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Pani-rosa [81]
3 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]3 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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Below is the required code.

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%% Newton Raphson Method

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Command Window:

(a) First Root:

Initial guess:

1.5

Condition of convergence:

0.01

x = -1.815662 and error = 3.315662 at iteration = 1.000000

x = -0.644115 and error = 1.171547 at iteration = 2.000000

x = 0.208270 and error = 0.852385 at iteration = 3.000000

x = 0.434602 and error = 0.226332 at iteration = 4.000000

x = 0.451631 and error = 0.017029 at iteration = 5.000000

x = 0.451732 and error = 0.000101 at iteration = 6.000000

The solution x = 0.451732

>>

Second Root:

Initial guess:

3.5

Condition of convergence:

0.01

x = 3.300299 and error = 0.199701 at iteration = 1.000000

x = 3.305650 and error = 0.005351 at iteration = 2.000000

The solution x = 3.305650

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(b) Guess x=0.5:

Initial guess:

0.5

Condition of convergence:

0.01

x = 0.450883 and error = 0.049117 at iteration = 1.000000

x = 0.451732 and error = 0.000849 at iteration = 2.000000

The solution x = 0.451732

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Guess x=1.75:

Initial guess:

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Condition of convergence:

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x = 227.641471 and error = 225.891471 at iteration = 1.000000

x = 218.000998 and error = 9.640473 at iteration = 2.000000

x = 215.771507 and error = 2.229491 at iteration = 3.000000

x = 217.692636 and error = 1.921130 at iteration = 4.000000

x = 216.703197 and error = 0.989439 at iteration = 5.000000

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The solution x = 216.971251

>>

Guess x=3.0:

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Condition of convergence:

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x = 3.309861 and error = 0.309861 at iteration = 1.000000

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The solution x = 3.305651

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Guess x=4.7:

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x = -1.916100 and error = 1.051861 at iteration = 240.000000

x = -0.748896 and error = 1.167204 at iteration = 241.000000

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x = 0.428332 and error = 0.265602 at iteration = 243.000000

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x = 0.451732 and error = 0.000187 at iteration = 245.000000

The solution x = 0.451732

>>

Explanation:

The two solutions are x =0.451732 and 3.305651 within the range 0 < x< 5.

The initial guess x = 1.75 fails to determine the solution as it's not in the range. So the solution turns to unstable with initial guess x = 1.75.

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