Chapter 8, Problem 83P

To determine

(a)

The rate of flow of oil through the funnel when diameter of pipe is doubled.

The funnel effectiveness when the diameter of the pipe is doubled.

Answer to Problem 83P

The rate of flow through the funnel is $61.284\times {10}^{-6}{\text{m}}^{\text{3}}/\text{s}$.

The funnel effectiveness is $5.68\%$.

Explanation of Solution

Given information:

The temperature of oil is $20°\text{C}$, the height of cylindrical reservoir is $20\text{cm}$, the diameter of pipe is $2\text{cm}$, the height pipe is $40\text{cm}$ and the entrance effects are negligible. The flow is steady and incompressible. The density of oil is $888.1\text{kg}/{\text{m}}^{\text{3}}$ and the viscosity is $0.8374\text{kg}/\text{m}\cdot \text{s}$.

Write the expression for the flow rate in the tunnel.

$\dot{\vee }=\frac{\rho g\left(H+h\right)\pi d^4}{128\mu L}$....... (I)

Here, the pipe length is $L$, height of the reservoir is $H$, height of the pipe is $h$, diameter of pipe is $d$, the gravitational acceleration is $g$, the density of oil is $\rho$ and the dynamic viscosity of oil is $\mu$.

The exit point is taken as reference level.

$\begin{array}{l}{z}_2=0\\ z_1=H+h\end{array}$....... (II)

Write the expression for the energy equation.

$\frac{P_1}{\rho g}+{\alpha }_1\frac{V_1^2}{2g}+z_1+h_{pump}=\frac{P_2}{\rho g}+{\alpha }_2\frac{V_2^2}{2g}+z_2+h_{turbine}+h_L$....... (III)

Here, the inlet pressure is $P_1$, the exit pressure is $P_2$, the inlet velocity is $V_1$, the exit velocity is $V_2$, datum at the inlet is $z_1$, the exit datum is $z_2$, the gravitational acceleration is $g$, the kinetic energy correction factor at inlet is ${\alpha }_1$, the kinetic energy correction factor at exit is ${\alpha }_2$, the head loss in the pump is $h_{pump}$, the head loss in the turbine is $h_{turbine}$, the density of water is $\rho$ and the head loss in the pipe is $h_L$.

The exit is taken as the reference level; there is no turbine or pump, the flow is frictionless, the correction factor of kinetic energy is unity and the head loss is zero.

The fluid is open to atmosphere at inlet and outlet.

$\begin{array}{l}{P}_1=P_{atm}\\ P_2=P_{atm}\end{array}$

Write the expression for the maximum flow rate.

${\dot{\vee }}_{\max }=V_2\left(\frac{\pi }{4}{d}^2\right)$....... (IV)

Write the expression for the funnel effectiveness.

$\epsilon =\frac{\dot{\vee }}{{\dot{\vee }}_{\max }}$....... (V)

Here, the given flow rate is $\dot{\vee }$.

Calculation:

Substitute $888.1\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $40\text{cm}$ for $L$, $2\text{cm}$ for $d$, $40\text{cm}$ for $h$, $20\text{cm}$ for $H$, $0.8374\text{kg}/\text{m}\cdot \text{s}$ for $\mu$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (I).

$\begin{array}{c}\dot{\vee }=\frac{\left(888.1\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(20\text{cm}+40\text{cm}\right)\left(\pi {\left(2\text{cm}\right)}^4\right)}{128\left(0.8374\text{kg}/\text{m}\cdot \text{s}\right)\left(40\text{cm}\right)}\\ =\frac{\left(\begin{array}{l}\left(888.1\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ \times \left(20\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)+40\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\\ \times \left(\pi {\left(1\text{cm}\left(\frac{2\text{m}}{100\text{cm}}\right)\right)}^4\right)\end{array}\right)}{128\left(0.8374\text{kg}/\text{m}\cdot \text{s}\right)\left(40\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}\\ =61.284\times {10}^{-6}{\text{m}}^{\text{3}}/\text{s}\end{array}$

Therefore, the rate through the funnel is $61.284\times {10}^{-6}{\text{m}}^{\text{3}}/\text{s}$.

Substitute $40\text{cm}$ for $h$ and $20cm$ for $H$ in Equation (II).

