FE Mechanical Practice Exam and Technical Study Guide Errata - … · 2019-11-21 · FE Mechanical...

FE Mechanical Errata -1 www.engproguides.com

FE Mechanical Practice Exam and Technical Study Guide Errata

This product has been updated to incorporate all changes shown in the comments on the webpage and email comments as of January, 10 2018. If you have purchased this product prior to this date and wish for the latest version then please email Justin Kauwale at [email protected] or you can use this document to see the changes that were made.

The following changes have recently been incorporated

December 30, 2018: Statics Problem & Solution 8 has been revised, see attached.

January 10, 2019:

Mathematics page 5 referenced the quadratic equation incorrectly. The reference should have been to the Pythagorean theorem. Please see attached.

Mechanics of materials pages 39 to 41, referenced the Machinery’s Handbook incorrectly and did not provide enough information in the problem. Please see attached.

Fluid of mechanics, pages 24 – 26, contained missing images. These pages have been re-printed, see attached.

January 11, 2019:

On page 4 of computational tools, see the below change in red.

On page 5 of computational, see the below change in red.


On page 44 of Fluid Mechanics, added the inner areas of schedule 40 steel pipes. The inner diameters are slightly different from the nominal diameters

On Solution 4 of Probability & Statistics, the incorrect value from the problem was used in the solution, see below.

On page 12 of computational tools, the last step was incorrect. This required the possible solutions to be changed.


On page 21 of Mechanics of Materials, the elastic and plastic definitions were switched.


On page 33 of Mechanics of Materials, clarified that the strain energy equation is for a simple, elastic bar.

On page 19 of Engineering Economics, clarified that the answer letter is (c) for practice problem 3.

On page 14 of Electricity & Magnetism, corrected the answer for the sample problem.

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Figure 6: This figure plots the function y = log(x).

The NCEES FE Reference Handbook shows all the applicable logarithmic laws that are used to solve equations with log functions. You should be familiar with how to use those laws on the FE exam. An example of one of those laws is the change of base law, which is shown below.

Change Base: In order to change the original base “b” of a log function, take the log of the value “a” with the new base “n” divided by the log of the original base “b” with the new base “n”.

𝐶ℎ𝑎𝑛𝑔𝑒 𝐵𝑎𝑠𝑒: log 𝑎 →log 𝑎log 𝑏

𝐸𝑥𝑎𝑚𝑝𝑙𝑒: log 1,000 →log 1,000log 100

32

log 1,000 1.5

2.5 NATURAL LOGARITHMIC EQUATIONS

Natural logarithmic equations follow the same rules and laws as logarithmic equations, except the only difference is that the base is “e”. The term “e” is a mathematical constant, founded by Euler. It is approximated as “2.71828”. Natural logarithmic equations are used in the rotational sections of mechanics and with pulleys. The natural log is not written as log base e, but written as ln and the base “e’ is implied.

𝑁𝑎𝑡𝑢𝑟𝑎𝑙 𝐿𝑜𝑔: log 𝑥 𝑦

𝑁𝑎𝑡𝑢𝑟𝑎𝑙 𝐿𝑜𝑔: ln 𝑥 𝑦

ln 𝑥 𝑦 ⇒ 𝑒 𝑥

2.6 TRIGONOMETRY

The main skills that you need to know for trigonometry are the right angle equations and the law of cosines & sines for all other triangles. These skills will help you to determine the angles and lengths of triangles for any situation.


“x” value Press your value for “x”

𝑅 → 𝑃 Press the Rectangular to Polar Button

“y” value Press your value for “𝑦”

= Press the equal sign

The first value shown will be your R -component

𝑅 → 𝜃 Press this button to show the y-component

The second value shown will be your 𝜃 -component

The Casio FX-115ES Plus/Casio FX-991EX ClassWiz and TI-36X Pro also has a complex mode which allows you to add/multiply complex numbers in various forms and easily converting the results between rectangular and polar. The stored lines of calculations and the various functions, including matrices and integrals make this calculator one of the favored among test takers.

5.4 VECTOR MATHEMATICAL OPERATIONS

Addition & Subtraction of Vectors: Vectors can be added or subtracted from one another by simply adding or subtracting their real components together and doing the same for their imaginary components. It is very easy to do this when vectors are presented in their Rectangular coordinates, however it is more difficult to do this by hand when the vectors are in polar form. For example, if you are adding 3 and 3j, then the answer is simply 3 + 3j. Another example would be if you were to add the vectors “2+2j” and “4+4j”, then the answer would be “6+6j”.

1st Example: You can complete this problem by inputting the polar vectors and adding them. If you are doing the equation by hand, then you should follow the recommended method by converting polar to rectangular and then adding the like terms to one another (real + real and imaginary + imaginary).

𝐶𝑜𝑛𝑣𝑒𝑟𝑡 𝑡𝑜 𝑅𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 → 3∠0° 3∠90° → 3 3𝑗

𝐶𝑜𝑛𝑣𝑒𝑟𝑡 𝑡𝑜 𝑃𝑜𝑙𝑎𝑟 → 3 3𝑗 → 3√2∠45°

2nd Example: This example is a little less intuitive, but since both vectors have the same angle, they can be added directly or with your calculator. If you are doing the equation by hand, then you should follow the recommended method by converting polar to rectangular and then adding the like terms to one another (real + real and imaginary + imaginary).

