Chapter 14.3, Problem 14.74P

(a)

To determine

Find the magnitude and line of action of the propulsive thrust developed by the engine when the speed of the airplane is $300\text{mi}/\text{h}$.

Answer to Problem 14.74P

The magnitude of airplane is $\underset{\_}{9,690\text{lb}}$.

The line of action of the propulsive thrust developed by the engine when the speed of the airplane is $300\text{mi}/\text{h}$ is $\underset{\_}{3.38\text{ft}}$.

Explanation of Solution

Given information:

The mass $\left(\frac{dm}{dt}\right)$ flow rate is $200\text{lb}/\text{s}$.

The initial velocity $\left(u_A\right)$ is $300\text{mi}/\text{h}$.

The velocity $\left(u_B\right)$ of engine is $2000\text{ft}/\text{s}$.

Calculation:

Sketch the free body diagram of airplane as shown in Figure 1.

Refer to Figure 1.

Here, F is force that the plane exerts on the air.

Express the force that the plane exerts on air by using impulse momentum principle.

$\begin{array}{l}\left(\Delta m\right)u_A+F\left(\Delta t\right)=\left(\Delta m\right)u_B\\ F=\frac{\Delta m}{\Delta t}\left(u_B-u_A\right)\\ F=\frac{dm}{dt}\left(u_B-u_A\right)\end{array}$ (1)

Take a moment about B,

$\begin{array}{c}-e\left(\Delta m\right)u_A+M_B\left(\Delta t\right)=0\\ M_B=e\frac{dm}{dt}u{A}_{}\end{array}$ (2)

Find the unit mass flow rate from $\text{lb}/\text{s}$ to $\text{slug}/\text{s}$.

$\frac{dm}{dt}=\frac{1}{g}\frac{dW}{dt}$ (3)

Here, $\frac{dW}{dt}$ is weight flow rate of the sand and g is acceleration due to gravity.

Substitute $200\text{lb}/\text{s}$ for $\frac{dW}{dt}$ and $32.2{\text{ft}/\text{s}}^2$ for g in Equation (3).

$\begin{array}{c}\frac{dm}{dt}=\frac{1}{\left(32.2{\text{ft}/\text{s}}^2\right)}\left(200\text{lb}/\text{s}\right)\\ =\left(6.211\times \frac{\text{lb}\cdot \text{s}}{\text{ft}}\right)\left(\frac{\text{s}}{\text{s}}\right)\\ =\left(6.211\times \frac{\text{lb}\cdot {\text{s}}^2}{\text{ft}}\right)\left(\frac{\text{1}}{\text{s}}\right)\\ =6.2112\text{slug}/\text{s}\end{array}$

Find the magnitude (F) of airplane using the relation as follows:

Substitute $6.2112\text{slug}/\text{s}$ for $\frac{dm}{dt}$, $300\text{mi}/\text{h}$ for $\left(u_A\right)$, and $2000\text{ft}/\text{s}$ for $\left(u_B\right)$ in Equation (1).

$\begin{array}{c}F=6.2112\left(2000-300\text{mi}/\text{h}\times \left(\frac{1\text{mi}/\text{s}}{3600\text{mi}/\text{h}}\times \frac{5280\text{ft}/\text{s}}{1\text{mi}/\text{s}}\right)\right)\\ =6.2112\left(2000-440\right)\\ =9,689.47\\ F\approx 9,690\text{lb}\end{array}$

Thus, the magnitude of airplane is $\underset{\_}{9,690\text{lb}}$.

Find the line of action of the propulsive thrust developed by the engine when the speed of the airplane is $300\text{mi}/\text{h}$ using the relation:

From Equation (2).

$\begin{array}{l}Fd=M_B\\ d=\frac{M_B}{F}\end{array}$ (4)

Substitute $e\frac{dm}{dt}u{A}_{}$ for $M_B$ and $\left(u_B-u_A\right)$ for F in Equation (4).

$d=\frac{eu{A}_{}}{\left(u_B-u_A\right)}$ (5)

Substitute $12\text{ft}$ for e, $300\text{mi}/\text{h}$ for $\left(u_A\right)$, and $2000\text{ft}/\text{s}$ for $\left(u_B\right)$ in Equation (5).

$\begin{array}{c}d=\frac{12\times \left(300\text{mi}/\text{h}\times \frac{1\text{mi}/\text{s}}{3600\text{mi}/\text{h}}\times \frac{5280\text{ft}/\text{s}}{1\text{mi}/\text{s}}\right)}{\left(2000-\left(\frac{1\text{mi}/\text{s}}{3600\text{mi}/\text{h}}\times \frac{5280\text{ft}/\text{s}}{1\text{mi}/\text{s}}\right)\right)}\\ =\frac{12\times 440}{2000-440}\\ =3.38\text{ft}\end{array}$

Thus, the line of action of the propulsive thrust developed by the engine when the speed of the airplane is $300\text{mi}/\text{h}$ is $\underset{\_}{3.38\text{ft}}$.

(b)

To determine

Find the magnitude and line of action of the propulsive thrust developed by the engine when the speed of the airplane is $600\text{mi}/\text{h}$.

Answer to Problem 14.74P

The magnitude of airplane is $\underset{\_}{6,960\text{lb}}$.

The line of action of the propulsive thrust developed by the engine when the speed of the airplane is $300\text{mi}/\text{h}$ is $\underset{\_}{9.43\text{ft}}$.

Explanation of Solution

Calculation:

Find the magnitude (F) of airplane using the relation as follows:

Substitute $6.2112\text{slug}/\text{s}$ for $\frac{dm}{dt}$, $600\text{mi}/\text{h}$ for $\left(u_A\right)$, and $2000\text{ft}/\text{s}$ for $\left(u_B\right)$ in Equation (1).

$\begin{array}{c}F=6.2112\left(2000-600\text{mi}/\text{h}\times \left(\frac{1\text{mi}/\text{s}}{3600\text{mi}/\text{h}}\times \frac{5280\text{ft}/\text{s}}{1\text{mi}/\text{s}}\right)\right)\\ =6.2112\left(2000-880\right)\\ =6,956.54\\ F\approx 6.960\text{lb}\end{array}$

Thus, the magnitude of airplane is $\underset{\_}{6,960\text{lb}}$.

Find the line of action of the propulsive thrust developed by the engine when the speed of the airplane is $600\text{mi}/\text{h}$ using the relation:

Substitute $12\text{ft}$ for e, $600\text{mi}/\text{h}$ for $\left(u_A\right)$, and $2000\text{ft}/\text{s}$ for $\left(u_B\right)$ in Equation (5).

$\begin{array}{c}d=\frac{12\times \left(600\text{mi}/\text{h}\frac{1\text{mi}/\text{s}}{3600\text{mi}/\text{h}}\times \frac{5280\text{ft}/\text{s}}{1\text{mi}/\text{s}}\right)}{\left(2000-\left(600\text{mi}/\text{h}\times \frac{1\text{mi}/\text{s}}{3600\text{mi}/\text{h}}\times \frac{5280\text{ft}/\text{s}}{1\text{mi}/\text{s}}\right)\right)}\\ =\frac{12\times 880}{2000-880}\\ =9.43\text{ft}\end{array}$

Thus, the line of action of the propulsive thrust developed by the engine when the speed of the airplane is $300\text{mi}/\text{h}$ is $\underset{\_}{9.43\text{ft}}$.