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adelina 88 [10]

2 years ago

10

A single-phase load is located 2800 ft from its source. The load draws a current of 86 A and operates on 480 V. The maximum volt

age drop at the load cannot be greater than 3%. What size aluminum conductors should be installed to operate this load?

1 answer:

Akimi4 [234]2 years ago

6 0

check photo for answer

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Show the ERD with relational notation with crowfoot. Your ERD must show PK, FKs, min and max cardinality, and correct line types

zhenek [66]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.

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3 years ago

Suppose you have two boxes in front of you. One box contains a Thevenin Equivalent (voltage source in series with a resistor) an

fomenos

Answer:

1. Measure the temperature of the boxes and leave them unconnected.

2. Norton reduces his circuit down to a single resistance in parallel with a constant current source. A real-life Norton equivalent circuit would be continuously wasting power (as heat) as the current source dumps energy into the resistor, even when externally unconnected, while a Thevenin equivalent circuit would sit there doing nothing.

3. The Norton equivalent box would get warm and eventually run out of power. The Thevenin equivalent box would stay at ambient temperature.

8 0

3 years ago

Water vapor at 100 psi, 500 F and a velocity of 100 ft./sec enters a nozzle operating at steady sate and expands adiabatically t

almond37 [142]

Answer:

a)exit velocity of the steam, V2 = 2016.8 ft/s

b) the amount of entropy produced is 0.006 Btu/Ibm.R

Explanation:

Given:

P1 = 100 psi

V1 = 100 ft./sec

T1 = 500f

P2 = 40 psi

n = 95% = 0.95

a) for nozzle:

Let's apply steady gas equation.

$h_1 + \frac{(v_1) ^2}{2} = h_2 + \frac{(v_2)^2}{2}$

h1 and h2 = inlet and exit enthalpy respectively.

At T1 = 500f and P1 = 100 psi,

h1 = 1278.8 Btu/Ibm

s1 = 1.708 Btu/Ibm.R

At P2 = 40psi and s1 = 1.708 Btu/Ibm.R

1193.5 Btu/Ibm

Let's find the actual h2 using the formula :

$n = \frac{h_1 - h_2*}{h_1 - h_2}$

$n = \frac{1278.8 - h_2*}{1278.8 - 1193.5}$

solving for h2, we have

$h_2 = 1197.77 Btu/Ibm$

Take Btu/Ibm = 25037 ft²/s²

Using the first equation, exit velocity of the steam =

$(1278.8 * 25037) + \frac{(100)^2}{2}= (1197.77*25037)+ \frac{(V_2)^2}{2}$

Solving for V2, we have

V2 = 2016.8 ft/s

b) The amount of entropy produced in BTU/ lbm R will be calculated using :

Δs = s2 - s1

Where s1 = 1.708 Btu/Ibm.R

At h2 = 1197.77 Btu/Ibm and P2 =40 psi,

S2 = 1.714 Btu/Ibm.R

Therefore, amount of entropy produced will be:

Δs = 1.714Btu/Ibm.R - 1.708Btu/Ibm.R

= 0.006 Btu/Ibm.R

3 0

3 years ago

Which two statements about professional technical jobs in the energy industry are correct?

Tanya [424]

The answer is both B and D

4 0

2 years ago

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An oxygen–nitrogen mixture consists of 35 kg of oxygen and 40 kg of nitrogen. This mixture is cooled to 84 K at 0.1 MPa pressure

Tamiku [17]

Answer:

The mass of oxygen in liquid phase = 14.703 kg

The mass of oxygen in the vapor phase = 20.302 kg

Explanation:

Given that:

The mass of the oxygen $m_{O_2}$ = 35 kg

The mass of the nitrogen $m_{N_2}$ = 40 kg

The cooling temperature of the mixture T = 84 K

The cooling pressure of the mixture P = 0.1 MPa

From the equilibrium diagram for te-phase mixture od oxygen-nitrogen as 0.1 MPa graph. The properties of liquid and vapor percentages are obtained.

i.e.

Liquid percentage of $O_2$ = 70% = 0.70

Vapor percentage of $O_2$ = 34% = 0.34

The molar mass (mm) of oxygen and nitrogen are 32 g/mol and 28 g/mol respectively

Thus, the number of moles of each component is:

number of moles of oxygen = 35/32

number of moles of oxygen = 1.0938 kmol

number of moles of nitrogen = 40/28

number of moles of nitrogen = 1.4286 kmol

Hence, the total no. of moles in the mixture is:

$N_{total} = 1.0938+1.4286$

$N_{total} = 2.5224 \ kmol$

So, the total no of moles in the whole system is:

$N_f + N_g = 2.5224 --- (1)$

The total number of moles for oxygen in the system is

$0.7 \ N_f + 0.34 \ N_g = 1.0938 --- (2)$

From equation (1), let N_f = 2.5224 - N_g, then replace the value of N_f into equation (2)

∴

0.7(2.5224 - N_g) + 0.34 N_g = 1.0938

1.76568 - 0.7 N_g + 0.34 N_g = 1.0938

1.76568 - 0.36 N_g = 1.0938

1.76568 - 1.0938 = 0.36 N_g

0.67188 = 0.36 N_g

N_g = 0.67188/0.36

N_g = 1.866

From equation (1)

$N_f + N_g = 2.5224$

N_f + 1.866 = 2.5224

N_f = 2.5224 - 1.866

N_f = 0.6564

Thus, the mass of oxygen in the liquid and vapor phases is:

$m_{fO_2} = 0.7 \times 0.6564 \times 32$

$m_{fO_2} = 14.703 \ kg$

The mass of oxygen in liquid phase = 14.703 kg

$m_{g_O_2} = 0.34 \times 1.866 \times 32$

$m_{g_O_2} = 20.302 \ kg$

The mass of oxygen in the vapor phase = 20.302 kg

8 0

3 years ago