𝐶𝑜𝑛𝑣𝑒𝑟𝑡 𝑡𝑜 𝑅𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 → 2∠45° 4∠45° → 1.414 1.414𝑗 2.828 2.828𝑗 → 4.24 4.24𝑗

𝐶𝑜𝑛𝑣𝑒𝑟𝑡 𝑡𝑜 𝑃𝑜𝑙𝑎𝑟 → 4.24 4.24𝑗 → 6∠45°


3rd Example: In this example, the answer cannot be seen as intuitively, but as long as you either use your calculator to add polar coordinates or convert to rectangular and then add the like terms, then you will be able to complete this problem.

3∠35° 5∠60°

𝐶𝑜𝑛𝑣𝑒𝑟𝑡 𝑡𝑜 𝑅𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 → 3∠35° 5∠60° → 2.46 1.72𝑗 2.5 4.33𝑗

𝐶𝑜𝑛𝑣𝑒𝑟𝑡 𝑆𝑢𝑚 𝑡𝑜 𝑃𝑜𝑙𝑎𝑟 → 4.96 6.05𝑗 → 7.82∠50.65°

This figure below shows how vectors can be added and subtracted. This figure also corresponds to the first example and it also shows how the answer would change if vectors A and B were subtracted or added.

Multiplication & Division of Vectors: The multiplication and division of vectors is easy for vectors in polar form but difficult for vectors in rectangular form. Your calculator should be able to divide and subtract in both rectangular form, but if you are not familiar with this function on your calculator, then you can also complete this by hand. In order to multiply or divide vectors by hand you must convert the vector to polar form and then multiply or divide the radius of the vectors and then add or subtract the angles. You add the vector angles for multiplication and subtract the vectors angles for division.

1st Example: In this example, the vectors are already in polar form, so you can simply multiply the radiuses and add the vector angles.

3∠0° 3∠0° 9∠0°

2nd Example: Again, the vectors are already in polar form, so multiply the radiuses and add the vector angles.

3∠0° 3∠45° 9∠45°

3rd Example: In this example, the vectors must first be converted to polar and then the radiuses can be divided and the vector angles can be subtracted.


2 2𝑗 4 4𝑗 → 2.828∠45° 5.66∠45°

12

∠0° → 0.5 0𝑗

6.0 DIFFERENTIAL EQUATIONS Differential equations are used in mechanical engineering to curve fit data. This type of math typically uses software, so it is unlikely that you will have a question on differential equations on the FE exam.

7.0 NUMERICAL METHODS Numerical methods are used with differential equations in mechanical engineering to curve fit data. This type of math typically uses software, so it is unlikely that you will have a question on differential equations on the FE exam.

Statics-37 www.engproguides.com

𝑦 242.8 𝑙𝑏𝑓

The correct answer is most nearly, (C) 243 lbf.

9.8 PRACTICE PROBLEM 8 – STATIC FRICTION

What is the force, “X” required to keep the block stationary. The block weighs 100 lbs and the static friction coefficient is 0.5.

(A) 15 lbf

(B) 90 lbf

(C) 100 lbf

(D) 150 lbf

9.8.1 Solution 8 – Static Friction In this problem, you first need to calculate the normal force, with the free body diagram shown below.


You can convert 100 pounds to pound force and you can calculate the normal force.

𝐹 100 𝑙𝑏𝑠 ∗32.2

𝑓𝑡𝑠

32.2𝑙𝑏 𝑓𝑡

𝑙𝑏𝑓 𝑠

100 𝑙𝑏𝑓

Next sum up the forces acting in the “x-direction”, which is pointing parallel to the slope of the ground.

𝐹

𝑁 ∗ 𝜇 𝑋 ∗ cos 35° 100 𝑙𝑏𝑓 ∗ sin 35° 0

𝑁 ∗ 0.5 𝑋 ∗ cos 35° 100 𝑙𝑏𝑓 ∗ sin 35° 0

𝑁 2 ∗ 57.4 0.82 ∗ 𝑋 → 𝑁 114.8 1.64𝑋

Next sum up the forces acting in the “y-direction”, which is pointing perpendicular to the slope of the ground.

𝐹 0

𝑁 𝑋 ∗ sin 35 100 𝑙𝑏𝑓 ∗ cos 35° 0

𝑁 𝑋 ∗ 0.57 81.92 𝑙𝑏𝑓 0

Next, plug in the equation for “N” from above.

114.8 1.64𝑋 𝑋 ∗ 0.57 81.92 𝑙𝑏𝑓 0

2.21 ∗ X 32.9 𝑙𝑏𝑓

𝑋 14.9 𝑙𝑏𝑓

𝑁 90.25 𝑙𝑏𝑓

The correct answer is most nearly, (A) 15 lbf.

9.9 PRACTICE PROBLEM 9 – CENTROID

Find the expression for the centroid, 𝑦, of the shaded area.


9.10 PRACTICE PROBLEM 10 – CENTROID

Find the centroid coordinates, (�̅�, 𝑦) for the shaded region of the following figure.

