Chapter 3, Problem 146P

An oil pipeline and a 1.3-m³ rigid air tank are connected to each other by a manometer, as shown in Fig. P3-146. If the tank contains 15 kg of air at 80 $°$ C. determine (a) the absolute pressure in the pipeline and (b) the change in $\Delta h$ when the temperature in the tank drops to 20 $°$ C. Assume the pressure in the oil pipeline to remain constant, and the air volume in the manometer to be negligible relative to the volume of the tank.

To determine

(a)

The absolute pressure in the pipeline.

Answer to Problem 146P

The absolute pressure in the pipeline is $\underset{\_}{1123067.7\text{N}/{\text{m}}^{\text{2}}}$.

Explanation of Solution

Given information:

Volume rigid air tank is $1.3{\text{m}}^{\text{3}}$, mass of air in the tank is $15\text{kg}$, temperature of air in the tank is $80°\text{C}$, specific gravity of oil in the pipeline is $2.68$, specific gravity of mercury is $13.6$.

Write the expression for absolute pressure of air in the tank.

  $P_{air}=\frac{mRT}{V}$...... (I)

Here, the mass of air in the tank is $m$, characteristic gas constant of air is $R$, temperature of air in the tank $T,$ and the volume of tank is $V$.

Write the expression for absolute pressure in the pipeline.

  $P_{oil}=-\left(S_{oil}{\rho }_{w}g{h}_{oil}\right)-\left(S_{Hg}{\rho }_{w}g{h}_{Hg}\right)+P_{air}$...... (II)

Here, specific gravity of oil is $S_{oil}$, density of water is ${\rho }_w$, acceleration due to gravity is $g$, height of oil in the manometer is $h_{oil}$, specific gravity of mercury is $S_{Hg},$ and height of mercury in the manometer is $h_{Hg}$.

Calculation:

Substitute $15\text{kg}$ for $m$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$, $80°\text{C}$ for $T,$ and $1.3{\text{m}}^3$ for $V$ in equation (I).

  $\begin{array}{l}{P}_{air}=\frac{15\text{kg}\times 0.287\text{kJ}/\text{kg}\cdot \text{K}\times 80°\text{C}}{1.3{\text{m}}^3}\\ =\frac{4.305\text{kJ}/\text{K}\times \left(80°\text{C+273}\text{.15}\right)\text{K}}{1.3{\text{m}}^3}\\ =\frac{1520.31\text{kJ}\left(\frac{\text{1}\text{kN}\cdot \text{m}}{\text{1}\text{kJ}}\right)}{1.3{\text{m}}^3}\\ =\frac{1520.31\text{kN}\cdot \text{m}}{1.3{\text{m}}^{\text{3}}}\end{array}$

  $=1169.469\text{kN}/{\text{m}}^{\text{2}}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_w$, $2.68$ for $S_{oil}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $13.6$ for $S_{Hg}$, $75\text{cm}$ for $h_{oil}$, $20\text{cm}$ for $h_{Hg}$ and $1169.469\text{kN}/{\text{m}}^{\text{2}}$ for $P_{air}$ in equation (II).

$P_{oil}=\left[\begin{array}{l}-\left(2.68\times 1000\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}\times 75\text{cm}\right)\\ -\left(13.6\times 1000\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}\times 20\text{cm}\right)+\left(1169.469\text{kN}/{\text{m}}^{\text{2}}\right)\end{array}\right]$

  $=\left[\begin{array}{l}-\left(2680\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}\times 75\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\\ -\left(13600\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}\times 20\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\\ +\left(1169.469\text{kN}/{\text{m}}^{\text{2}}\left(\frac{{10}^3\text{N}/{\text{m}}^{\text{2}}}{1\text{kN}/{\text{m}}^{\text{2}}}\right)\right)\end{array}\right]$

  $=\left[\begin{array}{l}-\left(26290.8\text{kg}/{\text{m}}^2\cdot {\text{s}}^2\times 0.75\text{m}\right)-\left(133416\text{kg}/{\text{m}}^2\cdot {\text{s}}^2\times 0.20\text{m}\right)+\\ \left(1169469\text{N}/{\text{m}}^{\text{2}}\right)\end{array}\right]$

  $=\left[\begin{array}{l}-\left(19718.1\text{kg}/\text{m}\cdot {\text{s}}^2\left(\frac{1\text{N}/{\text{m}}^{\text{2}}}{1\text{kg}/\text{m}\cdot {\text{s}}^2}\right)\right)-\left(26683.2\text{kg}/\text{m}\cdot {\text{s}}^2\left(\frac{1\text{N}/{\text{m}}^{\text{2}}}{1\text{kg}/\text{m}\cdot {\text{s}}^2}\right)\right)\\ +\left(1169469\text{N}/{\text{m}}^{\text{2}}\right)\end{array}\right]$

  $\begin{array}{c}{P}_{oil}=\left[-\left(19718.1\text{N}/{\text{m}}^{\text{2}}\right)-\left(26683.2\text{N}/{\text{m}}^{\text{2}}\right)+\left(1169469\text{N}/{\text{m}}^{\text{2}}\right)\right]\\ =1123067.7\text{N}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

The absolute pressure in the pipeline is $\underset{\_}{1123067.7\text{N}/{\text{m}}^{\text{2}}}$.

To determine

(b)

The change in $\Delta \text{h}$ when the temperature in the tank drops to $20°\text{C}$.

Answer to Problem 146P

The change in $\Delta \text{h}$ when the temperature in the tank drops to $20°\text{C}$ is $\underset{\_}{20.39\text{cm}}$.

