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A commercial facility has a three-phase, 277 volt, 1200 Amp service. What is the required nominal transformer size needed for th

alekssr [168]

Answer:

The answer is "Option e".

Explanation:

Given value:

$\to v_{ph}=277 \ \ \ \ \ \ \ I_{ph} =1200\\\\$

$vA= v_{Ph} I_{ph} \ \ \ \to for \ single \ phase\\\\=3v_{ph} I_{ph} \ \ \ \ \ \to for \ 3 - \phi \\\\s = 3\times 277 \times 1200 \\\\$

$=997.200 \ K VA\\\\= 1000 \ k VA$

7 0

3 years ago

A 35-ft³ rigid tank has propane at 25 psia, 540 R and is connected by a valve to another tank of 20 ft³ with propane at 40 psia,

gulaghasi [49]

Answer:

final pressure = 200KPa or 29.138psia

Explanation:

The detailed step by step calculations with appropriate conversion factors applied are as shown in the attachment.

8 0

3 years ago

PLEASE HELPPPPPPP!!!!,

raketka [301]

software engineers

hardware engineers

metallurgic engineers

biomechanical engineers

4 0

3 years ago

Read 2 more answers

What is tear test? What do we understand from this test? How?

Anvisha [2.4K]

Explanation:

The tear test determines the force required by a material to undergo complete failure when there is already a crack or tear present in it.

With this test we understand a material's resistance to failure when there is already a crack present.

The material which already has a crack is placed in a tensile testing or universal test machine. So, both sides of the material along the crack are pulled until material failure takes place.

6 0

4 years ago

Nancy ate a 500 Cal lunch. Neglecting efficiency issues (i.e., assuming 100% conversion of energy to work), to what height could

VladimirAG [237]

Answer:

$4265.04\ \text{m}$

$2.38\times 10^{10}\ \text{W}$

Explanation:

PE = Energy of food = 500 cal = $500\times4184=2.092\times10^6\ \text{J}$

m = Mass of object = 50 kg

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Potential energy of food is given by

$PE=mgh\\\Rightarrow h=\dfrac{PE}{mg}\\\Rightarrow h=\dfrac{2.092\times 10^6}{50\times 9.81}\\\Rightarrow h=4265.04\ \text{m}$

Nancy could raise the weight to a maximum height of $4265.04\ \text{m}$ .

Mass of $H_2$ used per year = $25\times 10^{9}\ \text{kg/year}$

Energy of $H_2$ = $\dfrac{30\times10^9}{1000}=30\times 10^6\ \text{J/kg}$

Power

$P=25\times 10^{9}\ \text{kg/year}\times 30\times 10^6\ \text{J/kg}\\\Rightarrow P=7.5\times 10^{17}\ \text{J/year}\\\Rightarrow P=\dfrac{7.5\times 10^{17}}{365.25\times 24\times 60\times 60}\\\Rightarrow P=2.38\times 10^{10}\ \text{W}$

The power requirement is $2.38\times 10^{10}\ \text{W}$ .

6 0

3 years ago

Need the answer now- Englishkailangan ang sagot ngayon-tagalogmag practice tayo!​

Need the answer now- Englishkailangan ang sagot ngayon-tagalogmag practice tayo!