(A) (6ft, 11ft) (B) (4ft, 10ft) (C) (4ft, 11.5ft) (D) (5ft, 11ft)

9.10.1 Solution 10 - Centroid Find the equation for the centroid of a circular segment in the Statics section of the NCEES FE Reference Handbook. The figure is rotated 90 degrees, so switch the x and y equations and shift the center point of the x centroid. The equations become

�̅�12𝑓𝑡 ∗ cos 60°

16𝑓𝑡

𝑦2 ∗ 𝑎

3sin 𝜃

𝜃 𝑠𝑖𝑛𝜃 ∗ 𝑐𝑜𝑠𝜃

In the above equation, a=12ft and 𝜃 = 90o-60o = 30o = 𝜋/6.

𝑦2 ∗ 12𝑓𝑡

3∗

sin 30°𝜋/6 sin 30° ∗ cos 30°

11.04𝑓𝑡

So, the centroid, (�̅�, 𝑦) = (6ft, 11ft)

The correct answer is (A) (6ft, 11ft)

Dynamics, Kinematics & Vibrations-34 www.engproguides.com

There are three unknowns, the two normal forces and the force F. The frictional force is a function of the normal force B. The three equations are the balanced equation in the X-direction, Y-direction and the moment about point B.

𝑋 𝐷𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 → 𝐹 𝐹 𝑁

𝑌 𝐷𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 → 𝑊 30𝑙𝑏𝑠 𝑁

𝑀𝑜𝑚𝑒𝑛𝑡 𝐵 → 0 𝑊 ∗ 1.5𝑓𝑡 𝑁 ∗ 3

Substitute into moment equation.

0 30𝑙𝑏𝑠 ∗ 1.5𝑓𝑡 𝑁 ∗ 3

𝑁 15 𝑙𝑏𝑠

Next, substitute NA and NB into the X-direction equation.

0.4𝑁 𝐹 𝑁

0.4 ∗ 30 𝐹 15

𝐹 3 𝐿𝐵𝑆

The answer is most nearly, (a) 3 LBS.

13.2 PRACTICE PROBLEM 2 - FRICTION

A braking system applies a force of 1,000 N (F) to a rotating part that has a moment of 100 N-m. The rotating part is pinned at O and is allowed to rotate about O. The coefficient of kinetic friction is 0.4. What is the reaction force in the X-direction at point O? The radius of the rotating part is R = 250 mm.


The answer is most nearly,

(a) 100 N

(b) 200 N

(c) 300 N

(d) 400 N

13.2.1 Solution 2 - Friction A braking system applies a force of 1,000 N (F) to a rotating part that has a moment of 100 N-m. The rotating part is pinned at O and is allowed to rotate about O. The coefficient of kinetic friction is 0.4. What is the reaction force in the X-direction at point O? The radius of the rotating part is R = 250 mm. Assume the weights of the components are negligible.

First draw a free body diagram. The brake force F is applied downward and the frictional force is opposing the rotating motion. The reactionary forces at O will be opposite the brake force F in the y-direction and opposite the frictional forces in the x-direction.

You should use the moment at O to solve for the reactionary force in the x-direction, which will simply be equal to the frictional force.

𝑀𝑜𝑚𝑒𝑛𝑡 𝑂 → 0 𝑀 𝐹 ∗ 0.25𝑚

𝑀𝑜𝑚𝑒𝑛𝑡 𝑂 → 0 100 𝑁 ∗ 𝑚 𝐹 ∗ 0.25𝑚

𝐹 𝐹 400 𝑁

The correct answer is most nearly, (d) 400 N.


13.5 PRACTICE PROBLEM 5 – WORK ENERGY

What is the velocity of the block after it has dropped 5 meters? Assume the block starts at a velocity of 1 m/s. There is an opposing frictional force that has been calculated to be as shown in the figure below.

(a) 1 m/s

(b) 3.3 m/s

(c) 9.4 m/s

(d) 14 m/s

13.5.1 Solution 5 – Work Energy In this problem, you can use the energy balance equation with potential energy, work and kinetic energy.

Initially, you can assume that the work is equal to zero because the net force has not acted over a distance yet. If you assume the block is 5 meters above the 0 meter point, then the potential energy will be “mg” multiplied by “5 meters”. The initial velocity is 1 m/s, so there will be some kinetic energy.

𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 → 𝑊 0; 𝑃𝐸 𝑚𝑔ℎ; 𝐾𝐸12

𝑚𝑣

𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 → 100 𝑘𝑔 ∗ 9.81𝑚𝑠

∗ 5 𝑚12

100 𝑘𝑔 ∗ 1 4,955 𝐽


At the final condition, at point 0 meters, there will have been 5 meters of work done by the net force and the potential energy will be equal to zero. The kinetic energy will account for the loss of potential energy.

𝐹𝑖𝑛𝑎𝑙 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 → 100𝑁 512

100 𝑘𝑔 ∗ 𝑣

Next equate the initial and final conditions.

𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝐹𝑖𝑛𝑎𝑙 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛

4,955 𝐽 100 512

100 𝑘𝑔 ∗ 𝑣

𝑣 9.43 𝑚/𝑠

Another method is to solve for the net force, which is equal to the frictional force minus the gravitational force. In this problem, you can assume that upward forces are positive and downward forces are negative.

