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

Answer 1

Answer:

Answer:

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

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:

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

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:

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

- Where, dV/dt = Volume flow rate:

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

- Then from Chain rule we have:

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

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

- Separate variables:

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

- Integrate both sides:

$\frac { 1 } { √(2*g) }*(4*y^1^.^5*tan(Q) ^2 + y^-^0^.^5d^2 + 4*d*y^0^.^5*tan(Q))= {-d^2} .dt\n\n\frac { 1 } { √(2*g) }*((8)/(5)*y^2^.^5*tan(Q) ^2 + 2*d^2*y^0^.^5 + (8)/(3)*d*y^1^.^5*tan(Q)) = -d^2*t + C\n\n\frac { 1 } { √(2*y*g) }*((8)/(5)*y^3*tan(Q) ^2 + 2*d^2*y + (8)/(3)*d*y^2*tan(Q)) = -d^2*t + C\n\n0.1204*y^3 + 0.000017638*y + 0.00217*y^2 = -0.0000390625*t + C\n\n3175.424*y^3 + 0.45153*y + 55.552*y^2 = -t + C$

- Evaluate @ t = 0 , y = 0.3 m

$3175.424*(0.15)^3 + 0.45153*(0.15) + 55.552*(0.15)^2 = 0 + C\n\nC = 12.0347055\n\n3175.424*y^3 + 0.45153*y + 55.552*y^2 = t + 12.0347055$

- Time taken from y = 300 to 150 mm:

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

- 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.

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Kim is working on the cost estimate and feasible design options for a building. Which stage of a construction plan is Kim working on now? A. design development B. schematic design C. mechanical D. structural

Answers

Kim is working on the cost estimate. The stage of a construction plan is Kim working on now is schematic design. The correct option is B.

What is a schematic design?

A schematic design is an outline of a house or a building or another construction thing. The schematic design makes the outline map of the exterior or interior of the building. It is the foremost phase of designing something.

The design expert discusses the project three-dimensionally at this point in the process. To define the character of the finished project and an ideal fulfillment of the project program, a variety of potential design concepts are investigated.

The schematic design consists of a rough sketch with markings and measurements.

Therefore, the correct option is B. schematic design.

To learn more about schematic design, refer to the link:

brainly.com/question/14959467

#SPJ5

Answer:

B. schematic design

Explanation:

This correct for Plato/edmentum

What is the difference between absolute and gage pressure?

Answers

Explanation:

Step1

Absolute pressure is the pressure above zero level of the pressure. Absolute pressure is considering atmospheric pressure in it. Absolute pressure is always positive. There is no negative absolute pressure.

The expression for absolute pressure is given as follows:

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Here, $P_(ab)$ is absolute pressure, $P_(g)$ is gauge pressure and $P_(atm)$ is atmospheric pressure.

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___________ are used in an automotive shop and can be harmful to the environment if not disposed of properly.Oil filters
Fluorescent lamps
Mercury-containing lamps
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Answers

Answer: D all above

Explanation:

Jus done it

Air is contained in a cylinder device fitted with a piston-cylinder. The piston initially rests on a set of stops, and a pressure of 300 kPa is required to move the piston. Initially, the air is at 100 kPa and 27°C and occupies a volume of 0.4 m^3. A) Determine the amount of heat transferred to the air, in kJ, while increasing the temperature to 1200 K. Assume air has constant specific heats evaluated at 300 K.

Answers

Answer:

The amount of heat transferred to the air is 340.24 kJ

Explanation:

From P-V diagram,

Initial temperature T1 = 27°C

Initial pressure P1 = 100 kPa

final pressure P3 = P2 = 300 kPa

volume at point 2, V2 = V1 = 0.4 m³

final temperature T2 = T3 = 1200 K

To determine the final pressure V3, use ideal gas equation

PV = mRT

Where R is the specific gas constant = 0.2870 KPa m³ kg K

But,

from initial condition, mass m = PV/RT

m = (P1*V1)/R*T1

T1 = 27+273 = 300K

m = (100*0.4)/(0.2870*300) = 0.4646 kg

Then;

Final volume V3 = mRT3/P3

V3 = (0.4646*0.2870*1200)/300

V3 = 0.5334 m³

Total work done W is determined where there is volume change which from point 2 to 3.

W = P3*(V3-V2)

W = 300*(0.5334-0.4) = 40.02 kJ

To get the internal energy, the heat capacity at room temperature Cv is 0.718 kJ/kg K

∆U = m*Cv*(T2-T1)

∆U = 0.4646*0.718(1200-300)

∆U = 300.22 kJ

The heat transfer Q = W + ∆U

Q = 40.02 + 300.22 = 340.24 kJ

Determine the amount of heat transferred to the air, in kJ, while increasing the temperature to 1200 K is 340.24 kJ

The attached file shows the Pressure - Volume relationship (P -V graph)

Calculate the "exact" alkalinity of the water in Problem 3-2 if the pH is 9.43.Calculate the "approximate" alkalinity (in mg/L as CaCO3 ) of a water containing 120 mg/L of bicarbonate ion and 15 mg/L of carbonate ion.

Answers

Answer:

A) approximate alkalinity = 123.361 mg/l

B) exact alkalinity = 124.708 mg/l

Explanation:

Given data :

A) determine approximate alkalinity first

Bicarbonate ion = 120 mg/l

carbonate ion = 15 mg/l

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B) calculate the exact alkalinity of the water if the pH = 9.43

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9.43 + pOH = 14. therefore pOH = 14 - 9.43 = 4.57

[OH^- ] = 10^-4.57 = 2.692*10^-5 moles/l

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A power plant burns natural gas to supply heat to a heat engine which rejects heat to the adjacent river. The power plant produces 800 MW of electrical power and has a thermal efficiency of 38%. Determine the heat transfer rates from the natural gas and to the river, in MW.

Answers

A. The heat transfer rate from natural gas is 2105.26 MW

B. The heat transfer rate to river is 1305.26 MW

Efficiency formula

Efficiency = (power output / power input) × 100

A. How to determine the heat transfer from natural gas

Efficiency = 38%
Power output = 800 MW

Power input =?

Power input = Power input / efficiency

Power input = 800 / 38%

Power input = 800 / 0.38

Power input = 2105.26 MW

Thus, the heat transfer from natural gas is 2105.26 MW

B. How to determine the heat transfer to the river

Total heat = 2105.26 MW
Heat used by plant = 800 MW
Heat to the river =?

Heat to the river = 2105.26 – 800

Heat to the river = 1305.26 MW

Learn more about efficiency:

brainly.com/question/2009210

Answer:

heat transfer from natural gas is 2105.26 MW

heat transfer to river is 1305.26 MW

Explanation:

given data

power output Wn = 800 MW

efficiency = 38%

solution

we know that efficiency is express as

$\eta = (Wn)/(Qin)$ ......................1

put here value we get

38% = $(800)/(Qin)$

Qin = 2105.26 MW

so heat supply is 2105.26

so we can say

Wn = Qin - Qout

800 = 2105.26 - Qout

Qout = 2105.26 - 800

Qout = 1305.26 MW

so heat transfer from natural gas is 2105.26 MW

and heat transfer to river is 1305.26 MW

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