Ratio Of Maximum Velocity To Average Velocity
Ratio Of Maximum Velocity To Average Velocity Assignment Help  Ratio Of Maximum Velocity To Average Velocity Homework Help
Ratio of Maximum Velocity to Average Velocity
The maximumvelocity occurs at centre where r = 0. Put in Equation
U_{max} = 1/4μ ∂p / ∂x R^{2}
Mean or Average velocity is obtained by dividing the discharge of the fluid across the corss sectional area of pipe (πr^{2}).
Te discharge (Q) across the section is obtained by considering the flow through a circular ring element of radius ‘r’ and thickness ‘dr’ as shown in.
dQ = Velocity at a radius r x Area of ring element
= u x 2 πr dr =  1/4μ ∂p / ∂x [ R^{2}  r^{2 }] x 2 πr dr
R R
Q = ∫ dQ =  1/4μ (∂p / ∂x ) x 2π ∫ (R^{2}  r^{2}) r dr
0 0
R
= 1/4μ (  ∂p / ∂x )2π ∫ ( R^{2} r  r^{3}) dr
0
= 1/4μ ( ∂p / ∂x )2π (R^{4} / 2  R^{4} / 4) = 1/4μ ( ∂p / ∂x )2π R^{4} / 4
Q = π / 8μ ( ∂p / ∂x ) R^{4}
Average velocity,

U_{ave } = Q / A = π / 8μ ( ∂p / ∂x ) R^{4} / πR^{2}

U_{ave} = 1 / 8μ ( ∂p / ∂x )R^{2}
Ratio of maximum and average velocity
U_{max } =  1 / 4μ (∂p / ∂x )R^{2} / 1 / 8μ ( ∂p / ∂x )R^{2}
U_{max } = 8/4 = 2

U
Ratio of maximum velocity to average velocity is 2.
For more help in Ratio of Maximum Velocity to Average Velocity click the button below to submit your homework assignment
U_{max} = 1/4μ ∂p / ∂x R^{2}
Mean or Average velocity is obtained by dividing the discharge of the fluid across the corss sectional area of pipe (πr^{2}).
Te discharge (Q) across the section is obtained by considering the flow through a circular ring element of radius ‘r’ and thickness ‘dr’ as shown in.
dQ = Velocity at a radius r x Area of ring element
= u x 2 πr dr =  1/4μ ∂p / ∂x [ R^{2}  r^{2 }] x 2 πr dr
R R
Q = ∫ dQ =  1/4μ (∂p / ∂x ) x 2π ∫ (R^{2}  r^{2}) r dr
0 0
R
= 1/4μ (  ∂p / ∂x )2π ∫ ( R^{2} r  r^{3}) dr
0
= 1/4μ ( ∂p / ∂x )2π (R^{4} / 2  R^{4} / 4) = 1/4μ ( ∂p / ∂x )2π R^{4} / 4
Q = π / 8μ ( ∂p / ∂x ) R^{4}
Average velocity,

U_{ave } = Q / A = π / 8μ ( ∂p / ∂x ) R^{4} / πR^{2}

U_{ave} = 1 / 8μ ( ∂p / ∂x )R^{2}
Ratio of maximum and average velocity
U_{max } =  1 / 4μ (∂p / ∂x )R^{2} / 1 / 8μ ( ∂p / ∂x )R^{2}
U_{max } = 8/4 = 2

U
Ratio of maximum velocity to average velocity is 2.
For more help in Ratio of Maximum Velocity to Average Velocity click the button below to submit your homework assignment