𝑁𝑒𝑡 𝐹𝑜𝑟𝑐𝑒 → 100 𝑘𝑔 ∗ 9.81𝑚𝑠

100 𝑁 881 𝑁 → 𝑎 8.8𝑚𝑠

;

𝑣 𝑣 2𝑎𝑥; 𝑣 9.43𝑚𝑠

;

The correct answer is most nearly (C) 9.4 m/s.

13.6 PRACTICE PROBLEM 6 – WORK ENERGY

At what distance will the spring compress such that it brings the block to rest in the figure below. Assume the spring is not compressed and k = 1,000 N/m. The block starts at an initial velocity of 5 m/s. Assume a frictionless surface.

(a) 1 m

(b) 3 m

(c) 5 m

(d) 7 m

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(c) 289 l

(d) 420 l

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Equation 6: Relationship between COP and EER

𝐶𝑂𝑃 𝐸𝐸𝑅

3.412

9.0 NON-REACTING MIXTURES OF GASES A non-reacting gas mixture is a mixture of two or more ideal gases that can be treated as a sum of the individual ideal gases. For example, the total mass of a mixture is just the sum of the gases.

𝑀𝑎𝑠𝑠 → 𝑚 𝑚 , 𝑚 , . . . 𝑚 ,

The number of particles of each type of gas will be dependent on the number of moles in the mixture. So the same equation can be re-written with the mole variable.

𝑀𝑜𝑙𝑒𝑠 → 𝑛 𝑛 , 𝑛 , . . . 𝑛 ,

These two equations can be combined with the molecular weight for each gas.

𝑚 𝑛 , ∗ 𝑀𝑊 , 𝑛 , ∗ 𝑀𝑊 , . . . 𝑛 , ∗ 𝑀𝑊 ,

The second concept that you should know for the non-reacting mixtures of gases is that the ideal gas law applies to this mixture of gases. This means that the volume of the gas mixture is equal to the sum of each volume of individual gas. The same is also true for pressures.

𝑉𝑜𝑙𝑢𝑚𝑒 → 𝑉 𝑉 , 𝑉 , . . . 𝑉 ,

𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 → 𝑃 𝑃 , 𝑃 , . . . 𝑃 ,

The last concept that you need to know is that the properties of the gas mixture is dependent on the individual properties of each gas and the mole fraction of each individual gas.

𝑀𝑜𝑙𝑒 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛 → 𝑋 ,𝑛 ,

𝑛

𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 → 𝑢 𝑢 , ∗ 𝑋 , 𝑢 , ∗ 𝑋 , . . . 𝑢 , ∗ 𝑋 ,

𝐸𝑛𝑡ℎ𝑎𝑙𝑝𝑦 → ℎ ℎ , ∗ 𝑋 , ℎ , ∗ 𝑋 , . . . ℎ , ∗ 𝑋 ,

𝐻𝑒𝑎𝑡 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 → 𝑐 𝑐,

∗ 𝑋 , 𝑐,

∗ 𝑋 , . . . 𝑐,

∗ 𝑋 ,

𝐸𝑛𝑡𝑟𝑜𝑝𝑦 → 𝑠 𝑠 , ∗ 𝑋 , 𝑠 , ∗ 𝑋 , . . . 𝑠 , ∗ 𝑋 ,

Thermodynamics - 81 http://www.engproguides.com

Fuel is simply any material that can be burned to release heat. But the most common fuels in the Thermal & Fluids application are the hydrocarbon fuels. This category includes fuels like coal, gasoline and natural gas. These fuels power the power plants and cars in the United States of America. The fuels are a combination of carbon and hydrogen.

𝐻𝑦𝑑𝑟𝑜𝑐𝑎𝑟𝑏𝑜𝑛 𝑓𝑢𝑒𝑙𝑠 𝐶 𝐻

𝑤ℎ𝑒𝑟𝑒 𝑥 𝑎𝑛𝑑 𝑦 𝑎 𝑞𝑢𝑎𝑛𝑡𝑖𝑦 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑒𝑙𝑒𝑚𝑒𝑛𝑡

For example, these are the formulas for gasoline, diesel and methane.

𝐺𝑎𝑠𝑜𝑙𝑖𝑛𝑒 𝐶 𝐻 ; 𝐷𝑖𝑒𝑠𝑒𝑙 𝐶 𝐻 ; 𝑀𝑒𝑡ℎ𝑎𝑛𝑒 𝐶 𝐻

12.2 AIR

Air is about 21% oxygen (by volume) and 78% nitrogen and about 1% other gases. The FE exam may ask you to calculate the amount of air used in a combustion process and air should take into account both oxygen and nitrogen.

𝐴𝑖𝑟 21%𝑂 79%𝑁

𝐴𝑖𝑟 𝑂 3.76𝑁

For example, when air is used in the propane combustion equation, the following equation is used.

12.3 STOICHIOMETRY

Stoichiometry of chemical reactions means that species react in exact proportions. The proportions that react are governed by the specific fuel’s combustion equation. When completing these types of problems, remember that the proportions that are in the equations are

based on moles and not weight. If a problem gives you weights, you need to first convert

the weight to moles before you can use the stoichiometry equations.

In order to convert weight to moles you need the molecular weight, so make sure you remember the periodic table in the NCEES FE Reference Handbook. Here are some of the common molecular weights used in the combustion equations.

