Momentum Thickness
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Momentum Thickness
It is the distance measured perpendicular to the solid boundary, by which boundary will be displaced to compensate for the reduction in momentum of the flowing fluid due to boundary layer formation. It is denoted by 'θ'.
Derivation:
Momentum of fluid/sec = Mass / Sec x Velocity
= (ρ. u b dy) u
= ρ u2 b. dy
Momentum of fluid if no plate is placed (u = V),
= (ρ. u b dy) U
Loss of momentum/sec through strip = ρ u b. dy U - ρ u2 b dy = ρ b u (U - u) dy
Total loss of momentum/sec through BC,
δ
= ∫ ρ b u (U - u) dy
0
Let ( ) be the distance by which plate is displaced, when fluid flowing with constant velocity.
Loss of momentum/sec at a distance θ with velocity U
= Mass of flow through θ x Velocity = (ρ x θ x b x U). U
= ρ θ b U2
Equating Equations
δ
= ρ θ b U2 = ∫ ρ b u (U - u) dy
0
δ
θ = ∫ u ( U - u) dy / U2
0
δ
θ = ∫ u / U (1- u/U) dy
0
For more help in Momentum Thickness click the button below to submit your homework assignment
= (ρ. u b dy) u
= ρ u2 b. dy
Momentum of fluid if no plate is placed (u = V),
= (ρ. u b dy) U
Loss of momentum/sec through strip = ρ u b. dy U - ρ u2 b dy = ρ b u (U - u) dy
Total loss of momentum/sec through BC,
δ
= ∫ ρ b u (U - u) dy
0
Let ( ) be the distance by which plate is displaced, when fluid flowing with constant velocity.
Loss of momentum/sec at a distance θ with velocity U
= Mass of flow through θ x Velocity = (ρ x θ x b x U). U
= ρ θ b U2
Equating Equations
δ
= ρ θ b U2 = ∫ ρ b u (U - u) dy
0
δ
θ = ∫ u ( U - u) dy / U2
0
δ
θ = ∫ u / U (1- u/U) dy
0
For more help in Momentum Thickness click the button below to submit your homework assignment