Chapter 14, Problem 14.112RP

The nozzle shown discharges a stream of water at the rate of 200 gal/min. Knowing that at both B and C the stream of water moves with velocity of magnitude 100 ft/s, and neglecting the weight of the vane, determine the force-couple system that must be applied at A to hold the vane in place $\left(1\text{f}{t}^3=7.48\text{gal}\right)\text{.}$

To determine

The force couple system that must be applied at A to hold the vane in plane.

Answer to Problem 14.112RP

The reaction force at A is $59.1\text{ lb}$ at $20°$clockwise from $x$ axis and resultant couple at A is $41.36\text{ lb-ft}$ in clockwise direction.

Explanation of Solution

Given:

Rate of discharge of water is $200\text{gal/min}$.

Magnitude of velocity of water is $100\text{ ft/s}$.

Concept used:

Write the expression for mass flow rate of water.

$\frac{\Delta m}{\Delta t}=\rho Q$ ...... (1)

Here, $\frac{\Delta m}{\Delta t}$ is the mass flow rate, $\rho$ is the density of water and $Q$ is the discharge rate of water.

Draw the impulse moment diagram for the vane as shown below:

Consider $x$ components of momentum.

$A_x\Delta t=\left(\Delta m\right)v_C\sin 40°$

Simplify the above expression for $A_x$.

$A_x=\frac{\Delta m}{\Delta t}{v}_c\sin 40°$ ...... (2)

Here, $A_x$ is the reaction at A in $x$ direction and $v_C$ is the velocity of flow at C.

Consider the $y$ component of momentum.

$\left(\Delta m\right)v_B+A_y\Delta t=\left(\Delta m\right)v_C\cos 40°$

Simplify the above expression.

$A_y=\frac{\Delta m}{\Delta t}\left(v_C\cos 40°-v_B\right)$ ...... (3) Here, $A_y$ is the $y$ component of reaction at A and $v_B$ is the velocity of flow at B.

Write the expression for the magnitude of reaction force at $A$.

$A=\sqrt{A_x+A_y}$ ...... (4)

Here, $A$ is the magnitude of reaction at A.

Write the expression for the direction of reaction at A.

$\alpha ={\tan }^{-1}\left(\frac{A_y}{A_x}\right)$ ...... (5)

Write the expression for momentum equilibrium at point A.

$\frac{\Delta m}{\Delta t}\left(3\text{ in}\right)v_C+M_A=\left(9\text{ in}\right)\left(\frac{\Delta m}{\Delta t}\right)\left(v_C\cos 40°\right)-\left(15\text{ in}\text{.}\right)\left(\frac{\Delta m}{\Delta t}\right)\left(v_C\sin 40°\right)$

Simplify the above expression for $M_A$.

$M_A=\left(9\text{ in}\right)\left(\frac{\Delta m}{\Delta t}\right)\left(v_C\cos 40°\right)-\left(15\text{ in}\text{.}\right)\left(\frac{\Delta m}{\Delta t}\right)\left(v_C\sin 40°\right)-\frac{\Delta m}{\Delta t}\left(3\text{ in}\right)v_B$ ...... (6)

Here, $M_A$ is the resultant moment at A.

Calculation:

Substitute $1.9378{\text{ slug/ft}}^3$ for $\rho$ and $200\text{ gal/min}$ for $Q$ in equation (1).

$\begin{array}{c}\frac{\Delta m}{\Delta t}=\left(1.9378{\text{ slug/ft}}^3\right)\left(200\text{ gal/min}\left(\frac{1{\text{ ft}}^3}{7.48\text{ gal}}\right)\left(\frac{1\text{ min}}{60\text{ s}}\right)\right)\\ =0.8636\text{ slug/s}\end{array}$

Substitute $100\text{ ft/s}$ for $v_C$ and $0.8636\text{ slug/s}$ for $\frac{\Delta m}{\Delta t}$ in equation (2).

$\begin{array}{c}{A}_x=\left(0.8636\right)\left(100\sin 40°\right)\\ =55.5\text{ lb}\end{array}$

Substitute $0.8636\text{ slug/s}$ for $\frac{\Delta m}{\Delta t}$, $100\text{ ft/s}$ for $v_C$ and $100\text{ ft/s}$ for $v_B$ in equation (3).

$\begin{array}{c}{A}_y=\left(0.8636\right)\left(100\cos 40°-100\right)\\ =-20.2\text{ lb}\end{array}$

Substitute $55.5\text{ lb}$ for $A_x$ and $-20.2\text{ lb}$ for $A_y$ in equation (4).

$\begin{array}{c}A=\sqrt{{\left(55.5\right)}^2+{\left(-20.2\right)}^2}\\ =59.1\text{ lb}\end{array}$

Substitute $55.5\text{ lb}$ for $A_x$ and $-20.2\text{ lb}$ for $A_y$ in equation (5).

$\begin{array}{c}\alpha ={\tan }^{-1}\left(\frac{-20.2}{55.5}\right)\\ =-20°\end{array}$

Substitute $100\text{ft/s}$ for $v_C$, $100\text{ ft/s}$ for $v_B$ and $0.8636\text{ slug/s}$ for $\left(\frac{\Delta m}{\Delta t}\right)$ in equation (6).

$\begin{array}{l}{M}_A=\left[\begin{array}{l}\left(9\text{ in}\right)\left(0.8636\right)\left(100\cos 40°\right)-\left(15\text{ in}\text{.}\right)\left(0.8636\right)\left(100\sin 40°\right)\\ -\left(0.8636\right)\left(3\text{ in}\right)\left(100\right)\end{array}\right]\\ =0.8636\left(900\cos 40°-1500\sin 40°-300\right)\\ =-496.32\text{}\text{lb-in}\left(\frac{1\text{ ft}}{12\text{ in}\text{.}}\right)\\ =-41.36\text{ lb-ft}\end{array}$

The reaction force at A is $59.1\text{ lb}$ at $20°$clockwise from $x$ axis and resultant couple at A is $41.36\text{ lb-ft}$ in clockwise direction.

Conclusion:

Thus, the reaction force at A is $59.1\text{ lb}$ at $20°$clockwise from $x$ axis and resultant couple at A is $41.36\text{ lb-ft}$ in clockwise direction.