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Isentropic Efficiencies of Turbines, Compressors and Nozzles

Isentropic Efficiencies of Turbines, Compressors and Nozzles. Lecture # 28 Nov. 3, 08. Announcements. Midterm-II: Friday, Nov. 7 th , 11 am, in class 10 problems from Chapters 4,5,6 4 from Ch. 4, 2 from Ch.5, 4 from Ch. 6

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Isentropic Efficiencies of Turbines, Compressors and Nozzles

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  1. Isentropic Efficiencies of Turbines, Compressors and Nozzles Lecture # 28 Nov. 3, 08

  2. Announcements Midterm-II: Friday, Nov. 7th, 11 am, in class 10 problems from Chapters 4,5,6 4 from Ch. 4, 2 from Ch.5, 4 from Ch. 6 Today’s material is NOT included. Entropy Balances for control volumes (HW # 10) is included. Total Points: 20 Review Session in Class on Wednesday Homework # 10 Solutions will be posted on Wed. morning Equation sheet will contain all Key Equations listed at end of chapters 1-6; Cover page with unit conversions will be provided. Where necessary property tables will be given

  3. Isentropic Processes A process for which the entropy is constant, is called an isentropic process. On the h-s diagram and on the T-s diagram, we can draw the processes, as shown in the next figures. The state at the end of the isentropic process can be obtained by matching the initial entropy and another property such as pressure or specific volume or temperature.

  4. Fig06_08 Fig06_08

  5. Fig06_09 Fig06_09

  6. Ideal Gas Model Let us now derive some analytical relations for an isentropic process for an ideal gas. We can write

  7. Constant Specific Heats Let us consider an isentropic process for an ideal gas when the specific heats are constants. Introducing the ideal gas relations,

  8. Constant Specific Heats We can then derive the following relations from the above equations: Substituting the above relation by eliminating the temperature ratio from the above relation, we can write: The process is shown in the following figure

  9. Fig06_10 Fig06_10

  10. Isentropic Efficiencies Isentropic efficiencies involve a comparison between the actual performance of a device and the performance that would be achieved under idealized circumstances for the same inlet state and the same exit pressure. Isentropic Turbine Efficiency For a turbine with no heat transfer and KE and PE effects neglected, we can write an expression for rate of work as:

  11. Fig06_E6 Fig06_E6

  12. Isentropic Efficiency, Turbine The maximum value for rate work is obtained for the smallest value of permitted h2 for the given exit pressure. This value is obtained using the second law constraint. Since there is no heat transfer, The smallest value of s2 corresponds to a reversible process, with s2 = s1. All sates to the left of the isentropic vertical lines are not possible. The corresponding enthalpy for isentropic process is h2s. .

  13. Turbine Isentropic Efficiency The turbine isentropic efficiency is defined as: Typical values of isentropic efficiencies are between 0.7 and 0.9.

  14. Isentropic Efficiency of Compressors For compressors, we want to the minimum amount of work. The actual work is given by: The minimum work corresponds to an isentropic process (since there is no heat transfer). Hence

  15. Fig06_12 Fig06_12

  16. Isentropic Efficiency of Compressors The isentropic efficiency of a compressor is defined as: Typical values for isentropic efficiency of a compressor is between 75 and 85%. Isentropic pump efficiency can be defined similarly.

  17. Nozzles Since nozzles produce kinetic energy from a drop in enthalpy, the nozzle isentropic efficiency is defined as: For well-designed nozzles, the isentropic efficiencies are around 95%, hence nozzles are fairly easy to design free of internal irreversibilities.

  18. Sample Problem 2 A steam turbine operates at steady state with inlet conditions of p1 = 5 bar, T1 = 320oC. Steam leaves the turbine at a pressure of 1 bar. There is no significant heat transfer between the turbine and surroundings. KE and PE effects are negligible. If the isentropic efficiency is 75%, determine the work developed per unit mass flowing through the turbine, in kJ/kg. Given: Turbine inlet and outlet conditions, isentropic efficiency. KE and PE effects are negligible, and no heat transfer.

  19. Fig06_E6 Fig06_E6

  20. Solution The work developed can be calculated from the isentropic efficiency and the isentropic work. Evaluate the following properties: h1 = 3105.6 kJ/kg. s1 = 7.5308 kJ/kg.K The exit pressure is 1 bar, and if for an isentropic process, the exit entropy is 7.5308 kJ/kg.K. The corresponding h2s = 2743.0 kJ/kg. Substituting the values for the inlet and isentropic enthalpy, the actual work is:

  21. Sample Problem 2 Steam enters a nozzle operating at steady state at p1 = 140 lbf/in2 and T1 = 600oF with a velocity of 100 ft/s. The pressure and temperature at the exit are p2 = 40 lbf/in2 and T2 = 350 oF. There is no significant heat transfer between nozzle and its surroundings and changes in PE between inlet and exit can be neglected. Determine the nozzle isentropic efficiency. Solution: The nozzle efficiency is defined as the ratio of actual Kinetic energy developed to the isentropic value

  22. Fig06_E6 Fig06_E6

  23. Solution For the actual operation, The inlet enthalpy for T1 = 600oF and p1 = 140lbf/in2 is 1326.4 Btu/lbm; s1 = 1.7191 Btu/lbm.oR T2 = 350oF, p2 = 40 lbf/in2, h2 = 1211.8 Btu/lb. Hence

  24. Solution Now we need to evaluate h2s for the isentropic process. Match pressure = 40 lbf/in2, and s2 = s1 = 1.7191 Btu/lbm.oR results in h2s = 1202.3 Btu/lbm

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