University of Johannesburg The Isentropic Efficiency Increases Paper

User Generated

Qnar96

Engineering

University of Johannesburg

Description

I need help with a Engineering question. All explanations and answers will be used to help me learn.
I did most of the work I just need someone to review and complete the answer and make it look slightly more presentable.

Attached in the word document is the assignment.

MUST provide evidence (with references) to support any decision you make in solving parts of the problem that require additional published information from Thermodynamics: An Engineering Approach, 8th or 9th Edition; by Cengal and Boles, McGraw-Hill

The project is basically done I just need the following changes to be made:

For section E it says (see below) but the answer given simply shows one answer without the rest of what is asked. Please make this correction.

(e) Calculate the turbine isentropic efficiencies associated with the

two temperatures determined in part (d). Repeat the calculations

for at least three more temperatures between the minimum and

the maximum. You are asked to present calculation results in a

clearly annotated Table or do the same in EXCEL and plot the

isentropic efficiency vrs. entering temperature, in EXCEL or

MathLab. Comment on the results. What are the “takeaways”?

I only need these changes to be made as I will not have the time to so this myself.

Unformatted Attachment Preview

Given saturated : mass Tann : art vapor steam = 200 kPa 2kg Is 20°C ' = water pumping water efficiency isentropic Alternative a 100hPa = 80% = e: 200hPa pump water I • , B I ET eoopnpa 20°C 2kg Is aoopnpa Z ) 13 saturated ca -33 isentropic vapor Alternatives : steam pump water I B. oiler I T • , > Delivery " saturated Z 100hPa 20°C turbine # -1 vapor 200hPa 2kg Is b P - 3 y • - . 3 n n 2 50 bar h T n 7 2 z or 50bar a • ✓ 2bar 1bar - A- - . - - - - - - - - • 4 2 7 A ^ a 20°C 7 4 • - - - - • . 4 a • n 1 C 4--84-01 kT/kg Sy 0.27964 - > s z >v bJ/kg gfh , )=h , + h; up = (h2-hdtO8( O 8 - P, = 1bar specific pump work = Wp = = Wp Actual Ideal Power = Power 84.01 = = (P . Pe) ha xo , = 0.1001 84.11001 batting (2-1)×102×0.001001 0.1001 mi (hi - - = RAH 2 Kgb x x = ha ( 84.14 84.04 - ( 84.11001 - 8401) = 0.250252mW =o.2rW (84.1109-84.01)+0.8/84.01) O ' AT h ,) Mi (ha ht) - = - t = 84.14 - 8 kJ/kg . d Pa B = at PL, 50 bar = 200hPa = hi hg (2bar) 2bar = , Sy = Sg (2bar ) - - 53--54 From steam Table hz T3 has ha hz Pa = h ha - - = = = has thy 2bar Sz , = = 2706 2546 - And kT1kg AT Ieng → - 62 Tz ha - hz GTI fag hi-hat h 3 Tz = = 2790.30 koTfag 164°C Lower bound : bound : h 3=3560 84-11001 t . y kJ/kg/k kpmg 1899 1-3=554 722°C = 7127 ATI kg 84.11001 Upper e) 7.127 = KJ 1kg 554.722 °C 2706.1899 = - , 3560.62 = = 2707 = = = = 166°C 2546 BST 84-11001-25462×100 8h 11001 . - 3560.82 = kg 70.81 % Pz = 50bar T ^ H u Q . I i s MPa 100hPa ⑧ - k•_$6 t - 200hPa , n > S choosing boiler SL, (isentropic ) = from Ss table so = exit 7.1269 temperature kT/kgk , he and = isentropic efficiency : 2706.2kt Hag Sasso from T so superheated table 5hPa ) > Bat (@ we TL choose , at 5hPa Rate 263.942 = h , h, = 54=7-0375 417.5 = he 3481.2 RTING where - 520°C = htuookpa) - , kJtkg K . = Wp (actual) - - - St (200hPa) ( Sfg ( zoo kpa)) TX Wp Cpw ( DT ) - 4.18 (100-20) 83.18516, G ( DP) - = Mp 54=55 7- 0375 to the boiler transfer of heat (0-00104315)×1500040%2 = = 6.3g O 8 fajypg - 7. 0375=1.5302 x 45 TX ( Sfg ( 200hPa)) hz - - 0.984 = hit WP = hfg Q 83 It . 6.39=89.49 ninth hz) GTI tag 213481.2-89.491=504.90 - - ht t X t he 0.984 ( 2201.51 Q - - - = 6783.43 kW heat transfer to the 2670.98 RT Hag y =hu , = ha 45 - 3481.2-2706.22=0.9565 3681.2 - Lf = 95.5% = Micha hd - 2670.98 WT ft - = 1550 fav = 213481.2 - 2766.2 ) boiler . g) 100hPa Vt @ Misers h ha 0.001 VlB = pump , h za 5000 1000 a cut, - tha - 100 865 =cwTz . 