Q： Rankine cycle pump work Integration

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Homework Statement

To determine pump work

Relevant Equations

Open system, work integral equation

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：

 

[haruspex]

$pV^k=c$ is for an adiabatic gaseous process. This stage of the Rankine cycle is with a liquid, so theoretically isovolumetric.

 

[Chestermiller]

$pV^k=c$ 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.