Chapter 8, Problem 133P

To determine

(a)

The flow rate of oil through pipe when the pipe is placed horizontally.

Answer to Problem 133P

The flow rate of oil through pipe when the pipe is placed horizontally is $7.46\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}$.

Explanation of Solution

Given Information:

The diameter of the pipe is $6\text{cm}$, the length of the pipe is

  $33\text{m}$, the inlet pressure of the oil is $745\text{kPa}$ and the pressure at outlet is $97\text{kPa}$.

Write the expression to calculate the area of the pipe.

  $A=\frac{\pi D^2}{4}$...... (I)

Here, the diameter of the pipe is $D$ and the area of the pipe is $A$.

Write the expression to calculate change in pressure.

  $\Delta P=P_1-P_2$...... (II)

Here, the pressure at inlet is $P_1$ and the pressure at outlet is $P_2$.

Write the expression to calculate the volume flow rate in an inclined pipe.

  $\dot{\vee }=\frac{\left[\Delta P-\rho gL\sin \theta \right]\pi D^4}{128\mu L}$...... (III)

Here, the volume flow rate is $\dot{\vee }$, the viscosity of the oil is $\mu$, the density of the oil is $\rho$, the angle of inclination or declination is $\theta$ and the length of the pipe is $L$.

Calculation:

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

  $\begin{array}{c}A=\frac{\pi {\left(6\text{cm}\right)}^2}{4}\\ =\frac{\pi {\left[\left(6\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{\text{cm}}\right)\right]}^2}{4}\\ =\frac{113.09\times {10}^{-4}{\text{m}}^2}{4}\\ =2.827\times {10}^{-3}{\text{m}}^{\text{2}}\end{array}$

Substitute $745\text{kPa}$ for $P_1$ and $97\text{kPa}$ for $P_2$ in Equation (II).

  $\begin{array}{c}\Delta P=745\text{kPa}-97\text{kPa}\\ =\text{648}\text{kPa}\\ =\left(\text{648}\text{kPa}\right)\left(\frac{{10}^3\text{Pa}}{\text{kPa}}\right)\\ =648\times {10}^3\text{Pa}\end{array}$

Refer to the Table A-7, "Properties of liquids" to obtain the value of $\rho$ for oil at $20°\text{C}$ as $888.1\text{kg}/{\text{m}}^{\text{3}}$ and $\mu$ for oil at $20°\text{C}$ as $0.8374\text{kg}/\text{m}\cdot \text{s}$.

Substitute $648\times {10}^3\text{Pa}$ for $\Delta P$, $888.1\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $0.8374\text{kg}/\text{m}\cdot \text{s}$ for $\mu$, $0°$ for $\theta$, $6\text{cm}$ for $D$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $33\text{m}$ for $L$ in Equation (III).

  $\begin{array}{c}\dot{\vee }=\frac{\left[\begin{array}{l}\left[648\times {10}^3\text{Pa}-\left(888.1\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(33\text{m}\right)\sin 0°\right]\\ \times \pi {\left(6\times {10}^{-2}\text{cm}\right)}^4\end{array}\right]}{128\left(0.8374\text{kg}/\text{m}\cdot \text{s}\right)\left(33\text{m}\right)}\\ =\frac{\left[648\times {10}^3\text{Pa}\left(\frac{\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}{\text{Pa}}\right)-0\right]\pi {\left[\left(6\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{\text{cm}}\right)\right]}^4}{128\left(27.6342\text{kg}/\text{s}\right)}\\ =0.00746{\text{m}}^{\text{3}}/\text{s}\\ =7.46\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\end{array}$

Conclusion:

The flow rate of oil through pipe when the pipe is placed horizontally is $7.46\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}$.

To determine

(b)

The flow rate of oil through pipe when the pipe is inclined $15°$ upward.

Answer to Problem 133P

The flow rate of oil through pipe when the pipe is inclined $15°$ upward is $6.59\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}$.

Explanation of Solution

Calculation:

Substitute $648\times {10}^3\text{Pa}$ for $\Delta P$, $888.1\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $0.8374\text{kg}/\text{m}\cdot \text{s}$ for $\mu$, $15°$ for $\theta$, $6\text{cm}$ for $D$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $33\text{m}$ for $L$ in Equation (III).

