## Info

The numbers in ordinary type are specifications, which also neglect pressure losses in the boiler and condenser. Values in bold type are obtained from the steam tables using the input specifications. The underlined numbers are determined by application of the 1st Law to the values of the previous two categories. Specifically:

Turbine The heat loss from a large turbine is negligible compared to the shaft work produced. Therefore, this component can be assumed to operate adiabatically, or qi_2 = 0. With these eliminations, Eq (4.9) reduces to: w1-2 = h1 - h2, the shaft work produced by the turbine.

Condenser The condenser is responsible for supplying the reject heat of the cycle to the low-temperature reservoir. The reject heat is obtained by condensing the exhaust steam from the turbine and subcooling the liquid. (This is necessary because the pump cannot handle a two-phase mixture). The First law for the condenser is q2-3 = qL = h2 - h3

Pump Cycle specifications include the work required to operate the pump. The First law utilizes this number to determine the enthalpy at the pump outlet h4 = h3 + w3-4

Boiler This component receives the heat input to the cycle from the hot reservoir. Inside this unit, the inlet subcooled liquid is heated to saturation, completely vaporized, and the steam is superheated. The First law for the boiler is q4-1 = qH = h1 - h4

Cycle Efficiency The net (shaft) work produced by the cycle is:

wS = w2-3 - w4-1 = 658 kJ/kg and the efficiency of the Rankine cycle, as defined by Eq (4.4), is:

qH 2851.9

Use of the 2nd law to calculate efficiencies of work-producing and work-consuming devices

Even in cases where an adiabatic flow component such as a pump or a turbine is not reversible, Eq (4.12) provides a means of calculating its efficiency, or the ratio of the actual work produced or consumed to that for perfectly reversible operation.

Turbine The first step is to determine the exit steam quality if the turbine were reversible. If this were so, the entropy of the inlet superheated steam (location 1) should be equal to the entropy of the exhaust steam, or (s2)rev = s1 = 6.77 J/kg-K. Together with the prescribed 15 kPa outlet pressure and the associated saturated liquid/vapor entropies, the outlet quality would be:

A reversible turbine with the same inlet steam conditions and the same exhaust pressure condenses 7% more steam than does the actual turbine. In the reversible situation, the exit enthalpy is evaluated from xrev and the enthalpies of the saturated phases at location 2:

(h2)rev = hf + xrevhfg = 225.9 + 0.83x2373.1 = 2192 kJ/kg and the reversible work is:

(w1-2)rev = h1 - (h2)rev = 3025 - 2192 = 833 kJ/mole

Fig. 4.10 A steam turbine

Turbine rotor

Fig. 4.10 A steam turbine

Using the actual work calculated previously in conjunction with the above reversible work gives the efficiency of the turbine:

wi o 661 turbine efficiency =-1-2— =-= 0.79 (79%)

Additional problems involving steam turbines in tandem with other components of a power generation system are Nos. 4.6 and 4.13.

## Getting Started With Solar

Do we really want the one thing that gives us its resources unconditionally to suffer even more than it is suffering now? Nature, is a part of our being from the earliest human days. We respect Nature and it gives us its bounty, but in the recent past greedy money hungry corporations have made us all so destructive, so wasteful.

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