycle 1 uses R: Cooling rate of 00 Btu/s at 55 o F. Flow rate of R m 1 (cooling per second) / (cooling per pound) (00 Btu/s) / ( C ) 00 / lbm R/s (ans) C, as calculated above, has units of Btu/lbm. ow are the two cycles connected? By the heat exchanger between the two. The heat released from the condenser in cycle 1 must equal the heat used by the evaporator in cycle. Or m 1,1 m C, (this way, the units are correct: this-many Btu/s equals this-many Btu/s) Flow rate of tetrafluoroethane m m 1 (,1 / C, ) lbm tetrafluoroethane/s (ans) Power requirement of each compressor: Cycle 1: Power m 1 W c,1 (4.015 lbm R/s)(9.1 Btu/lbm R) 41.5 Btu/s (ans) Cycle : Power m W c, (4.015 lbm tetrafluoro./s)(4.6 Btu/lbm tetrafluoro.) Btu/s (ans) eat output from 1 m 1,1 (0.64 lbm R/s)(66.6 Btu/lbm R) 41.5 Btu/s (ans) Overall COP (m 1 C,1 ) / (m 1 W c,1 + m W c, ) (00 Btu/s) / (41.5 Btu/s Btu/s) 1.4 (ans)

8 5) a) One adiabatic turbine arrangement W t Reactor T o F P psia T? P 0 psia The maximum work that can be produced is when the turbine is reversible 0 To compute the maximum work, let us look at the case of reversible turbine. (ee previous problem sets for more detailed solution to turbine/compressor problems). T o F, P psia. This is supersaturated steam because P > P sat (500 o F). upersaturated steam table gives for this condition: Btu/lbm, Btu/lbm R (These numbers are important because we keep using them for the latter part of the problem) Btu/lbm R. When we look in the steam table for P 0 psia we see that this entropy value lies in between 0.58 Btu/lbm R and sat vap 1.70 Btu/lbm R. Therefore the outlet of the turbine must be a mixture of liquid and vapor. To find the quality: x sat liq ( ) Btu/lbm R ( ) Btu/lbm R x fraction of vapor sat vap The quality (x) allows us to calculate : x + (1 x) (0.804)(1156. Btu/lbm) + ( )(196.7 Btu/lbm) sat vap These enthalpy values from steam table saturated condition at 0 psia. (1-x) Btu/lbm sat vap sat vap ince the turbine is adiabatic, W s + W s W s 1 ( ) Btu/lbm -47. Btu/lbm (ans) W s < 0 for turbine because the system produces work b) Two-turbine arrangement W t,1 W t, Reactor T o F P psia T? P P * T 500 o F P P* T 4? P 4 0 psia R There must be some heat R that gets transferred from reactor to the steam to raise its temperature from T (unknown for now) to 500 o F again. The process, we re told, occurs at constant P P*.

9 c) P* 10 psia. Our strategy is the same as before, except with different P s and two turbines. First turbine We have the same inlet as before, so 1 and 1 are same as in part a) Btu/lbm R team table for P 10 psia: 0.59 Btu/lbm R and sat vap Btu/lbm R. Therefore the outlet of the turbine must be a mixture of liquid and vapor. sat liq ( ) Btu/lbm R x This all looks kind of ( ) Btu/lbm R familiar, doesn t it? sat vap W t,1 1 ( ) Btu/lbm Btu/lbm econd turbine Inlet now is at P 10 psia and T 500 o F. Looking up in steam table for this condition: Btu/lbm and Btu/lbm R Btu/lbm R. team table for P 4 0 psia: 0.58 Btu/lbm R and sat vap 1.70 Btu/lbm R. Therefore the outlet of the turbine must be a mixture of liquid and vapor. 4 sat liq ( ) Btu/lbm R x ( ) Btu/lbm R sat vap x sat vap + (1 x)sat liq (0.947)(10. Btu/lbm) + ( )(97. Btu/lbm) Btu/lbm x sat vap + (1 x)sat liq (0.881)(1156. Btu/lbm) + ( )(196.7 Btu/lbm) Btu/lbm W t, 4 ( ) Btu/lbm Btu/lbm The total work W t,1 + W t, (-55.5 Btu/lbm) + (-14.5 Btu/lbm) Btu/lbm (ans) Jumping ahead to problem e) for P* 10 psia e) What is the heat absorbed during the reheating process? The heat absorbed ( R, see diagram in part b) will be because no work is done during reheating: R + W s R Btu/lbm Btu/lbm Btu/lbm (ans)

10 d) P* 5 psia. Our strategy is the same as before, except with different P s and we have two turbines. First turbine We have the same inlet as before, so 1 and 1 are same as in part a) Btu/lbm R team table for P 5 psia: Btu/lbm R and sat vap Btu/lbm R. Therefore the outlet of the turbine must be a mixture of liquid and vapor. sat liq ( ) Btu/lbm R x 0.85 ( ) Btu/lbm R sat vap x sat vap + (1 x)sat liq (0.85)(1167.1Btu/lbm) + (1-0.85)(8.0 Btu/lbm) Btu/lbm W t,1 1 ( ) Btu/lbm -1.8 Btu/lbm econd turbine Inlet now is at P 5 psia and T 500 o F. Looking up in steam table for this condition: Btu/lbm and Btu/lbm R Btu/lbm R. team table for P 4 0 psia: 0.58 Btu/lbm R and sat vap 1.70 Btu/lbm R. 4 > sat vap (0 psia) The outlet of the turbine must be vapor. Looking up in steam table at P 0 psia and for 1.70 Btu/lbm R, we find it lies between 50 o F and 400 o F. Using linear interpolation (very similar to the way we find quality), we find that the temperature T 4 is approximately 8 o F. And at this condition, Btu/lbm W t, 4 ( ) Btu/lbm Btu/lbm The total work W t,1 + W t, (-1.8 Btu/lbm) + (-54.8 Btu/lbm) Btu/lbm (ans) e) What is the heat absorbed during the reheating process? We calculated for the case of P* 10 psia, that R Btu/lbm imilarly for the case of P* 5 psia, (see previous page). The heat absorbed ( R, see diagram in part b) will be because no work is done during reheating: R + W s R Btu/lbm Btu/lbm -8.4 Btu/lbm (ans) f) and g) The work produced by the turbine can be calculated by measuring the change in enthalpy for the vertical lines. Naturally, we should get the same answer even if we use the Mollier diagram.