Chapter 3, Problem 3.177P

To determine

(a)

The formula for the viscosity of the liquid.

Answer to Problem 3.177P

The formula for the viscosity of the liquid is $\left(H+L\right)\left(\frac{\pi \rho g{d}^4}{128LQ}\right)-\left(\frac{\rho Q{\alpha }_2}{16\pi L}\right)$.

Explanation of Solution

Given information:

The flow rate is $Q$ . The flow is laminar. The head loss in the pipe is $h_f=\frac{32\mu LV}{\rho g{d}^2}$ . The density is $\rho$.

The Figure-(1) shows the section (1) and the section (2) for the given system.

Figure-(1)

Write the expression for the Bernoulli’s equation between the section (1) and section (2) of the given system.

$\begin{array}{l}\frac{p_1}{\rho g}+{\alpha }_1\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+{\alpha }_2\frac{V_2^2}{2g}+z_2+h_f\\ \left(\frac{p_1-p_2}{\rho g}\right)+\left({\alpha }_1\frac{V_1^2}{2g}-{\alpha }_2\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=h_f\end{array}$ …… (I)

Here, the pressure at inlet is $p_1$, the velocity at inlet is $V_1$, the height of the inlet from the datum is $z_1$, the pressure at outlet is $p_2$, the velocity at outlet is $V_2$, the height of the outlet from the datum is $z_2$, the density of water is $\rho$, the head loss due to friction is $h_f$, the kinetic energy correction factor at section (1) is ${\alpha }_1$, the kinetic energy correction factor at section (2) is ${\alpha }_2$ and the acceleration due to gravity is $g$.

Write the expression for the head loss due to friction in the pipe region.

$h_f=\frac{32\mu L{V}_2}{\rho g{d}^2}$

Here, the lead loss due to friction is $h_f$, dynamic viscosity is $\mu$, the length of the pipe is $L$, the flow velocity is $V_2$, density is $\rho$, the gravitational acceleration is $g$, the diameter of the pipe is $d$.

Since the pressure at the inlet and outlet are same so,

$p_1=p_2$

Since the fluid is stationary at section (1), so the velocity of fluid at section (1) is zero.

$V_1=0$

Write the expression for the height of section (2) from datum.

$z_1=H+L$

Substitute $p_2$ for $p_1$, $0$ for $V_1$, $0$ for $z_2$, $\frac{32\mu L{V}_2}{\rho g{d}^2}$ for $h_f$ and $H+L$ for $z_1$ in Equation (I).

$\begin{array}{l}\left(\frac{p_2-p_2}{\rho g}\right)+\left({\alpha }_1\frac{0}{2g}-{\alpha }_2\frac{V_2^2}{2g}\right)+\left\{\left(H+L\right)-0\right\}=\frac{32\mu L{V}_2}{\rho g{d}^2}\\ 0+0+\left(H+L\right)-{\alpha }_2\frac{V_2^2}{2g}=\frac{32\mu L{V}_2}{\rho g{d}^2}\\ {\alpha }_2\frac{V_2^2}{2g}+\frac{32\mu L{V}_2}{\rho g{d}^2}-\left(H+L\right)=0\end{array}$ …… (II)

Write the expression for the flow rate at section (2).

$Q=A_2{V}_2$ …… (III)

Here, the flow rate is $Q$ and the area is $A_2$.

Write the expression for the area at section (2).

$A_2=\frac{\pi }{4}{d}^2$

Substitute $\frac{\pi }{4}{d}^2$ for $A_2$ in Equation (III).

$\begin{array}{l}Q=\frac{\pi }{4}{d}^2{V}_2\\ V_2=\frac{Q}{\frac{\pi }{4}{d}^2}\\ V_2=\frac{4Q}{\pi d^2}\end{array}$

Substitute $\frac{4Q}{\pi d^2}$ for $V_2$ in Equation (II).

$\begin{array}{l}{\alpha }_2\frac{{\left(\frac{4Q}{\pi d^2}\right)}^2}{2g}+\frac{32\mu L\left(\frac{4Q}{\pi d^2}\right)}{\rho g{d}^2}-\left(H+L\right)=0\\ \frac{8Q^2{\alpha }_2}{{\pi }^2{d}^{4}g}+\frac{128\mu LQ}{\pi \rho g{d}^4}-\left(H+L\right)=0\\ \frac{128\mu LQ}{\pi \rho g{d}^4}=\left[\left(H+L\right)-\frac{8Q^2{\alpha }_2}{{\pi }^2{d}^{4}g}\right]\\ \mu =\frac{\pi \rho g{d}^4}{128LQ}\left[\left(H+L\right)-\frac{8Q^2{\alpha }_2}{{\pi }^2{d}^{4}g}\right]\end{array}$

Further solve the above expression.

