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Unformatted text preview: 1 non ideal turbine State 1: T = 500 C P = ? w State 2: P = 50 kPa saturated vapor FIG. 1: Non ideal turbine problem. Nonideal turbine with an unknown inlet state Consider an adiabatic turbine with an isentropic efficiency of t = 0 . 8 and an exit pressure of 50 kPa (or 0.5 bar). For a given inlet temperature T 1 , calcu- late the inlet pressure P 1 which will yield a saturated vapor exit state. Make a plot of P 1 vs. T 1 . This problem has some relevance. Often the inlet temperature of a turbine is fixed due to material limitations. The exit pressure will also be fixed by the condenser conditions, and the exit should also be a saturated vapor to prevent liquid damage of the turbine blades. The unknown variable is the inlet pressure which can be controlled by throttling the steam from the boiler. This problem is trivial to solve if the turbine is isentropic: s 1 will be equal to s 2 , and s 1 , T 1 will fix P 1 . The problem is a good deal more complicated if the turbine is nonideal, as is the case here. The isentropic efficiency of the turbine is defined by T = w a w s = h 1- h 2 a h 1- h 2 s (1) in which state 2 s represents the state with s 2 s = s 1 and P 2 s = P 2 = 50 kPa. The actual exit enthalpy is fixed (sat. vapor at 50 kPa), so h 2 a = 2646 kJ/kg. If Eq. (1) is rearranged, and the independent variables noted, we get h 1 ( T 1 ,P 1 ) = 1 1- T ( h 2 a- T h 2 s ( P 2 ,s 2 s = s 1 )) (2) This is one equation for two unknowns: P 1 and s 1 . A second equation is simply s 1 = s 1 ( T 1 ,P 1 ) (3) i.e., s 1 is a function of the two properties T 1 and P 1 . Combining the equations together, we get a single nonlinear equation for P 1 ; that is, we have something like F ( P 1 ) = 0 T 1 = 500 C T s P 1 ' P 2 = 50 kPa 2s' 2a' 2a s s initial guess revised estimate P 1 FIG. 2: Iteration strategy....
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- Spring '11