Drawing Water Pressure Diagram Along Pipeline

A

,  H

t  l / c

, at BB will be:

V

B

t  l / c

B

,  H

t  l / c

. Thus, at each cross-section, the water pressure value changes as follows:

At AA:  H A

 H A c ( VA ​V A

) (*)

t  l/ c

t g t

t  l/ c

At BB:

 H B

 H B c ( V B V B

) (**)

t  l/ c

t g t

t  l/ c

In the two formulas above, the right side has a minus sign because this is a reverse wave. If we omit

Through friction loss, the water is transmitted without deformation, so:

 H B  H A and

tt  l/ c

V B V A

Transform and add the two formulas (*) and (**) together and rearrange them to get:

tt  l/ c

 H A  H B c ( VA V B )

(14-9)

tt  l/ c g t

t  l / c

Equation (14-9) is the equation for retrograde wave propagation from BB to AA.

* . Consider the case of a forward wave from A to B (Figure 14-5,b):

Figure (14-5,b) is a diagram to determine the equation of the forward water wave chain. On this diagram, line (1) is lower than line (2) because this is a forward pressure increasing wave. The symbols and explanations to establish the formula are similar to the case of the reverse wave, except that the direction is reversed. There is also a change in the water pressure at each cross-section A and B as follows:

At AA:

 H A  H A 

c ( VA ​V A

) (*')

t  l / ct

g tt  l / c

At BB:  H B  H B c ( V B V B

) (**')

t  l/ct gt

t  l/ c

In the two formulas above, the right side has a plus sign because this is a forward wave. Also omit

Through friction loss, the water is transmitted without deformation, so:

 H A  H B and

tt  l/ c

V A V B

. Transform and add the two formulas (*') and (**') together and rearrange them to get

tt  l/ c

The forward wave propagation equation (14-10) from AA to BB is as follows:

 H B  H A

( V B V A

) (14-10)

c

tt  l / c

g tt  l / c

b. Indirect water wave equation, relative value

For ease of calculation, the absolute water impact values ​​(14-9) and (14-10) are converted to relative water impact values ​​(dimensionless) in the following way:

Set  H

H 0

called relative water pressure value; H 0 is the static water column without water impact;

VQ

v q

is the relative velocity and flow;

V max

  c Q max

2g H 0 F

Q max

cV max

2g H 0

is the pipeline inertia coefficient.

Divide both sides of equations (14-9) and (14-10) by

H 0 and the right side multiplied by

score

V max , finally by placing the dimensionless quantities above we have the equation

V max

The relative values ​​for the two cases of water wave propagation are as follows:

- Equation of retrograde wave transmission from B to A:

 A  B 2 ( v A v B )

(11-14)

tt  l / ctt  l / c

- Equation of forward wave transmission from A to B:

 B  A 2 ( v B v A )

(14-12)

tt  l / ctt  l / c

General application to the transmission of wave chains on pipelines,

The half-water phase difference is 

= l/c. So we have the system of equations for chain wave propagation

n 

n 

v

B

Reverse wave propagation:  A

B



(n  1) 

2 ( v A

(n  1) )

(14-13)

Forward wave transmission:

THREE





n (n  1) 

2 ( v B

(n  1) )

(14-14)

n 

v

A

Equations (14-13) and (14-14) are called wave propagation equations, or chain equations because based on them we can determine the water pressure values ​​at successive half-phases when knowing the boundary conditions and initial conditions.

XIV. 2. 2. Calculation of water impact by analytical method

The absolute or relative water wave propagation equations can be used in combination with boundary conditions or initial conditions to calculate the water pressure. Here we present how to use the relative wave propagation equations (14-13) and (14-14) to calculate with two methods: analytical and graphical. In this section we use the analytical method to solve.

1. Boundary conditions and initial conditions

To solve the water impact problem by analytical method, we first determine the

Boundary conditions and initial conditions at two cross sections AA and BB of the pipe.

n 

- At cross-section BB, where it contacts the pressure tank or large reservoir, the water surface oscillation is almost constant, so it is considered that there is no water pressure, meaning:  B 0 ;

- At section AA in front of the turbine: to find the exact boundary conditions here, it is necessary to study the opening and closing mode of the flow direction mechanism or needle valve over time:

2gH A

* For impulse turbines , the opening and closing rules follow the relationship

Q A 

, so the relative velocity at AA at the end of the first phase will be:

Q A 2 2g( H 0  H A )

1  A

2 

1  A

1

v

A 2  2  Q max

2 

2g H 0

max 2 

 1

And similarly we have the boundary condition at AA at the end of any nth phase will be:

Q A

1  A

2n 

1  A

n

v A 2n 

  2n 

 n

, or:

2n Q max

n

v A  n

(14-15)

1  A

n

n

In which: v A

- relative velocity at AA at the end of phase n;

n

 A - relative water pressure at AA at the end of phase n.

 n - relative opening of the needle valve at the end of phase n.

* For counter-attack turbine : the on-off rule at the end of phase n will be:

1 2n 

H 0 H A

Q A Q ' D 2

H 0

v

A

2n 

2n 

Q max

12n 

Q

D

' 2

1 max 1

, flow through the turbine during the process

The transition is very complicated, and it is necessary to rely on the characteristic curve of the turbine to determine the flows.

