October 30

Exam Question 29 Pump Horsepower

What is the minimum size of motor needed for a 2 x 1 pump to deliver 100 gallons per minute (gpm) of water to a water skimmer that is located 50 feet above the pump and has an operating pressure of 35 psig. The water pressure on the suction side of the pump is 5 in Hg vacuum. Assume standard wall thickness.
A) 100
B) 75
C) 25
D) 50

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June 27

Exam Problem #25 (The Reynolds Number)


Which of the following statements about the Reynolds Number Are FALSE?

I Is a dimensionless number
II If the Reynolds Number is great than 2000, the flow is laminar
III As the hydraulic diameter of the pipe increases so does the Reynolds Number
IV The greater the viscous forces, the greater the Reynolds Number

A) All of the statements are false
B) II & IV
C) I & IV


Today’s engineering exam question looks at what the Reynolds Number is and what factors can and can not change the Reynolds Number.


The general equations are:

Re = \frac {\rho v D_h}{\mu} = \frac {v D_h}{\nu } = \frac {Q D_h}{\nu A}


ν = kinematic viscosity
μ = dynamic viscosity
Dh = Hydraulic Diameter (also known as equivalent diameter)
v = velocity
A = Area
ρ = density


The Reynolds Number (Re) is a dimensionless number that is the ratio of inertia forces to viscous forces. When this number is less than 2,000, the flow is laminar. When the number is greater than 4,000 the flow is considered turbulent.
I – The Reynolds number is a dimensionless number. This statement is true.
II – When the Reynolds number is less that 2000 the flow is laminar. This statement is false
III – Since the hydraulic diameter is in the numerator. An increase will result in a larger Reynolds Number. This statement is true.
IV – As the viscous forces increase the Reynolds number decreases. This statement is false.
Since II and IV are false, the correct answer is B
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June 19

Exam Problem #19 (Gas Velocity in a Pipe)


Your client specs state that the velocity of the gas shall not exceed 60 feet per second. You are sizing a line that has a design capacity of 3,000,000 cubic feet per day with a pressure of 125 psig. When the temperature of the gas is at 80 F the compressibility of the 0.966.
What is the smallest diameter schedule 40 diameter pipe that can be used?
A) 6 Inch
B) 3 Inch
C) 2 Inch
D) 4 Inch


Today’s PE/EIT question of the day calculates the minimum size a pipe can be used to keep the speeds below 60 feet per second by calculating the gas velocity in a pipe. This is done by using the gas velocity equation in API RP 14e


Gas Velocity Equation

V\ =\ \frac{60\ Z\ Q_g\ T}{d^2\ P_{inlet}}

Where V is the velocity in feet per second, Q is the flow rate in mmscfd, d is the inner diameter of the pipe, T is the temperature in Rankine, P is the pressure in psia, and Z is the compressibility. 60 is a conversion factor that adjust for the differences in units and also takes into account the base temperature and pressure of 60F and 14.7 psia


Temperature (T) 80 F
Pressure (P) 125 psig
Compressibility (Z) 0.966
Flow Rate (Q) 3 mmscfd
Max Velocity (V) 60 ft/s


First we need to rewrite the equation to solve for the inner diameter

d\ =\ \sqrt{\frac{60\ Z\ Q_g\ T}{V\ P_{inlet}}}

d\ =\ \sqrt{\frac{60 (0.966) (3)(540)}{(60)(125+14.7)}} = 3.35\ inches

The smallest Sch 40 pipe that has an inner diameter of 3.35 inches is a 4 inch diameter.
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Choice D is correct

June 18

Exam Problem #18 (Gas Flow Rate)


Natural gas (SG = 0.7) flows through a 20 inch diameter schedule 80 pipe. The pipe is 75 miles long with an inlet pressure of 1400 psig and an outlet pressure of 400 psig. The average temperature of the gas is 60º F. If a compressibility (Z) of 0.85 and a Moody Friction Factor of 0.013 is assumed.
What is the daily flow rate?
A) 228.2 MMSCFD
B) 158.6 MMSCFD
C) 173.8 MMSCFD
D) 1460 MMSCFD


Today’s EIT/PE exam problem comes from the oil and gas industry and shows how to get an estimate of the gas flow rate of natural gas through a pipeline. This problem uses the General Pressure Drop Equation from API 14e (equation 2.6) which was based on of Bernoulli’s equation.


General Pressure Drop Equation

P^2_1 - P^2_2 = 25.2 \frac{(SG)(Q^2)ZTLf}{d^5}

Rearranged to solve for flow rate

Q = 0.199 \sqrt {\frac {d^5(P^2_1 - P^2_2)}{ZTL(SG)f}}

d is the inner diameter of the pipe in inches, P1 and P2 are the inlet and outlet pressures in psia, Z is compressibility, f is the moody friction factor, T is the temperature in Rankine, L is the length of pipe in feet, Q is the flow rate in MMSCFD,


Specific Gravity (SG) 0.7
Pipe inner Diameter (d) 17.938 inches
Inlet Pressure (P1) 1400 psig
Outlet Pressure (P2) 400 psig
Average Temperature (T) 60º F
Compressibility (Z) 0.85
Moody Friction Faction (f) 0.013
Pipe Length (l) 75 miles


Q = 0.199 \sqrt {\frac {(17.938)^5((1400+14.7)^2 -(400+14.7)^2)}{(0.85)(60+460)(75*5280)(.7)(0.013)}}=173.8\ MMSCFD

There is about 173.8 MMSCFD flowing through the pipeline. Choice C is correct
Some people may try using either the Panhandle equation or the Weymouth Equation. Under certain conditions these may be used. Though in this sample question we are given a friction factor, which neither the Panhandle or Weymouth equations use.
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