Chapter 8, Problem 152P

To determine

The coefficient of discharge of the orifice.

Answer to Problem 152P

The coefficient of discharge is $0.237$.

Explanation of Solution

Given information:

The diameter of tank 1 is $30\text{cm}$, diameter of tank 2 is $12\text{cm}$, diameter of orifice is $5\text{mm}$. The fluid height in tank 1 is $45\text{cm}$ and the fluid height in tank 2 is $15\text{cm}$. The time taken to stop the flow is $200\text{s}$. The loss of head due to friction is negligible.

Assume the free surface for tank as $1$ and the exit of orifice as $0$.

Write the expression to calculate the energy equation for the system.

  $\frac{P_1}{\rho g}+{\alpha }_1\frac{V_1^2}{2g}+z_1+h_{pump}=\frac{P_0}{\rho g}+{\alpha }_0\frac{V_0^2}{2g}+z_0+h_{turbine}+h_L$  ......(I)

Here, the pressure in tank is $P_1$, datum head in tank is $z_1$, the pressure in orifice is $P_0$, datum head in orifice is $z_0$, the kinetic energy correction factor for tank 1 is ${\alpha }_1$, the kinetic energy correction factor for orifice is ${\alpha }_0$, the turbine head is $h_{turbine}$, the pump head is $h_{pump}$, the head loss due to the friction is $h_L$, the density is $\rho$ and the gravitational acceleration is $g$.

The surface of the tank is open to atmosphere.

  $P_1=P_{atm}$

Here, the atmospheric pressure is $P_{atm}$.

Write the expression for the pressure in orifice.

  $P_0=P_{atm}+\rho g{h}_2$

Here, the fluid height in tank 2 is $h_2$.

Write the expression for the net fluid height in the system.

  $h=h_1-h_2$  ......(II)

Here, the height in tank 1 is $h_1$.

Write the Equation (II) in derivative form.

  $\begin{array}{l}dh=d{h}_1-d{h}_2\\ d{h}_2=d{h}_1-dh\end{array}$

Here, the fluid height in tank is $h_1$.

Write the expression for the volume flow rate through the orifice.

  $\dot{V}=C_d{V}_0{A}_0$   ...... (III)

Here, the coefficient of discharge is $C_d$, the area of orifice is $A_0$ and the velocity at the exit of the orifice is $V_0$.

Write the expression for the volume flow through orifice in terms of time derivatives.

  $\begin{array}{c}\dot{V}=A_2\frac{d{h}_2}{dt}\\ \dot{V}=-A_1\frac{d{h}_1}{dt}\\ -A_1\frac{d{h}_1}{dt}=A_2\frac{d{h}_2}{dt}\\ d{h}_2=-\frac{A_1}{A_2}d{h}_1\end{array}$  ......(IV)

Here, the area of tank 2 is $A_2$ and the area of tank 1 is $A_1$.

Write the expression for the area of tank 1.

  $A_1=\frac{\pi }{4}{D}_1^2$

Here, the diameter of tank 1 is $D_1$.

Write the expression for the area of tank 2.

  $A_2=\frac{\pi }{4}{D}_2^2$

Here, the diameter of tank 2is $D_2$.

Write the expression for the area of orifice.

  $A_0=\frac{\pi }{4}{D}_0^2$

Here, the diameter of orifice is $D_0$.

Calculation:

Substitute $0$ for $V_1$, $P_{atm}$ for $P_1$, $h_1$ for $z_1$, $0$ for $h_{turbine}$, $0$ for $h_{pump}$, $0$ for $h_L$, $P_{atm}+\rho g{h}_2$ for $P_0$, $0$ for $z_2$, $1$ for ${\alpha }_1$ and $1$ for ${\alpha }_0$ in Equation (I).

  $\begin{array}{l}\frac{P_{atm}}{\rho g}+\left(1\right)\frac{{\left(0\right)}^2}{2g}+h_1+0=\frac{P_{atm}+\rho g{h}_2}{\rho g}\left(1\right)\frac{V_0^2}{2g}+0+0+0\\ h_1=\frac{\rho g{h}_2}{\rho g}+\frac{V_0^2}{2g}\\ h_1=h_2+\frac{V_0^2}{2g}\\ V_0=\sqrt{2g\left(h_1-h_2\right)}\end{array}$   ....... (V)

Substitute $h$ for $\left(h_1-h_2\right)$ in Equation (V).

  $V_0=\sqrt{2gh}$   ....... (VI)

Substitute $d{h}_1-dh$ for $d{h}_2$ in Equation (IV).

  $\begin{array}{l}d{h}_1-dh=-\frac{A_1}{A_2}d{h}_1\\ dh=d{h}_1\left(1+\frac{A_1}{A_2}\right)\\ d{h}_1=\frac{dh}{\left(1+\frac{A_1}{A_2}\right)}\end{array}$

Substitute $-A_1\frac{d{h}_1}{dt}$ for $\dot{V}$ in Equation (III).