Element Molecular Weight (g/mol) Carbon [C] 12 Nitrogen [N] 14 Oxygen [O] 16 Sulfur [S] 16 Hydrogen [H] 1

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Mechanics of Materials-49 www.engproguides.com

13.9.1 Solution 9 – Shear and Moment Diagram First conduct the shear diagram. This shear diagram shows that the moment will increase linearly.

You also know that the ends of the moment diagram will have moment values equal to M1 and M2. So this takes out B & D since the moments are equal to zero at the right ends. Since the force is a point force, then the moment will increase linearly as opposed to in a curve as shown in C.

The correct answer is most nearly (A).

13.10 PRACTICE PROBLEM 10 – STRESS TRANSFORMATIONS

A compressive shear stress of 10 psi and a shear stress of 5 psi acts upon a material. What are the principal stresses?

(a) -.4 psi & 14 psi

(b) -20 psi & 10 psi

(c) -30 psi & 30 psi

(d) -31 psi & 1 psi

13.10.1 Solution 10 – Stress Transformations The principal shear stresses are found with the below equation. Compressive stresses are negative and tensile stresses are positive.

𝜎 ,𝜎 𝜎

2𝜎 𝜎

2 𝜏

Mechanics of Materials-50 www.engproguides.com

𝜎 ,10 𝑝𝑠𝑖

210 𝑝𝑠𝑖

25 𝑝𝑠𝑖

𝜎 , 7.5 𝑝𝑠𝑖 7.1 𝑝𝑠𝑖

𝜎 0.4 𝑝𝑠𝑖 𝑎𝑛𝑑 14.6 𝑝𝑠𝑖

The correct answer is most nearly (A) -0.4 psi & 15 psi

13.11 PRACTICE PROBLEM 11 – STRESS FROM AXIAL LOAD

A force, F, of 45 kips is applied to a two segment steel rod as shown below. What is the stress experienced at segment BC?

(a) 6 kpsi

(b) 14 kpsi

(c) 18 kpsi

(d) 25 kpsi

13.11.1 Solution 11 – Stress from Axial Load The internal force at segment AB and the internal force at segment BC will both equal the force, F. Solve for the stress at segment BC, with diameter equal to 1.5”.

𝜎 𝑃/𝐴

𝜎45 𝑘𝑖𝑝𝑠

𝜋 ∗1.5𝑖𝑛

2

25 𝑘𝑝𝑠𝑖

The correct answer is most nearly (D).

Fluid Mechanics - 8 http://www.engproguides.com

2.4 HEAT CAPACITY

Heat capacity describes the amount of energy required to raise a fluid’s or solid’s temperature by 1 degree. This term will be discussed more in Section 12.0 Thermodynamics.

There are two types of heat capacity as shown below. The cp term is used for processes involving constant pressure and the cv term is used for processes involving constant volume. The cp term is used to calculate the change in enthalpy of a fluid or solid as a function of a change in temperature of the fluid or solid. Enthalpy includes the pressure-volume work. For liquids and solids, the terms are very near to one another, because liquids and solids are nearly incompressible. However, for gases the terms will be more different.

2.5 SPECIFIC HEAT RATIO

The specific heat ratio for gases is the ratio of the specific heat capacity at constant pressure and specific heat capacity at constant volume. This is value is used calculate isentropic relationships.

𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝐻𝑒𝑎𝑡 𝑅𝑎𝑡𝑖𝑜, 𝑘𝑐𝑐

𝑘 1.4

3.0 FLUID STATICS The paragraphs after fluid statics involve fluids in movement. The majority of Section 11 Fluid Mechanics is on moving fluids and not fluids at rest, but there may be a few questions on the FE exam that involve fluid statics. The main concepts that you must understand are the pressure due to fluid height, manometers and buoyancy.

3.1 PRESSURE DUE TO A STATIC FLUID

The first concept that you need to understand is that the static pressure at certain points within a fluid will vary based on the height of the fluid above the point. Static pressure is the pressure acting upon a body or point due to a fluid, when a fluid is at rest.

For example, in the figure below the pressure at point 5A will be greater than the pressure at point 2A, which will be greater than the pressure at point 1A. Another concept that you need to understand is that any container that is open to the atmosphere will have a pressure equal to the atmospheric pressure at that location acting upon the open parts of the container. For example, points 1A and 1B, will have a static pressure equal to the atmospheric pressure and if the location of this container is at sea level, then that pressure will be equal to 14.7 psia. The next concept that you need to understand is that all points at the same elevation will have the same static pressure, for example, points 2A, 2B, 3A, 3B, 4A and 4B will all be at the same static pressure. It may be easier to see how 2A, 3A, 4A and 4B at the same pressure, but maybe not 2B and 3B. It appears that there is not that much fluid acting upon those points, but you need to think about how 2B, 3B and 4B are all at the same pressure. If these points were not at the same pressure

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13.4.1 Solution 4 – Pressure Drop 500 GPM of an unknown fluid flows through a 6" schedule 40 pipe. If the pipe has a relative roughness value of .0004 and a Reynolds number of 250,000, then what is the pressure drop through 100 ft of pipe?

