Chapter 5, Problem 116P

A diffuser in a pipe flow is basically a slow expansion of the pipe diameter, which slows down the fluid velocity and increases the pressure (the Bernoulli effect). Water at room temperature flows with a volume flow rate of 00250 m³/s through a horizontal diffuser in which the pipe diameter increases gradually from D₁= 6.00 to D₂= 11.00 cm. The irreversible head loss through the diffuser is estimated to be 0.450 in. The flow is turbulent, and the kinetic energy correction factors at both the inlet and outlet of the diffuser are assumed to be 1.05.

(a) Calculate the pressure difference P₂- P₁in units of kPa using the energy equation.
(b) Repeat using the Bernoulli equation (ignore irreversible head losses and ignore kinetic energy correction factors-in other words, set the kinetic energy correction factors to 1). Calculate the percentage error in the result due to the Bernoulli approximation. and explain why (or why not) Bernoulli is applicable here.
(c) It may be surprising that the answer to part (a) is positive. i.e., the pressure rises downstream. How is this possible? Explain by calculating the change i n energy grade line EGL and the change in hydraulic grade line .HGL from the upstream to the downstream location. In particular, does EGL go up or down, and does HGL go up or down?

To determine

(a)

The pressure difference between outlet to inlet of diffuser.

Answer to Problem 116P

The pressure difference between outlet to inlet of diffuser is $32.98\text{kPa}$.

Explanation of Solution

Given information:

The volume flow rate of diffuser is $0.0250{\text{m}}^{\text{3}}/\text{s}$, inlet diameter of diffuser is $6\text{cm}$, outlet diameter of diffuser is $11\text{cm}$,irreversible head loss is $0.45\text{m}$ and energy correction factor is $1.05$ pipe is horizontal and flow is steady and irreversible.

Write the expression for area of pipe.

  $A=\frac{\pi }{4}{d}^2$   ....... (I)

Here diameter of section is $d$.

Write the expression for velocity.

  $v=\frac{\dot{V}}{A}$   ....... (II)

Here volume flow rate is $\dot{V}$.

Consider inlet section as $1$ and outlet section as $2$.

Write the expression for energy equation in control volume.

  $\frac{P_1}{\rho g}+{\alpha }_1\frac{v_1^2}{2g}+z_1=\frac{P_2}{\rho g}+{\alpha }_2\frac{v_2^2}{2g}+z_2+h_l$   ....... (III)

Here, density is $\rho$, acceleration due to gravity is $g$, pressure at inlet is $P_1$, velocity at inlet is $v_1$,energy correction factor at inlet is ${\alpha }_1$, elevation at inlet is $z_1$ and pressure at outlet is $P_2$,velocity at outlet is $v_2$,energy correction factor at outlet is ${\alpha }_2$, elevation at outlet is $z_2$ loss in pipe is $h_l$.

Substitute $z$ for $z_1$ and $z_2$, $\alpha$ for ${\alpha }_1$ and ${\alpha }_2$ in Equation (III).

  $\begin{array}{l}\frac{P_1}{\rho g}+\alpha \frac{v_1^2}{2g}+z=\frac{P_2}{\rho g}+\alpha \frac{v_2^2}{2g}+z+h_l\\ \frac{P_2-P_1}{\rho g}=\alpha \frac{v_1^2-v_2^2}{2g}-h_l\\ P_2-P_1=\rho \alpha \frac{v_1^2-v_2^2}{2}-\rho g{h}_l\end{array}$   ....... (IV)

Calculation:

Substitute $6\text{cm}$ for $d$ in Equation (I).

  $\begin{array}{c}{A}_1=\frac{\pi }{4}{\left(6\text{cm}\right)}^2\\ =\frac{\pi }{4}{\left(6\text{cm}\times \frac{\text{1}\text{m}}{100\text{cm}}\right)}^2\\ =\frac{\pi }{4}{\left(0.06\text{m}\right)}^2\\ =0.002826{\text{m}}^{\text{2}}\end{array}$

Substitute $0.002826{\text{m}}^{\text{2}}$ for $A_1$ and $0.0250{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$ in Equation (II)

  $\begin{array}{c}{v}_1=\frac{0.0250{\text{m}}^{\text{3}}/\text{s}}{0.002826{\text{m}}^{\text{2}}}\\ =8.846\text{m}/\text{s}\end{array}$

Substitute $11\text{cm}$ for $d$ in Equation (I).

