Chapter 2, Problem 2.11P

To determine

The elevation of the liquid level in the open piezometer tube $B$.

The elevation of the liquid level in the open piezometer tube $C$.

Answer to Problem 2.11P

The elevation of the liquid level in the open piezometer tube $B$ is $2.73\text{m}$.

The elevation of the liquid level in the open piezometer tube $C$ is $1.93\text{m}$.

Explanation of Solution

Calculation:

Write the hydrostatic formula for tube $B$.

$\left[p_A+{\gamma }_{\text{air}}\left(z_{\text{air}}\right)+{\gamma }_{\text{gasoline}}\left(z_{{}_{\text{gasoline}}}\right)-{\gamma }_{\text{gasoline}}\left(z_B\right)-p_{B_{\text{gage}}}\right]=0$ …… (I)

Here, the gage pressure of tube $B$ is $p_{B_{\text{gage}}}$, the pressure of gage $A$ is $p_A$, the specific weight of air is ${\gamma }_{\text{air}}$, the distance above the air-gasoline interface is $z_{\text{air}}$, the specific weight of gasoline is ${\gamma }_{\text{gasoline}}$, the distance between the glycerin-gasoline interface to the air-gasoline interface is $z_{\text{gasoline}}$, the distance above the glycerin interface of the tube $B$ is $z_B$ and

Calculate the liquid level above the glycerin interface.

$z_{BT}=\left(z_B\right)+\left(z_{\text{glycerin}}\right)$ …… (II)

Here, the distance of liquid level above the glycerin interface for tube $B$ is $z_{BT}$ and the distance between glycerin interface to glycerin-gasoline interface is $z_{\text{glycerin}}$.

Write the hydrostatic formula for tube $C$.

$\left[\begin{array}{l}{p}_A+{\gamma }_{\text{air}}\left(z_{\text{air}}\right)+{\gamma }_{\text{gasoline}}\left(z_{{}_{\text{gasoline}}}\right)+{\gamma }_{\text{glycerin}}\left(z_{\text{glycerin}}\right)\\ -{\gamma }_{\text{glycerin}}\left(z_C\right)-p_{C_{\text{gage}}}\end{array}\right]=0$ …… (III)

Here, the gage pressure of tube $C$ is $p_{C_{\text{gage}}}$, the specific weight of glycerin is ${\gamma }_{\text{glycerin}}$ and the distance of liquid level above the glycerin interface for tube $C$ is $z_C$.

Substitute $1.5\text{kPa}$ for $p_A$, $12\text{N}/{\text{m}}^{\text{3}}$ for ${\gamma }_{\text{air}}$, $2\text{m}$ for $z_{\text{air}}$, $6670\text{N}/{\text{m}}^{\text{3}}$ for ${\gamma }_{\text{gasoline}}$, $1.5\text{m}$ for $z_{{}_{\text{gasoline}}}$ and $0\text{kPa}$ for $p_{B_{\text{gage}}}$ in Equation (I).

$\begin{array}{l}\left[\begin{array}{l}\left(1.5\text{kPa}\right)+\left(12\text{N}/{\text{m}}^{\text{3}}\right)\left(2\text{m}\right)+\left(6670\text{N}/{\text{m}}^{\text{3}}\right)\left(1.5\text{m}\right)\\ -\left(6670\text{N}/{\text{m}}^{\text{3}}\right)\left(z_B\right)-\left(0\text{kPa}\right)\end{array}\right]=0\\ \left[\begin{array}{l}\left(1.5\text{kPa}\right)\left(\frac{1000\text{N}/{\text{m}}^2}{1\text{kPa}}\right)+\left(24\text{N}/{\text{m}}^2\right)+\left(10005\text{N}/{\text{m}}^2\right)\\ -\left(6670\text{N}/{\text{m}}^{\text{3}}\right)\left(z_B\right)\end{array}\right]=0\\ \left(11529\text{N}/{\text{m}}^2\right)=\left(6670\text{N}/{\text{m}}^{\text{3}}\right)\left(z_B\right)\\ z_B=\frac{\left(11529\text{N}/{\text{m}}^2\right)}{\left(6670\text{N}/{\text{m}}^{\text{3}}\right)}\end{array}$

Further simplify the above value.

$\begin{array}{c}{z}_B=1.7284\text{m}\\ \approx 1.73\text{m}\end{array}$

Substitute $1.73\text{m}$ for $z_B$ and $1\text{m}$ for $z_{\text{glycerin}}$ in Equation (II).

$\begin{array}{c}{z}_{BT}=\left(1.73\text{m}\right)+\left(1\text{m}\right)\\ =2.73\text{m}\end{array}$

Thus, the liquid level above the glycerin interface for tube $B$ is $2.73\text{m}$.

Substitute $1.5\text{kPa}$ for $p_A$, $12\text{N}/{\text{m}}^{\text{3}}$ for ${\gamma }_{\text{air}}$, $2\text{m}$ for $z_{\text{air}}$, $6670\text{N}/{\text{m}}^{\text{3}}$ for ${\gamma }_{\text{gasoline}}$, $1.5\text{m}$ for $z_{\text{gasoline}}$, $0\text{kPa}$ for $p_{C_{\text{gage}}}$, $12360\text{N}/{\text{m}}^{\text{3}}$ for ${\gamma }_{\text{glycerin}}$ and $1\text{m}$ for $z_{\text{glycerin}}$ in Equation (III).

$\begin{array}{l}\left[\begin{array}{l}\left(1.5\text{kPa}\right)+\left(12\text{N}/{\text{m}}^{\text{3}}\right)\left(2\text{m}\right)+\left(6670\text{N}/{\text{m}}^{\text{3}}\right)\left(1.5\text{m}\right)+\left(12360\text{N}/{\text{m}}^{\text{3}}\right)\left(1\text{m}\right)\\ -\left(12360\text{N}/{\text{m}}^{\text{3}}\right)\left(z_C\right)-\left(0\text{kPa}\right)\end{array}\right]=0\\ \left[\left(1.5\text{kPa}\right)\left(\frac{1000\text{N}/{\text{m}}^2}{1\text{kPa}}\right)+\left(22389\text{N}/{\text{m}}^2\right)\right]=\left(12360\text{N}/{\text{m}}^{\text{3}}\right)\left(z_C\right)\\ \left(23889\text{N}/{\text{m}}^2\right)=\left(12360\text{N}/{\text{m}}^{\text{3}}\right)\left(z_C\right)\\ z_C=\frac{\left(23889\text{N}/{\text{m}}^2\right)}{\left(12360\text{N}/{\text{m}}^{\text{3}}\right)}\end{array}$

Further simplify the above value.

$\begin{array}{c}{z}_C=1.9327\text{m}\\ \approx 1.93\text{m}\end{array}$

Conclusion:

Thus, the liquid level above the glycerin interface for tube $C$ is $1.93\text{m}$.