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3 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]3 years ago

4 0

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

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A composite wall is composed of 20 cm of concrete block with k = 0.5 W/m-K and 5 cm of foam insulation with k = 0.03 W/m-K. The

wariber [46]

Answer:

4.8°C

Explanation:

The rate of heat transfer through the wall is given by:

$q=\frac{Ak}{L}dT$

$\frac{q}{A}=\frac{k}{L}dT$

Assumptions:

1) the system is at equilibrium

2) the heat transfer from foam side to interface and interface to block side is equal. There is no heat retention at any point

3) the external surface of the wall (concrete block side) is large enough that all heat is dissipated and there is no increase in temperature of the air on that side

${k_{fi}= 0.03 W/m.K$

${L_{fi}= 5 cm = 0.05 m$

${T_{fi}= 25 \°C$

${k_{cb} = 0.5 W/m.K$

${L_{cb}= 20 cm = 0.20 m$

${T_{cb}= 0 \°C$

${T_{m}= ? \°C =$ temperature at the interface

Solving for ${T_{m}$ will give the temperature at the interface:

$\frac{q}{A}=\frac{k_{fi} }{L_{fi} }(T_{fi} -T_{m})=\frac{k_{cb} }{L_{cb} }(T_{m} -T_{cb})$

$\frac{0.03}{0.05 }(25 -T_{m})=\frac{0.5}{0.2}(T_{m} -0})$

$15 -0.6T_{m}=2.5T_{m}$

$3.1T_{m}=15$

$T_{m}=4.8$

3 0

4 years ago

A bakery wants to determine how many trays of doughnuts it should prepare each day. Demand is normal with a mean of 15 trays and

shutvik [7]

The number of trays that should be prepared if the owner wants a service level of at least 95% is; 7 trays

<h3>How to utilize z-score statistics?</h3>

We are given;

Mean; μ = 15

Standard Deviation; σ = 5

We are told that the distribution of demand score is a bell shaped distribution that is a normal distribution.

Formula for z-score is;

z = (x' - μ)/σ

We want to find the value of x such that the probability is 0.95;

P(X > x) = P(z > (x - 15)/5) = 0.95

⇒ 1 - P(z ≤ (x - 15)/5) = 0.95

Thus;

P(z ≤ (x - 15)/5) = 1 - 0.95

P(z ≤ (x - 15)/5) = 0.05

The value of z from the z-table of 0.05 is -1.645

Thus;

(x - 15)/5 = -1.645

x ≈ 7

Complete Question is;

A bakery wants to determine how many trays of doughnuts it should prepare each day. Demand is normal with a mean of 15 trays and standard deviation of 5 trays. If the owner wants a service level of at least 95%, how many trays should he prepare (rounded to the nearest whole tray)? Assume doughnuts have no salvage value after the day is complete. 6 5 4 7 unable to determine with the above information.

Read more about Z-score at; brainly.com/question/25638875

#SPJ1

4 0

2 years ago

Consider a modification of the air-standard Otto cycle in which the isentropic compression and expansion processes are each repl

Ulleksa [173]

Answer:

The answers to the question are

(1) Process 1 to 2

W = 295.16 kJ/kg

Q = -73.79 kJ/kg

(2) Process 2 to 3

W = 0

Q = 1135.376 kJ/kg

(3) Process 3 to 4

W = -1049.835 kJ/kg

Q = 262.459 kJ/kg

(4) Process 4 to 3

W=0

Q = -569.09 kJ/kg

(b) The thermal efficiency = 49.9 %

(c) The mean effective pressure is 9.44 bar

Explanation:

(a) Volume compression ratio $\frac{v_1}{v_2} = 10$

Initial pressure p₁ = 1 bar

Initial temperature, T₁ = 310 K

cp = 1.005 kJ/kg⋅K

Temperature T₃ = 2200 K from the isentropic chart of the Otto cycle

For a polytropic process we have

$\frac{p_1}{p_2} = (\frac{v_2}{v_1} )^n$ Therefore p₂ = p₁ ÷ $(\frac{v_2}{v_1} )^n$ = (1 bar) ÷ $(\frac{1}{10} )^{1.3}$ = 19.953 bar

Similarly for a polytropic process we have

$\frac{T_1}{T_2} = (\frac{v_2}{v_1} )^{n-1}$ or T₂ = T₁ ÷ $(\frac{v_2}{v_1} )^{n-1}$ = $\frac{310}{0.1^{0.3}}$ = 618.531 K

The molar mass of air is 28.9628 g/mol.

