Chapter 5 - 6

Question 1

Hl due to sudden expansion?

Answer 1

-Use the Bernoulli equation
- momentum conservation
-sub the Bernoulli equation into momentum to get
hL = (V1 - V2)^2 / (2g)
k = ( 1 - A1/A2 )^2 (minor loss coefficient)
hl = k * (V1^2) / (2g)

Question 2

Hl due to sudden contraction?

Answer 2

-Use the Bernoulli equation
- momentum conservation
-sub the Bernoulli equation into momentum to get
hL = (V1 - V2)^2 / (2g)
k = ( 1/C1 - 1) (Where C is the contraction coefficient)
hL = k * (V2^2) / (2g)

Question 3

What are the values of the minor loss coefficient in a:
90 elbow?
T bend?
Y bend?
Valve?

Answer 3

90, k = 0.2

T bend, k = 0.2 straight & k= 1 perpendicular

Y, both k = 0.4

Valve, k (1/4 closed) = 0.26, k (1/2) = 2.1, k (3/4) = 17 &
k (closed) = infinity.

Question 4

What is:
the total energy gradient?
the Hydraulic gradient?

Answer 4

Total energy gradient is the line of which the total head of the water in the pipe has: P/(pg) + v^2/(2g) + Z

The hydraulic gradient is just the elevation and pressure head of the water in the pipe: P/(p*g) + Z

Question 5

How do pipes in the parallel act?

Answer 5

Head loss in the pipes parallel to each other is always equal no matter what the dimensions are.

The sum of the flowrates in parallel equals the overall flowrate.

Question 6

How do siphons work?

What needs to happen for them to work?

Answer 6

Water travels upwards in the pipe to get over a hump and then back down into a different reservoir. The pressure is lower at the top of the hump so water flows there. Then when it gets to the top it can only flow down to the lower reservoir.

The pipe has to be full constantly and there is a maximum height the hump can be for the siphon to work. (usually lower than theoretical as you lose pressure from water flowing through a pipe.)

Question 7

What is the Hardy Cross method?

Answer 7

It’s a trial and error method to find the flow rates in a network of pipes. Hl = R * Q * |Q|
Where R = (8YL) / (pi^2 * g * D^5).

Question 8

How do you use the hardy cross method?

Answer 8

• number pipes and circuits and assume sign convention.
• draw table
• Obtain Ho from H = R * Q * |Q|
• Calculate Ho/Qo
• Calculate Sum of Ho
• Calculate the Sum of Ho/Qo
• Calculate dqo = - (sum Ho) / (2 * sum Ho/Qo)
• Repeat for further circuits.
• Apply correction factors to Q (e.g. Q1 = Q0 + dq, add both correction factors to the pipe that shares two systems).
• repeat the process until applied corrections are insignificant.