8/10/2019 Eat 239 (Pipe Network) Example 4 Solution

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8/10/2019 Eat 239 (Pipe Network) Example 4 Solution

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3 

Notes:

The flow rate in each pipe above is the assumed value and we wanted the ACTUAL Q. The

correction for the assumed value can be calculated by using following formula:

∑

∑

 

Thus, we need to find summation of h for each pipe and the ratio of h/Q.

Also,

The signage value of δQ obtained is referred to magnitude. Meaning, if you are getting a –ve

value of δQ. Then, the ACTUAL value of Q is actually less than your assumed value. 

DO NOT CONFUSE with Q. 

***************************************************************************

STEP 2: Calculate the K for all pipe in both Loop 1 and 2. K can be calculated from

following formula: 

K= fL3.03D5

 

Comment: I think calculation of K is straight forward.

***************************************************************************

STEP 3: Calculate energy losses due to friction, h by applying following formula:

h=KQ 

Comment: Bit tricky as you already successfully making me confused. Do rememberthat, -ve value of Q is referring to direction of Q (anti-clock wise) and thus, for –Q, the

value of h also would be –ve. We can say that head loss, h are the same direction with

flow (That why –Q would have –h). Also, the unit for h is m & thus Q needed to be in

m3 /s.

***************************************************************************

STEP 3: Calculate the ratio h/Q.

Comment: All should be +ve value. As h and Q for EACH pipe should be in the same

direction.

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4 

Step 4: Started with Iteration 1 

By now, you should be able to developed table something like this:

LOOP  PIPE 

D  L 

K 

Assumed Q   h 

h/Q   Q new 

mm  m  L/s  m 

1 

AC  500  3000  317  400  50.7  126.7  420.1 

CD  350  2000  1257  50  3.1  62.0  94.0 

DB  350  3000  1885  ‐150  ‐42.4  282.7  129.9 

BA  500  2000  211  ‐400  ‐33.8  84.5  379.9 

SUMMATION:  ‐22.4  556.8 

2 

CE  250  3000  10139  150  228.1  1520.8  126.1 

EF  250  2000  6759  50  16.9  338.0  26.1 

FD  250  3000  10139  ‐100  ‐101.4  1013.9  123.9 

DC  350  2000  1257  ‐50  ‐3.1  62.8  ‐94.0 

SUMMATION: 

140.6  2935.4 

Notes:

1) 

The corrected assumed value, Q NEW = Q (initially assumed) + δQ 

2)  The equation above applicable for all pipes in Loop 1 and 2, except for pipe CD (or

DC). For this pipe, the corrected value δQ has to be considered for both loops as the

 pipe was in both loops. The formula for calculation as follow:

δQCD = δQ1 - δQ2 

3) 

Then you should proceed with the Second Iteration by using your new Q obtained inthe First Iteration as your newly assumed Q.

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5 

4)  The method above should be repeated until δQ = 0 for both loop (or at least 3 iteration

during the final). Your friend asked me, what happened if after 1st iteration, we already

got δQ=0 for both loops. Answer: It is quite impossible for us to correctly guess the

flow rate on our first assumption. BUT, if this happened, I would do at least up to 2

iterations.

Kindly refer to the following for marking scheme.

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7 

2 

CE  250  3000  10139  150  228.1  1520.8  126.1 

EF  250  2000  6759  50  16.9  338.0  26.1 

FD  250  3000  10139  ‐100  ‐101.4  1013.9  123.9 

DC  350  2000  1257  ‐50  ‐3.1  62.8  ‐94.0 

SUMMATION:  140.6  2935.4 

Iteration 2

LOOP  PIPE 

D  L 

K 

Q   h 

h/Q   Qnew 

mm  m  L/s  m 

1 

AC  500  3000  317  420.1  55.9  133.1  416.0 

CD  350  2000 

1257  94.0  11.1  118.2  89.7 

DB 

350 

3000 

1885 

129.9  ‐

31.8 

244.9  ‐

134.0 

BA  500  2000  211  379.9  ‐30.5  80.2  ‐384.0 

4.7  576.4 

2 

CE  250  3000  10139  126.1  161.1  1278.2  126.3 

EF  250  2000  6759  26.1  4.6  176.2  26.3 

FD 

250 

3000 

10139  123.9  ‐155.7  1256.5  ‐123.7 

DC 

350 

2000 

1257  ‐94.0  ‐11.1  118.2  ‐89.7 

‐1.1  2829.0 

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