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Pani-rosa [81]
2 years ago
10

Technicians have to determine the flow rate of water in a pipe with the aid of a venturi installation and a mercury au.oaeter. T

he expected water welocity is about 1.5 a/s. The pipe diameter is 40 mm and the venturi throat has a diameter of 20 mm. The venturi meter has a discharge coefficient C = 0.97. Determine what level difference can be expected in the manometer.
Engineering
1 answer:
Schach [20]2 years ago
7 0

Answer:

Manometric difference x=142.85 mm.

Explanation:

Given :

 Pipe diameter d_1=40 mm

venturi meter d_2=20 mm

We can know that discharge through venturi meter is given as

Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt {A_1^2-A_2^2}}

A_1=1.24\times 10^{-3},A_2=3.12\times 10^{-4}

Q=A_1V_1

Q=1.24\times 10^{-3}\times 1.5=0.00186 m^3/s

0.00186=0.97\dfrac{1.24\times 10^{-3}\times 3.12\times 10^{-4} \sqrt{2gh}}{\sqrt {(1.24\times 10^{-3})^2-(3.12\times 10^{-4})^2}}

h=1.8 m

We know that h=x\left (\dfrac{\rho_{hg}}{\rho_w}-1\right )

Where x is the manometric deflection

⇒ 1.8=x\left (\dfrac{13600}{1000}-1\right )

So x=14.28 mm

Manometric difference x=142.85 mm.

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Answer:

  as soon as there is a design to improve

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Then, I started that step again when my boss told me, "make it better."

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The more interesting question is, "when do you <em>stop</em> the design improvement step?" (Judging by the constant barrage of software updates, that answer is, "never.")

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Answer:

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Explanation:

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dW = F*dr

For pressure the force F equials the pressure multiplied by the area of the piston, and considering dx as the displacement:

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dW = - P*dV

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Considering the gas as ideal, the pressure can be calculated as P = \frac{n*R*T}{V}, so:

\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV

Replacion this and solving equation (1) between state 1 and 2:

\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV

Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})

Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}

The internal energy depends only on the temperature of the gas, so there is no internal energy change U_{2} - U_{1} = 0, so the heat exchanged to the system equals the work done by the system:

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Answer:

Answer explained below

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--------------------------------------------------------------------------------

Psuedo Code:

countSort(int numList[]) {

int count[10000];

count[i] = 0; for all i;

for(int num in numList){

count[num]+= 1;

}

return count;

}

--------------------------------------------------------------------------------

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E.g. Lets say we want to search if 3 was present in the original list.

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