Ratio Maximum Velocity To Average Velocity
Ratio Maximum Velocity To Average Velocity Assignment Help  Ratio Maximum Velocity To Average Velocity Homework Help
Ratio of Maximum Velocity to Average Velocity
The velocity will be maximum to y = ½.
U_{max} =  1/2μ ∂p / ∂x [ t x t / 2  (t / 2)^{2}] =  1/2μ ∂p / ∂x [ t^{2 }/ 2  t^{2} / 4 ]
U_{max} =  1/ 8μ ∂p / ∂x t^{2}
Average velocity u = Q/A
dQ = Velocity at a distance y x Area of strip
=  1/2μ ∂p / ∂x [ ty  y^{2} ] x dy x 1
Integrating from 0 to t
t t
Q = ∫ dQ = ∫  1/2μ ∂p / ∂x [ ty  y^{2} ] dy x 1
0 0
Q =  1/2μ ∂p / ∂x [ t ( y^{2} / 2)_{0}^{t}  (y^{3} / 3)_{0}^{1} ] =  1/2μ ∂p / ∂x [ t^{3} / 2  t^{3 }/ 3 ]
Q =  1/2μ ∂p / ∂x t^{3 }/ 6
Q =  1/ 12μ ∂p / ∂x t^{3}
u = Q / A =  1/12μ . ∂p / ∂x . t3 / t x 1
u =  1/ 12μ ∂p / ∂x t^{2}
Ratio Umax to u
U_{max } =  1 / 8μ (∂p / ∂x) t^{2 }/ 1/ 12μ ∂p / ∂x t^{2}

u
U_{max } = 3 / 2

u
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U_{max} =  1/2μ ∂p / ∂x [ t x t / 2  (t / 2)^{2}] =  1/2μ ∂p / ∂x [ t^{2 }/ 2  t^{2} / 4 ]
U_{max} =  1/ 8μ ∂p / ∂x t^{2}
Average velocity u = Q/A
dQ = Velocity at a distance y x Area of strip
=  1/2μ ∂p / ∂x [ ty  y^{2} ] x dy x 1
Integrating from 0 to t
t t
Q = ∫ dQ = ∫  1/2μ ∂p / ∂x [ ty  y^{2} ] dy x 1
0 0
Q =  1/2μ ∂p / ∂x [ t ( y^{2} / 2)_{0}^{t}  (y^{3} / 3)_{0}^{1} ] =  1/2μ ∂p / ∂x [ t^{3} / 2  t^{3 }/ 3 ]
Q =  1/2μ ∂p / ∂x t^{3 }/ 6
Q =  1/ 12μ ∂p / ∂x t^{3}
u = Q / A =  1/12μ . ∂p / ∂x . t3 / t x 1
u =  1/ 12μ ∂p / ∂x t^{2}
Ratio Umax to u
U_{max } =  1 / 8μ (∂p / ∂x) t^{2 }/ 1/ 12μ ∂p / ∂x t^{2}

u
U_{max } = 3 / 2

u
For more help in Ratio of Maximum Velocity to Average Velocity click the button below to submit your homework assignment