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3 years ago

Which statement best describes the functions of a power transformer?

Engineering

2 answers:

pav-90 [236]3 years ago

7 0

........The answer is D

valkas [14]3 years ago

6 0

Answer:

Explanation:

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1. (5 pts) An adiabatic steam turbine operating reversibly in a powerplant receives 5 kg/s steam at 3000 kPa, 500 °C. Twenty per

KiRa [710]

Answer:

temperature of first extraction 330.8°C

temperature of second extraction 140.8°C

power output=3168Kw

Explanation:

Hello!

To solve this problem we must use the following steps.

1. We will call 1 the water vapor inlet, 2 the first extraction at 100kPa and 3 the second extraction at 200kPa

2. We use the continuity equation that states that the mass flow that enters must equal the two mass flows that leave

m1=m2+m3

As the problem says, 20% of the flow represents the first extraction for which 5 * 20% = 1kg / s

solving

5=1+m3

m3=4kg/s

we find the enthalpies and temeperatures in each of the states, using thermodynamic tables

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties

4.we find the enthalpy and entropy of state 1 using pressure and temperature

h1=Enthalpy(Water;T=T1;P=P1)

h1=3457KJ/kg

s1=Entropy(Water;T=T1;P=P1)

s1=7.234KJ/kg

remembering that it is a reversible process we find the enthalpy and the temperature in the first extraction with the pressure 1000 kPa and the entropy of state 1

h2=Enthalpy(Water;s=s1;P=P2)

h2=3116KJ/kg

T2=Temperature(Water;P=P2;s=s1)

T2=330.8°C

5.we find the enthalpy and the temperature in the second extraction with the pressure 200 kPav y the entropy of state 1

h3=Enthalpy(Water;s=s1;P=P3)

h3=2750KJ/kg

T3=Temperature(Water;P=P3;s=s1)

T3=140.8°C

Finally, to find the power of the turbine, we must use the first law of thermodynamics that states that the energy that enters is the same that must come out.

For this case, the turbine uses a mass flow of 5kg / s until the first extraction, and then uses a mass flow of 4kg / s for the second extraction, taking into account the above we infer the following equation

W=m1(h1-h2)+m3(h2-h3)

W=5(3457-3116)+4(3116-2750)=3168Kw

7 0

3 years ago

A well-insulated, rigid vessel contains 3 kg of saturated liquid water at 40°C. The vessel also contains an electrical resistor

fenix001 [56]

Answer:

The final temperature in the vessel after the resistor has been operating for 30 min is 111.67°C

Explanation:

given information:

mass, m = 3 kg

initial temperature, T₁ = 40°C

current, I = 10 A

voltage, V = 50 V

time, t = 30 min = 1800 s

Heat for the system because of the resistance is

Q = V I t

where

V = voltage (V)

I = current (A)

t = time (s)

Q = heat transfer to the system (J)

so,

Q = V x I x t

= 50 x 10 x 1800

= 900000

= 9 x 10⁵ J

the heat transfer in the closed system is

Q = ΔU + W

where

U = internal energy

W = work done by the system

thus,

Q = ΔU + W

9 x 10⁵ = ΔU + 0, W = 0 because the tank is a well-insulated and rigid.

ΔU = 9 x 10⁵ J = 900 kJ

then, the energy change in the system is

ΔU = m c ΔT

ΔT = ΔU / m c, c = 4.186 J/g°C

= 900 / (3 x 4.186)

= 71.67°C

so,the final temperature (T₂)

ΔT = T₂ - T₁

T₂ = ΔT + T₁

= 71.67°C + 40°C

= 111.67°C

7 0

3 years ago

A positively charged electric has a charge of 2 Coulombs. If 12.5x10^18 free electrons are added to it, what will be the net cha

vitfil [10]

The net charge on the said dielectric is 0

1C = 6.25 * 10¹⁸ electrons; hence:

-12.5x10¹⁸ free electrons = (-12.5x10¹⁸ free electrons * (1C/ 6.25 * 10¹⁸ electrons))

-12.5x10¹⁸ free electrons = -2 C

A positively charged electric has a charge of 2 Coulombs, hence:

Q₁ = 2C

12.5x10¹⁸ free electrons are added to it, hence:

Q₂ = -2C

$Net\ charge(Q_{net})=Q_1+Q_2=2C+(-2C)=0$

Hence the net charge on the said dielectric is 0

Find out more at: brainly.com/question/10464929

3 0

3 years ago

An element has two naturally occurring isotopes, isotope 1 with an atomic weight of 78.918 amu and isotope 2 with an atomic weig

Sidana [21]

Answer:

The fraction of isotope 1 is 50.7 % and fraction fraction of isotope 1 is 49.2 %.