  $\begin{array}{c}{A}_2=\frac{\pi }{4}{\left(11\text{cm}\right)}^2\\ =\frac{\pi }{4}{\left(11\text{cm}\times \frac{\text{1}\text{m}}{100\text{cm}}\right)}^2\\ =\frac{\pi }{4}{\left(0.11\text{m}\right)}^2\\ =0.0094985{\text{m}}^{\text{2}}\end{array}$

Substitute $0.0094985{\text{m}}^{\text{2}}$ for $A_2$ and $0.0250{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$ in Equation (II)

  $\begin{array}{c}{v}_2=\frac{0.0250{\text{m}}^{\text{3}}/\text{s}}{0.0094985{\text{m}}^{\text{2}}}\\ =2.631\text{m}/\text{s}\end{array}$

Refer to table "Properties of saturated water" to obtain density of water as $998\text{kg}/{\text{m}}^{\text{3}}$.

Substitute, $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $8.846\text{m}/\text{s}$ for $v_1$, $2.631\text{m}/\text{s}$ for $v_2$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$

  $1.05$ for $\alpha$ and $0.45\text{m}$ for $h_l$ in Equation (IV).

  $\begin{array}{c}{P}_2-P_1=\left[\begin{array}{l}\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(1.05\right)\frac{{\left(8.846\text{m}/\text{s}\right)}^2-{\left(2.631\text{m}/\text{s}\right)}^2}{2}-\\ \left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.45\text{m}\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(523.95\text{kg}/{\text{m}}^{\text{3}}\right)\left(71.329{\text{m}}^2/{\text{s}}^2\right)-\\ \left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(4.4145{\text{m}}^2/{\text{s}}^{\text{2}}\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(37372.8295\text{kg}/\text{m}\cdot {\text{s}}^2\times \frac{\text{kPa}}{1000\text{kg}/\text{m}\cdot {\text{s}}^2}\right)-\\ \left(4405.671\text{kg}/\text{m}\cdot {\text{s}}^2\times \frac{\text{kPa}}{1000\text{kg}/\text{m}\cdot {\text{s}}^2}\right)\end{array}\right]\\ =32.98\text{kPa}\end{array}$

Conclusion:

The pressure difference between outlet to inlet of diffuser is $32.98\text{kPa}$.

To determine

(b)

The pressure difference using Bernoulli equation.

The percent error due to Bernoulli approximation.

Answer to Problem 116P

The pressure difference using Bernoulli equation is $35.61\text{kPa}$.

The percent error due to Bernoulli approximation is $7.38\%$.

Explanation of Solution

Given information:

Ignore irreversible head loss, kinetic energy correction factor $1$.

Write the expression for Bernoulli equation.

  $\frac{P_1}{\rho g}+\frac{v_1^2}{2g}+z_1=\frac{P_2}{\rho g}+\frac{v_2^2}{2g}+z_2$   ....... (V)

Write the expression for error percentage.

  $\text{error}=\frac{\Delta P_{\text{B}}-\Delta P_{\text{E}}}{\Delta P_{\text{B}}}\times 100\%$   ....... (VI)

Here, pressure difference using energy equation is $\Delta P_{\text{E}}$ and pressure difference using Bernoulli equation is $\Delta P_{\text{B}}$.

Calculation:

Substitute $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $8.846\text{m}/\text{s}$ for $v_1$, $2.631\text{m}/\text{s}$ for $v_2$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $z$ for $z_1$, $z_2$ in Equation (V).

  $\begin{array}{l}\frac{P_1}{\rho g}-\frac{P_2}{\rho g}=\frac{v_2^2}{2g}-\frac{v_1^2}{2g}\\ P_2-P_1=998\text{kg}/{\text{m}}^{\text{3}}\times \frac{{\left(8.846\text{m}/\text{s}\right)}^2-{\left(2.631\text{m}/\text{s}\right)}^2}{2}\\ P_2-P_1=\left(35610\text{kg}/\text{m}\cdot {\text{s}}^2\times \frac{\text{kPa}}{1000\text{kg}/\text{m}\cdot {\text{s}}^2}\right)\\ P_2-P_1=35.61\text{kPa}\end{array}$

Substitute $32.976\text{kPa}$ for $\Delta P_{\text{E}}$ and $35.61\text{kPa}$ for $\Delta P_{\text{B}}$ in Equation (VI).

