Force on a plate due to air jet

AI Thread Summary
The force exerted on a plate by an air jet is calculated using the formula F = ρAv², where ρ is the air density, A is the plate area, and v is the jet velocity. The discussion clarifies that the term m d(v)/dt equals zero because the velocity of the jet remains constant as it strikes the plate. By applying conservation of linear momentum, the rate of change of momentum in the jet is equated to the force on the plate. The analysis shows that the change in density over time can be expressed in terms of mass flow rate and area, leading back to the original force equation. This comprehensive approach confirms the relationship between the air jet's properties and the force on the plate.
ank160
Messages
21
Reaction score
0
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.
 
Physics news on Phys.org
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
 
comparing a flat solar panel of area 2π r² and a hemisphere of the same area, the hemispherical solar panel would only occupy the area π r² of while the flat panel would occupy an entire 2π r² of land. wouldn't the hemispherical version have the same area of panel exposed to the sun, occupy less land space and can therefore increase the number of panels one land can have fitted? this would increase the power output proportionally as well. when I searched it up I wasn't satisfied with...
Back
Top