(A) 1.7 ft of head

(B) 3.3 ft of head

(C) 4.2 ft of head

(D) 4.9 ft of head

Use the Darcy Weisbach equation to find the pressure drop.

ℎ𝑓𝐿𝑣2𝐷𝑔

𝐷𝑎𝑟𝑐𝑦 𝑊𝑒𝑖𝑠𝑏𝑎𝑐ℎ 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛

𝑤ℎ𝑒𝑟𝑒 ℎ 𝑓𝑡 𝑜𝑓 ℎ𝑒𝑎𝑑; 𝑓 𝐷𝑎𝑟𝑐𝑦 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟; 𝑣 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑓𝑡

𝑠𝑒𝑐,

𝐷 𝑖𝑛𝑛𝑒𝑟 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑓𝑡 , 𝑔 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 32.2𝑓𝑡

𝑠𝑒𝑐

First, you need to use the Moody Diagram in the FE Handbook to find the friction factor.

𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑟𝑜𝑢𝑔𝑛𝑒𝑠𝑠 0.0004 𝑎𝑛𝑑 𝑅𝑒𝑦𝑛𝑜𝑙𝑑 𝑠 𝑁𝑢𝑚𝑏𝑒𝑟 250,000


13.9.1 Solution 9 – Impulse Momentum

Water flows through a 50 cm pipe at a flow rate of 190 kg/s and pressure of 4kPa. As the water flows, it hits a 90 degree elbow. What is the force needed to restrain the elbow?

(a) 35 N

(b) 70 N

(c) 706 N

(d) 986 N

Use the momentum equation for steady flow, where the subscript 1 designates fluid entering the control volume and 2 designates fluid leaving the control volume.

𝐹 𝑄 𝜌 𝑣 𝑄 𝜌 𝑣

The velocity entering and leaving the elbow is the same and is calculated as:

𝑣 𝑣 190𝑘𝑔𝑠

∗1𝑚

1000𝑘𝑔∗

1

𝜋 ∗0.5 𝑚

2

0.97𝑚𝑠

Separate the forces into the x and y components.


𝐹 𝐹 , 𝐹 𝐹 , 𝑃 𝐴

𝐹 , 𝑃 𝐴 0 𝑄 𝜌 𝑣

Since,

𝑄 𝜌 𝑚

then,

𝐹 , 𝑚 𝑣 𝑃 𝐴 190𝑘𝑔𝑠

∗ 0.97𝑚𝑠

∗1𝑁

1 𝑘𝑔 ∗ 1𝑚𝑠

4000𝑁

𝑚∗ 𝜋 ∗

0.5𝑚2

969 𝑁

𝐹 𝐹 , 𝑄 𝜌 𝑣

𝐹 , 𝑚 𝑣 190𝑘𝑔𝑠

∗ 0.97𝑚𝑠

∗1𝑁

1 𝑘𝑔 ∗ 1𝑚𝑠

184 𝑁

Then, find the total reactive force.

𝐹 𝐹 , 𝐹 , 969 𝑁 184 𝑁 986 𝑁

The answer is (D) 986 N.

Fluid Mechanics - 52 www.engproguides.com 10-15 out of 110 problems

13.13 PRACTICE PROBLEM 13 – SCALING LAWS

A pump is sized for 200 GPM at 150 ft of head. If the pump speed is decreased by 25%, what is the new flow of the pump? Assume the diameter remains the same.

(a) 100 GPM

(b) 115 GPM

(c) 150 GPM

(d) 200 GPM

13.14 PRACTICE PROBLEM 14 – EXTERNAL FLOW

Water flows around a 4cm diameter pipe that is placed across a large channel. Water flows in the 20m long, 10m deep channel at a rate of 10m/s and a temperature of 20oC. What is the drag force on the pipe? Assume a drag coefficient of 0.4.

(a) 8 kN

(b) 16 kN

(c) 20 kN

(d) 25 kN


𝑊𝑒 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑐𝑜𝑛𝑣𝑒𝑟𝑡 4 𝑝𝑠𝑖 𝑡𝑜 𝑓𝑒𝑒𝑡 ℎ𝑒𝑎𝑑 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑

𝑓𝑡 𝑜𝑓 ℎ𝑒𝑎𝑑 4 ∗2.31𝑆𝐺

𝑓𝑡 𝑜𝑓 ℎ𝑒𝑎𝑑 9.24

𝑄 50𝑔𝑎𝑙𝑙𝑜𝑛𝑠𝑚𝑖𝑛𝑢𝑡𝑒

∗. 134 𝑓𝑡𝑔𝑎𝑙𝑙𝑜𝑛

∗1 𝑚𝑖𝑛60 𝑠

0.112𝑓𝑡

𝑠

𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑤𝑜𝑟𝑘9.24 150 ∗ 0.112 ∗ 62.4 ∗ 1.75

. 69 ∗ .9

𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑤𝑜𝑟𝑘 3,136 𝑓𝑡𝑙𝑏𝑓

𝑠

𝐶𝑜𝑛𝑣𝑒𝑟𝑡 𝑡𝑜 𝐻𝑃 → 3,136𝑓𝑡 𝑙𝑏𝑓

𝑠∗ 1.818𝑥10 5.7 𝐻𝑃

The answer is most nearly, (C), 7.5 HP.

