ChemE 260 Isentropic Efficiency Dr. William Baratuci Senior Lecturer Chemical Engineering Department University of Washington TCD 8: C CB 6: 11 & 12 May 13, 2005

Isentropic Efficiency: S Compare work input or output of a real device to to that of an isentropic device. Adiabatic Turbines: Adiabatic Nozzles: Adiabatic Compressors: We will apply the concept of isentropic efficiency to turbines, compressors, pumps and nozzles. Pumps are not listed here because the equation is the same as the one for the compressor. An isentropic turbine produces the most work possible beginning from the real inlet state and ending at the same outlet pressure as the real turbine. An isentropic compressor requires the least work possible beginning from the real inlet state and ending at the same outlet pressure as the real compressor. We define the isentropic efficiency of a turbine as the ratio of the actual work output to the work output of an isentropic turbine with the same inlet conditions and same outlet pressure. IF the REAL turbine is also adiabatic, then The 1st Law is: WS,act = mdot ( Hin – Hout,act ) The isentropic efficiency is the ratio of the change in the specific enthalpy for the actual or real turbine to the change in the specific enthalpy of the isentropic turbine. We define the isentropic efficiency of a compressor as the ratio of the isentropic work input to the actual work input of a compressor with the same inlet conditions and same outlet pressure. IF the REAL compressor is also adiabatic, then The isentropic efficiency is the ratio of the change in the specific enthalpy for the isentropic compressor to the change in the specific enthalpy of the actual or real compressor. Isentropic efficiency is a bit odd for nozzles because there is no shaft work involved ! The goal of a nozzle is to increase the kinetic energy of the flowing fluid. So, we define the isentropic efficiency of a nozzle as the ratio of the specific kinetic energy of the effluent for an actual or real nozzle to the specific kinetic energy of the effluent from an isentropic nozzle. Real nozzles are not isentropic primarily because of friction within the fluid and between the fluid and the wall of the nozzle. IF the the outlet velocity is MUCH MUCH greater than the inlet velocity, then we get a simpler form of the equation for the isentropic efficiency of a nozzle in terms of the enthalpy of the fluid. Baratuci ChemE 260 May 13, 2005

TS & HS Diagrams for Turbines Only states to the right of path 1-2 are “accessible” states. TS & HS Diagrams look deceptively similar for turbines, compressors pumps and nozzles. If the turbine is isentropic, then S^ cannot decrease because that would result in Sgen < 0. Therefore, the effluent of an adiabatic turbine can only occupy states to the right of S1^. The isentropic path is the best case. The isothermal path is the WORST case. The enthalpy actually goes UP ! How does that happen ? Irreversibilities within the turbine are so bad that you have to put shaft work INTO the system ! It isn’t really a turbine anymore. But isn’t really a compressor either. It is not a device you would want to buy. It converts shaft work and PV work into relatively low value internal energy. Baratuci ChemE 260 May 13, 2005

PV Diagram for a Compressor An isentropic compressor requires less shaft work input than an actual, adiabatic compressor. But, if we could build an isothermal compressor, that was also adiabatic, it would require even LESS work input than an isentropic compressor. I hope that seems suspicious to you. Isothermal process requires less shaft work for an adiabatic compressor. Baratuci ChemE 260 May 13, 2005

HS Diagram for a Compressor The problem is that an a compressor that is BOTH isothermal AND ADIABATIC is not possible. Its process path moves into states that are not “accessible”. An isothermal, adiabatic compressor leads to negative values for Sgen and that is a violation of the 2nd Law. Unlike turbines, we DON’T WANT our compressor to be adiabatic. It seems weird that we actually WANT our compressor to LOSE heat so that its performance and efficiency improve. But we do ! IN the BEST case, we can RECOVER this heat and use it somewhere else in our car/airplane/factory/system. We can let our compressors lose heat to the air that surrounds them. We can give them a “cooling jacket” with cool water circulating through it. Or we can use multiple compressors and cool the working fluid between the compression steps. Isothermal process for an adiabatic compressor violates the 2nd Law ! Baratuci ChemE 260 May 13, 2005

Multi-Stage Compression: PV Diagram It is easy to visualize the savings in shaft work when a 2-stage compressor train is used instead of 1 big compressor. The intercooler keeps the specific volume of the gas lower. 3-stage compression trains are common when you must get a gas up to a very high pressure. Baratuci ChemE 260 May 13, 2005 Thermo-CD © B-Cubed 2005

Multi-Stage Compression: HS Diagram Remember that the increase in enthalpy is how much shaft work you must put into the compression process. This HS Diagram makes it very clear why multi-stage compression can dramatically reduce the shaft work requirement for a compression process. Isothermal Efficiency Only used for multi-stage compressors. Compare a multi-stage compressor to one isothermal compressor. Baratuci ChemE 260 May 13, 2005 Thermo-CD © B-Cubed 2005

2-Stage Compressor Power Requirement #1 P1 #2 Intercooler (HEX) P2 PX , T1 Assume: Each compressor is isentropic (polytropic with  = ) The working fluid is an ideal gas in all states in the process. The heat capacities of the working fluid are constant. Intercooler returns the fluid to T1. Compressor #1 With a few assumptions we can determine the shaft work requirement and the total shaft work requirement for the system. The equation can be simplified if we use the intercooler to return the fluid temperature to the temperature of the feed, T1. The big question is how to decide how much shaft work to put into the 1st compressor and how much to put into the 2nd one ? More to the point, what is PX ? Do we make PX = average of P1 and P2 ? Why ? How do we decide ? . Baratuci ChemE 260 May 13, 2005 Compressor #2

2-Stage Compressor Design Calculus to the rescue ! Solve for PX ! (check the 2nd derivative to be sure you found a minimum and not a maximum !) Solution: or: We determine the value of PX that minimizes the specific shaft work requirement for the entire process by Taking the derivative of the specific shaft work requirement for the entire process with respect to PX… Setting the derivative equal to zero… And solving the resulting equation for PX. We need to make sure that the extreme point that we found is a minimum and not a maximum by making sure that the sign of the 2nd derivative is positive. The results are nice and simple. That seems surprising. We do NOT set PX at the algebraic average of P1 and P2 . We set PX at the GEOMETRIC average of P1 and P2 . The square-root of P1 times P2 is called the geometric average. Real compressors are not isentropic and frequently the gases are not ideal, but the geometric average pressure is a good starting point for choosing the intermediate pressure in a 2-stage compressor system. Are there 3-stage compressor trains ? Yes, they are fairly common. Are there 4-stage compressor trains ? Very few. The costs of al the extra equipment and its maintenance usually outweigh the savings due to improved efficiency. Baratuci ChemE 260 May 13, 2005

Next Class … Reversible Work Lost Work Work requirement for a reversible process operating between the actual inlet and outlet states. Lost Work Extra work that we put in that gets wasted by irreversibilities Work that we don’t get out because it gets wasted by irrversibilities. Where does this lost work go ? Actual Work = Reversible Work – Lost Work 2nd Law Efficiency Compare Actual Work to Reversible Work (instead of comparing to isentropic work) Baratuci ChemE 260 May 13, 2005

Example Problem Air is compressed by an internally reversible compressor from 100 kPa and 300 K to a pressure of 900 kPa. Calculate the specific shaft work requirement for the compressor for each of the following cases: Isentropic compression with  = 1.4 Polytropic compression with  = 1.3 Isothermal compression Internally reversible 2-stage compression with intercooling,  = 1.3. Calculate the isothermal efficiency of this 2-stage compressor.