Chapter 1.11, Problem 57P

To determine

Express the gage pressure and absolute pressures in $\text{kPa}$, $\text{kN}/{\text{m}}^{\text{2}}$, $\text{lbf}/\text{in}{\text{.}}^{\text{2}}$, $\text{psi}$ and $\text{mmHg}$.

Answer to Problem 57P

The absolute pressures in $\text{kPa}$,$\text{kN}/{\text{m}}^{\text{2}}$, $\text{lbf}/\text{in}{\text{.}}^{\text{2}}$, $\text{psi}$ and $\text{mmHg}$ are $18\text{kPa}$, $18\text{kN}/{\text{m}}^{\text{2}}$, $2.61\text{lbf}/\text{in}{\text{.}}^{\text{2}}$,$\text{2}\text{.61}\text{psi}$ and $\text{135}\text{mmHg}$ respectively.

Explanation of Solution

Write the formula to calculate the gage pressure $\left(P_{\text{gage}}\right)$.

$P_{\text{gage}}=P_{\text{abs}}-P_{\text{atm}}\left(\text{for pressures above}{P}_{\text{atm}}\right)$

Here, absolute pressure is $P_{\text{abs}}$ and atmospheric pressure is $P_{\text{atm}}$.

Write the formula to calculate the absolute pressure $\left(P_{\text{abs}}\right)$.

$P_{\text{abs}}=P_{\text{atm}}-P_{\text{vac}}\left(\text{for pressures below}{P}_{\text{atm}}\right)$ (I)

Here, vacuum pressure is $P_{\text{vac}}$.

Write the expression of conversion relation for the pressure from $\text{kPa}$ to $\text{kN}/{\text{m}}^{\text{2}}$.

$\begin{array}{c}P\left(\text{kPa}\right)=P\left(\text{kPa}\right)\times \frac{1\text{kN}/{\text{m}}^{\text{2}}}{1\text{kPa}}\\ =P\left(\text{kN}/{\text{m}}^{\text{2}}\right)\end{array}$

Write the expression of conversion relation for the pressure from $\text{kPa}$ to $\text{lbf}/\text{in}{\text{.}}^{\text{2}}$.

$\begin{array}{c}P\left(\text{kPa}\right)=P\left(\text{kPa}\right)\times \frac{1\text{lbf}/\text{in}{\text{.}}^{\text{2}}}{6.895\text{kPa}}\\ =P\left(\text{lbf}/\text{in}{\text{.}}^{\text{2}}\right)\end{array}$

Write the expression of conversion relation for the pressure from $\text{kPa}$ to $\text{psi}$.

$\begin{array}{c}P\left(\text{kPa}\right)=P\left(\text{kPa}\right)\times \frac{1\text{psi}}{6.895\text{kPa}}\\ =P\left(\text{psi}\right)\end{array}$

Write the expression of conversion relation for the pressure from $\text{kPa}$ to $\text{mmHg}$.

$\begin{array}{c}P\left(\text{kPa}\right)=P\left(\text{kPa}\right)\times \frac{1\text{mmHg}}{0.1333\text{kPa}}\\ =P\left(\text{mmHg}\right)\end{array}$

Conclusion:

The gage pressure cannot be found for the pressures below the atmospheric pressure.

Substitute $\text{98}\text{kPa}$ for $P_{\text{atm}}$ and $\text{80}\text{kPa}$ for $P_{\text{vac}}$ in Equation (I).

$\begin{array}{c}{P}_{\text{abs}}=\text{98}\text{kPa}-\text{80}\text{kPa}\\ =1\text{8}\text{kPa}\end{array}$

Convert the unit of pressure from $\text{kPa}$ to $\text{kN}/{\text{m}}^{\text{2}}$.

$\begin{array}{c}{P}_{\text{abs}}=18\text{kPa}\\ =18\text{kPa}\times \frac{1\text{kN}/{\text{m}}^{\text{2}}}{1\text{kPa}}\\ =18\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Convert the unit of pressure from $\text{kPa}$ to $\text{lbf}/\text{in}{\text{.}}^{\text{2}}$.

$\begin{array}{c}{P}_{\text{abs}}=18\text{kPa}\\ =18\text{kPa}\times \frac{1\text{lbf}/\text{in}{\text{.}}^{\text{2}}}{6.895\text{kPa}}\\ =2.61\text{lbf}/\text{in}{\text{.}}^{\text{2}}\end{array}$

Convert the unit of pressure from $\text{kPa}$ to $\text{psi}$.

$\begin{array}{c}{P}_{\text{abs}}=18\text{kPa}\\ =18\text{kPa}\times \frac{1\text{psi}}{6.895\text{kPa}}\\ =2.61\text{psi}\end{array}$

Convert the unit of pressure from $\text{kPa}$ to $\text{mmHg}$.

$\begin{array}{c}{P}_{\text{abs}}=18\text{kPa}\\ =18\text{kPa}\times \frac{1\text{mmHg}}{0.1333\text{kPa}}\\ =135\text{mmHg}\end{array}$

Thus, the absolute pressures in $\text{kPa}$,$\text{kN}/{\text{m}}^{\text{2}}$, $\text{lbf}/\text{in}{\text{.}}^{\text{2}}$, $\text{psi}$ and $\text{mmHg}$ are $18\text{kPa}$, $18\text{kN}/{\text{m}}^{\text{2}}$, $2.61\text{lbf}/\text{in}{\text{.}}^{\text{2}}$,$\text{2}\text{.61}\text{psi}$ and $\text{135}\text{mmHg}$ respectively.