Force on a plate due to air jet

[ank160]

Suppose there is a plate with area A, on which jet of air with velocity 'v' is striking, then force on it will be = (density) * A* v^2

But if we go by basics i.e. F = d(mv)/dt = v d(m)/dt + m d(v)/ dt

here i am not able to figure out that why the second term i.e. m d(v)/ dt is zero, because when it becomes zero then only we will get the formula that i have written in second line.

Plz help.

 

[boneh3ad]

You are also assuming that the jet is the same size as the plate (or larger than the plate). If it was smaller than the plate, then the A term would represent the area of the jet.

Anyway, look it a different way: as a conservation of linear momentum in a control volume. Initially, the rate of change of the momentum in the jet coming into this control volume is

$$\frac{d(\rho v)}{dt} = \rho \frac{dv}{dt} + v \frac{d \rho}{dt} = v \frac{d \rho}{dt} = \frac{F_{\textrm{plate}}}{V}$$

Where $\frac{dv}{dt}=0$ because of the fact that $v$ coming into the control volume is not changing (the jet is coming at the plate at a constant $v$). This rate of change of momentum per volume $V$ is equal to the force on the plate $F_{\textrm{plate}}$ per unit volume because inside this control volume, the flow is decelerated to zero x-velocity (or whatever direction you choose to be normal to the plate).

So the next step is to look at $\frac{d \rho}{dt}$.

$$\frac{d \rho}{dt} = \frac{1}{V}\frac{d m}{dt} = \frac{1}{V}\rho v A$$

Plugging that back into the original momentum balance gives you

$$\frac{1}{V}\rho v^2 A = \frac{F_{\textrm{plate}}}{V}$$

Which simplifies to

$$F_{\textrm{plate}} = \rho v^2 A$$