1-2=294.465 K Tz=2a46 → h) s gen To E - - = = = Sagen - 3467 - = 407K - can be for heat transfer to occur , Tz 827.72K = = 607k . k¥7 ) .az kJ/bg (f % = - ¥7 ( 971.38 since temperature tout) Tef 9911 3417.99 = - minimum tf ) 1762.74 = Ein 134°C = QU = out (Ein ÷ 9 Tu h , @ 2oz 6.175 = 1230 = - ¥ =L h, - za = - entropy 971.38 kT/Gg 1762.74=-1.944 be positive ( entropy change 2nd law of thermodynamics this violates the generation negative ) , must . We will first start by calculating the cost of each component. In our system, there are three main components: the pump, the boiler, and the turbine. There are also costs associated with the piping system and the running electricity costs of the pump. We will assume the installation cost of a 200W pump to be of 1000$ (source https://www.homeadvisor.com/cost/plumbing/replace-a-wellpump/ ). We have previously calculated that the power needed for the pump is 0.25025 kW, which is approximately 250 W. If the pump runs one hour, it will consume 250 kWh of electricity. The average price a US resident pays is 13.19 cents per kWh (source https://www.electricchoice.com/electricity-prices-by state/#:~:text=The%20average%20electricity%20rate%20is,is%2013.31%20cents%20per%20k Wh). For 250 kWh, the cost to be paid in dollars is 32.97$ per hour or 3297 cents. Now, for the boiler, the average installation cost extends to 5,677$ on average (source https://www.homeadvisor.com/cost/heating-and-cooling/install-aboiler/#:~:text=Installing%20a%20new%20boiler%20costs,AFUE)%20cost%20%246%2C000% 20to%20%2411%2C000). A small turbine would cost 2000$ (https://iea-etsap.org/ETechDS/PDF/E06-hydropower-GS-gct_ADfina_gs.pdf ). The piping costs around 10$ per square foot (https://www.thumbtack.com/p/pipe-installation-cost ). With all of these in mind the total cost is the initial cost of each component with the running costs of the pump. (a) Draw a neat and clearly labelled schematic of the alternative for supplying the 2kg/s of saturated (vapor) steam at 200 kPa. (b) Represent the alternative on T-s, P-v and h-s diagrams with appropriate points to correspond with those on your clearly labelled schematic. Connect the points on the diagrams with arrows to show the directions of the processes. (c) Determine the ideal power (kW) required by the pump and hence determine the actual power input to the pump. (d) Based on the conditions given in the problem, determine lower (minimum)and upper (maximum) bounds on the specific entropy of the fluid entering the turbine. Determine the equivalent temperatures associated with the lower and upper bound entropies. Provide details and justifications to your answers. (e) Calculate the turbine isentropic efficiencies associated with the two temperatures determined in part (d). Repeat the calculations for at least three more temperatures between the minimum and the maximum. You are asked to present calculation results in a clearly annotated Table or do the same in EXCEL and plot the isentropic efficiency vrs. entering temperature, in EXCEL or MathLab. Comment on the results. What are the “takeaways”? (f) With an acceptable (provide justification for your choices) value of boiler exit temperature and turbine isentropic efficiency, calculate the rate of heat transfer per unit time to the boiler (assuming from an external heat source of course) and the work output per unit time of the turbine. (g) Since the temperature of the heat source is not known, what MUST be the minimum value of the heat source temperature for energy to be transferred to the boiler? Justify your answer. (h) By selecting an appropriate control volume(s), estimate the value of the entropy generation rate of this alternative. Does the process violate the second law? Explain your answer. (i) Write at least a one-page assessment (typed, 1.5 spacing, Times Roman) of this alternative as a Project (and Cost) Engineer who must “sell” this idea to the CEO of EKG & Associates. Clearly document all interpolations/extrapolations in Appendices and attach appropriately to your solution.
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Explanation & Answer

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Attached. Please let me know if you have any questions or need revisions.

Temperature Effieciency
164
251.02
313.7
400.17
475.13
554.72

70.81
74.07
77.18
84.05
87.63
100

Temp vs Eff
120

100

80

60

40

20

0
0

100

200...


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