Chapter 2, Problem 2.1P

Interpretation Introduction

Interpretation:

The altitude of the airplane above the sea level is to be determined as accurately as possible.

Concept Introduction:

The hydrostatic approximation combined with ideal gas law is:

  $\frac{dp}{dz}=-\frac{pg}{RT}$ ............(1)

Here, p is the pressure on the fluid at height z at temperature T . R is the universal gas constant and g is the acceleration due to gravity.

Answer to Problem 2.1P

The altitude of the airplane above the sea level is $9219.8\text{ ft}$ .

Explanation of Solution

Given information:

The biometric pressure at the sea level is measured to be $30.1\text{ in}\text{.Hg}$ with temperature of ${70}^{\circ }\text{F}$ . In the indicator of the airplane in flight, the gage pressure is $10.6\text{ psia}$ and the air temperature is ${46}^{\circ }\text{C}$ .

First, assume that the change in the temperature of air is linear with the height of the airplane. Sea level is indicated by ‘0’ and airplane level is indicated by ‘1’. If the height of the airplane from the sea level be ‘ h’ , then

  $\begin{array}{c}{T}_0={70}^{°}\text{F}=530\text{ R}\\ T_1={46}^{°}\text{F}=506\text{ R}\\ T=T_0+\frac{T_1-T_0}{z_1-z_0}\left(z\right)\\ T=530+\frac{506-530}{h-0}\left(z\right)\\ T=530-\frac{24}{h}\left(z\right)\end{array}$

Use this relationship in equation (1) and integrate it to get an expression in terms of pressure and height as:

  $\begin{array}{c}\frac{dp}{dz}=-\frac{pg}{RT}\\ \frac{dp}{dz}=-\frac{pg}{R\left(530-\frac{24z}{h}\right)}\\ {\displaystyle \underset{p_0}{\overset{p_1}{\int }}\frac{dp}{p}}=-\frac{g}{R}{\displaystyle \underset{0}{\overset{h}{\int }}\frac{dz}{\left(530-\frac{24z}{h}\right)}}\\ \ln \frac{p_1}{p_0}=-\frac{gh}{24R}\ln \frac{530}{506}\end{array}$

Substitute the given values of the variables as:

  $\begin{array}{c}{p}_0=30.1\text{ in}\text{. Hg}=14.78\text{ psia}\\ p_1=10.6\text{ psia}\\ R=0.3704\frac{\text{psia}\cdot {\text{ft}}^3}{\text{lbm}\cdot \text{R}}\\ g=32.17\frac{\text{ft}}{{\text{s}}^2}\end{array}$

Calculate the value of h as:

  $\begin{array}{c}\ln \frac{p_1}{p_0}=-\frac{gh}{24R}\ln \frac{530}{506}\\ \ln \frac{10.6}{14.78}=-\frac{\left(32.17\right)h}{24\left(0.3704\right)}\ln \frac{530}{506}\\ h=1.99\frac{\cancel{{\text{ft}}^{\text{2}}}\cdot \cancel{{\text{s}}^{\text{2}}}}{\cancel{\text{lbm}}}\frac{\cancel{\text{lbf}}}{\cancel{\text{in}{\text{.}}^2}}\times \frac{144\cancel{\text{in}{\text{.}}^2}}{\cancel{{\text{ft}}^{\text{2}}}}\times \frac{32.174\frac{\text{ft}\cdot \cancel{\text{lbm}}}{\cancel{{\text{s}}^2}}}{\cancel{\text{lbf}}}\\ h=9219.8\text{ ft}\end{array}$