- #1

tracker890 Source h

- 90

- 11

- 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：