Chapter 14.3, Problem 14.80P

While cruising in level flight at a speed of 570 mi/h, a jet airplane scoops in air at a rate of 240 Ib/s and discharges it with a velocity of 2200 ft/s relative to the airplane. Determine (a) the power actually used to propel the airplane. (b) the total power developed by the engine, (c) the mechanical efficiency of the airplane.

To determine

(a)

The power used to propel the airplane.

Answer to Problem 14.80P

power = $15450hp$

Explanation of Solution

Given information:

Flight speed $v=570mi/h=836ft/s$

Exhaust relative to the plane $u=2200ft/s$

$dm=150lb/s$

Mass flow rate = $\frac{dm}{dt}=\frac{150lb/s}{32.2ft/s^2}$

$\frac{dm}{dt}=7.4534slug/s$

From the relation of principle of impulse and moment:

$\begin{array}{l}F=\left(\frac{\Delta m}{\Delta t}\right)u-\left(\frac{\Delta m}{\Delta t}\right)vor,\\ \frac{\Delta m}{\Delta t}=\frac{dm}{dt}=\frac{F}{u-v}or,\\ F=\frac{dm}{dt}\left(u-v\right)\end{array}$

$\begin{array}{l}F=7.4534\left(2200-836\right)\\ F=10,166lb\end{array}$

Calculation:

Power used to propel the airplane:

$\begin{array}{l}{P}_1=Fv\\ P_1=10,166(836)\\ P_1=8.499\times {10}^{6}ft.lb/sor,\\ P_1=15,450hp\end{array}$

To determine

(b)

The total power developed by the engine.

Answer to Problem 14.80P

Total Power = $28060hp$

Explanation of Solution

Given information:

Flight speed $v=570mi/h=836ft/s$

Exhaust relative to the plane $u=2200ft/s$

$dm=150lb/s$

Mass flow rate = $\frac{dm}{dt}=\frac{150lb/s}{32.2ft/s^2}$

$\frac{dm}{dt}=7.4534slug/s$

From the relation of principle of impulse and moment:

$\begin{array}{l}F=\left(\frac{\Delta m}{\Delta t}\right)u-\left(\frac{\Delta m}{\Delta t}\right)vor,\\ \frac{\Delta m}{\Delta t}=\frac{dm}{dt}=\frac{F}{u-v}or,\\ F=\frac{dm}{dt}\left(u-v\right)\end{array}$

$\begin{array}{l}F=7.4534\left(2200-836\right)\\ F=10,166lb\end{array}$

Calculation:

Power used to propel the airplane:

$\begin{array}{l}{P}_1=Fv\\ P_1=10,166(836)\\ P_1=8.499\times {10}^{6}ft.lb/sor,\\ P_1=15,450hp\end{array}$

Power of kinetic energy of the exhaust:

$\begin{array}{l}{P}_2=\frac{1}{2}\Delta m{\left(u-v\right)}^2\\ P_2=\frac{1}{2}\frac{dm}{dt}{\left(u-v\right)}^2\\ P_2=\frac{1}{2}\times 7.4534{\left(2200-836\right)}^2\\ P_2=6.934\times {10}^{6}ft.lb/s\end{array}$

Now, total power of the airplane, $P=P_1+P_2$

$\begin{array}{l}P=8.499\times {10}^6+6.934\times {10}^6\\ P=15.433\times {10}^{6}ft.lb/sor,\\ P=28,060hp\end{array}$

To determine

(c)

The mechanical efficiency of the airplane.

Answer to Problem 14.80P

Total Power = $28060hp$

Explanation of Solution

Given information:

Flight speed $v=570mi/h=836ft/s$

Exhaust relative to the plane $u=2200ft/s$

$mass=150lb/s$

Mass flow rate = $\frac{dm}{dt}=\frac{150lb/s}{32.2ft/s^2}$

$\frac{dm}{dt}=7.4534slug/s$

From the relation of principle of impulse and moment:

$\begin{array}{l}F=\left(\frac{\Delta m}{\Delta t}\right)u-\left(\frac{\Delta m}{\Delta t}\right)vor,\\ \frac{\Delta m}{\Delta t}=\frac{dm}{dt}=\frac{F}{u-v}or,\\ F=\frac{dm}{dt}\left(u-v\right)\end{array}$

$\begin{array}{l}F=7.4534\left(2200-836\right)\\ F=10,166lb\end{array}$

Calculation:

Power used to propel the airplane:

$\begin{array}{l}{P}_1=Fv\\ P_1=10,166(836)\\ P_1=8.499\times {10}^{6}ft.lb/sor,\\ P_1=15,450hp\end{array}$

Power of kinetic energy of the exhaust:

$\begin{array}{l}{P}_2=\frac{1}{2}\Delta m{\left(u-v\right)}^2\\ P_2=\frac{1}{2}\frac{dm}{dt}{\left(u-v\right)}^2\\ P_2=\frac{1}{2}\times 7.4534{\left(2200-836\right)}^2\\ P_2=6.934\times {10}^{6}ft.lb/s\end{array}$

Now, total power of the airplane, $P=P_1+P_2$

$\begin{array}{l}P=8.499\times {10}^6+6.934\times {10}^6\\ P=15.433\times {10}^{6}ft.lb/sor,\\ P=28,060hp\end{array}$

Mechanical efficiency to propel the airplane $=\frac{Powerused}{TotalPower}$

Mechanical efficiency $=\frac{P_1}{P}$

Mechanical efficiency $\begin{array}{l}=\frac{8.499\times {10}^6}{15.433\times {10}^6}\\ =0.5507\approx 0.551\end{array}$