Question

Problem 130 pts): a. Explain the primary purpose of regeneration in a Rankine power plant. Sketch a T-s diagram using either open or closed feedwater heater to explain your response. b. Explain the advantages and disadvantages of open and closed feedwater heaters. c. Assume steady-flow, no heat transfer, no work, one entrance, and one exit and neglect elevation changes. Show that the stagnation enthalpy is constant throughout a nozzle. d. Recall that the pressure ratio that maximizes the net work for the simple Brayton cycle and makes T4 = T2. What happens if the regenerative Brayton cycle operates at a pressure ratio larger than this ratio? The pressure ratio that maximizes the net work is given by: 120-12 T e. Explain why the assumption that the water vapor is an ideal gas is valid when the mixture temperature is below 50°C. I

please answer all parts and show work please

Answer 1

wers T-s diagram Of Rankirse cycle Used iu Rankine power plant as Shown, in which feed water entry point is @ and exit point 4 s is 6. 4 S Primary purpose of R@genfatim in a Rankine cycle - To increase the mean temperature of Heat addition which leads to increased amount of heat suppléed at high temperature, also Regeratim reduces the heat rejection which is more efficient. feed water teater Advantages disadvantages Open pump is required for each feedwater High rate of Heat transfer Inexpensive Simple heater . . closed No need of pump for each feedwater heater because of different pressure Less heat transfer • More expensive More complex,

Nozzle V2 hoz hos bi ecca hz vi Assumptims ㅜ • Steady flow stagnant No heat transfer enthapy No work transfer Neglecting [Eleva tion] potential energies change. . + 7,² - Apply first law per unit mass for one entrance, me emit Inet that What tho tv² the one between inlet and stagnatim state so subscribled by ho=h, tv,? 2 OD Also we b This term ho is called os Stagnatim enthapy. know that → for high speed flow enthalpy and kinetic energy are combined into stagnatim enthalpy ho I now, htu No coin Ein = out hi + v,2 = hz + V2² 2 2 I ha, hoz Therefore stagnatim enthapy remains constant during Steady flow process.

2(k-1) rp = з TI ) Whet 2- is the pressure Ratio that maximizes the Network for Simple Bragtes cycle and makes Ty=72 with given diagram/curve of Whet - rp Spopt. spmax rp if To =Tmax las it is) T1=Tmin к 2(k-1) (Tmax Tmin receptinum) if operates beyond rpopt. Wnet will start to decrease this Assumption is completely wrong first of all it is not in gas state (mixture below 50°c] → also we can not cassume that because it is not is gaseacy form. Second for Any gas → → By definitim of Ideal gas ů (this Assumption) does not fullfill the Basic Criteria → So this Assumption is wrong.