...Let me know when you get to the rankine cycle problem and determining the efficiency. I think its like #61? I don't understand their efficiency equations, or at least I don't see why it is different from what's in the MERM.
Hi namod65
It is #63. They just neglected the pump work.
Let's see with detail what's going on. Define the following states:
1 = condenser discharge = pump suction
2 = pump discharge = boiler inlet
3 = boiler discharge = turbine inlet
4 = turbine discharge = condenser inlet
Then, the thermal efficiency would be
eta = (Wturbine - Wpump)/Qboiler
where
Wpump=
h2 -
h1. In terms of enthalpies, this equation is:
eta = [(h3 - h4) - (h2 - h1)] / (h3 - h2)
Neglecting the pump work means saying that
h2 =
h1. If you insert that in the above equation you get what they used in the solution.
Here's why we can neglect the pump work: Remember that for compressed liquids, the change in enthalpy is usually approximated as:
h2 - h1 = v1 (P2 - P1)
so we can actually calculate it as:
h2 -
h1 = (0.01623 ft3/lbm) x (700 - 2) lbf/in2 x
[144in2/ft2] x
[32.2 lbm ft/lbf/s2] x
[ 1 (Btu/lbm)/ (25037 ft2/s2) ] = 2.1 Btu/lbm
So if
h1 =94.02 Btu/lbm then
h2 = 96.12 Btu/lbm. (they are indeed very close)
If we use the original equation to calculate the efficiency we get:
eta = [(h3 - h4) - (h2 - h1)] / (h3 - h2) = [(1625.9 - 1027.4) - (96.12 - 94.02)]/(1625.9 - 96.12) = 0.389 = 39%
In the basic Rankine cycle, it is
usually a safe bet to neglect the pump work. You would have to be dealing with a HUGE pressure rise across the pump for it to make a difference. That is in fact the point of vapor power cycles: performing the work consuming step (isentropic compression) when the fluid is liquid, which requires much less energy per unit mass than what the work producing step (turbine expansion) produces per unit mass for the same pressure levels.
