Chapter 8, Problem 61P

A horizontal pipe has an abrupt expansion from $D_1=5$ cm to $D_2=10$ cm. The water velocity in the smaller section is 8 m/s and the how is turbulent. The pressure in the smaller section is P₁ = 410 kPa. Taking the kinetic energy correction factor to be 1.06 at both the inlet and the outlet, determine the downstream pressure P₂, and estimate the error that would have occurred if Bernoulli’s equation had been used.
Answers: 423 kPa, 17.3 kPa

To determine

The value of the downstream pressure.

The error which occurs when the Bernoulli's equation is used.

Answer to Problem 61P

The value of the downstream pressure $P_2$ is $439.9785\text{kPa}$.

The error in downstream pressure due to Bernoulli's equation is $21.58\text{kPa}$.

Explanation of Solution

Given information:

The velocity of water in the smaller section is $8\text{m}/\text{s}$ and the flow is found to be turbulent. The pressure at the smaller section is $410\text{kPa,}$ and the kinetic energy correction factor is $1.06$.

Write the expression for the cross sectional flow area for smaller section.

  $A_1=\frac{\pi }{4}{D}_1^2$........... (I)

Here, the diameter of the small section is $D_1$.

Write the expression for the cross sectional flow area for bigger section.

  $A_2=\frac{\pi }{4}{D}_2^2$........... (II)

Here, the diameter of the big pipe is $D_2$.

Write the expression for the velocity at bigger section using continuity equation.

  $\begin{array}{l}\rho V_1{A}_1=\rho V_2{A}_2\\ V_2=\frac{A_1}{A_2}{V}_1\end{array}$........... (III)

Here, the velocity at the small section is $V_1$.

Write the expression for the loss coefficient for expansion based on the velocity in the smaller section.

  $K_L=\alpha {\left(1-\frac{D_1^2}{D_2^2}\right)}^2$........... (IV)

Here, the correction factor for the kinetic energy is $\alpha$.

Write the expression for the head loss due to sudden expansion from smaller section to bigger section.

  $h_L=K_L\frac{V_1^2}{2g}$........... (V)

Here, the gravitational acceleration is $g$.

Write the expression for the Bernoulli's equation.

  $\frac{P_1}{\rho g}+\alpha \frac{V_1^2}{2g}+z_1=\frac{P_2}{\rho g}+\alpha \frac{V_2^2}{2g}+z_2+h_L$........... (VI)

Here, the smaller section pressure is $P_1$, the bigger section pressure is $P_2$, the datum at bigger section is $z_2$, the datum at smaller section is $z_1$, the gravitational acceleration is $g$ and the density of water is $\rho$.

Write the expression for the real Bernoulli's equation.

  $\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{{P^{\prime }}_2}{\rho g}+\frac{V_2^2}{2g}+z_2$........... (VII)

Here, the downstream pressure using Bernoulli's equation is ${P^{\prime }}_2$.

Write the expression for the error in pressure due to use of Bernoulli's equation.

  $E={P^{\prime }}_2-P_2$........... (VIII)

Calculation:

Substitute $5\text{cm}$ for $D_1$ in Equation (I).

  $\begin{array}{c}{A}_1=\frac{\pi }{4}{\left(5\text{cm}\right)}^2\\ =\frac{\pi }{4}{\left(5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\\ =1.96\times {10}^{-3}{\text{m}}^{\text{2}}\end{array}$

Substitute $10\text{cm}$ for $D_1$ in Equation (I).

  $\begin{array}{c}{A}_2=\frac{\pi }{4}{\left(10\text{cm}\right)}^2\\ =\frac{\pi }{4}{\left(10\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\\ =7.85\times {10}^{-3}{\text{m}}^{\text{2}}\end{array}$

Substitute $1.96\times {10}^{-3}{\text{m}}^{\text{2}}$ for $A_1$, $7.85\times {10}^{-3}{\text{m}}^{\text{2}}$ for $A_2$ and $8\text{m}/\text{s}$ for $V_1$ in Equation (III).

  $\begin{array}{c}{V}_2=\frac{\left(1.96\times {10}^{-3}{\text{m}}^{\text{2}}\right)}{\left(7.85\times {10}^{-3}{\text{m}}^{\text{2}}\right)}\left(8\text{m}/\text{s}\right)\\ =0.24\times 8\text{m}/\text{s}\\ =1.99\text{}\text{m}/\text{s}\end{array}$

Substitute $10\text{cm}$ for $D_1$, $5\text{cm}$ for $D_1$ and $1.06$ for $\alpha$ in Equation (IV).

