Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 4, Problem 4.33P
To determine

(a)

The expression for radial flow velocity.

Expert Solution
Check Mark

Answer to Problem 4.33P

The expression for radial flow velocity is Vr=4Qπrb.

Explanation of Solution

Given information:

The volume flow rate of incompressible flow towards a drain is Q, the area of the wedge is A, the effect of friction and gravity are neglected, the width of the wedge is b, the angle of the wedge is 45°, the flow is assumed to be purely radial.

Write the expression for the velocity of radial flow.

vr=QA ... (I)

Here, volume flow rate of fluid is Q, area of the wedge is A.

Write the expression for the area of the wedge

A=θ2π2πrb ... (II)

Here, angle of the wedge is θ, the distance from the drain where the velocity is calculated is r, the width of the wedge is b.

Calculation:

Substitute π4 for θ in Equation (II)

A= π 4 2π2πrbA=π4rb

Substitute π4rb for A in Equation (I).

vr=Qπ4 rb=4Qπ rb

Conclusion:

The radial velocity at distance r from the wedge is vr=4Qπrb.

To determine

(b)

The value of viscous term in the momentum equation in r direction is zero.

Expert Solution
Check Mark

Explanation of Solution

Write the expression for the velocity of radial flow in r direction

vr=4Qπrb

Substitute C=4Qπb in Equation (II).

vr=Cr

Write the expression for viscous term from momentum Equation.

V=v2vr v r r 2 2 r 2 v θ θ=1r r r v r r v r r 2 2 r 2 v θ θ ... (III)

Here, the viscous term is V, velocity in r direction is vr, the velocity in θ direction is vθ.

Calculation:

Substitute Cr for vr, 0 for 2r2vθθ in Equation (III)

V=1r r r C r r +C r 3 0=1r r r C r r +C r 3 0=1r r r C r 2 +C r 3 0=1r r C r +C r 3 0

V=C r 3 +C r 3 0=0

Conclusion:

Thus, the viscous term in momentum equation in r direction is 0.

(c)

To determine

The pressure distribution p(r) at boundary condition p=pο, r=R.

(c)

Expert Solution
Check Mark

Answer to Problem 4.33P

The pressure distribution p(r) at boundary condition p=pο, r=R is p=pο+ρC221R21r2.

Explanation of Solution

Given information:

The boundary condition for the pressure distribution is p=pο, the boundary condition for distance from the drain is r=R.

Write the expression for maximum normal shear stress.

vrt+vrvrr1rvθ2=1ρpr+gr+V (IV)

Here the viscous term is V, The change in radial velocity with respect to time is vrt, The change in radial velocity with respect to r distance is vrr, acceleration due to gravity in r direction is gr, the change in pressure with respect to distance r from sink is 1ρpr.

Calculation:

Substitute 0 for V, 0 for gr, 0 for vrt, 0 for 1rvθ2 in Equation (IV).

vrvrr=1ρpr ... (V)

Substitute Cr for vr in Equation (V).

Cr C r r=1ρprρCr C r r=prpr=ρCr×C r 2 p=ρC2r3r ... (VI)

Integrate Equation (VI).

dp=ρ C 2 r 3 rdp=ρC2 1 r 3 rp=ρC21 2 r 2 p=ρC22r2+K ... (VII)

Here, K is integrating constant

Substitute boundary condition pο for p, R for r in Equation (VII)

pο=ρC22R2+KK=pο+ρC22R2

Substitute K in Equation (VII)

p=ρC22r2+pο+ρC22R2=pο+ρC22R2ρC22r2=pο+ρC221 R 2 1 r 2

Conclusion:

The pressure distribution p(r) at boundary condition p=pο, r=R is p=pο+ρC221R21r2 ..

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Chapter 4 Solutions

Fluid Mechanics

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