Chapter 14.3, Problem 14.74P

The jet engine shown scoops in air at A at a rate of 200 lb/s and discharges it at B with a velocity of 2000 ft/s relative to the airplane. Determine the magnitude and line of action of the propulsive thrust developed by the engine when the speed of the airplane is (a) 300 mi/h. (b) 600 mi/h.

To determine

(a)

The magnitude and line of action of propulsive thrust developed by the engine when the speed of the air-plane is 300 mi/h.

Answer to Problem 14.74P

Magnitude $F=9690lb.$

Distance $d=3.38ft.$

Explanation of Solution

Given information:

$mass=200lb/s.$

Velocity of B $u_B=2000ft/s.$

Velocity of A $u_A=300mi/h=440ft/s.$

$e=12ft$

To determine the line of action of propulsive thrust use a frame of reference moving with the plane. Let, F be the force exerts by the plane on the air.

Now, applying the impulse moment principle:

$\left(\Delta m\right)u_A+F\left(\Delta t\right)=\left(\Delta m\right)u_B$

$\begin{array}{l}F=\frac{\Delta m}{\Delta t}\left(u_B-u_A\right)\\ F=\frac{dm}{dt}\left(u_B-u_A\right)\end{array}$

Now, $\frac{dm}{dt}=\frac{200}{32.2}=6.2112slug/s$

Hence,

$\begin{array}{l}F=6.2112\left(2000-440\right)\\ F=9690lb\end{array}$

Again, moment about B,

$\begin{array}{l}-e\left(\Delta m\right)u_A+M_B\left(\Delta t\right)=0\\ M_B=-e\frac{\Delta m}{\Delta t}{u}_{A}or,\\ M_B=-e\frac{dm}{dt}{u}_A\end{array}$

$M_B=-e\frac{dm}{dt}{u}_A$

Let d be the distance,

$\begin{array}{l}Fd=M_B\\ d=\frac{M_B}{F}\end{array}$

$\begin{array}{l}d=\frac{e\frac{dm}{dt}{u}_A}{\frac{dm}{dt}\left(u_B-u_A\right)}or,\\ d=\frac{e{u}_A}{\left(u_B-u_A\right)}\end{array}$

$\begin{array}{l}d=\frac{\left(12\right)\left(440\right)}{\left(2000-440\right)}\\ d=\frac{5280}{1560}\\ d=3.38ft\end{array}$

To determine

(b)

The magnitude and line of action of propulsive thrust developed by the engine when the speed of the air-plane is 600 mi/h.

Answer to Problem 14.74P

Magnitude $F=6960lb.$

Distance $d=9.43ft.$

Explanation of Solution

Given information:

$mass=200lb/s.$

Velocity of B $u_B=2000ft/s.$

Velocity of A $u_A=600mi/h=880ft/s.$

$e=12ft.$

To determine the line of action of propulsive thrust use a frame of reference moving with the plane, let, F be the force exerts by the plane on the air.

Now, applying the impulse moment principle:

$\left(\Delta m\right)u_A+F\left(\Delta t\right)=\left(\Delta m\right)u_B$

$\begin{array}{l}F=\frac{\Delta m}{\Delta t}\left(u_B-u_A\right)\\ F=\frac{dm}{dt}\left(u_B-u_A\right)\end{array}$

Now, $\frac{dm}{dt}=\frac{200}{32.2}=6.2112slug/s$

Hence,

$\begin{array}{l}F=6.2112\left(2000-880\right)\\ F=6960lb\end{array}$

Again, moment about B,

$\begin{array}{l}-e\left(\Delta m\right)u_A+M_B\left(\Delta t\right)=0\\ M_B=-e\frac{\Delta m}{\Delta t}{u}_{A}or,\\ M_B=-e\frac{dm}{dt}{u}_A\end{array}$

$M_B=-e\frac{dm}{dt}{u}_A$

Let d be the distance,

$\begin{array}{l}Fd=M_B\\ d=\frac{M_B}{F}\end{array}$

$\begin{array}{l}d=\frac{e\frac{dm}{dt}{u}_A}{\frac{dm}{dt}\left(u_B-u_A\right)}or,\\ d=\frac{e{u}_A}{\left(u_B-u_A\right)}\end{array}$

$\begin{array}{l}d=\frac{\left(12\right)\left(880\right)}{\left(2000-880\right)}\\ d=\frac{10560}{1120}\\ d=9.428\approx 9.43ft\end{array}$