Question

A conical funnel of half-angle θ = 30 drains through a small hole of diameter d = 6:25 mm at the vertex. The speed of the liquid leaving the funnel is V= √ 2gy where y is the height of the liquid free surface above the hole. The funnel initially is filled to height y0 = 300 mm. Obtain an expression for the time, t, for the funnel to completely drain, and evaluate. Find the time to drain from 300 mm to 150 mm (a change in depth of 150 mm), and from 150 mm to completely empty (also a change in depth of 150 mm). Can you explain the discrepancy in these times?

Answer 1

`3175.424*y^3 + 0.45153*y + 55.552*y^2 = t + 12.0347055`

t = 12.03

t = 81.473

velocity of the fluid decreases with level of fluid. Hence, in later stages the time taken is greater than in earlier stages

Step-by-step explanation:

Given:

- The half angle θ = 30°

- The diameter of the small hole d = 6.25 mm

- The flow rate out of the funnel Q = A*√ 2gy

- The volume of frustum of cone is given by:

`V = (\pi )/(12)*y*(D^2 + d^2 + d*D})`

Where,

D: Is the larger diameter of the frustum

d: Is the smaller diameter of the frustum

y: The height of the liquid free surface from small diameter d base.

Find:

- Obtain an expression for the time, t, for the funnel to completely drain, and evaluate.

- Find the time to drain from 300 mm to 150 mm (a change in depth of 150 mm) and from 150 mm to completely empty (also a change in depth of 150 mm)

- Can you explain the discrepancy in these times?

Solution:

- We will use rate of change analysis by considering the rate of change of volume, and then apply the chain rule.

- The Volume of the frustum is a function of d , D , y. V = f ( d , D , y ). In our case the diameter of base d remains constant. Then we are left with:

V = f ( D , y )

- Determine a relationship between y and d. We will use half angle θ to determine the relationship between D and y by applying trigonometric ratios:

tan ( θ ) = D / 2*y

D = 2*y*tan ( θ )

- We developed a relationship between D and y in terms of half angle which remains constant. So the Volume is now only a function of one variable y:

V = f ( y )

- The volume of frustum of the cone can be written as:

`V = (\pi )/(12)*y*(D^2 + d^2 + d*D})`

Substituting the relationship for D in terms of y we have:

`V = (\pi )/(12)*y*(4*y^2*tan(Q) ^2 + d^2 + 2*d*y*tan(Q)})`

- Now by rate of change of Volume analysis we have:

dV / dt = [dV / dy] * [dy / dt]

- Computing dV / dy, where V = f(y) only:

`V = (\pi )/(12)*(4*y^3*tan(Q) ^2 + y*d^2 + 2*d*y^2*tan(Q)})\\\\(dV)/(dy) = (\pi )/(12)*(12*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q)})\\\\`

- Where, dV/dt = Volume flow rate:

`(dV)/(dt) = - Q\\(dV)/(dt) = - A*V\\(dV)/(dt) = - (\pi*d^2 )/(4) *√(2*g*y)`

- Then from Chain rule we have:

[dy / dt] = [dV / dt] / [dV / dy]

`(dy)/(dt) = (- (\pi*d^2 )/(4) *√(2*g*y))/((\pi )/(12)*(12*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q))\\\\} \\(dy)/(dt) = (-d^2 *√(2*g*y))/((4*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q))\\\\}`

- Separate variables:

`\frac {(4*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q)} {√(2*g*y)} .dy = {-d^2} .dt\\\\\frac { 1 } { √(2*g) }*(4*y^1^.^5*tan(Q) ^2 + y^-^0^.^5d^2 + 4*d*y^0^.^5*tan(Q))= {-d^2} .dt\\`

- Integrate both sides:

`\frac { 1 } { √(2*g) }*(4*y^1^.^5*tan(Q) ^2 + y^-^0^.^5d^2 + 4*d*y^0^.^5*tan(Q))= {-d^2} .dt\\\\\frac { 1 } { √(2*g) }*((8)/(5)*y^2^.^5*tan(Q) ^2 + 2*d^2*y^0^.^5 + (8)/(3)*d*y^1^.^5*tan(Q)) = -d^2*t + C\\\\\frac { 1 } { √(2*y*g) }*((8)/(5)*y^3*tan(Q) ^2 + 2*d^2*y + (8)/(3)*d*y^2*tan(Q)) = -d^2*t + C\\\\0.1204*y^3 + 0.000017638*y + 0.00217*y^2 = -0.0000390625*t + C\\\\3175.424*y^3 + 0.45153*y + 55.552*y^2 = -t + C`

- Evaluate @ t = 0 , y = 0.3 m

`3175.424*(0.15)^3 + 0.45153*(0.15) + 55.552*(0.15)^2 = 0 + C\\\\C = 12.0347055\\\\3175.424*y^3 + 0.45153*y + 55.552*y^2 = t + 12.0347055`

- Time taken from y = 300 to 150 mm:

`3175.424*0.15^3 + 0.45153*0.15 + 55.552*0.15^2 = -t + 90.871587\\\\t = 0 - -12.0347055\\\\t = 12.0347055 s\\`

- Time taken from y = 150 to 0 mm:

t = t_(0-300) - ( t_(0-150) = 90.871587 - 12.0347055 = 81.4721 s

- The discrepancy is time can be explained by the velocity of fluid coming out of bottom is function of y. The velocity of the fluid decreases as the level of fluid decreases hence the time taken from y =150 mm to 0 mm is larger than y = 300 mm to y = 150 mm.