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4. Two elevated tanks supply water to a demand node with a valve at its outlet (Figure 4). The elevated tanks are cylindrical and have diameters of 5 m. The initial water level in Tank 1 is 96 m and the initial water level in Tank 2 is 89 m. The valve is half open and has a discharge coefficient of 0.15 m^5/2/s. The initial steady-state flow through the valve is 350 L/s. The valve discharges to the atmosphere. Both pipes have a Hazen-Williams 'C' factor of 100, an internal diameter of 250 mm, and a length of 300 m. Assuming quasi-steady conditions in the system, determine the pressure head at the demand node and the flow in the pipes in the first three time steps of the simulation. Use a time step of 15 seconds to carry out the quasi-steady state simulation. Tank 1 HGL Tank 2 96 m 89 m H L=300 m L=300 m Q

          4. Two elevated tanks supply water to a demand node with a valve at its outlet (Figure 4). The elevated tanks are cylindrical and have diameters of 5 m. The initial water level in Tank 1 is 96 m and the initial water level in Tank 2 is 89 m. The valve is half open and has a discharge coefficient of 0.15 m^5/2/s. The initial steady-state flow through the valve is 350 L/s. The valve discharges to the atmosphere. Both pipes have a Hazen-Williams 'C' factor of 100, an internal diameter of 250 mm, and a length of 300 m. Assuming quasi-steady conditions in the system, determine the pressure head at the demand node and the flow in the pipes in the first three time steps of the simulation. Use a time step of 15 seconds to carry out the quasi-steady state simulation.

Tank 1
HGL
Tank 2
96 m
89 m
H
L=300 m
L=300 m
Q

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4. Two elevated tanks supply water to a demand node with a valve at its outlet (Figure 4). The elevated tanks are cylindrical and have diameters of 5 m. The initial water level in Tank 1 is 96 m and the initial water level in Tank 2 is 89 m. The valve is half open and has a discharge coefficient of 0.15 m^5/2/s. The initial steady-state flow through the valve is 350 L/s. The valve discharges to the atmosphere. Both pipes have a Hazen-Williams 'C' factor of 100, an internal diameter of 250 mm, and a length of 300 m. Assuming quasi-steady conditions in the system, determine the pressure head at the demand node and the flow in the pipes in the first three time steps of the simulation. Use a time step of 15 seconds to carry out the quasi-steady state simulation.

Tank 1
HGL
Tank 2
96 m
89 m
H
L=300 m
L=300 m
Q

Added by Elizabeth L.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sri K

02/26/2023

Step 1

First, we need to find the head loss in each pipe using the Hazen-Williams equation: $H_f = 10.67 \times \frac{L}{C^2 \times D^{4.87}} \times Q^{1.85}$ where $H_f$ is the head loss, $L$ is the length of the pipe, $C$ is the Hazen-Williams factor, $D$ is the Show more…

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Two elevated tanks supply water to a demand node with a valve at its outlet (Figure 4). The elevated tanks are cylindrical and have diameters of 5 m. The initial water level in Tank 1 is 96 m and the initial water level in Tank 2 is 89 m. The valve is half open and has a discharge coefficient of 0.15 m^3/s. The initial steady-state flow through the valve is 350 L/s. The valve discharges to the atmosphere. Both pipes have a Hazen-Williams factor of 100, an internal diameter of 250 mm, and a length of 300 m. Assuming quasi-steady conditions in the system, determine the pressure head at the demand node and the flow in the pipes in the first three time steps of the simulation. Use a time step of 15 seconds to carry out the quasi-steady state simulation. Tank 2 Tank 1 HGL 89 m 96 m L-300 m L-300 m

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Transcript

00:01 As in the given question, according to the question which we have f of x is equal to sine x.

00:09 So here sign h x is 1 to 1 function and it has an inverse inverse which is sine h inverse x is equal to m within the bracket we have x plus square root of x squared plus 1 so we will take this as our equation first.

00:47 So at the first step to solve our first part of the question we will take y is equal to sine h x and therefore sine h x is a gate route passing through the point which are x comma y and which is equal to 0 .0 the orison.

01:14 So therefore we can construct our diagram accordingly.

01:19 These are the coordinate x -axis, and here we have y -axis.

01:26 So our first curve will be like this.

01:34 And secondly, i'm sorry, this is not a curve.

01:37 Okay, now fine.

01:39 So this is our first curve.

01:44 Let me draw again like this.

01:49 And on the other hand our second line will be like this way.

01:56 So here we have our value for y is equal to sine hx and this is for y is equal to sine inverse x.

02:07 And similarly we have here y is equal to sine h inverse x and this is y is equal to some h x.

02:17 This is a point of a reason.

02:19 So or we can say that sign h x is equal to e to the power of x minus e to the power of minus x divided by 2.

02:31 When x is greater than 0 then e to the power of minus x will be equal to or similar to 1.

02:43 Difference between e to the power of minus x and 1 is less than 1 always.

02:49 So this way we will get sine h x similar to e to the power of x...

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