Explanation:

Given that

Weight of isotope 1 = 78.918 amu

Weight of isotope 2 = 80.916 amu

Average atomic weight= 79.903 amu

Lets take fraction of isotope 1 is x then fraction of isotope 2 will be 1-x.

The total weight will be summation of these two isotopes

79.903 = 78.918 x + 80.916(1-x)

By solving above equation

80.916 - 79.903 = (80.916-78.918) x

x=0.507

So the fraction of isotope 1 is 50.7 % and fraction fraction of isotope 1 is 49.2 %.

4 0

4 years ago

Determine the real roots of f (x) = −0.6x2 + 2.4x + 5.5:(a) Graphically.(b) Using the quadratic formula.(c) Using three iteratio

zubka84 [21]

The three methods used to find the real roots of the function are,

graphically, the quadratic formula, and by iteration.

The correct vales are;

(a) Graphically, the roots obtained are; x ≈ -1.629, and 5.629

(b) Using the quadratic formula, the real roots of the given function are; x ≈ -1.62589, x ≈ 5.62859

Reasons:

The given function is presented as follows;

f(x) = -0.6·x² + 2.4·x + 5.5

(a) The graph of the function is plotted on MS Excel, with increments in the

x-values of 0.01, to obtain the approximation of the x-intercepts which are

the real roots as follows;

$\begin{array}{|c|cc|}x&&f(x)\\-1.63&&-0.00614\\-1.62&&0.03736\\5.62&&0.03736 \\5.63&&-0.00614\end{array}\right]$

Checking for the approximation of x-value of the intercept, we have;

$x = -1.63 + \dfrac{0 - (-0.00614)}{0.0376 - (-0.00614)} \times (-1.62-(-1.63)) \approx -1.629$

Therefore, based on the similarity of the values at the intercepts, the x-

values (real roots of the function) at the x-intercepts (y = 0) are;

x ≈ -1.629, and 5.629

(b) The real roots of the quadratic equation are found using the quadratic

formula as follows;

The quadratic formula for finding the roots of the quadratic equation

presented in the form f(x) = a·x² + b·x + c, is given as follows;

$x = \mathbf{ \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}}$

Comparison to the given function, f(x) = -0.6·x² + 2.4·x + 5.5, gives;

a = -0.6, b = 2.4, and c = 5.5

Therefore, we get;

$x = \dfrac{-2.4\pm \sqrt{2.4^{2}-4\times (-0.6)\times 5.5}}{2\times (-0.6)} = \dfrac{-2.4\pm\sqrt{18.96} }{-1.2} = \dfrac{12 \pm\sqrt{474} }{6}$

Which gives

The real roots are; x ≈ -1.62859, and x ≈ 5.62859

$x_l$ = 5, and $x_u$ = 10

The first iteration is therefore;

$x_r = \dfrac{5 + 10}{2} = 7.5$

$Estimated \ error , \ \epsilon _a = \left|\dfrac{10- 5}{10 + 5} \right | \times 100\% = 33.33\%$

$True \ error, \ \epsilon _t = \left|\dfrac{5.62859 - 7.5}{5.62859} \right | \times 100\% = 33.25\%$

f(5) × f(7.5) = 2.5 × (-10.25) = -25.625

The bracket is therefore; $x_l$ = 5, and $x_u$ = 7.5

Second iteration:

$x_r = \dfrac{5 + 7.5}{2} = 6.25$

$Estimated \ error , \ \epsilon _a = \left|\dfrac{7.5- 5}{7.5+ 5} \right | \times 100\% = 20\%$

$True \ error, \ \epsilon _t = \mathbf{\left|\dfrac{5.62859 - 6.25}{5.62859} \right | \times 100\%} \approx 11.04\%$

f(5) × f(6.25) = 2.5 × (-2.9375) = -7.34375

The bracket is therefore; $x_l$ = 5, and $x_u$ = 6.25

Third iteration

$x_r = \dfrac{5 + 6.25}{2} = 5.625$

$Estimated \ error , \ \epsilon _a = \left|\dfrac{5.625- 5}{5.625+ 5} \right | \times 100\% = 5.88\%$

$True \ error, \ \epsilon _t = \mathbf{\left|\dfrac{5.62859 - 5.625}{5.62859} \right | \times 100\%} \approx 6.378 \times 10^{-2}\%$

f(5) × f(5.625) = 2.5 × (0.015625) = 0.015625

Therefore, the bracket is $x_l$ = 5.625, and $x_u$ = 6.25

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6 0

3 years ago