Therefore R = $\frac{8.3145}{28.9628}$ = 0.287 kJ/kg⋅K

cp = 1.005 kJ/kg⋅K Therefore cv = cp - R = 1.005- 0.287 = 0.718 kJ/kg⋅K

1). For process 1 to 2 which is polytropic process we have

W = $\frac{R(T_2-T_1)}{n-1}$ = $\frac{0.287(618.531-310)}{1.3 - 1}$ = 295.16 kJ/kg

Q = $(\frac{n-\gamma}{\gamma - 1} )W$ = $(\frac{1.3-1.4}{1.4-1} ) 295.16 kJ/kg$ = -73.79 kJ/kg

W = 295.16 kJ/kg

Q = -73.79 kJ/kg

2). For process 2 to 3 which is reversible constant volume heating we have

W = 0 and Q = cv×(T₃ - T₂) = 0.718× (2200-618.531) = 1135.376 kJ/kg

W = 0

Q = 1135.376 kJ/kg

3). For process 3 to 4 which is polytropic process we have

W = $\frac{R(T_4-T_3)}{n-1}$ = Where T₄ is given by $\frac{T_4}{T_3} = (\frac{v_3}{v_4} )^{n-1}$ or T₄ = T₃ × $0.1^{0.3}$

= 2200 × $0.1^{0.3}$ T₄ = 1102.611 K

W = $\frac{0.287(1102.611-2200)}{1.3 - 1}$ = -1049.835 kJ/kg

and Q = 262.459 kJ/kg

W = -1049.835 kJ/kg

Q = 262.459 kJ/kg

4). For process 4 to 1 which is reversible constant volume cooling we have

W = 0 and Q = cv×(T₁ - T₄) = 0.718×(310 - 1102.611) = -569.09 kJ/kg

W=0

Q = -569.09 kJ/kg

(b) The thermal efficiency is given by

$\eta = 1-\frac{T_4-T_1}{T_3-T_2}$ = $1-\frac{1102.611-310}{2200-618.531}$ = 0.499 or 49.9 % Efficient

(c) The mean effective pressure is given by

$p_{m} = \frac{p_1r[(r^{n-1}-1)(r_p-1)]}{ (n-1)(r-1)}$ where r = compression ratio and $r_p$ = $\frac{p_3}{p_2}$

However p₃ = $\frac{p_2T_3}{T_2}$ = $\frac{(19.953)(2200)}{618.531}$ =70.97 atm

$r_p$ = $\frac{p_3}{p_2}$ = $\frac{70.97}{19.953} = 3.56$

Therefore $p_m =\frac{1*10*[(10^{0.3}-1)(3.56-1)]}{0.3*9}$ = 9.44 bar

Please find attached generalized diagrams of the Otto cycle

8 0

3 years ago

A solid cylindrical workpiece made of 304 stainless steel is 150 mm in diameter and 100 mm is high. It is reduced in height by 5

goblinko [34]

Answer:

45.3 MN

Explanation:

The forging force at the end of the stroke is given by

F = Y.π.r².[1 + (2μr/3h)]

The final height, h is given as h = 100/2

h = 50 mm

Next, we find the final radius by applying the volume constancy law

volumes before deformation = volumes after deformation

π * 75² * 2 * 100 = π * r² * 2 * 50

75² * 2 = r²

r² = 11250

r = √11250

r = 106 mm

E = In(100/50)

E = 0.69

From the graph flow, we find that Y = 1000 MPa, and thus, we apply the formula

F = Y.π.r².[1 + (2μr/3h)]

F = 1000 * 3.142 * 0.106² * [1 + (2 * 0.2 * 0.106/ 3 * 0.05)]

F = 35.3 * [1 + 0.2826]

F = 35.3 * 1.2826

F = 45.3 MN

7 0

3 years ago

Emergency plans are being formulated so that rapid action can be taken in the event of an equipment failure. It is predicted tha

chubhunter [2.5K]

Answer:

Ammonia gas a hazardous gas to our health, when we are exposed to it for a long time. The gas is lighter than air, that means it's high concentration may not be noticed at the point of leakage, because it flows with the wind direction. Ammonia gas detector are used to determine the concentration of the gas at a particular place. We can use the dispersion modelling software program to know the exact position, where we can place the gas detector, which would be where evacuation is needed.

During evacuation, when the concentration of the gas has increased, a self-contained breathing apparatus should be used for breathing, and an encapsulated suit should be worn to prevent ammonia from reacting with our sweat or any other chemical burn. A mechanic ventilation will also be needed in the place of evacuation, so that the ammonia concentration in that area can be dispersed.

7 0

3 years ago

Read 2 more answers