  $\begin{array}{c}{K}_L=1.06{\left(1-\frac{{\left(5\text{cm}\right)}^2}{{\left(10\text{cm}\right)}^2}\right)}^2\\ =1.06{\left(1-\frac{{\left(5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}{{\left(10\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}\right)}^2\\ =1.06\left(1-\frac{0.0025\text{m}}{0.01\text{m}}\right)\\ =0.795\end{array}$

Substitute $8\text{m}/\text{s}$ for $V_1$, $9.81\text{m}/{\text{s}}^{\text{2}}$ and $0.795$ for $K_L$, for $g$ in Equation (V).

  $\begin{array}{c}{h}_L=\left(0.795\right)\times \frac{{\left(8\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ =0.795\times 3.26\text{m}\\ \text{=2}\text{.59}\text{m}\end{array}$

Substitute $410\text{kPa}$ for $P_1$, $0$ for $z_1$, $0$ for $z_2$, $8\text{m}/\text{s}$ for $V_1$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $1.99\text{m}/\text{s}$ for $V_2$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $1.06$ for $\alpha ,$ and $2.59\text{m}$ for $h_L$ in Equation (VI).

$\left(\begin{array}{l}\frac{410\text{kPa}}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ +1.06\frac{{\left(8\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ +0\end{array}\right)=\left(\begin{array}{l}\frac{P_2}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ +1.06\frac{{\left(1.99\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ +0+2.59\text{m}\end{array}\right)$

$\frac{P_2}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}=\left(\begin{array}{l}\frac{410\text{kPa}\left(\frac{1000\text{kg}/{\text{ms}}^{\text{2}}}{1\text{kPa}}\right)}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ +1.06\frac{{\left(8\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ -1.06\frac{{\left(1.99\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ -2.59\text{m}\end{array}\right)$

$\begin{array}{l}{P}_2=\left(41.79\text{m}+3.45\text{m}-0.21\text{m}-2.59\text{m}\right)\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ P_2=416336.4\text{kg}/{\text{ms}}^{\text{2}}\end{array}$

  $\begin{array}{c}{P}_2=416336.4\text{kg}/{\text{ms}}^{\text{2}}\left(\frac{1\text{kPa}}{1000\text{kg}/{\text{ms}}^{\text{2}}}\right)\\ =416.3364\text{kPa}\end{array}$

Therefore, the downstream pressure $P_2$ is $418.396\text{kPa}$.

Substitute $410\text{kPa}$ for $P_1$, $0$ for $z_1$, $0$ for $z_2$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $8\text{m}/\text{s}$ for $V_1$ and $1.99\text{m}/\text{s}$ for $V_2$ in Equation (VII).

$\left(\begin{array}{l}\frac{410\text{kPa}}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ +\frac{{\left(8\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ +0\end{array}\right)=\left(\begin{array}{l}\frac{{P^{\prime }}_2}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ +\frac{{\left(1.99\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ +0\end{array}\right)$

$\frac{{P^{\prime }}_2}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}=\left(\begin{array}{l}\frac{410\text{kPa}\left(\frac{1000\text{kg}/{\text{ms}}^{\text{2}}}{1\text{kPa}}\right)}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ +\frac{{\left(8\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ -\frac{{\left(1.99\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ -2.59\text{m}\end{array}\right)$

$\begin{array}{l}{P^{\prime }}_2=\left(41.79\text{m}+3.26\text{m}-0.20\text{m}\right)\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ {P^{\prime }}_2=439978.5\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\end{array}$

  $\begin{array}{c}{P^{\prime }}_2=439978.5\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\left(\frac{1\text{kPa}}{1000\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}\right)\\ =439.9785\text{kPa}\end{array}$

Substitute $418.396\text{kPa}$ for $P_2$ and $439978.5\text{kPa}$ for ${P^{\prime }}_2$ in Equation (VIII).

  $\begin{array}{c}E=439.9785\text{kPa}-418.396\text{kPa}\\ =\text{21}\text{.58}\text{kPa}\end{array}$

The error in downstream pressure due to Bernoulli's equation is $21.58\text{kPa}$.

Conclusion:

The value of downstream pressure $P_2$ is $439.9785\text{kPa}$.

The error in downstream pressure due to Bernoulli's equation is $21.58\text{kPa}$.