Engineering, 26.02.2020 03:29 Jofish5815

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?

Answers: 1

Show answers

Answers

Answer from: gianna61

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

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:

$V = \frac{\pi }{12}*y*(D^2 + d^2 + d*D})$

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:

$V = \frac{\pi }{12}*y*(D^2 + d^2 + d*D})$

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

$V = \frac{\pi }{12}*y*(4*y^2*tan(Q) ^2 + d^2 + 2*d*y*tan(Q)})$

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

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

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

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

- Where, dV/dt = Volume flow rate:

$\frac{dV}{dt} = - Q\\\frac{dV}{dt} = - A*V\\\frac{dV}{dt} = - \frac{\pi*d^2 }{4} *\sqrt{2*g*y}$

- Then from Chain rule we have:

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

$\frac{dy}{dt} = \frac{- \frac{\pi*d^2 }{4} *\sqrt{2*g*y}}{\frac{\pi }{12}*(12*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q)}\\\\} \\\frac{dy}{dt} = \frac{-d^2 *\sqrt{2*g*y}}{(4*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q)}\\\\}$

- Separate variables:

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

- Integrate both sides:

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

- 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.

Answer from: Quest

answer:

200k

explanation:

Answer from: Quest

answer:

Romans

Explanation:

Hope it helps

Another question on Engineering

Engineering, 04.07.2019 18:10

Steel is coated with a thin layer of ceramic to protect against corrosion. what do you expect to happen to the coating when the temperature of the steel is increased significantly? explain.

Answers: 1

Answer

Engineering, 04.07.2019 18:10

Afluid flows with a velocity field given by v=(x/t)i.. determine the local and convective accelerations when x=3 and t=1.

Answers: 2

Answer

Engineering, 04.07.2019 18:10

Which of the following ziegler nichols tuning methods the response of the controller to a step input should exhibit an s-shaped curve? a)-open loop mode b)-closed loop mode c)-both modes (open & closed) d)-none of the modes (open & closed)

Answers: 3

Answer

Engineering, 04.07.2019 18:10

Courses that are developed by subject matter experts, internal or extemal to the college or university. these programs are marketed by the school (clo2) marks a)-vocational schools b)-vendor training c)-colleges & universities d)-continuing education programs

Answers: 2

Answer

You know the right answer?

A conical funnel of half-angle θ = 30 drains through a small hole of diameter d = 6:25 mm at the ver...

Questions