Enthelpy and Isentropic compression/expansion

In summary, the specific work done by an isentropic compressor/pump in an ideal Rankine cycle is given by two different equations: Wpump = h2 - h1 and Wpump = v(P2 - P1), where v = v1. However, when these calculations are carried out between any two states, there can be a significant discrepancy in the results. This is because the process is isentropic and the water is actually a compressed liquid at state 2, rather than a saturated liquid. This difference in the state of the water can lead to vastly different answers when using the two equations for specific work.
  • #1
RoRoRo
2
0
In a certain thermodynamics textbook, specific work done by an isentropic compressor/pump in an ideal rankine cycles, is given by the following;

Wpump = h2 - h1
Wpump = v(P2 - P1), where v = v1

When I carry out these two calculations between any two states, I get vastly different answers.

For example, compressor in a rankine cycle operating between the following conditions for saturated water;

P1 = 20 kPa
v1 = 0.001017 m3/kg
h1 = 251.42 kJ/kg

P2 = 500 kPa
h2 = 640.09 kJ/kg

Wpump = h2 - h1 = 640090 - 251420 = 388 670 J/kg

Wpump in = v(P2 - P1) = 0.001017 ( 500000 - 20000 ) = 488.16 J/kg

I assume that I'm missing something. Is there any explanation for why I'm seeing such a huge discrepancy in results for two supposedly equivalent expressions?
 
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  • #2
Is the liquid water saturated in both states?
 
  • #3
I was assuming that the water was a saturated liquid at both states but because the process is isentropic, its actually a compressed liquid at state 2. I think...
 
  • #4
RoRoRo said:
I was assuming that the water was a saturated liquid at both states but because the process is isentropic, its actually a compressed liquid at state 2. I think...
Correct.
 

1. What is Enthalpy?

Enthalpy is a thermodynamic property that describes the total energy of a system, including its internal energy and the work required to change its volume or pressure.

2. How is Enthalpy related to Isentropic compression/expansion?

Enthalpy is a key factor in determining the efficiency of isentropic compression or expansion. In an isentropic process, the enthalpy remains constant, meaning that no energy is lost to heat transfer. This allows for a more efficient conversion of energy.

3. What is Isentropic compression?

Isentropic compression is a thermodynamic process in which a gas is compressed without any heat transfer, resulting in an increase in pressure and temperature. This process is reversible and does not involve any energy losses due to friction or heat transfer.

4. What is Isentropic expansion?

Isentropic expansion is the opposite of isentropic compression. It is a thermodynamic process in which a gas expands without any heat transfer, resulting in a decrease in pressure and temperature. This process is reversible and does not involve any energy losses due to friction or heat transfer.

5. How is Enthalpy calculated in Isentropic processes?

In an isentropic process, the change in enthalpy is equal to the work done on or by the system. This can be calculated using the equation: ΔH = W = PΔV, where ΔH is the change in enthalpy, W is the work done, P is the pressure, and ΔV is the change in volume.

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