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Investigation Into Maximising the Efficiency of Rankine and Brayton Cycles - Lab Report Example

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This report underlines the efficiency of the two respective cycles that can be improved to increase the productivity of power plants. In order to increase the efficiency of the Rankine cycle, multiple feed heaters, reheaters, superheaters, and turbines each work to prevent the erosion of the steam turbines. …
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Investigation Into Maximising the Efficiency of Rankine and Brayton Cycles
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PART Rankine cycle Modification with superheats Superheats maximize the efficiency of rankine cycle, by preventing the occurrence of water droplets since it triggers the generation of drier steam as shown in process 3’-4’ involving expansion. The limitation of the modification process may arise if the expander, obtained from the manufacturer, is a high speed turbine (Turns 60). This usually triggers erosion of the blades. The following hardware and cycle diagrams describe the principles under which modification with superheat works: Equation and calculation of efficiency Case scenario : A refrigerant enters the turbine as a saturated vapor at 1.6 MPa and leaves at 0.7 MPa. The mass flow rate of the refrigerant is 6 kg/s Geting the efficiency, requires the computation of the enthalpy values: At state 1: Assumption is that the properties are similar to the saturated liquid in the condenser: Hnece h1 = hf(0.7 MPa) = 88.82 kJ/kg and v1 = 0.0008331 m3/kg with pressure of 1600 kPa: this gives We then find h2 = h1 + |wP1| = 86.78 kJ/kg + 0.75 kJ/kg = 89.57 kJ/kg. h3 = hg(1.6 MPa) = 277.863 kJ/kg. s3 = sg(1.6 MPa) = 0.90784 kJ/kg∙K. For state 4, P4 P4 = 700 kPa and x4 = 0.979. We thus find the enthalpy from the quality as h4 = hf(P4 = 700 kPa) + x4 hfg(P4 = 700 kPa) = 88.82 kJ/kg + (0.979)(176.212 kJ/kg) or h4 = 261.41 kJ/kg. From this, the following graph arises: The heat intput to the steam generator, qh = h3 – h2 = 277.86 kJ/kg – 89.54 kJ/kg = 188.3 kJ/kg The condenser heat rejection, ql = |h1 – h4| = |88.62 kJ/kg – 261.41 kJ/kg| = 172.6 kJ/kg. The net work, w = qh - |qL| =188.3 kJ/kg| - 172.6 kJ/kg = 15.70 kJ/kg. The efficiency = w / qH = (15.70 kJ/k ) / (188.3 kJ/kg) or h = 8.3%. when using superheat: = xout = 0.979 Hence the efficiency when superheating is 97.9% which is far much higher than 8.3% when without. This shows the effect of superheating on increasing the dryness of the steam to about 97%. Without superheating, 8.3% efficiency shows that more water droplets are released leading to increased erosion of the turbine. Reheat Reheat involves allowing elevation in temperature, during the addition of heat to the saturated liquid, while still maintaining the dryness of the steam moving from the turbine )Tester et al 39). In which case, the vapor reheat process cut down on the initial superheat applied while raising the temperatura of heat addition. The following images shows the hardware and the cycle diagrams : The turbine and pumps are reversible and adiabatic meaning that there is no heat transfer or entropy change. Thus the first law for each device involves only one inlet and one outlet is q = w + hout – hin. Equation for efficiency: Scenario case: high-pressure turbine at 10 MPa and 550oC and leaves at 0.8 MPa while net output power is 80 mw. Rest of the steam is reheated to 500oC. condenser pressure is 10kPa Assumpition: State 1, 2 are unchanged, h1 = hf(10 kPa) = 191.81 kJ/kg and v1 = 0.001010 m3/kg. The pumps are isentropic and therefofre calculate the work of the first pump as follows: |wp1| = v1(P2 – P1). Thus, h2 = h1 + |wP1| = 191.81 kJ/kg + 0.80 kJ/kg = 192.61 kJ/kg. h3 = hf(800 kPa) = 720.87 kJ/kg and v3 = 0.001115 m3/kg. isentropic pump work for the second pump. h4 = h3 + |wP2| = 720.87 kJ/kg + 10.26 kJ/kg = 731.13 kJ/kg. h5 = h(10 MPa, 550oC) = 3502.0 kJ/kg. s5 = s(1 MPa, 550oC) =6.7585 kJ/kg∙K. h6 = h(P = PFWH = 800 kPa, s6 = s5). This state is in the gas region so we have to find h6 by interpolation between the first two rows at 800 kPa. This gives h6 = 2812,8 kJ/kg. h7 = h(0.8 MPa, 500oC) = 3481.3 kJ/kg. s7 = s(0.8 MPa, 500oC) =7.8692 kJ/kg∙K. h8 = h(P = Pcond = 10 kPa, s8 = s7). h8 = hf(P8 = 10 kPa) + x8 