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3 years ago

What the minimum wire size for a general residential application on a 20 A circuit

Engineering

1 answer:

prisoha [69]3 years ago

7 0

Answer:

The minimum wire size for a 20 A circuit would be 14 gauge wire. However modern fire and building (in the United States) code dictates that 12 gauge wire be used for 20 A. This is for added safety because there is a chance that 14 gauge wire can be overloaded by 20 A if the situation is right.

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What is the aim of reviewing a research paper?

qaws [65]

Answer:

Purpose of review papers

They carefully identify and synthesize relevant literature to evaluate a specific research question, substantive domain, theoretical approach, or methodology and thereby provide readers with a state-of-the-art understanding of the research topic.

3 0

3 years ago

Read 2 more answers

How has dissection used in engineering?

NeTakaya

Answer:

Product dissection has been widely deployed in engineering education as a means to aid in student's understanding of functional product elements, development of new concept ideas, and their preparation for industry.

Explanation:

I hope this helps :) have a wonderful day!

6 0

4 years ago

Read 2 more answers

The absolute pressure in water at a depth of 9 m is read to be 185 kPa. Determine: a. The local atmospheric pressure b. The abso

ser-zykov [4K]

Answer:

a)Patm=135.95Kpa

b)Pabs=175.91Kpa

Explanation:

the absolute pressure is the sum of the water pressure plus the atmospheric pressure, which means that for point a we have the following equation

Pabs=Pw+Patm(1)

Where

Pabs=absolute pressure

Pw=Water pressure

Patm= atmospheric pressure

Water pressure is calculated with the following equation

Pw=γ.h(2)

where

γ=especific weight of water=9.81KN/M^3

H=depht

Solving using ecuations 1 y 2

Patm=Pabs-Pw

Patm=185-9.81*5=135.95Kpa

Solving using ecuations 1 y 2, and atmospheric pressure

Pabs=0.8x5x9.81+135.95=175.91Kpa

8 0

4 years ago

A sign structure on the NJ Turnpike is to be designed to resist a wind force that produces a moment of 25 k-ft in one direction.

Jlenok [28]

Solution :

Finding the cohesion of the soil(c) using the relation:

$c = \frac{q_u}{2}$

Here, $q_u$ is the unconfined compression strength of the soil;

$c = \frac{800}{2}$

= 400 psf

∴ The cohesion value is greater than 0

So the use of the angle of internal friction is 0

Referring to the table relation between bearing capacity factors and angle of internal friction.

For the angle of inter friction $0^\circ$

$N_c = 5.14$

$N_q = 1.0$

$N_r = 0$

Therefore,

$q_{ult} = (400 \times 5.14 )+(110 \times 3 \times 1.0)+(0.5 \times 100 \times 13 \times 0)$

= 2386 psf

∴ Allowable bearing capacity $q_{a} = \frac{Q_{allow}}{A}$

$=\frac{30}{B^2}$

∴ $q_a = \frac{q_{ult}}{F.O.S}$

$\frac{30}{B^2} = \frac{2386}{3}$

∴ B = 0.2 ft

Therefore, the dimension of the square footing is 0.2 ft x 0.2 ft

$=0.04 \ ft^2$

7 0

3 years ago

The elementary liquid-phase series reaction

liraira [26]

Answer:

Concentration of A: $\frac{C_{A} }{C_{Ao} } =e^{-k_{1}t }$

Concentration of B: $\frac{C_{B} }{C_{Ao} } =\frac{k_{1} }{k_{2}-k_{1} } (e^{-k_{1}t } -e^{-k_{2}t } )$

Concentration of C: $\frac{C_{C} }{C_{Ao} } =1+\frac{k_{1} }{k_{2}-k_{1} } e^{-k_{2}t } -\frac{k_{2} }{k_{2}-k_{1} } e^{-k_{1} t}$

the image shows the graphs of the three concentrations

Explanation:

We have the reaction:

A ------->k1--------->B------------->k2--------->C

Each reaction:

$r_{A} =-k_{1} C_{A} \\r_{B} =k_{1} C_{A} -k_{2} C_{B} \\r_{C} =k_{2} C_{C}$

Where Cn is the concentration of each specie (A,B,C)

The mass balance for A:

$-\frac{dC_{A} }{dt} =-r_{A} \\-\frac{dC_{A} }{dt}=k_{1} C_{A} \\-\int\limits^y_x {\frac{dC_{A} }{dt} } \,=k_{1} t\\\frac{C_{A} }{C_{Ao} } =e^{-k_{1}t }$

Where x=CAo and y=CA

The mass balance for B:

$-\frac{dC_{B} }{dt} =-r_{B} \\-\frac{dC_{B} }{dt}=k_{2} C_{B} -k_{1} C_{A} \\\frac{dC_{B} }{dt}+k_{2} C_{B}=k_{1} C_{A}\\\frac{C_{B} }{C_{Ao} } =\frac{k_{1} }{k_{2}-k_{1} } (e^{-k_{1}t }-ex^{-k_{2}t } )$

The mass balance for C:

$\frac{C_{C} }{C_{Ao} } =1-\frac{C_{A} }{C_{Ao} } -\frac{C_{B} }{C_{Ao} } \\\frac{C_{C} }{C_{Ao} }=1+\frac{k_{1} }{k_{2}-k_{1} } e^{-k_{2} t}-\frac{k_{2} }{k_{2}-k_{1} } e^{-k_{1}t }$

The maximum concentration of C is:

$C_{Cmax} =C_{Ao} (\frac{k_{2} }{k_{1} } )^{\frac{k_{2} }{k_{2}-k_{1} }} =1.6(\frac{0.01}{0.4} )^{\frac{0.01}{0.01-0.4} } =1.76mol/dm^{3}$

and the maximum time is:

$t_{max} =\frac{ln\frac{k_{2} }{k_{1} } }{k_{2}-k_{1} } =\frac{ln\frac{0.01}{0.4} }{0.01-0.4} =9.4 h$

6 0

3 years ago