  $\begin{array}{c}\dot{\vee }=\frac{\left[\begin{array}{l}\left[648\times {10}^3\text{Pa}-\left(888.1\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(33\text{m}\right)\sin 15°\right]\\ \times \pi {\left(6\times {10}^{-2}\text{cm}\right)}^4\end{array}\right]}{128\left(0.8374\text{kg}/\text{m}\cdot \text{s}\right)\left(33\text{m}\right)}\\ =\frac{\left[648\times {10}^3\text{Pa}\left(\frac{\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}{\text{Pa}}\right)-74411.6\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right]\pi {\left[\left(6\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{\text{cm}}\right)\right]}^4}{128\left(27.6342\text{kg}/\text{s}\right)}\\ =0.00659{\text{m}}^{\text{3}}/\text{s}\\ =6.59\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\end{array}$

Conclusion:

The flow rate of oil through pipe when the pipe is inclined $15°$ upward is $6.59\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}$.

To determine

(c)

The flow rate of oil through pipe when the pipe is inclined $15°$ downward and the verification of flow to be laminar.

The value of Reynolds number is 159.71 and the flow of the oil is laminar.

Answer to Problem 133P

The flow rate of oil through pipe when the pipe is inclined $15°$ downward is $7.22\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}$.

Explanation of Solution

Write the expression to calculate the velocity of flow.

  $V=\frac{\dot{\vee }}{A}$...... (IV)

Here, the velocity of the flow is $V$.

Write the expression to calculate Reynolds's number.

  $\mathrm{Re}=\frac{\rho VD}{\mu }$...... (V)

Here, the Reynolds's number is $\mathrm{Re}$.

Calculation:

Substitute $648\times {10}^3\text{Pa}$ for $\Delta P$, $888.1\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $0.8374\text{kg}/\text{m}\cdot \text{s}$ for $\mu$, $-15°$ for $\theta$, $6\text{cm}$ for $D$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $33\text{m}$ for $L$ in Equation (III).

  $\begin{array}{c}\dot{\vee }=\frac{\left[\begin{array}{l}\left[648\times {10}^3\text{Pa}-\left(888.1\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(33\text{m}\right)\sin \left(-15°\right)\right]\\ \times \pi {\left(6\text{cm}\right)}^4\end{array}\right]}{128\left(0.8374\text{kg}/\text{m}\cdot \text{s}\right)\left(33\text{m}\right)}\\ =\frac{\left[648\times {10}^3\text{Pa}\left(\frac{\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}{\text{Pa}}\right)-\left(-74411.6\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right)\right]\pi {\left[\left(6\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{\text{cm}}\right)\right]}^4}{128\left(27.6342\text{kg}/\text{s}\right)}\\ =0.00722{\text{m}}^{\text{3}}/\text{s}\\ =7.22\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\end{array}$

Substitute $7.22\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}$ for $\dot{\vee }$ and $2.827\times {10}^{-3}{\text{cm}}^{\text{2}}$ for $A$ in Equation (IV).

  $\begin{array}{c}V=\frac{7.22\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}}{2.827\times {10}^{-3}{\text{m}}^{\text{2}}}\\ =2.51\text{m}/\text{s}\end{array}$

Substitute $888.1\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $0.8374\text{kg}/\text{m}\cdot \text{s}$ for $\mu$, $2.51\text{m}/\text{s}$ for $V$, $6\text{cm}$ for $D$ and $33\text{m}$ for $L$ in Equation (V).

  $\begin{array}{c}\mathrm{Re}=\frac{888.1\text{kg}/{\text{m}}^{\text{3}}\times 2.51\text{m}/\text{s}\times 6\text{cm}}{0.8374\text{kg}/\text{m}\cdot \text{s}}\\ =\frac{2.229\times {10}^3\text{kg}/{\text{m}}^2\cdot \text{s}\times \left(6\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{\text{cm}}\right)}{0.8374\text{kg}/\text{m}\cdot \text{s}}\\ =159.71\end{array}$

Conclusion:

The flow rate of oil through pipe when the pipe is inclined $15°$ downward is

  $7.22\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}$.

The value of Reynolds number is 159.71, therefore the flow of the oil is laminar.