Chapter 2, Problem 87RQ

To determine

The gage pressure of the gasoline line.

Explanation of Solution

Given:

Gage pressure $\left(P_{\text{gage}}\right)$ is $180\text{kPa}$.

Height of the water in the manometer $\left(h_{\text{w}}\right)$ is $45\text{cm}$.

Height of the mercury in the manometer $\left(h_{\text{Hg}}\right)$ is $10\text{cm}$.

Height of the oil in the manometer $\left(h_{\text{oil}}\right)$ is $50\text{cm}$.

Height of the gasoline in the manometer $\left(h_{\text{gasoline}}\right)$ is $22\text{cm}$.

Density of the water $\left({\rho }_{\text{w}}\right)$ is $1000\text{kg}/{\text{m}}^{\text{3}}$.

Specific gravity of the oil $\left({\text{SG}}_{\text{oil}}\right)$ is $0.79$.

Specific gravity of the mercury $\left({\text{SG}}_{\text{Hg}}\right)$ is $13.6$.

Specific gravity of the gasoline $\left({\text{SG}}_{\text{gasoline}}\right)$ is $0.70$.

Acceleration due to gravity $\left(g\right)$ is $9.81\text{m}/{\text{s}}^{\text{2}}$.

Calculation:

Calculate the density of the oil $\left({\rho }_{\text{oil}}\right)$.

  $\begin{array}{c}{\rho }_{\text{oil}}=\left({\rho }_{\text{w}}\right)\left({\text{SG}}_{\text{oil}}\right)\\ =\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(\text{0}\text{.79}\right)\\ =790\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Calculate the density of the mercury $\left({\rho }_{\text{Hg}}\right)$.

  $\begin{array}{c}{\rho }_{\text{Hg}}=\left({\rho }_{\text{w}}\right)\left({\text{SG}}_{\text{Hg}}\right)\\ =\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(\text{13}\text{.6}\right)\\ =13,600\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Calculate the density of the gasoline $\left({\rho }_{\text{gasoline}}\right)$.

  $\begin{array}{c}{\rho }_{\text{gasoline}}=\left({\rho }_{\text{w}}\right)\left({\text{SG}}_{\text{Hg}}\right)\\ =\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(\text{0}\text{.70}\right)\\ =700\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Calculate the gage pressure of the gasoline line $\left(P_{\text{gasoline}}\right)$.

  $P_{\text{gage}}-{\rho }_{\text{w}}g{h}_{\text{w}}+{\rho }_{\text{oil}}g{h}_{\text{oil}}-{\rho }_{\text{Hg}}g{h}_{\text{Hg}}-{\rho }_{\text{gasoline}}g{h}_{\text{gasoline}}=P_{\text{gasoline}}$

  $\begin{array}{c}{P}_{\text{gasoline}}=P_{\text{gage}}-{\rho }_{\text{w}}g{h}_{\text{w}}+{\rho }_{\text{oil}}g{h}_{\text{oil}}-{\rho }_{\text{Hg}}g{h}_{\text{Hg}}-{\rho }_{\text{gasoline}}g{h}_{\text{gasoline}}\\ =P_{\text{atm}}-g\left({\rho }_{\text{w}}{h}_{\text{w}}-{\rho }_{\text{oil}}{h}_{\text{oil}}+{\rho }_{\text{Hg}}{h}_{\text{Hg}}+{\rho }_{\text{gasoline}}{h}_{\text{gasoline}}\right)\\ =180\text{kPa}-\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left[\begin{array}{l}\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(45\text{cm}\right)-\left(790\text{kg}/{\text{m}}^{\text{3}}\right)\left(50\text{cm}\right)+\\ \left(13,600\text{kg}/{\text{m}}^{\text{3}}\right)\left(10\text{cm}\right)+\left(700\text{kg}/{\text{m}}^{\text{3}}\right)\left(22\text{cm}\right)\end{array}\right]\\ =180\text{kPa}-\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left[\begin{array}{l}\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.45\text{m}\right)-\left(790\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.50\text{m}\right)+\\ \left(13,600\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.10\text{m}\right)+\left(700\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.22\text{m}\right)\end{array}\right]\\ =180\text{kPa}-\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left[\begin{array}{l}\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\\ \left(0.45\text{m}\right)-\\ \left(790\text{kg}/{\text{m}}^{\text{3}}\right)\\ \left(0.50\text{m}\right)+\\ \left(13,600\text{kg}/{\text{m}}^{\text{3}}\right)\\ \left(0.10\text{m}\right)+\\ \left(700\text{kg}/{\text{m}}^{\text{3}}\right)\\ \left(0.22\text{m}\right)\end{array}\right]\left(\frac{1\text{kN}}{1000\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{\text{1}\text{kPa}}{1000\text{kN}/{\text{m}}^{\text{2}}}\right)\\ =164.6\text{kPa}\end{array}$

Thus, the gage pressure of the gasoline line is $\underset{\_}{164.6\text{kPa}}$.