Derivative amount

Q 'corresponding to the CCHD apertures a o

corresponding to specific water columns

1

(or

n ' specifically). To solve analytically in a very approximate way, the Italian scientist

1

Levi hypothesized that: the change in the flow rate

Q ' is proportional to the flap opening

1

Q ' a 0 2n 

a o current direction , that is:12n 2n  n

(*); where  n is the relative aperture

Q

'

max

a 0 max

for the flow direction mechanism at the end of the nth phase. So the formula (*) can be in the following form of

1  A

n

The boundary condition at AA at the end of the nth phase returns to formula (14-15) above:

nn

v A 

(14-15)

- Initial condition: at time t = 0, water has not yet hit, so the characteristics H, Q at the pipe cross-sections are all in stable mode. If we ignore hydraulic losses, then

The pressure line is horizontal, which means  H = 0 so

 A  B and

v A v B . On the other hand when

0 0

0 0

t

The wave arrives at BB at a very close time t = L/c, so we can consider it as v A v B v B .

v B has not had time to change,

0 0 t

2. Analytical method for simple pipelines

A simple pipe is a pipe whose diameter, thickness and pipe material are constant throughout the length of the pipe and do not branch. We use the chained equations (14-3) and (14-4) and the initial and boundary conditions (14-15) to solve in turn to determine the relative water pressure at the end of the pipe at the end of each phase:

2 

 1  A ,

 2   A ,

 3  A , ...,

A





n 2n 

when knowing the needle valve opening and closing mode

4 

6 

(t) or the vane opening a 0 (t) (Figure 14-6).

Figure 14-6. Relative opening and closing mode of water direction flap.

2 

a. Determine the relative water pressure at the end of the first phase (n = 1)  1  A :

Write the equation for forward wave propagation from A to B (equation 14-3):

 B A2 ( v  Bv A)

(*)



2 2 

Based on the boundary conditions at BB we have

 B 0

and the boundary condition at AA we have

v A 2 

; Based on the initial condition at B we have

v B v A  0

 0 ,

1  A

2 

1  A

2 

2 0

1  A

0

2 

So replace them in (*) we have:  A

2 (  0

2 

) and use the symbol at the end

Phase 1: Derive the equation to determine the relative water pressure solution at the end of the first phase:

1  1



1 0

1

2 

(14-16)

4 

b. Determine the relative water pressure at the end of phase n.

3 

4 

- Find the water pressure at the end of the second phase (n = 2)

 2   A : we write the equation

Forward wave from A to B we have:

THREE

1  A

4 





v

3 

3 4 

2 ( v B

v A

) , based on boundary conditions at B

and A we have:

 B 0 and

A   4 

. Need to find more

B write equation

3 

v

4 

Transmitting the wave back from B to A we have:

AB

1  A

2 





3 



2 3 

2 ( v A

v B

) . With

 B 0 and

2 

3 

3 

2 

Boundary condition at A has: v A

  2 

, substitute and extract v B

we have solution:

1   2





2 1

(14-17)

2 0 2 

- In a similar way, determine the solution for phases 3, 4, ... We have

2n 

general solution formula for the end of nth phase (  n  A

) any:

1  n

 n 1n  1

 n 0 2 

 i i  1

(14-18)

c. Water enters the primary phase and the limiting phase

In calculating and installing pressure pipes, the issue people are concerned with is determining

determine the maximum positive water pressure value to calculate the pipe strength and value

Minimum negative water pressure to check pipe placement to avoid vacuum in

So here we look at this practical issue.

Through actual calculation and operation of the pressure pipeline of the power plant, it is seen that the maximum water pressure  max falls into one of two cases: falls at the end of the first phase, that is,  max  1 (figure 14-7,a) or falls into the final phase, called the limiting phase, i.e.

 max  m

(Figure 14-7,b). Next we will determine the water values ​​and these two phases.

Water and first phase  max  1

When opening and closing the valve or the flow deflector, the water pressure reaches its maximum value at the end of the first phase, the following phases have smaller values. Therefore, we only need to calculate the water pressure value at the end of this phase. From the solution equation at the end of the first phase (14-16) square 2

and solve we have the solution: 

2 ⎡(  02 ) 

⎤(*)

(  02 ) 2( 22 )

1

0 1

1 ⎢ ⎣1 ⎥ ⎦

Need to choose the sign of the root of (*). If you choose the root with the (+) sign in (*) then if

turbine closing from initial relative opening

 0 1 (full load) to fully closed (  1 0 ) then

  2    2   c V max   H max  

This is absurd. So the solution would be:

1 0 g H 0 H 0

(  02 ) 2( 22 )

1

0 1

  2  ⎡ (  0    2 ) 

max

⎤(14-19)

1 ⎢ ⎣1 ⎥ ⎦

The first phase of water collision usually occurs in high water column (usually H  150 - 250 m).

Figure 14-7. Diagram of water entering the first phase and the limiting phase.

Water and phase limit  max  m

During the turbine opening and closing process, the water pressure increases gradually and reaches its maximum value.

into the final phase

 max (figure 14-7,b), that is

 max  m , called water and the limiting phase. To

To establish the formula for calculating water pressure and limiting phase, we use the solution (14-18) written for phase (m-1) and phase m as follows:

1  m  1

 m  1

1m 2

 m  1

 0 

2   

 i i  1

(*)

1  m

 m 1m  1

 m  0 

2   

 i i  1

(**)

From figure (14-7,b) we can see that

 m  1  m

and take the corresponding sides of (*) and (**)

Subtracting each other we have:

(  m  m  1 )

1 

1  m

m

. Consider the opening and closing process according to the rules.

linear law then relative aperture difference



m  1

 m 