$\begin{array}{c}{z}_1=20\text{cm}+40\text{cm}\\ \text{=}20\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)+40\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.6\text{m}\end{array}$

Substitute $1$ for ${\alpha }_1$, $1$ for ${\alpha }_2$, $P_{atm}$ for $P_1$, $P_{atm}$ for $P_2$, $0$ for $z_2$, $0$ for $V_1$, $0$ for $h_{pump}$, $0$ for $h_{turbine}$ and $0$ for $h_L$ in Equation (III).

$\begin{array}{l}\frac{P_{atm}}{\rho g}+1\frac{0}{2g}+z_1+0=\frac{P_{atm}}{\rho g}+1\frac{V_2^2}{2g}+0+0+0\\ z_1=\frac{V_2^2}{2g}\\ V_2=\sqrt{2g{z}_1}\end{array}$....... (VI)

Substitute $0.6\text{m}$ for $z_1$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (VI).

$\begin{array}{c}{V}_2=\sqrt{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.6\text{m}\right)}\\ =\sqrt{11.772{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ =3.431\text{m}/\text{s}\end{array}$

Substitute $3.431\text{m}/\text{s}$ for $V_2$ and $2\text{cm}$ for $d$ in Equation (IV).

$\begin{array}{c}{\dot{\vee }}_{\max }=\left(3.431\text{m}/\text{s}\right)\left(\frac{\pi }{4}{\left(2\text{cm}\right)}^2\right)\\ =\left(3.431\text{m}/\text{s}\right)\left(\frac{\pi }{4}{\left(2\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\right)\\ =1.078\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\end{array}$

Substitute $61.284\times {10}^{-6}{\text{m}}^{\text{3}}/\text{s}$ for $\dot{\vee }$ and $1.078\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}$ for ${\dot{\vee }}_{\max }$ in Equation (V).

$\begin{array}{c}\epsilon =\frac{61.284\times {10}^{-6}{\text{m}}^{\text{3}}/\text{s}}{1.078\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}}\\ =0.0568\\ =5.68\%\end{array}$

Therefore, the funnel effectiveness is $5.68\%$.

Conclusion:

Therefore, the rate of flow through the funnel is $61.284\times {10}^{-6}{\text{m}}^{\text{3}}/\text{s}$.

Therefore, the funnel effectiveness is $5.68\%$.

To determine

(b)

The rate of flow of oil via the funnel when the length of pipe is tripled and diameter is maintained the same.

The funnel effectiveness when the length of pipe is tripled and diameter is maintained the same.

Answer to Problem 83P

The rate of flow through the funnel is $2.979\times {10}^{-6}{\text{m}}^{\text{3}}/\text{s}$.

The funnel effectiveness is $0.72\%$.

Explanation of Solution

Given information:

The temperature of oil is $20°\text{C}$, the height of cylindrical reservoir is $20\text{cm}$, the diameter of pipe is $1\text{cm}$, the height pipe is $120\text{cm}$ and the entrance effects are negligible.

Write the expression for the flow rate in the tunnel.

$\dot{\vee }=\frac{\rho g\left(H+h\right)\pi d^4}{128\mu L}$....... (I)

Here, the pipe length is $L$, height of the reservoir is $H$, height of the pipe is $h$, diameter of pipe is $d$, the gravitational acceleration is $g$, the density of oil is $\rho$ and the dynamic viscosity of oil is $\mu$.

The exit point is taken as reference level.

$\begin{array}{l}{z}_2=0\\ z_1=H+h\end{array}$....... (II)

Write the expression for the energy equation.

$\frac{P_1}{\rho g}+{\alpha }_1\frac{V_1^2}{2g}+z_1+h_{pump}=\frac{P_2}{\rho g}+{\alpha }_2\frac{V_2^2}{2g}+z_2+h_{turbine}+h_L$....... (III)

Here, the inlet pressure is $P_1$, the exit pressure is $P_2$, the inlet velocity is $V_1$, the exit velocity is $V_2$, datum at the inlet is $z_1$, the exit datum is $z_2$, the gravitational acceleration is $g$, the kinetic energy correction factor at inlet is ${\alpha }_1$, the kinetic energy correction factor at exit is ${\alpha }_2$, the head loss in the pump is $h_{pump}$, the head loss in the turbine is $h_{turbine}$, the density of water is $\rho$ and the head loss in the pipe is $h_L$.