hfg(P8 = 10 kPa) = 191.81 kJ/kg + (0.9627)(2392.1 kJ/kg) or h8 = 2494.7 kJ/kg. The T-S diagram obtained: In this cycle there are three distinct mass flow rates at different points in the cycle. These are shown in the equations below. (Here, represents the mass flow into the feedwater heater.) Substituting the enthalpy values found above to compute this ratio. Substituting the values found for the enthalpies in the cycle and the mass flow rate ratio gives the net work per unit mass flowing through the steam generator as follows: From this specific work, we can find the mass flow rate required for a power output of 80 MW. From the equations for the efficiency and the net work, we see that we can use the computed value of work to simplify the efficiency calculation. h = 44.4% As compared to the simple Rankine cycle, the modification by reheat increases the efficiency from 8.3% to 44.4%. Multiple feed heaters The multiple feedheaters preheat the working liquid by use of hot vapor from the expander. This helps in increasing the overall cycle efficiency becaus of the heat provided before the liquid reaches the boiler. The open feed water involves mixing of the extracted hot vapor with a compressed liquid in mixing chamber, usually insulate. The mixing leads to condensation of the vapor, leading to transfer of the heat of vaporization to the fluid (Wu 40). Limitation arises if the chamber, obtained from the manufacture is small thereby limiting the condensation process required before the liquid leaves for the system piping. In closed feedwater heater involves condensation of the vapor acquired from the turbine as it flow on the Shell side. The condensation process releases heat of vaporization which is then taken by the compressed fluid. The following figure shows the hardware and cycle diagram for the multiple feeders: Recomendaciones  at atmospheric pressure and temperature, the fluid used should always be in liquid state to ensure ease of handling.  the lowest ambient temperature used in operation should be higher than the freezing point to ensure production of more dryer steam. Part 2: Brayton cycle Brayton cycle with intercooling Intercooling in conjunction with reheating and regeneration helps in enhancing the back work ratio of within the gas turbine cycle especially in cases where there is use of multistage compression. In order to achieve best performance with intercooling, the pressure is always kept equal as stated in the equation below: Figure showing the hardware diagram Figure showing cycle diagram a) T-s diagram. Equations and assumptions Referring to the T-s diagram, the regenerator effectiveness is given as: Considering the cold air-standard assumptions, equation (5) reduces to: (7) Assumptions Changes in the kinetic and potential energy are insignificant Presence of steady operating conditions. Utilizing the cold air-standard assumptions, we use the following equation get the termal efficiency Case scenario: An ideal gas-turbine cycle with two stage compression and expansion. The air enters each stage of the compressor at 300K and each stage of the turbine at 1200K. In the absence of any regeneration, the back work ratio and the thermal efficiency are determined as follows: Given: T1 = 300K From Table A-17: h1 = 300.19 kJ/kg Pr1 = 1.386 Pr2 = (P2/P1)*Pr1 = (√3)(1.386) = 2.401 T2 = 351K H2 = 351.39 kJ/kg Given: T6 = 1200K From Table A-17: h6 = 1277.79 kJ/kg Pr6 = 238 Pr7 = (P7/P6)*Pr6 = (1/√3)(238) = 137.4 T7 = 1048K H7 = 1100.75 kJ/kg Then wcomp, in = 2*(wcomp, in, 1 ) = 2*(h2 - h1) = 2*(351.39 – 300.19) = 102.4 kJ/kg wturb, out = 2*(wturb, out, 1) = 2*(h6 – h7) = 2*(1277.79 – 1100.75) = 354.08 kJ/kg wnet = wturb, out – wcomp, in = 354.08 – 102.4 = 251.68 kJ/kg qin = qprimary + qreheat = (h6 – h2) + (h6 – h7) = (1277.79 – 351.39) + (1277.79 – 1100.75) = 1103.44 kJ/kg Thus, rbw = wcomp, in/ wturb, out = 102.4/354.08 = 0.289 or 28.9% ηth = wnet, in/qin = 251.68/1103.44 = 0.228 or 22.8% The calculations shows that the back work ratio is 28.9% while the thermal efficiency is 22.8% Brayton cycle with regeneration Recuperation is used in the Brayton cycle to transfer heat from the turbine to the the compressor since the temperature at the former is higher than that found in the latter. In which case, there is use of counterblow heat exchanger to transfer the heat that is then used to heat the steam