PE Exam

[cantaloup]

You are working for United Developers, Inc. (UDI). UDI just acquired a tract of land at the dead-end of Sunny Rd. which connects to Lamar Rd., a main road. The length of Sunny Rd is 650 ft. There is a 12-in water main along Lamar Rd. An existing 6-in water main connects the 12-in water main to a fire hydrant located 250 ft from the dead-end of Sunny Rd. The length of the 6-in water main section is 400 ft; a section of 4-in water line runs from the fire hydrant to the dead-end. The UDI vice-president wants to build a duplex on the newly acquired tract of land. Here comes the problem: According to the previous owner of the tract of land, the fire flow test data on the fire hydrant shows that the fire hydrant delivers an adequate fire flow of 1000 gpm at a pressure of 24.60 psi during summer peak hour. The town ordinance requires that a fire flow of 1500 gpm at 20 psi minimum must be provided if UDI wants to build a duplex on the tract of land. With a flow of 1500 gpm at the fire hydrant, the pressure is dropping below 20 psi; therefore UDI vice-president wants you to calculate the length of 6-in pipe (the section connected to 12-in pipe on Lamar Rd) which must be replaced with 8-in pipe so that fire flow will be adequate; given C=100, pipe contraction coefficient minor loss: K=0.5, no change in elevation along the route.

The length (in feet) of 8-in pipe to replace the existing 6-in pipe is most nearly:

A/ 216

B/ 230

C/ 242

D/ 253

PE_question_pipe_in_series_find_L_2_.pdf

 

[3gorgesdam]

Is this for afternoon module?

 

[cantaloup]

Last PE exam people said that the WR module is easy! Yeah right! Attached is a file for the same pipe problem which has been simplified. If you guys want to take the PE exam with WR depth exam in Oct. and don't understand this problem then need to study ch17 CERM again. Answer will be posted in 48 hrs.

PE_question_only_pipe_in_series_find_L_2__simplified.pdf

 

[cantaloup]

You guys gave up? Heh heh heh

Here is the answer:

SOLUTION:

This is ?pipe in series? problem; Q is the same through out, total head loss is the sum of head loss of each section.

Note: For regular water, one psi of pressure is equivalent to 2.31 ft of head

Q1 = 1000gpm * 0.002228 = 2.228 cfs; Q2 = 1500 gpm * 0.002228 = 3.34 cfs ;

Head @ Point A = head at point C plus head loss along existing 6-in pipe.

Applying H-W formula: 24.6psi*2.31 + 4.72 * 2.23^1.85 * 400 / (100^1.85 * 0.5^4.87) = 105.40 ft

Velocity in the 6-in pipe, section BC: VBC = 3.34^2/(0.5^2*3.14/4) = 17.02 fps

Minor loss of pipe contraction: HL = 0.5*V^2/2*g = 0.5*17.02^2/2*32.2 = 2.25 ft

Total friction loss allowed with min. pressure: HT = 105.40ft ? 20psi*2.31-2.25 = 56.97ft

From H-W formula applied to both sections:

X is the length of 8-in section and 400-X is the length of 6-in section

HT = 4.72 * Q^1.85 * X / (C^1.85 * D1^4.87) + 4.72 * Q ^1.85 * (400 ? X) / (C^1.85 * D2^4.87)

D1 = 8? = 0.67 ft ; D2 = 6? = 0.50 ft

Flow in both pipe: Q2 = 3.34 cfs

Plug & chug => X = 234.1 ft .Answer is C (length must be > 234.1ft).

QUITE EASY, RIGHT?

NOTE: Hydraulics modeling such as EPANET can be used to find answer.

 

[Blu1913]

Where did:

For regular water, one psi of pressure is equivalent to 2.31 ft of head

Come from?

And the is an afternoon example, yes?

 

[JasonT33]

If you dont know that 1 psi = 2.308 ft of water, you may want to consider taking another afternoon module other than WR.

Pressure varies linearly with depth. This is fundamental. Relationship between pressure and hydrostatic head for fluid (incompressible)

p = y*h y = weight density

 

[cantaloup]

Regular water means municipal water at 60 F, specific weight gamma = 62.37lbf/ft3

The energy equation: P/gamma + V squared /2*g + z = ?.

If P is given in psi then 1 psi = 144 psf , then one psi = 144psf / 62.37 lbf /ft3 = 2.31 ft

They would give this in WR afternoon, not likely for morning.

Remember though, PE exam questions are created by engineers in working, not by professors in university, this is the problem I encounter once in a while.

In practice, for municipal water distribution system, engineers use H-W formula more often, for storm drainage they use Manning formula. D-W formula is used for general cases dealing with all liquid at a wide range of temp and density.

 

[cantaloup]

Another note: If in the question they give value of f (friction factor) instead of C value, you guys must use the formula :

Head loss = f x L/D x V2/(2*g) instead of the H-W formula

You guy should know it by now , PE exam is only a few weeks away.