14.12 SOLUTION 12 – PERFORMANCE CURVE

Background: A fan has been selected at the design point shown on the below fan curve. What is the minimum number of fans required to be placed in series, to achieve a flow rate of 3,000 CFM at a pressure of 4.0 in. wg?

If fans are placed in series, then their pressures are added. (3) Fans are required to achieve a pressure of 5.4 in. wg.

14.13 SOLUTION 13 – SCALING LAWS

A pump is sized for 200 GPM at 150 ft of head. If the pump speed is decreased by 25%, what is the new flow of the pump? Assume the diameter remains the same.

Use the affinity laws.

𝑄

𝑄 𝑁

𝑁 ; 𝑖𝑓 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑖𝑠 ℎ𝑒𝑙𝑑 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

200𝑄

𝑥0.75𝑥


200 ∗ 0.75 ∗ 𝑥 𝑄 ∗ 𝑥

𝑄 150 𝑔𝑝𝑚

14.14 SOLUTION 14 – EXTERNAL FLOW

The find the drag force on the pipe, use the following equation for immersed objects in a large body of water.

𝐹𝐶 𝜌𝑣 𝐴

2

From the NCEES FE Reference Handbook, the density of water at 20oC is 998.2 kg/m3 (997 at 25). Solve for the front face area of the pipe that the fluid flow will see.

𝐴 𝐷 ∗ 𝐶ℎ𝑎𝑛𝑛𝑒𝑙 𝑊𝑖𝑑𝑡ℎ 4𝑐𝑚 ∗1𝑚

100𝑐𝑚∗ 20𝑚 0.8𝑚

Solve for the drag force.

𝐹0.4 ∗ 998.2𝑘𝑔/𝑚 10𝑚/𝑠 0.8𝑚

2∗

1𝑁1𝑘𝑔 ∗ 𝑚/𝑠

𝐹 15,970𝑁

The answer is most nearly (B) 16kN.

14.15 SOLUTION 15 – COMPRESSIBLE FLOW

The stagnation pressure is the static pressure of the air, when all the kinetic energy is isentropically converted to pressure energy. It is calculated from the pressure in the duct using the following isentropic relationship.

𝑃𝑃

1𝑘 1

2∗ 𝑀𝑎

Where k is the specific heat ratio, cp/cv and Po is the stagnation pressure. The specific heat ration of air is 1.4. The pressure in the above equation is given in absolute values.

𝑃 21𝑝𝑠𝑖𝑔 14.7 35.7𝑝𝑠𝑖𝑎

Solve for the stagnation pressure.

𝑃 𝑃 1𝑘 1

2∗ 𝑀𝑎 35.7𝑝𝑠𝑖𝑎 ∗ 1

1.4 12

∗ 0.4

..

𝑃 39.9𝑝𝑠𝑖𝑎


13.12 PRACTICE PROBLEM 12 – POWER AND EFFICIENCY

A hydraulic pump must pump 50 GPM of a fluid with a specific gravity 1.75 from one reservoir to the next reservoir. The pump is located at the same elevation as the first reservoir and the second reservoir is located 150 feet above the first reservoir. Assume that the friction loss in the piping is 4 psi. What is the minimum motor horsepower required of the pump. Assume the pump has an efficiency of 69% and the motor has an efficiency of 90%.

The answer is most nearly,

(a) 3 HP

(b) 5 HP

(c) 7.5 HP

(d) 10 HP

13.12.1 Solution 12 – Power and Efficiency A hydraulic pump must pump 50 GPM of a fluid of specific gravity 1.75 from one reservoir to the next reservoir. The pump is located at the same elevation as the first reservoir and the second reservoir is located 150 feet above the first reservoir. Assume that the friction loss in the piping is 4 psi. What is the minimum motor horsepower required of the pump. Assume the pump has an efficiency of 69% and the motor has an efficiency of 90%.

𝑅𝑒𝑓𝑒𝑟 𝑡𝑜 𝑝𝑢𝑚𝑝 𝑝𝑜𝑤𝑒𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝐹𝐸 𝐻𝑎𝑛𝑑𝑏𝑜𝑜𝑘

𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑤𝑜𝑟𝑘∆𝐻 ∗ 𝑄 ∗ 𝐷𝑒𝑛𝑠𝑖𝑡𝑦

𝑃𝑢𝑚𝑝 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 ∗ 𝑀𝑜𝑡𝑜𝑟 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦

∆𝐻 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑟𝑜𝑝 𝑖𝑛 𝑓𝑒𝑒𝑡 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑; 𝑄 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑖𝑛 𝑓𝑡 /𝑠

𝑊𝑒 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑐𝑜𝑛𝑣𝑒𝑟𝑡 4 𝑝𝑠𝑖 𝑡𝑜 𝑓𝑒𝑒𝑡 ℎ𝑒𝑎𝑑 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑

𝑓𝑡 𝑜𝑓 ℎ𝑒𝑎𝑑 4 ∗2.31𝑆𝐺

𝑓𝑡 𝑜𝑓 ℎ𝑒𝑎𝑑 5.28

𝑄 50𝑔𝑎𝑙𝑙𝑜𝑛𝑠𝑚𝑖𝑛𝑢𝑡𝑒

∗. 134 𝑓𝑡𝑔𝑎𝑙𝑙𝑜𝑛

∗1 𝑚𝑖𝑛60 𝑠

0.112𝑓𝑡

𝑠

𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑤𝑜𝑟𝑘5.28 150 ∗ 0.112 ∗ 62.4 ∗ 1.75

. 69 ∗ .9

𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑤𝑜𝑟𝑘 3,058 𝑓𝑡𝑙𝑏𝑓

𝑠


𝐶𝑜𝑛𝑣𝑒𝑟𝑡 𝑡𝑜 𝐻𝑃 → 3,058𝑓𝑡 𝑙𝑏𝑓

𝑠∗ 1.818𝑥10 5.65 𝐻𝑃

The answer is most nearly, C, 7.5 HP.