before it reaches the combustion chamber. The limitation is that whenever the cycle is used to operate a perfectly reversible refrigerators, there is less heat extracted from the hot reservoir as compared to that released by refrigerator exhausts. This is usually forbidden by the 2nd law which assert that a refrigerator should not have a larger coefficient of performance. The following is a hardware diagram and cycle illustratign the working principle of recuperation: The equations used in calculating thermal efficiency: and For an ideal regenerator in which we have: T5=T4 We get: Therefore, (I) Using isentropic relations, we have: and therefore If we replace in (I) And finally; Using the above scenario: Pr7 = (P7/P6)*Pr6*.75 = (1/√3)(238)(.75) = 103.057 From Table A-17: T5 = 975K H5 = 1017.32 kJ/kg qin = qprimary + qreheat = (h6 – h5) + (h6 – h7) = (1277.79 – 1017.32) + (1277.79 – 100.75) = 437.51 kJ/kg and ηth = wnet/ qin = 251.68/437.51 = 0.575 or 57.5% Analysis As can be seen from the calculation, the absence of regeneration makes intercooling less efficient, even though the back ratio is significantly reduced. The presence of regenerator helps in the reduction of the exhaust gases obtained from the turbine consequently increasing the efficiency of the multistage compressor. Intuitively, intercooling should not be used in gas-turbine power plants unless there is intervention of regeneration. Part 3: Combination of the two cycles The combination of the two cycles will significantly help in improving the thermal efficiency of a power plant as shown in the calculations below. The two parts above have proved that Brayton and Rankine cycles can be used to increase thermal efficiency of turbines; consequently, the combination is most likely to be productive (Ghosh et al 120). The mechanism behind the combination is to utilize the high temperature released from the hot exhaust gases, in the gas turbine, in the Rankine steam cycle. The hight temperatures are used to heat the steam found in the boiler of the steam turbine. As shown in the figure, the heat not used in the topping cycle (Brayton cycle) is utilized as heat input in the bottoming cycle (Rankine cycle): Figure depicting the combination The T-s diagram From the discussion, the thermal efficiency of the combination is the summation of the individual efficiencies less the product of the efficiencies, as shown in the following formula: ηCC= (WB+ WT)/Qin= ηT+ ηB –ηT*ηB For instance, given Brayton cycle efficiency = 0.40 and Rankine cycle efficiency = 0.30 ηCC = 0.4 + 0.3 – (0.4 * 0.3) = 0.58 As shown in the example above, the combination of the two proves to be more efficient than the individuals. This justifies wide use of the combination aspect in power plants. The efficiency of the aspect arises since a drawback realized in one cycle is used as a benefit in the other cycle. Summary and conclusión It is true that the efficiency of the two respective cycles can be improved to increase the productivity of power plants. In order to increase the efficiency of the Rankine cycle, multiple feedheaters, reheaters, superheaters, and turbines each work to prevent the erosion of the steam turbines through triggering the reléase of drier steam. For the Brayton cycle, presence of regenerator helps in the reduction of the exhaust gases obtained from the turbine consequently increasing the efficiency of the multistage compressor. Intercooling in conjunction with reheating and regeneration helps in enhancing the back work ratio of within the gas turbine cycle especially in cases where there is use of multistage compression. Further, the combination of the two cycle proves to be even more efficient. Efficiency of the aspect arises since a drawback realized in one cycle is used as a benefit in the other cycle. Work cited Ghosh, Tushar K, and Mark A. Prelas. Energy Resources and Systems. Dordrecht: Springer, 2009. Print. Turns, Stephen R. Thermodynamics: Concepts and Applications. Cambridge [u.a.: Cambridge Univ. Press, 2006. Print. Wu, Chih. Thermodynamics and Heat Powered Cycles: A Cognitive Engineering Approach. New York: Nova Science Publishers, 2006. Print. Tester, Jefferson W, Elisabeth M. Drake, and Michael J. Driscoll. Sustainable Energy: Choosing Among Options. Cambridge, Mass: MIT Press, 2005. Print. Read More
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