Q: Rankine cycle pump work Integration

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    Themodynamics
In summary, the process from state 1 to state 2 is both an isentropic and isovolumic process. However, the input work of the pump, represented by (A) and (B) in the equations, is not the same. This is because (A) represents an isovolumic process while (B) represents an ideal gas adiabatic process.
  • #1
tracker890 Source h
90
11
Homework Statement
To determine pump work
Relevant Equations
Open system, work integral equation
1688721171384.png

Why is the process from state 1 to state 2 an isentropic and isovolumic process, but the input work of the pump is (A) not (B)?
isovolumic process:
$$ w_{pump,in}=\ \int_{1}^{2}{vdp\ =\ v\left(p_2-p_1\right)}\ldots\ldots\ldots\ldots..\left(A\right) $$

isentropic process:
$$ \because pv^k=c\ \ \therefore c\ =\ \left( \frac{c}{p} \right) ^k $$
$$w_{pump,in}=\ \int_{1}^{2}{vdp\ =\int_{1}^{2}{{(\frac{c}{p})}^kdp}=\ \frac{c^k}{1-k}\left[p_2^{1-k}-p_1^{1-k}\right]}=\frac{1}{1-k}\left[{\left({p_2}^k{v_2}^k\right)p}_2^{1-k}-\left({p_1}^k{v_1}^k\right)p_1^{1-k}\right]=\frac{1}{1-k}\left[p_2{v_2}^k-p_1{v_1}^k\right]\ldots\ldots\ldots\ldots..\left(B\right)$$

reference
 
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  • #2
##pV^k=c## is for an adiabatic gaseous process. This stage of the Rankine cycle is with a liquid, so theoretically isovolumetric.
 
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  • #3
haruspex said:
##pV^k=c## is for an adiabatic gaseous process. This stage of the Rankine cycle is with a liquid, so theoretically isovolumetric.
Not only an adiabatic gaseous process; an ideal gas adiabatic process.
 
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