Chapter 14.3, Problem 14.75P

A jet airliner is cruising at a speed of 900 km/h with each of its three engines discharging air with a velocity of 800 m/s relative to the plane. Determine the speed of the airliner after it has lost the use of (a) one of its engines. (b) two of its engines. Assume that the drag due to air friction is proportional to the square of the speed and that the remaining engines keep operating at the same rate.

To determine

(a)

The speed of the air-liner after it has lost the use of one of its engine.

Answer to Problem 14.75P

Speed $v_1=760{\rm K}m/h.$

Explanation of Solution

Given information:

Cruising speed

$v_0=900{\rm K}m/h=250m/s.$

Discharge relative velocity

$u=800m/s.$

Thrust formula for engine is,

$F=\frac{dm}{dt}\left(u_A-v_B\right)$

And drag force $D=k{v}_A{}^2$

When all the three engine working the cruising speed is $v_0$. Then, the rate of mass factor is,

$\Rightarrow 3F-D=0$

$\Rightarrow 3\frac{dm}{dt}\left(u-v_0\right)-k{v}_A^2=0$

$\Rightarrow \frac{dm}{dt}=\frac{k{v}_A^2}{3\left(u-v_0\right)}or,$

$\begin{array}{l}\Rightarrow \frac{dm}{dt}=\frac{k{\left(250\right)}^2}{3\left(800-250\right)}\\ \Rightarrow \frac{dm}{dt}=\frac{62500k}{1650}\\ \Rightarrow \frac{dm}{dt}=37.87k\\ \end{array}$

Calculation:

When one engine fails two are working then,

Cruising speed = $v_1$

$\Rightarrow 2F-D=0$

$\Rightarrow 2\frac{dm}{dt}\left(u-v_1\right)-k{v}_1^2=0$

$\begin{array}{l}\Rightarrow 2\left(37.87k\right)\left(800-v_1\right)-k{v}_1^2=0\\ \Rightarrow 6059k-75.74k{v}_1-k{v}_1^2=0\\ \Rightarrow v_1^2+75.74v_1-6059=0\end{array}$

By solving the above equation we get,

$\begin{array}{l}{v}_1=211.20m/sor,\\ v_1=760{\rm K}m/h\end{array}$

To determine

(b)

The speed of the air-liner two of its engine fails.

Answer to Problem 14.75P

Speed $v_2=562{\rm K}m/h.$

Explanation of Solution

Given information:

Cruising speed

$v_0=900{\rm K}m/h=250m/s.$

Discharge relative velocity

$u=800m/s.$

Thrust formula for engine is,

And drag force $D=k{v}_A{}^2$

When all the three engine working the cruising speed is $v_0$. Then, the rate of mass factor is,

$\Rightarrow 3F-D=0$

$\Rightarrow 3\frac{dm}{dt}\left(u-v_0\right)-k{v}_A^2=0$

$\Rightarrow \frac{dm}{dt}=\frac{k{v}_A^2}{3\left(u-v_0\right)}or,$

$\begin{array}{l}\Rightarrow \frac{dm}{dt}=\frac{k{\left(250\right)}^2}{3\left(800-250\right)}\\ \Rightarrow \frac{dm}{dt}=\frac{62500k}{1650}\\ \Rightarrow \frac{dm}{dt}=37.87k\\ \end{array}$

Calculation:

When two of its engine fails then only one is working then,

Cruising speed $=v_2$

$\Rightarrow F-D=0$

$\Rightarrow \frac{dm}{dt}\left(u-v_2\right)-k{v}_2^2=0$

$\begin{array}{l}\Rightarrow \left(37.87k\right)\left(800-v_2\right)-k{v}_2^2=0\\ \Rightarrow 30269k-37.87k{v}_2-k{v}_2^2=0\\ \Rightarrow v_2^2+37.87v_2-30269=0\end{array}$

By solving the above equation we get,

$\begin{array}{l}{v}_2=156.17m/sor,\\ v_2=562{\rm K}m/h\end{array}$