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Contents:
  1. Quick Start
  2. Water Service Pipe Lines
  3. Water Pipeline Engineer jobs

The maximum amount which can be put in block BM is determined b y drawing a closed loop u s i n g occupied blocks as corners see the dotted c i r c u i t i n Table 3. Now f o r each u n i t which i s added to block BM, one u n i t would have to be subtracted from block BN, added to block AN and subtracted from block AM to keep the yields and requirements consistent. In this case the maximum allocation to BM is 10, since this would evacuate block AM.

After making the best new allocation, re-calculate the e v a l u a t i o n number and evaluation sums as in Table 3. Allocate resource to the most p r o f i t a b l e block and repeat the r e - d i s t r i b u t i o n procedure until there is no further possible cost improvement, indicated by the fact that there i s no e v a l u a t i o n sum g r e a t e r than the cost coef- ficient in any block.

In our example we arrived at the optimum distribution in two steps, but more compl icated patterns involving more sources and consumers may need many more attempts. The example can o n l y serve to introduce the subject of t r a n s p o r t a - tion programming.

Quick Start

There are many other conditions which are dealt with in textbooks on the subject of mathematical optimization techn- niques such as Van der Veen and Dantzig and this example o n l y serves as an introduction. For instance, if two blocks in the table happened to be evacuated simultaneously, one of the blocks could be allocated a very small quantity denoted by 'e' say. Computat ions then proceed as before and the q u a n t i t y 'e' disregarded at the end.

L i n e a r programming i s one of the most powerful optimization tech- niques. The use of a computer is normally essential for complex systems, although the simple example given here is done by hand.

Water Service Pipe Lines

The technique may o n l y be used if the r e l a t i o n s h i p between v a r i a b l e s is linear, so it is restricted in application. Linear programming cannot be used for optimizing the design of pipe networks with closed loops without resort to successive approximations. It can be used to design trunk mains or tree-like networks where the flow in each branch is known. Since the relationships between flow, head loss, diameter and cost are non-l inear, the following technique is used to render the system linear: For each b r a n c h o r main p i p e , a number of pre-selected diameters a r e allowed and the length of each p i p e of different diameter is treated as the v a r i a b l e.

The program will indicate that some diameter pipes h a v e zero lengths, thereby i n effect e l i m i n a t i n g them. Any other type of linear constraint can be treated in the analysis. I t may be r e q u i r e d to m a i n t a i n the pressure at certain points in the network above a f i x e d minimum a I inear inequality of the greater-than-or-equal-to-type or within a certain range.

The example concerns a trunk main w i t h two drawoff points Fig. There are thus four v a r i a b I es , X,, X2, X3 and X4 which are the lengths of pipe of dif- ferent diameters.

This simple example could be optimized b y manual comparison of the costs of all alternatives giving the correct head loss, but linear programming is used here to demonstrate the tech- nique. The linear constraints on the system are expressed in equation form below and the coefficients of the equations are tabulated in Table 3. Lengths a r e expressed i n h u n d r e d metres.


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The computations proceed by setting all real variables to zero, so it i s necessary to introduce artificial slack variables into each equation to satisfy the equality. To i n i t i a t e the solution, the slack variables a, b a n d c a r e assigned the values 5, 4 and 5 respective- l y see the t h i r d column of Table 3. The X2 column is now designated the key column. The key column is that which shows the lowest opportunity value i n the cost mini- mization case.

To determine the maximum amount of the key column variable which may be introduced, calculate the replacement ratios for each row as follows:- Divide the amount of the program variable for each row by the corresponding number i n the key column. The lowest p o s i t i v e replace- ment ratio is selected as that is the maximum amount which could be introduced without v i o l a t i n g any of the constraints. The row with the lowest positive replacement ratio is designated the key row a n d the number at the intersection of the key column and key row, the key number.

After introducing a new variable, the matrix is rearranged T a b l e 3. The program v a r i a b l e and i t s cost coefficient in the key row a r e replaced by the new variable and its cost coefficient. From each number in a non-key row, subtract the corresponding number i n the key row m u l t i p l i e d b y the r a t i o of the o l d row number in the key column divided by the key number. The procedure of studying opportunity values and replacement ratios and revising the table is repeated until there is no further negative opportunity value.

