The perpendicular force to flow exerted by a jet on stationary inclined flat plate = _____ , when velocity of jet = 20m/s, angle between jet and plate = 60° and area of cross section of jet = 0.01 m²

Concept:

Vr = relative velocity of the jet with respect to the plate

• When the plate is moving (u) in the direction of the jet (V) i.e. away from the jet: Vr = V - U
• When the plate is moving (U) in the opposite direction of the jet (V) i.e. towards the jet: Vr = V + U

Force exerted by the jet on the plate in the direction normal to the plate:

\(\begin{array}{l} {F_n} = \dot m\left[ {{V_{1n}} - {V_{2n}}} \right] = \rho A{V_r}\left[ {{V_r}\sin \theta - 0} \right]\\ {F_n} = \rho AV_r^2\sin \theta \end{array}\)

Force exerted by the jet on the plate in the direction of the jet (Fx) and is perpendicular to the direction of flow (Fy)

\(\begin{array}{l} {F_x} = {F_n}\sin \theta = \rho AV_r^2{\sin ^2}\theta \\ {F_y} = {F_n}\cos \theta = \rho AV_r^2\sin \theta \cos \theta \end{array}\)

Calculation:

Given, velocity of jet = 20 m/s 

Angle between jet and plate = 60° and area of cross-section of jet = 0.01 m²

Flate plate is stationary, Then

Relative velocity of the jet with respect to the plate(V_r) = 20 - 0 = 20 m/s

The perpendicular force to flow exerted by a jet on a stationary inclined flat plate,

Fy = ρAV_r2 sin θ cos θ

F_y = 1000 × 0.01 × 20² × sin 60 × cos 60

F_y = 1732.05 N.