$\begin{array}{c}\mu =\left[\left(\frac{\pi \rho g{d}^4}{128LQ}\right)\left(H+L\right)-\left(\frac{\pi \rho g{d}^4}{128LQ}\right)\left(\frac{8Q^2{\alpha }_2}{{\pi }^2{d}^{4}g}\right)\right]\\ =\left[\left(H+L\right)\left(\frac{\pi \rho g{d}^4}{128LQ}\right)-\left(\frac{\rho }{16L}\right)\left(\frac{Q{\alpha }_2}{\pi }\right)\right]\\ =\left(H+L\right)\left(\frac{\pi \rho g{d}^4}{128LQ}\right)-\left(\frac{\rho Q{\alpha }_2}{16\pi L}\right)\end{array}$

Conclusion:

The formula for the viscosity of the liquid is $\left(H+L\right)\left(\frac{\pi \rho g{d}^4}{128LQ}\right)-\left(\frac{\rho Q{\alpha }_2}{16\pi L}\right)$.

To determine

(b)

The viscosity of the fluid.

Answer to Problem 3.177P

The viscosity of the fluid is $0.001925\text{kg}/\text{m}\cdot \text{s}$.

Explanation of Solution

Given information:

The diameter of the pipe is $\text{2}\text{mm}$ . The density is $800\text{kg}/{\text{m}}^{\text{3}}$ . The length of the pipe is $95\text{cm}$ . The height of the water level is $30\text{cm}$ . The flow rate is $760{\text{cm}}^3/\text{h}$.

Write the expression for the viscosity.

$\mu =\left(H+L\right)\left(\frac{\pi \rho g{d}^4}{128LQ}\right)-\left(\frac{\rho Q{\alpha }_2}{16\pi L}\right)$ …… (IV)

Calculation:

Substitute $2\text{mm}$ for $d$, $800\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $95\text{cm}$ for $L$, $760{\text{cm}}^3/\text{h}$ for $Q$, $9.81{\text{m}/\text{s}}^2$ for $g$ and $30\text{cm}$ for $H$ in Equation (IV).

Further solve the above expression.

$\begin{array}{c}\mu =1.925395\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\\ =0.001925395\text{kg}/\text{m}\cdot \text{s}\\ \approx 0.001925\text{kg}/\text{m}\cdot \text{s}\end{array}$

Conclusion:

The viscosity of the fluid is $0.001925\text{kg}/\text{m}\cdot \text{s}$.

To determine

(c)

The name of the fluid having viscosity $0.001925\text{kg}/\text{m}\cdot \text{s}$.

Answer to Problem 3.177P

The name of the fluid having viscosity $0.001925\text{kg}/\text{m}\cdot \text{s}$ is

Explanation of Solution

Given information:

The viscosity of the fluid is $0.001925\text{kg}/\text{m}\cdot \text{s}$.

Refer to the viscosity table for liquids to obtain the fluid having viscosity $0.001925\text{kg}/\text{m}\cdot \text{s}$ as propyl alcohol.

${\mu }_p=0.00192\text{kg}/\text{m}\cdot \text{s}$

Here, the dynamic viscosity of propyl alcohol is ${\mu }_p$.

${\mu }_p\approx \mu$

Thus, the fluid could be propyl alcohol.

Conclusion:

The name of the fluid having viscosity $0.001925\text{kg}/\text{m}\cdot \text{s}$ is propyl alcohol.

To determine

(d)

Whether the Reynolds number is less than $2000$ or not.

Answer to Problem 3.177P

Yes, the Reynolds number is less than $2000$.

Explanation of Solution

Given information:

The diameter of the pipe is $\text{2}\text{mm}$ . The density is $800\text{kg}/{\text{m}}^{\text{3}}$ . The length of the pipe is $95\text{cm}$ . The height of the water level is $30\text{cm}$ . The flow rate is $760{\text{cm}}^3/\text{h}$.

Write the expression for the Reynolds number.

$\mathrm{Re}=\frac{\rho V_{2}d}{\mu }$ …… (V)

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

Write the expression for the velocity at section (2).

$V_2=\frac{4Q}{\pi d^2}$

Substitute $\frac{4Q}{\pi d^2}$ for $V_2$ in Equation (V).

$\begin{array}{c}\mathrm{Re}=\frac{\rho \left(\frac{4Q}{\pi d^2}\right)d}{\mu }\\ =\frac{4\rho Q}{\pi \mu d}\end{array}$ …… (VI)

Calculation:

Substitute $2\text{mm}$ for $d$, $800\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $760{\text{cm}}^3/\text{h}$ for $Q$, $9.81{\text{m}/\text{s}}^2$ for $g$ and $0.001925\text{kg}/\text{m}\cdot \text{s}$ for $\mu$ Equation (VI).

$\begin{array}{c}\mathrm{Re}=\frac{4\left(800\text{kg}/{\text{m}}^{\text{3}}\right)\left(760{\text{cm}}^3/\text{h}\right)}{\pi \left(0.001925\text{kg}/\text{m}\cdot \text{s}\right)\left(2\text{mm}\right)}\\ =\frac{4\left(800\text{kg}/{\text{m}}^{\text{3}}\right)\left(760{\text{cm}}^3/\text{h}\right)\left(\frac{1{\text{m}}^3}{1\times {10}^6{\text{cm}}^3}\right)\left(\frac{1\text{h}}{3600\text{s}}\right)}{\pi \left(0.001925\text{kg}/\text{m}\cdot \text{s}\right)\left(2\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}\\ =55.85\\ \approx 56\end{array}$

Thus, the Reynolds number is $56$ which is less than $2000$.

$\mathrm{Re}<2000$

Conclusion:

Yes, the Reynolds number is less than $2000$.