In the example Table 3. Van der Veen, a n d D a n t z i g f o r a full des- cription of the technique. There are many other cases which can be mentioned below:- If the constraints are of the 5 less-than-or-equal-to type and not just equations, slack variables with zero cost coefficients are introduced into the 1. The artificial slack variables with high cost coef- f i c i e n t s a r e then omitted.

If the constraints are of the 2 greater-than-or-equal-to type, introduce artificial slack variables with high cost coefficients into the 1. If the objective function is to be minimized, the opportunity value with the highest negative value is selected, but if the function is to be maximized, the opportunity value with the h i g h e s t p o s i t i v e v a l u e i s selected.

The opportunity values represent shadow values of the corres-. If two replacement ratios a r e equal, whichever row i s selected, the amount of program variable in the other row will be zero when the matrix is rearranged. Merely assume it to have a very s m a l l v a l u e a n d proceed as before. Non-linear programming and search methods are also discussed. For instance. Buras, N. Cross, H. Isaacs, L. Lam, C. Wood, D. Transient pressures caused by a change of flow r a t e in conduits are often the cause of bursts. Transients in closed conduits are normally classed into two cate- gories: Slow motion mass oscillation of the fluid is referred to a surge, whereas rapid change in flow accompanied by elastic strain of the fluid and conduit is referred to as w a t e r hammer.

For slow or small changes in flow rate or pressure the two theories yield t h e same r e s u l ts. It i s normally easier to a n a l y s e a system b y r i g i d column theory whenever the theory is applicable than by elastic theory. A pressure difference applied across the ends of the column produces an instantaneous acceleration. The basic equa- tion relating the head d i f f e r e n c e between t h e e n d s of the water col- umq in a uniform bore conduit to the rate of change in velocity i s d e r i v e d from Newton's b a s i c l a w o f motion, a n d is.

Valves in Pipeline - Gate, Pressure relief valves etc. - Environmental Engineering

The equation is useful for calculating the head rise associated with slow deceleration of a water column. I t may be used f o r calcu- lating the water level variations in a surge shaft following power trip or starting up in a pumping line, or power load changes in a hydro-electric installation fed by a pressure pipeline. The equa- tion may be solved in steps of At b y computer, in tabular form o r graphical ly.

Example A m l o n g penstock w i t h a c r o s s s e c t i o n a l a r e a , A,, o f 1 m2 is protected against water hammer b y a surge s h a f t a t the tur- bine, with a cross sectional area, A2, of 2 m2 and an unrestricted orifice. Calculate the maximum rise in water level in the surge shaft neglecting friction. Then from Equ. Following the stopping of a pump at the upstream end of a pumping line, the pressure frequently drops sufficiently to cause vaporization at peaks along the line.

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In such cases the water column beyond the vapour pocket will decelerate slowly and r i g i d column theory is s u f f i c i e n t l y accurate for analysis. Thus if a valve at the d i s c h a r g e end of a pipeline is shut the fluid upstream of the gate will pack against the gate, causing a pressure rise. The pressure will rise sufficiently to stop the liquid in accordance with the momentum l a w. The amount o f water stopped p e r u n i t time depends o n the amount of water required to replace the volume created by the compression of w a t e r a n d e x p a n s i o n o f t h e p i p e.

The pressure wave caused by the valve closure referred to earlier, thus travels upstream, superimposed on the static head, as illustrated in Fig. When the wave front reaches the open reservoir end, the pressure in the pipe forces water backwards into the reservoir, so that the velocity now reverses and pressure drops b a c k to s t a t i c r e s e r v o i r p r e s s u r e a g a i n. A negative wave 2 thus travels downstream from the r e s e r v o i r. That wave front will in t u r n reach the closed end. Now the velocity in the entire pipeline is -v where v was the original velocity. The variation in head at the v a l v e will be as indicated i n Fig.

In the case of pumping lines, the most violent change i n flow conditions i s normally associated with a pump trip.

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Water Pipeline Engineer jobs

The water down- stream of the pump is suddenly decelerated, resulting in a sudden reduction in pressure. The negative wave travels towards the dis- charge end Fig. The pressure at the pump alter- nates from a pressure drop to a pressure r i s e.