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2 years ago

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

Engineering

1 answer:

prisoha [69]2 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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Identify factors that can cause a process to become out of control. Give several examples of such factors.

Oliga [24]

Answer:

Explained

Explanation:

This situation can occur because of various factors such as:

Gradual deterioration of lubrication and coolant.
change of environmental condition such as temperature, humidity, moisture, etc.
Change in the properties of incoming raw material
An increase or decrease in the temperature of the heat treating operation
Debris interfering with the manufacturing process.

4 0

2 years ago

The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet

alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

$T_{h_i}=200^{\circ}C$

$T_{h_o}=120^{\circ}C$

$T_{c_i}=100^{\circ}C$

$T_{c_o}=125^{\circ}C$

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

$m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]$

$\frac{m_hc_{ph}}{m_cc_{pc}}$ = $\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )$

we can see that heat capacity of hot fluid is minimum

Also from energy balance

$Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )$

$NTU=\frac{UA}{\left ( mc_p\right )_{h}}$ = $\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}$

$T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}$

$T_m=41.63^{\circ}C$

NTU=1.921

$And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}$

$\epsilon =\frac{1-0.2669}{1-0.0834}$

$\epsilon =0.799$

5 0

3 years ago

Why is it important to keep hand tools free of oil and grease?

Bezzdna [24]

Answer:

keeps it from rusting

Explanation:

5 0

3 years ago

Can someone draw a electric circuit with electrical symbols for this

erastovalidia [21]

Answer:

i cant draw it

i am sry

Explanation:

4 0

2 years ago

Read 2 more answers

A 8.0-m-diameter parachute of a new design is to be used to transport a load from flight altitude to the ground with an average

S_A_V [24]

Answer:

drag coefficient for the parachute is Cd = 1.84177

Explanation:

given data

diameter = 8.0 m

average vertical speed = 3 m/s

total weight of load = 200 N

to find out

drag coefficient for the parachute

solution

we will apply here drag coefficient formula that is express as here

drag coefficient Cd = $\frac{2w}{\rho v^2A}$ ......................1

here w is total load and A is area and v is speed

and here ρ = 1.22 kg/cu

put here value we get

Cd = $\frac{2*4*200}{1.229*5^2*\pi * 3^2}$

Cd = 1.84177

drag coefficient for the parachute is Cd = 1.84177

3 0

2 years ago