  $\begin{array}{l}-A_1\frac{d{h}_1}{dt}=C_d{V}_0{A}_0\\ -dt=A_1\left(\frac{d{h}_1}{C_d{V}_0{A}_0}\right)\end{array}$   ....... (VII)

Substitute $\frac{dh}{\left(1+\frac{A_1}{A_2}\right)}$ for $d{h}_1$ and $\sqrt{2gh}$ for $V_0$ in Equation (VII).

  $\begin{array}{c}-dt=A_1\left(\frac{\left(\frac{dh}{\left(1+\frac{A_1}{A_2}\right)}\right)}{C_d\left(\sqrt{2gh}\right)A_0}\right)\\ -dt=A_1\left(\frac{dh}{C_d\left(1+\frac{A_1}{A_2}\right)\left(\sqrt{2gh}\right)A_0}\right)\\ -dt=\frac{A_1{A}_{2}dh}{C_d{A}_0\left(A_1+A_2\right)\left(\sqrt{2gh}\right)}\end{array}$   ...... (VIII)

Integrate Equation (VIII) with limits $0$ to $t$ in left- hand side and limits $z_1$ to $z$ in right-hand side.

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{t}{\int }}-dt}={\displaystyle \underset{h_1}{\overset{h}{\int }}\frac{A_1{A}_{2}dh}{C_d{A}_0\left(A_1+A_2\right)\left(\sqrt{2gh}\right)}}\\ -{\left(t\right)}_0^t=\frac{A_1{A}_2}{C_d{A}_0\left(A_1+A_2\right)\left(\sqrt{2g}\right)}{\left(\frac{h^{\frac{1}{2}}}{\frac{1}{2}}\right)}_{h_1}^h\\ -\left(t-0\right)=\frac{A_1{A}_2}{C_d{A}_0\left(A_1+A_2\right)\left(\sqrt{2g}\right)}\left(\frac{h^{\frac{1}{2}}}{\frac{1}{2}}-\frac{h_1^{\frac{1}{2}}}{\frac{1}{2}}\right)\\ C_d=-\frac{2A_1{A}_2}{A_0\left(A_1+A_2\right)\left(t\right)\left(\sqrt{2g}\right)}\left(h^{\frac{1}{2}}-h_1^{\frac{1}{2}}\right)\end{array}$   ....... (IX)

Substitute $\frac{\pi }{4}{D}_0^2$ for $A_0$, $\frac{\pi }{4}{D}_1^2$ for $A_1$, and $\frac{\pi }{4}{D}_2^2$ for $A_2$ in Equation (IX).

  $\begin{array}{l}{C}_d=-\frac{2\left(\frac{\pi }{4}{D}_1^2\right)\left(\frac{\pi }{4}{D}_2^2\right)}{\left(\frac{\pi }{4}{D}_0^2\right)\left(\frac{\pi }{4}{D}_1^2+\frac{\pi }{4}{D}_2^2\right)\left(t\right)\left(\sqrt{2g}\right)}\left(h^{\frac{1}{2}}-h_1^{\frac{1}{2}}\right)\\ C_d=\frac{2\left(D_1^2\right)\left(D_2^2\right)}{\left(D_0^2\right)\left(D_1^2+D_2^2\right)\left(t\right)\left(\sqrt{2g}\right)}\left(h_1^{\frac{1}{2}}-h_{}^{\frac{1}{2}}\right)\end{array}$   ....... (X)

Substitute $45\text{cm}$ for $h_1$, $0$ for $h$, $30\text{cm}$ for $D_1$, $5\text{mm}$ for $D_0$, $12\text{cm}$ for $D_2$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $200\text{s}$ for $t$ in Equation (X).

  $C_d=\left(\begin{array}{l}\frac{2{\left(30\text{cm}\right)}^2{\left(12\text{cm}\right)}^2}{\left({\left(5\text{mm}\right)}^2\right)\left({\left(30\text{cm}\right)}^2+{\left(12\text{cm}\right)}^2\right)\left(200\text{s}\right)\left(\sqrt{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\right)}\\ \times \left({\left(45\text{cm}\right)}^{\frac{1}{2}}-{\left(0\right)}_{}^{\frac{1}{2}}\right)\end{array}\right)$

  $=\left(\begin{array}{c}\frac{2{\left(30\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2{\left(12\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}{\left({\left(5\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\right)\left(\begin{array}{l}{\left(30\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\\ +{\left(12\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\end{array}\right)\left(200\text{s}\right)\left(\sqrt{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\right)}\\ \times \left({\left(45\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^{\frac{1}{2}}-{\left(0\right)}_{}^{\frac{1}{2}}\right)\end{array}\right)$

  $=0.237$

Conclusion:

The coefficient of discharge is $0.237$.