13.11 PRACTICE PROBLEM 11 – COMPRESSOR

Problem removed

13.11.1 Solution 11 – Compressor Solution removed

Heat Transfer - 30 http://www.engproguides.com

𝑅𝑒𝐷 ∗ 𝑣

𝜈

0.33 𝑓𝑡 ∗ 6.5 𝑓𝑡𝑠

0.609 x 10 𝑓𝑡

𝑠

355,775

10.4 PROBLEM 4: CALCULATE THE CONVECTIVE HEAT TRANSFER COEFFICIENT

250 GPM of 120F water flows through a 4” diameter pipe that is located in the ceiling of a building. Calculate the convective heat transfer at the fluid inside the pipe. Assume the water in the pipe is not under extreme pressure.

120F, water properties: 𝜇 1.36 ∗

, 𝑐 0.999 ∗℉

, 𝑘 0.378 ∗

∗ ∗℉, 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑖𝑡𝑦

a) 556∗ ∗℉

b) 5,182∗ ∗℉

c) 6,738∗ ∗℉

d) 12,784∗ ∗℉

10.4.1 Solution 4: Calculate the Convective Heat Transfer Coefficient This is a forced convective heat transfer problem through a pipe and the objective is to solve for the convective heat transfer, ℎ . Use the Reynolds number in Problem 3 to determine whether the flow is laminar or turbulent. Because the Reynolds number is greater than 2,300, the flow is turbulent

The pipe is within the ceiling, exposed to ambient temperatures, meaning the temperature of the pipe wall temperature is lower than the fluid temperature.

Therefore, Equation 4.1.2 can be used to solve for the convective heat transfer coefficient, hconv.

𝑁𝑢𝐷 ∗ ℎ

𝑘 .023 ∗ 𝑅𝑒. ∗ 𝑃𝑟.

Solve for the Nusselt number, Nu, by first finding the Prandtl number. The Prandtl number, Pr, is calculated with the following equations.

𝑃𝑟𝜇 ∗ 𝑐

𝑘

Measurement, Instrumentation & Controls - 14 http://www.engproguides.com

6.0 PRACTICE EXAM PROBLEMS

6.1 PRACTICE PROBLEM 1 – ROUTH TEST

A characterizing equation is shown below. The variable “C” must be greater than what minimum value in order to ensure the equation stable?

𝑡 1000𝑡 𝐶 10 𝑡 100𝐶 0

(a) -9

(b) 0

(c) 9

(d) 1,000

6.1.1 Solution 1 – Routh Test First, plug in the values to the first two rows of the matrix below.

𝑡 1000𝑡 𝐶 10 𝑡 100𝐶 0 0

⎝

⎜⎛

𝐴 𝐶 𝐸𝐵 𝐷 0𝑏 𝑏 0𝑐 𝑐 0𝑑 𝑑 0⎠

⎟⎞

⎝

⎜⎛

1 𝐶 101000 100𝐶

100 ∗ 𝐶 10 100𝐶100

0

100𝐶 𝑐 ⎠

⎟⎞

All values in the first column must be greater than 0. Check the third row.

1000 ∗ 𝐶 10 100𝐶100

0

100𝐶 1,000 𝐶 0

99𝐶 1,000

𝐶 10.1

Now check the fourth row.

100𝐶 0

Mechanical Design & Analysis-104 www.engproguides.com

18.14 PRACTICE PROBLEM 14 – HYDRAULIC COMPONENTS

A pump is sized for 200 GPM at 150 ft of head. If the speed is decreased by 10%, what is the new design pressure of the pump? Assume the diameter remains the same.

(a) 100 ft of head

(b) 122 ft of head

(c) 135 ft of head

(d) 150 ft of head

Mechanical Design & Analysis-105 www.engproguides.com

18.14.1 Solution 14 – Hydraulic Components A pump is sized for 200 GPM at 150 ft of head. If the speed is decreased by 10%, what is the new design pressure of the pump? Assume the diameter remains the same.

Use the affinity laws.

𝑄

𝑄 𝑁

𝑁 ; 𝑖𝑓 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑖𝑠 ℎ𝑒𝑙𝑑 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

150𝑄

𝑥0.9𝑥

150 ∗ 0.9 ∗ 𝑥 𝑄 ∗ 𝑥

𝑄 121.5 𝑓𝑡 ℎ𝑒𝑎𝑑

FE Mechanical Practice Exam and Technical Study Guide Errata - … · 2019-11-21 · FE Mechanical...

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Transcript of FE Mechanical Practice Exam and Technical Study Guide Errata - … · 2019-11-21 · FE Mechanical...