  $\begin{array}{c}\text{error}=\frac{35.61\text{kPa}-32.976\text{kPa}}{35.61\text{kPa}}\times 100\%\\ =0.0738\times 100\%\\ =7.38\%\end{array}$

Bernoulli equation is not applicable here because of pressure head loss and kinetic energy correction factor as greater $1$.

Conclusion:

The pressure difference using Bernoulli equation is $35.61\text{kPa}$.

The percent error due to Bernoulli approximation is $7.38\%$.

To determine

(c)

The nature of Energy Grade line.

The nature of Hydraulic Grade line.

Answer to Problem 116P

The Energy Grade Line (EGL) decrease from the inlet to the exit and Energy grade Line goes down.

The Hydraulic Grade Line (HGL) increase from the inlet to the exit and Hydraulic grade Line goes up.

  

  Figure-(1)

The Figure (1) shows the nature of Energy Grade Line (EGL) and Hydraulic Grade Line (HGL).

Explanation of Solution

Given information:

Write the expression for net Energy Grade Line (EGL).

  $\Delta EGL=\left(\frac{P_1}{\rho g}+{\alpha }_1\frac{v_1^2}{2g}+z_1\right)-\left(\frac{P_2}{\rho g}+{\alpha }_2\frac{v_2^2}{2g}+z_2\right)$   ....... (VII)

Substitute $z$ for $z_1$, $z$ for $z_2$, $\alpha$ for ${\alpha }_1$ and $\alpha$ for ${\alpha }_2$ in Equation (VII).

  $\begin{array}{c}\Delta EGL=\frac{P_1-P_2}{\rho g}+\alpha \frac{v_1^2-v_2^2}{2g}+z-z\\ =\frac{P_1-P_2}{\rho g}+\alpha \frac{v_1^2-v_2^2}{2g}\end{array}$   ....... (VIII)

Write the expression for net Hydraulic Grade Line (HGL).

  $\Delta HGL=\left(\frac{P_1}{\rho g}+z_1\right)-\left(\frac{P_2}{\rho g}+z_2\right)$   ....... (IX)

Substitute $z$ for $z_1$ and $z$ for $z_2$ in Equation (IX).

  $\begin{array}{c}\Delta EGL=\frac{P_1-P_2}{\rho g}+z-z\\ =\frac{P_1-P_2}{\rho g}\end{array}$........(X)

Calculation:

Substitute $998\text{Kg}/{\text{m}}^{\text{3}}$ for $\rho ,$

  $8.846\text{m}/\text{s}$ for $v_1$, $2.631\text{m}/\text{s}$ for $v_2$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $1.05$ for $\alpha$

  $-32.98\text{kPa}$ for $P_1-P_2$ in equation (VIII).

  $\begin{array}{c}\Delta EGL=\frac{-32.98\text{kPa}}{\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+1.05\frac{{\left(8.846\text{m}/\text{s}\right)}^2-{\left(2.631\text{m}/\text{s}\right)}^2}{2\times 9.81\text{m}/{\text{s}}^{\text{2}}}\\ =\frac{-32.98\text{Pa}\times \frac{\text{1000}\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}{\text{1}\text{Pa}}}{\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+1.05\frac{{\left(8.846\text{m}/\text{s}\right)}^2-{\left(2.631\text{m}/\text{s}\right)}^2}{2\times 9.81\text{m}/{\text{s}}^{\text{2}}}\\ =-3.36\text{m}+3.81\text{m}\\ =0.45\text{m}\end{array}$

  $\Delta EGL$ is positive ,so Energy Grade Line will decrease.

Substitute $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $-32.98\text{kPa}$ for $P_1-P_2$ in equation (X).

  $\begin{array}{c}\Delta EGL=\frac{-32.98\text{kPa}}{\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ =\frac{-32.98\text{kPa}\times \frac{\text{1000}\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}{\text{1}\text{Pa}}}{\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ =-3.36\frac{\text{1}\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}{\text{1}\text{kg}/{\text{m}}^2\cdot {\text{s}}^{\text{2}}}\\ =-3.36\text{m}\end{array}$

  $\Delta HGL$ is negative, so Hydraulic Grade Line will increase.

Conclusion:

The Energy Grade Line (EGL) decrease from the inlet to the exit and Energy grade Line goes down.

The Hydraulic Grade Line (HGL) increase from the inlet to the exit and Hydraulic grade Line goes up.

  

  Figure (1)

The Figure (1) shows the nature of Energy Grade Line and Hydraulic Grade Line.