Ask question

Ask question

All categories

2 years ago

7

After earning a bachelor's degree, one must do which of the following before taking the PE examination to receive a Professional

Engineering license?

1 answer:

san4es73 [151]2 years ago

4 0

FUNDAMENTALS OF ENGINEERING (FE) Exam and get the EIT license

You might be interested in

A concentrated load P is applied to the upper end of a 1.47-m-long pipe. The outside diameter of the pipe is D = 112 mm and the

myrzilka [38]

Answer:

Pmax = 38251.73 N

Explanation:

Given info

L = 1.47 m

D = 112 mm ⇒ R = D/2 = 112/2 mm = 56 mm

d = 101 mm ⇒ r = D/2 = 101/2 mm = 50.5 mm

a) We can apply the following equation in order to get Q (First Moment of Area):

Q = 2*(A₁*y₁-A₂*y₂)

where

A₁ = π*R² = π*(56 mm)² = 3136 π mm²

y₁ = 4*R/(3*π) = 4*56/(3*π) mm = 224/(3*π) mm

A₂ = π*r² = π*(50.5 mm)² = 2550.25 π mm²

y₂ = 4*r/(3*π) = 4*50.5/(3*π) mm = 202/(3*π) mm

then

Q = 2*(3136 π mm²*224/(3*π) mm-2550.25 π mm²*202/(3*π) mm)

⇒ Q = 62437.833 mm³

b) If τallow = 83 MPa = 83 N/mm²

P = ?

We can use the equation

τ = V*Q / (t*I) ⇒ V = τ*t*I / Q

where

t = D - d = 112 mm - 101 mm = 11 mm

I = (π/64)*(D⁴-d⁴) = (π/64)*((112 mm)⁴-(101 mm)⁴) = 2615942.11 mm⁴

Q = 62437.833 mm³

we could also use this equation in order to get Q:

Q = (4/3)*(R³-r³)

⇒ Q = (4/3)*((56 mm)³-(50.5 mm)³) = 62437.833 mm³

then we have

V = (83 N/mm²)*(11 mm)*(2615942.11 mm⁴) / (62437.833 mm³)

⇒ V = 2942.255 N

Finally Pmax = V = 38251.73 N

6 0

3 years ago

Find a negative feedback controller with at least two tunable gains that (1) results in zero steady state error when the input i

IceJOKER [234]

Answer:

Gc(s) = $\frac{0.1s + 0.28727}{s}$

Explanation:

comparing the standard approximation with the plot attached we can tune the PI gains so that the desired response is obtained. this is because the time requirement of the setting is met while the %OS requirement is not achieved instead a 12% OS is seen from the plot.

attached is the detailed solution and the plot in Matlab

8 0

3 years ago

An ideal vapor-compression refrigeration cycle operates at steady state with Refrigerant 134a as the working fluid. Saturated va

Ksju [112]

Explanation:

Note: Refer the diagram below

Obtaining data from property tables

State 1:

$\left.\begin{array}{l}P_{1}=1.25 \text { bar } \\\text { Sat - vapour }\end{array}\right\} \begin{array}{l}h_{1}=234.45 \mathrm{kJ} / \mathrm{kg} \\S_{1}=0.9346 \mathrm{kJ} / \mathrm{kgk}\end{array}$

State 2:

$\left.\begin{array}{l}P_{2}=5 \text { bor } \\S_{2}=S_{1}\end{array}\right\} \quad h_{2}=262.78 \mathrm{kJ} / \mathrm{kg}$

State 3:

$\left.\begin{array}{l}P_{3}=5 \text { bar } \\\text { Sat }-4 q\end{array}\right\} h_{3}=71-33 \mathrm{kJ} / \mathrm{kg}$

State 4:

Throttling process $h_{4}=h_{3}=71.33 \mathrm{kJ} / \mathrm{kg}$

(a)

Magnitude of compressor power input

$\dot{w}_{c}=\dot{m}\left(h_{2}-h_{1}\right)=\left(8 \cdot 5 \frac{\mathrm{kg}}{\min } \times \frac{1 \mathrm{min}}{\csc }\right)(262.78-234 \cdot 45)\frac{kj}{kg}$

$w_{c}=4 \cdot 013 \mathrm{kw}$

(b)

Refrigerator capacity

$Q_{i n}=\dot{m}\left(h_{1}-h_{4}\right)=\left(\frac{g \cdot s}{60} k_{0} / s\right) \times(234 \cdot 45-71 \cdot 33) \frac{k J}{k_{8}}$

$Q_{i n}=23 \cdot 108 \mathrm{kW}\\1 ton of retregiration =3.51 k \omega$

$\ Q_{in} =6 \cdot 583 \text { tons }$

(c)

Cop:

$\beta=\frac{\left(h_{1}-h_{4}\right)}{\left(h_{2}-h_{1}\right)}=\frac{Q_{i n}}{\omega_{c}}=\frac{23 \cdot 108}{4 \cdot 013}$

$\beta=5 \cdot 758$

3 0

2 years ago

The alignment readings for the front of a vehicle are shown above. Camber and toe are within specification, caster is not. Techn

dlinn [17]

Answer:

B. B only

Given Information:

1. Camber and toe are within specification

2. Caster is not within specification

Technician A says that with the current settings, the left front tire tread may wear on the inside edge.

Technician B says that with the current settings, the vehicle may pull to the left

Explanation:

Lets discuss the effects of Camber, toe and caster misalignment

Effects of Camber and Toe misalignment:

Camber is the inward or outward tilt of the fron tires and is used to distribute load across the tread. Any misalignment causes uneven loading on the tires which results in tire wear on one edge.

The most common cause of tire wear on the inside edge is due to the camber misalignment which results in premature tire wear.

Another reason is of tire wear is vehicle’s toe. A slight misalignment of the toe reduces the life of the tire.

Since it is given that camber settings and toe settings are within specification therefore, tire tread wear on the inside edge cannot happen if camber and toe are within specification.

Technician A cannot be right.

Effects of Caster misalignment:

Whenever there is a misalignment of the castor then the vehicle will not be able to go in straight line rather it will pull to either left or right side. Caster misalignment also causes heavy or light steering depending upon the positive or negative misalignment of caster.

Since it is given that caster settings are not within specification therefore, the vehicle may pull to the left due to the caster misalignment.

Technician B must be right.

4 0

3 years ago

I need help!!!!!!!!!

katovenus [111]

Answer:

buy a new one

Explanation:

go on newegg and you will most likely find one there

8 0

2 years ago