Chapter 2, Problem 49P

Reconsider Prob. 2-48. Assuming a bear pressure increase during the compression, estimate the energy needed to compress the water isothermally.

To determine

The energy needed to compress the water isothermally.

Answer to Problem 49P

The energy needed to compress the water isothermally is $2.47\text{kJ}$.

Explanation of Solution

Given information:

The water contains by a piston cylinder is $10\text{kg}$, the atmospheric temperature is $20°\text{C}$, the pressure inside the cylinder is $100\text{atm}$ and the coefficient of compressibility of water remains constant.

Write the expression for the final volume of water.

  $V_1=V_0\left(1-\alpha \left(P_1-P_0\right)\right)$   ...... (I)

Here, the initial volume of water is $V_0$, final volume of water is $V_1$, the final pressure is $P_1$, initial pressure is $P_0$ and the compressibility of water is $\alpha$.

Write the expression for the volume of water.

  $V_0=\frac{m_w}{{\rho }_w}$   ...... (II)

Here, the mass of water is $m_w$ and the density of water is ${\rho }_w$.

Write the expression for the linear pressure increase.

  $P=a-bV$   ...... (III)

Here, the pressure is $P$, the volume is $V$ and the constants are $a$ & $b$.

Write the expression for the work done on water.

  $W=m{\displaystyle \int PdV}$   ...... (IV)

Here, the work done on water is $W$.

Calculation:

Substitute $10\text{kg}$ for $m_w$ and $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_w$ in Equation (II).

  $\begin{array}{c}{V}_0=\frac{10\text{kg}}{1000\text{kg}/{\text{m}}^{\text{3}}}\\ =\frac{10{\text{m}}^3}{1000}\\ =0.01{\text{m}}^3\end{array}$

Substitute $0.01{\text{m}}^3$ for $V_0$, $4.8\times {10}^{-5}/\text{atm}$ for $\alpha$, $100\text{atm}$ for $P_1$ and $1\text{atm}$ for $P_0$ in Equation (I).

  $\begin{array}{c}{V}_1=0.01{\text{m}}^3\left(1-\left(4.8\times {10}^{-5}/\text{atm}\right)\left(100\text{atm}-1\text{atm}\right)\right)\\ =0.01{\text{m}}^3\times 0.9952\\ =9.952\times {10}^{-3}{\text{m}}^3\end{array}$

Substitute $1\text{atm}$ for $P$ and $0.01{\text{m}}^3$ for $V$ in Equation (III).

  $\begin{array}{l}1\text{atm}=a-b\times \left(0.01{\text{m}}^3\right)\\ a=\left(1\text{atm}\times \frac{101.325\text{kPa}}{1\text{atm}}\right)+\left(b\times \left(0.01{\text{m}}^3\right)\right)\\ a=\left(101.325\text{kPa}\right)+\left(b\times \left(0.01{\text{m}}^3\right)\right)\end{array}$   ...... (V)

Substitute $100\text{atm}$ for $P$, $\left(101.325\text{kPa}\right)+\left(b\times \left(0.01{\text{m}}^3\right)\right)$ for $a$ and $9.952\times {10}^{-3}{\text{m}}^3$ for $V$ in Equation (III).

  $\begin{array}{l}100\text{atm}=\left(101.325\text{kPa}\right)+\left(b\times \left(0.01{\text{m}}^3\right)\right)-\left(b\times \left(9.952\times {10}^{-3}{\text{m}}^3\right)\right)\\ \left(4.8\times {10}^{-5}{\text{m}}^3\right)\times \text{b}=\left(100\text{atm}\times \frac{101.325\text{kN}/{\text{m}}^2}{1\text{atm}}\right)-\left(101.325\text{kPa}\times \frac{1\text{kN}/{\text{m}}^2}{1\text{kPa}}\right)\\ b=208982812.5\text{kN}/{\text{m}}^5\end{array}$

Substitute $208982812.5{\text{kN}/\text{m}}^5$ for $b$ in Equation (V).

  $\begin{array}{c}a=\left(101.325\text{kPa}\right)+\left(\left(208982812.5\text{kN}/{\text{m}}^5\right)\times 0.01{\text{m}}^3\right)\\ =\left(101.325\text{kPa}\times \frac{1\text{kN}/{\text{m}}^2}{1\text{kPa}}\right)+\left(2089828.125\text{kN}/{\text{m}}^2\right)\\ =2089929.45\text{kN}/{\text{m}}^2\end{array}$

Substitute $a-bV$ for $P$ in Equation (IV).

  $\begin{array}{c}W=m{\displaystyle \underset{V_0}{\overset{V_1}{\int }}\left(a-bV\right)dV}\\ =m{\left[aV-b\frac{V^2}{2}\right]}_{V_0}^{V_1}\\ =m\left[a\left(V_1-V_0\right)-\frac{b}{2}\left(V_1^2-V_0^2\right)\right]\end{array}$   ...... (VI)

Substitute $2089929.45\text{kN}/{\text{m}}^2$ for $a$, $208982812.5\text{kN}/{\text{m}}^5$ for $b$, $0.01{\text{m}}^3$ for $V_0$, $9.952\times {10}^{-3}{\text{m}}^3$ for $V_1$ and $10\text{kg}$ for $m$ in Equation (VI).

  $\begin{array}{c}W=\left(10\text{kg}\right)\left[\begin{array}{l}\left(2089929.45\text{kN}/{\text{m}}^2\right)\times \left(0.01{\text{m}}^3-9.952\times {10}^{-3}{\text{m}}^3\right)\\ -\frac{\left(208982812.5\text{kN}/{\text{m}}^5\right)}{2}\left({\left(0.01{\text{m}}^3\right)}^2-{\left(9.952\times {10}^{-3}{\text{m}}^3\right)}^2\right)\end{array}\right]\\ =\left(10\text{kg}\right)\left[\left(100.317\text{kN}\cdot \text{m}\right)-100.07\text{kN}\cdot \text{m}\right]\\ =2.47\text{kJ}\end{array}$

From work energy theorem, the net work is equal to the energy.

Conclusion:

The energy needed to compress the water isothermally is $2.47\text{kJ}$.