Chapter 5, Problem 136P

Liquid ethanol (p = 783 kg/m³) at a pressure of 230 kPa enters a 10-cm-diameter pipe at a rate of
2.8 kg/s. Ethanol leaves the pipe at 100 kPa at 15 m above the inlet level. Take the correction factor to be 1. If the diameter of the pipe at the exit is 12 cm, the irreversible head loss in this pipe is
(a) 0.95 m
(b) 1.93 m
(c) 1.23 m
(d4.11 m
(e)2.86m

To determine

The head loss in the pipe is

Answer to Problem 136P

The head loss in the pipe is $\text{1}\text{.929}\text{m}$.

Explanation of Solution

Given information:

The pressure at the entry is $230\text{kPa}$, pressure at the exit is $100\text{kPa}$, diameter of the pipe at the entry is $\text{10}\text{cm}$, diameter of the pipe at the exit is $\text{12}\text{cm}$, flow rate at the entry $\text{2}\text{.8}\text{kg}/\text{s}$, density of the liquid ethanol is $\text{783}\text{kg}/{\text{m}}^{\text{3}}$, datum height at exit of pipe is $\text{15}\text{m,}$ and correction factor is 1.

Write the expression for the Bernoulli Equation

  $\frac{p_1}{\rho g}+{\alpha }_1\frac{v_1^2}{2g}+z_1=\frac{p_2}{\rho g}+{\alpha }_2\frac{v_2^2}{2g}+z_2+h_l$  .......(I)

Here, pressure at entry of the pipe is $p_1$, pressure at point at exit of the pipe is $p_2$, velocity at entry of the pipe is $v_1$, velocity at exit of the pipe is $v_2$, datum height at entry of the pipe is $z_1$, datum height at exit of pipe is $z_2$, acceleration due to gravity is $g$, Density of water is $\rho$, head loss between the entry and exit of the pipe is $h_l$, the correction factor is $\alpha$

Write the expression for the mass flow rate.

  $\dot{m}=\rho A_1{v}_1$  .......(II)

Here, at the entry $A_1$, density of the liquid is $\rho$, and flow rate at the entry is $v_1$.

Write the expression for the area at the entry of pipe.

  $A_1=\frac{\pi D_1^2}{4}$  .......(III)

Here, diameter at the entry of the pipe is $D_1$.

Write the expression for the area at the entry of pipe.

  $A_2=\frac{\pi D_2^2}{4}$  .......(IV)

Here, diameter at the exit of the pipe is $D_2$.

Write the expression of the continuity equation.

  $v_2{A}_2=v_1{A}_1$  .......(V)

Calculation:

Substitute $\text{10}\text{cm}$ for $D_1$ in equation (III).

  $\begin{array}{c}{A}_{\text{1}}\text{=}\frac{\text{π}}{\text{4}}\left({\text{10}}^{\text{2}}{\text{cm}}^{\text{2}}\right)\\ \text{=}\frac{\text{π}}{\text{4}}\left({\text{10}}^{\text{2}}{\text{cm}}^{\text{2}}\text{×}\left(\frac{\text{1}{\text{m}}^{\text{2}}}{{\text{100}}^{\text{2}}{\text{cm}}^{\text{2}}}\right)\right)\\ \text{=0}\text{.00785}{\text{m}}^{\text{2}}\end{array}$

Substitute $\text{2}\text{.8}\text{kg}/\text{s}$ for $\dot{m}$, $\text{0}\text{.00785}{\text{m}}^{\text{2}}$ for $A_1$, and $\text{783}\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ in equation (II).

  $\begin{array}{l}2.8\text{kg}/\text{s}=783\text{kg}/{\text{m}}^{\text{3}}\times 0.00785{\text{m}}^2\times v_1\\ v_1=\frac{2.8\text{kg}/\text{s}}{783\text{kg}/{\text{m}}^{\text{3}}\times 0.00785{\text{m}}^2}\\ v_1=0.4554\text{m}/\text{s}\end{array}$

Substitute $\text{12}\text{cm}$ for $D_2$ in equation (IV).

  $\begin{array}{c}{A}_{\text{1}}\text{=}\frac{\text{π}}{\text{4}}\left({\text{12}}^{\text{2}}{\text{cm}}^{\text{2}}\right)\\ \text{=}\frac{\text{π}}{\text{4}}\left({\text{12}}^{\text{2}}{\text{cm}}^{\text{2}}\text{×}\left(\frac{\text{1}{\text{m}}^{\text{2}}}{{\text{100}}^{\text{2}}{\text{cm}}^{\text{2}}}\right)\right)\\ \text{=0}\text{.01131}{\text{m}}^{\text{2}}\end{array}$

Substitute $\text{0}\text{.4554}\text{m}/\text{s}$ for ${\text{v}}_{\text{1}}$, $\text{0}\text{.00785}{\text{m}}^{\text{2}}$ for $A_1$, $\text{0}\text{.01131}{\text{m}}^{\text{2}}$ for $A_2$ in equation (V).

  $\begin{array}{l}0.01131{\text{m}}^2\times v_2=0.00785{\text{m}}^{\text{2}}\times 0.4554\text{m}/\text{s}\\ v_2=\frac{0.00785{\text{m}}^{\text{2}}\times 0.4554\text{m}/\text{s}}{0.01131{\text{m}}^{\text{2}}}\\ v_2=0.316\text{m}/\text{s}\end{array}$

Substitute $\text{230}\text{kPa}$ for $p_1$, $\text{100}\text{kPa}$ for $p_2$, $\text{0}\text{.4554}\text{m}/\text{s}$ for $v_1$, $\text{0}\text{.316}\text{m}/\text{s}$ for $v_2$, $\text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}$ for $g$, $783\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $1$ for ${\alpha }_1$, $1$ for ${\alpha }_2$, $0$ for $z_1$ and $\text{15}\text{m}$ for $z_2$ in equation (I).

  $\begin{array}{l}\left[\begin{array}{l}\frac{\text{230}\text{kPa}}{\text{783}\text{kg}/{\text{m}}^{\text{3}}\text{×9}\text{.81}\text{m}/{\text{s}}^{\text{2}}}\\ +\left(1\right)\frac{{0.4554}^2{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{2\times 9.81\text{m}/{\text{s}}^{\text{2}}}+0\end{array}\right]=\left[\begin{array}{l}\frac{100\text{kPa}}{783\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}}\\ +(1)\frac{{0.316}^2{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{2\times 9.81\text{m}/{\text{s}}^{\text{2}}}+15+h_l\end{array}\right]\\ \frac{230\text{kPa}\times \left(\frac{1000\text{Pa}}{1\text{kPa}}\right)}{783\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}}+0.01057\text{m}=\frac{100\text{kPa}\times \left(\frac{\text{1000}\text{Pa}}{\text{1}\text{kPa}}\right)}{783\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}}+0.00508\text{m}+15\text{m}+h_l\\ 29.94\text{3}\text{m}+\text{0}\text{.01057}\text{m}=13.018\text{m}+0.00508\text{m}+15\text{m}+h_l\\ h_l=29.953\text{m}-\text{28}\text{.023}\text{m}\end{array}$

  $h_{\text{l}}\text{=1}\text{.929}\text{m}$

Conclusion:

The head loss in the pipe is $\text{1}\text{.929}\text{m}$.