Explanation of Solution

Given information:

Temperature of air in the tank is $20°\text{C}$.

The figure below shows the new schematic diagram of manometer.

  

Figure-(2)

Write the expression for absolute pressure of air in the tank.

  $P_{air}=\frac{mRT}{V}$...... (III)

Here, the mass of air in the tank is $m$, characteristic gas constant of air is $R$, temperature of air in the tank $T,$ and the volume of tank is $V$.

Write the expression for equation of balancing the volumes in both arms of the manometer.

  $\begin{array}{c}\pi {\left(3D\right)}^{2}c=\pi D^{2}b\\ 9D^{2}c=D^{2}b\\ b=9c\end{array}$...... (IV)

Here, the diameter of limb is $D$, the rise in height of mercury in left limb is $c,$ and the fall in height of mercury in right limb is $b$.

Write the expression for vertical height corresponding to $b$ in the right limb.

  $d=b\cos 50$...... (V)

Substitute $9c$ for $b$ in equation (V).

  $\begin{array}{c}d=9c\cos 50\\ =5.785c\end{array}$...... (VI)

Write the expression for equation of pressure head balance in manometer.

  $\left(\frac{P_{oil}}{{\rho }_{w}g}\right)+S_{oil}\times \left(h_{oil}+c\right)+S_{Hg}\times \left(h_{Hg}-d\right)=\left(\frac{P_{air}}{{\rho }_{w}g}\right)$...... (VII)

Substitute $5.785c$ for $d$ in equation (VII).

  $\left(\frac{P_{oil}}{{\rho }_{w}g}\right)+S_{oil}\times \left(h_{oil}+c\right)+S_{Hg}\times \left(h_{Hg}-5.785c\right)=\left(\frac{P_{air}}{{\rho }_{w}g}\right)$...... (VIII)

Calculation:

Substitute $15\text{kg}$ for $m$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$, $20°\text{C}$ for $T$ and $1.3{\text{m}}^3$ for $V$ in equation (III).

  $\begin{array}{l}{P}_{air}=\frac{15\text{kg}\times 0.287\text{kJ}/\text{kg}\cdot \text{K}\times 20°\text{C}}{1.3{\text{m}}^3}\\ =\frac{4.305\text{kJ}/\text{K}\times \left(20°\text{C+273}\text{.15}\right)\text{K}}{1.3{\text{m}}^3}\\ =\frac{1262.01\text{kJ}\left(\frac{\text{1}\text{kN}\cdot \text{m}}{\text{1}\text{kJ}}\right)}{1.3{\text{m}}^3}\\ =\frac{1262.01\text{kN}\cdot \text{m}}{1.3{\text{m}}^{\text{3}}}\end{array}$

  $=971\text{kN}/{\text{m}}^{\text{2}}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_w$, $2.68$ for $S_{oil}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $13.6$ for $S_{Hg}$, $75\text{cm}$ for $h_{oil}$, $20\text{cm}$ for $h_{Hg}$, $1123067.7\text{N}/{\text{m}}^{\text{2}}$ for $P_{oil}$ and $971\text{kN}/{\text{m}}^{\text{2}}$ for $P_{air}$ in equation (VIII).

  $\begin{array}{l}\left[\begin{array}{l}\left(\frac{1123067.7\text{N}/{\text{m}}^{\text{2}}}{1000\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}}\right)\\ +2.68\times \left(75\text{cm}+c\right)+\\ 13.6\times \left(20\text{cm}-5.785c\right)\end{array}\right]=\left(\frac{971\text{kN}/{\text{m}}^{\text{2}}}{1000\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}}\right)\\ \left(\begin{array}{l}\frac{1123067.7\text{N}/{\text{m}}^{\text{2}}}{1000\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}}-\\ \left(\frac{971\text{kN}/{\text{m}}^{\text{2}}\left(\frac{{10}^3\text{N}/{\text{m}}^{\text{2}}}{1\text{kN}/{\text{m}}^{\text{2}}}\right)}{1000\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}}\right)\end{array}\right)=-\left[\begin{array}{l}2.68\times \left(75\text{cm}+c\right)\\ +13.6\times \left(20\text{cm}-5.785c\right)\end{array}\right]\\ \left(\frac{\text{114}\text{.481}\text{N}}{\text{1}\text{kg}/{\text{s}}^{\text{2}}}\right)-\left(\frac{\text{98}\text{.98}\text{N}}{\text{1}\text{kg}/{\text{s}}^{\text{2}}}\right)=-\left(201\text{cm}+2.68c+272\text{cm}-78.676c\right)\\ \frac{15.501\text{N}\left(\frac{\text{1}\text{kgm}/{\text{s}}^{\text{2}}}{\text{1}\text{N}}\right)}{\text{1}\text{kg}/{\text{s}}^{\text{2}}}=75.996c-473\text{cm}\end{array}$

  $\begin{array}{l}\frac{15.501\text{kgm}/{\text{s}}^{\text{2}}}{\text{1}\text{kg}/{\text{s}}^{\text{2}}}+473\text{cm}=\text{75}\text{.996}c\\ 15.501\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)+473\text{cm}=\text{75}\text{.996}c\\ 1550.1\text{cm}=75.995c\\ c=20.39\text{cm}\end{array}$

Conclusion:

The change in $\Delta \text{h}$ when the temperature in the tank drops to $20°\text{C}$ is $\underset{\_}{20.39\text{cm}}$.