The exit is taken as the reference level; there is no turbine or pump, the flow is frictionless, the correction factor of kinetic energy is unity and the head loss is zero.

The fluid is open to atmosphere at inlet and outlet.

$\begin{array}{l}{P}_1=P_{atm}\\ P_2=P_{atm}\end{array}$

Write the expression for the maximum flow rate.

${\dot{\vee }}_{\max }=V_2\left(\frac{\pi }{4}{d}^2\right)$....... (IV)

Write the expression for the funnel effectiveness.

$\epsilon =\frac{\dot{\vee }}{{\dot{\vee }}_{\max }}$....... (V)

Here, the given flow rate is $\dot{\vee }$.

Calculation:

Substitute $888.1\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $1\text{cm}$ for $d$, $120\text{cm}$ for $h$, $20\text{cm}$ for $H$, $0.8374\text{kg}/\text{m}\cdot \text{s}$ for $\mu$, $120\text{cm}$ for $L$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (I).

$\begin{array}{c}\dot{\vee }=\frac{\left(888.1\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(20\text{cm}+120\text{cm}\right)\left(\pi {\left(1\text{cm}\right)}^4\right)}{128\left(0.8374\text{kg}/\text{m}\cdot \text{s}\right)\left(120\text{cm}\right)}\\ =\frac{\left(\begin{array}{l}\left(888.1\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ \times \left(20\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)+120\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\\ \times \left(\pi {\left(1\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^4\right)\end{array}\right)}{128\left(0.8374\text{kg}/\text{m}\cdot \text{s}\right)\left(120\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}\\ =2.979\times {10}^{-6}{\text{m}}^{\text{3}}/\text{s}\end{array}$

Therefore, the flow rate through the funnel is $2.979\times {10}^{-6}{\text{m}}^{\text{3}}/\text{s}$.

Substitute $120\text{cm}$ for $h$ and $20cm$ for $H$ in Equation (II).

$\begin{array}{c}{z}_1=20\text{cm}+120\text{cm}\\ =20\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)+120\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =1.4\text{m}\end{array}$

Substitute $1$ for ${\alpha }_1$, $1$ for ${\alpha }_2$, $P_{atm}$ for $P_1$, $P_{atm}$ for $P_2$, $0$ for $z_2$, $0$ for $V_1$, $0$ for $h_{pump}$, $0$ for $h_{turbine}$ and $0$ for $h_L$ in Equation (III).

$\begin{array}{l}\frac{P_{atm}}{\rho g}+1\frac{0}{2g}+z_1+0=\frac{P_{atm}}{\rho g}+1\frac{V_2^2}{2g}+0+0+0\\ z_1=\frac{V_2^2}{2g}\\ V_2=\sqrt{2g{z}_1}\end{array}$....... (VI)

Substitute $1.4\text{m}$ for $z_1$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (VI).

$\begin{array}{c}{V}_2=\sqrt{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(1.4\text{m}\right)}\\ =\sqrt{27.468{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ =5.241\text{m}/\text{s}\end{array}$

Substitute $5.241\text{m}/\text{s}$ for $V_2$ and $1\text{cm}$ for $d$ in Equation (IV).

$\begin{array}{c}{\dot{\vee }}_{\max }=\left(5.241\text{m}/\text{s}\right)\left(\frac{\pi }{4}{\left(1\text{cm}\right)}^2\right)\\ =\left(5.241\text{m}/\text{s}\right)\left(\frac{\pi }{4}{\left(1\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\right)\\ =0.412\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\end{array}$

Substitute $2.979\times {10}^{-6}{\text{m}}^{\text{3}}/\text{s}$ for $\dot{\vee }$ and $0.412\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}$ for ${\dot{\vee }}_{\max }$ in Equation (V).

$\begin{array}{c}\epsilon =\frac{2.979\times {10}^{-6}{\text{m}}^{\text{3}}/\text{s}}{0.412\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}}\\ =0.0072\times 100\\ =0.72\%\end{array}$

Conclusion:

The rate of flow through the funnel is $2.979\times {10}^{-6}{\text{m}}^{\text{3}}/\text{s}$.

The funnel effectiveness is $0.72\%$.