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2 years ago

BHE:FLI:JPM next sequence

Engineering

1 answer:

Natasha_Volkova [10]2 years ago

0 0

Answer:

NTQ

Explanation:

The given sequence is

BHE : FLI : JPM

If is clear that, alphabets on first places are B, F, J. Difference between their place vales is 4.

B+4=F,F+4=J ; so alphabet on first place of next term of sequence is J+4=N.

Similarly, alphabets on second places are H, L, P. Difference between their place vales is 4.

H+4=L,L+4=P ; so alphabet on second place of next term of sequence is P+4=T.

Alphabets on third places are E, I, M. Difference between their place vales is 4.

E+4=I,I+4=M ; so alphabet on third place of next term of sequence is M+4=Q.

Therefore, the next term is NTQ.

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Answer: Heat transfer (Q) is 521 kJ.

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PV= $\frac{m}{M}$ RT, where:

P is pressure;

V is volume;

m is mass;

M is molar mass;

R is a constant: R = 8.314. $10^{-5}$ m³bar.K⁻¹.mol⁻¹;

T is temperature;

Using this equation, find the intial temperature:

PV = $\frac{m}{M}$ RT

1.2 = $\frac{2}{16}$ .8.314. $10^{-5}$ .T

T = 1.924. $10^{5}$ K

To determine the final temperature, use Combined Gas Law:

$\frac{P_{i} . V_{i} }{T_{i} }$ = $\frac{P.V }{T}$ , in which, the left side of the equality is related to the initial values and the right side, to the final values.

As pressure is constant:

$\frac{V_{i} }{T_{i} } =\frac{V}{T}$

T = $\frac{V.T_{i} }{V_{i} }$

T = $\frac{4.1.924.10^{-5} }{2}$

T = 3.85. $10^{5}$ K

With the temperatures, calculate the heat transfer of the process:

Q = m.k.ΔT, where:

k is heat constant

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The heat transfer in the process is 521 kJ.

6 0

2 years ago

Read 2 more answers

g The function below takes a single string parameter: sentence. Complete the function to return everything but the middle 10 cha

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Answer:

def processString(sentence):
middle = len(sentence) // 2
output = sentence[0: middle - 5] + sentence[middle+5:]
return output
print(processString("I have a dream"))

Explanation:

Create a function processString that take sentence as input paratemer (Line 1).

Next create a variable middle to hold the value of middle index of sentence string (Line 2)

Build the output string by slicing the input sentence from first character to character middle - 5th and from middle + 5th till the end of the string (Line 3).

Test the function using a sample sentence and we shall get the output "I am"

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2 years ago

A three-phase line has an impedance of 0.4 j2.7 ohms per phase. The line feeds two balanced three-phase loads that are connected

Viktor [21]

Answer:

a. The magnitude of the line source voltage is

Vs = 4160 V

b. Total real and reactive power loss in the line is

Ploss = 12 kW

Qloss = j81 kvar

Sloss = 12 + j81 kVA

c. Real power and reactive power supplied at the sending end of the line

Ss = 540.046 + j476.95 kVA

Ps = 540.046 kW

Qs = j476.95 kvar

Explanation:

a. The magnitude of the line voltage at the source end of the line.

The voltage at the source end of the line is given by

Vs = Vload + (Total current×Zline)

Complex power of first load:

S₁ = 560.1 < cos⁻¹(0.707)

S₁ = 560.1 < 45° kVA

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S₂ = P₂×1 (unity power factor)

S₂ = 132×1

S₂ = 132 kVA

S₂ = 132 < cos⁻¹(1)

S₂ = 132 < 0° kVA

Total Complex power of load is

S = S₁ + S₂

S = 560.1 < 45° + 132 < 0°

S = 660 < 36.87° kVA

Total current is

I = S*/(3×Vload) ( * represents conjugate)

The phase voltage of load is

Vload = 3810.5/√3

Vload = 2200 V

I = 660 < -36.87°/(3×2200)

I = 100 < -36.87° A

The phase source voltage is

Vs = Vload + (Total current×Zline)

Vs = 2200 + (100 < -36.87°)×(0.4 + j2.7)

Vs = 2401.7 < 4.58° V

The magnitude of the line source voltage is

Vs = 2401.7×√3

Vs = 4160 V

b. Total real and reactive power loss in the line.

The 3-phase real power loss is given by

Ploss = 3×R×I²

Where R is the resistance of the line.

Ploss = 3×0.4×100²

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Qloss = 3×X×I²

Where X is the reactance of the line.

Qloss = 3×j2.7×100²

Qloss = j81000 var

Qloss = j81 kvar

Sloss = Ploss + Qloss

Sloss = 12 + j81 kVA

c. Real power and reactive power supplied at the sending end of the line

The complex power at sending end of the line is

Ss = 3×Vs×I*

Ss = 3×(2401.7 < 4.58)×(100 < 36.87°)

Ss = 540.046 + j476.95 kVA

So the sending end real power is

Ps = 540.046 kW

So the sending end reactive power is

Qs = j476.95 kvar

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2 years ago

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Answer:

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Explanation:

We know,

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From the above formulas we can conclude that the thermal resistance of a substance mainly depends upon heat transfer coefficient,whereas radius has negligible effects on heat transfer coefficient.

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Answer:

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2 years ago