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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July 2

Exam Problem #28 (Thermal Efficiency)


A steam turbine produces 1.21 gigawatts of electricity. The load on the cooling system is measured to be 520,000 tons. For this power generation system what is the thermal efficiency?
A) 55%
B) 43%
C) 30%
D) 34%


Today’s fundamental of engineering sample problem looks at calculating the thermal efficiency when given units in the metric and English systems


Thermal Efficiency

\eta = \frac {Q_{in} - Q_{out}}{Q_{in}}


Qin = 520,000 tons
Qout 1.21 gigawatts (gw)


First step is to convert the units to the same type. Since there are only two variables you can use either metric and convert the problem into watts or English and convert everything into tons. In this problem we are going to convert the units of watts into tons (1 ton = 12,000 btuh)

(\frac {1\ ton}{12,000\ btu})(\frac {3412\ btu}{1\ kilowatt}) = 0.283 \frac {ton}{kw}

Now to put everything back into the thermal efficiency equation

\eta = \frac {520,000 - (1,210\ mw)(1,000\ kw/mw)(0.283\ ton/kw)}{520,000} = 0.3415


The correct answer is D with a thermal efficiency of 34%

July 1

Exam Problem #27 (Refrigeration)


A refrigeration process produces 1,300 kJ/kg of cooling. A reactor requires a minimum of 110 tons of cooling in order to maintain a safety factor of 2.5  In order for the refrigeration process to provide 110 tons of cooling what must the mass flow rate be?
A) 100 kg/hr
B) 350 kg/hr
C) 107 kg/hr
D) 1015 kg/hr


Today’s fundamentals of engineering exam question looks at how the flow rate required to maintain a level of refrigeration.


\dot{m} = \frac{refrigeration\ rate} {cooling\ capacity}


Cooling capacity = 1,300 kJ/kg
Refrigeration rate = 110 tons
Ton of Refrigeration = 12,000 btu/hour


First we need to convert a ton of refrigeration to joules

Joules\ =\ 12,000 \frac {btu}{hr}(1,055 \frac {joule}{btu}) = 12,660,000\ J

Next we find the mass flow rate

\dot{m} = \frac{(1,266 kJ)(110\ Tons)} {1300 kj/kg} = 107\ \frac {kg}{hr}

The correct answer is C
For the previous fundamentals of engineering sample question click here

June 30

Exam Problem #26 (Adiabatic Process)


Which is the following is TRUE. For a adiabatic, internally reversible process, what is the net change in entropy?

A) Entropy goes to infinity and beyond
B) The entropy in the system decreases
C) The entropy in the system increases when reversed
D) The starting entropy and ending entropy are equal


Today PE/FE sample problem looks at entropy and how an adiabatic reversible process changes the net amount of entropy

Since there is no heat added or lost and the processes is fully reversible so the work in equals the work out the net effect on entropy is. The initial state of entropy would equal the final state of entropy, there will be no change. Choice D is correct.

If heat was taken or withdrawn that would have changed the amount of entropy in the system. The same is if there would have been any work done that was not reversible.

Click here to view the previous PE/FE engineering sample question

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
Click here to view the previous PE/FE engineering sample question