A problem on Impulse & Impulsive Forces

Question

A jet of water leaves a nozzle of $1$ inch diameter at a speed of $50$ ft/sec and impinges on a plate fixed at right angles to its direction. What pressure is exerted on the plate?

If the nozzle be drawn backwards with a velocity of $20$ ft/sec in the line of the jet, what pressure is then exerted on the plane?

Answer: $26.64~ pounds-wt$; $9.59~pounds-wt$

Attempt: Cross-sectional area of the nozzle=$\pi(\frac{1}{24})^2~sq-ft$

Mass of water issued from jet per sec =Vol. of water per sec $\times$ 62.5 =$\pi(\frac{1}{24})^2\times 50 \times 62.5 =\frac{275}{1008}\times 62.5 ~pounds$

We have $Ft=momentum \implies F =\frac{Mass \times velocity}{t}=\frac{\frac{275}{1008}\times 62.5\times 50 }{1}=852.5545635$

I am not getting the desired result for the first part and unable to solve the 2nd part. Please help me to solve. Thanks in advance.

Answer 1

Notice, the force exerted on the fixed plate in the direction of the jet line$$=\text{rate of change of momentum of water striking the plate}$$$$=\text{mass/sec} \times\text{(velocity before striking-velocity after striking)}=(\rho a V)(V-0)=\rho aV^2$$ Where, density of water, $\rho=1000\ kg/m^3$

Speed of water jet, $V=50\times \frac{30.48}{100}=15.24\ m/sec$

area of cross-section of nozzle, $a=\frac{\pi}{4}(2.54)^2\times 10^{-4}=5.0671\times 10^{-4}\ m^2$ Now, setting all the values in the above expression, we get the force (F) exerted by the jet on the fixed plate $$F=1000\times 5.0671\times 10^{-4}\times (15.24)^2$$ $$=\color{blue}{117.6872 \ \text{N (newtons)}}$$ $$=117.6872\times 0.2248 $$ $$=\color{blue}{26.4561 \ \text{lbf (pounds-wt)}}$$ We can calculate pressure $$=\frac{\text{force}}{\text{area of plate}}=\frac{F}{A}$$

Now, if the nozzle is drawn backwards then the relative velocity of the water jet w.r.t. fixed plate $$V=\text{velocity of jet}-\text{velocity of nozzle}=50-20=30\ ft/sec$$$$=30\times 0.3048=9.144\ m/sec$$ Now, setting the value of velocity of jet w.r.t. plate in the above expression, we get force exerted on the plate $F$ $$F=1000\times 5.0671\times 10^{-4}\times (9.144)^2$$$$=\color{blue}{42.3674 \ \text{N (newtons)}}$$$$=42.3674\times 0.2248$$$$=\color{blue}{9.5242 \ \ \text{lbf (pounds-wt)}}$$ We can calculate pressure $$=\frac{\text{force}}{\text{area of plate}}=\frac{F}{A}$$

Answer 2

The main source of confusion here is the definition of a "pound".

You say the mass of water that strikes the plate each second is $$(\text{volume of water per second}) \times 62.5\ \text{pounds}.$$

But actually that is the weight of the water. The mass of the water, which you need in order to correctly compute $\frac{\Delta mv}{t}$, is the $m$ in the formula $F = ma$; when discussing the weight of water, the weight is a force given by $F = mg$. Hence $m = \frac{62.5}{g}\ \text{slugs}$.

The exact solution depends on the values one takes for various constants; $62.5$ is slightly high for the density of water in pounds per cubic foot, and the answer you are "supposed" to get seems to depend on converting lbm to slugs using the approximation $g \approx 32$, whereas $32.174$ would be more accurate.

At any rate, if you assume $g = 32 \ \text{ft}/\text{s}^2$, then the answer you calculated would be off by a factor of $32$, having failed to divide $62.5$ by $g$ when calculating mass. Indeed, $852.5545635 / 32 = 26.64$ to four significant digits, so that explains what happened to your calculation.

All this business with lbf vs. lbm vs. slugs is (to me) is a good argument for doing the whole problem in metric, and only converting from N to lbf at the end, much more compelling than the notion that the conversion factors are much easier in metric (all being powers of $10$ rather than other numbers such as $12$ inches per foot).