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2 years ago

9

A single-phase transformer circuit feeds a motor and lighting load of 50 kilowatts. At a power factor of .8, the KVA rating of t

he step down transformer would be

1 answer:

AveGali [126]2 years ago

3 0

The KVA rating of the step down transformer at the given power factor would be 62.5 kVA.

<h3>What is power factor of a transformer?</h3>

Power factor (PF) is the ratio of working power, measured in kilowatts (kW), to apparent power, measured in kilovolt amperes (kVA).

PF = working power / apparent power

PF = kW/kVA

kVA = kW/PF

kVA = 50 kW/0.8

kVA = 62.5 kVA

Thus, the KVA rating of the step down transformer at the given power factor would be 62.5 kVA.

Learn more about power factor here: brainly.com/question/7956945

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Compute the thermal efficiency for an ideal gas turbine cycle that operates with a pressure ratio of 6.75 and uses helium gas.

pychu [463]

Solution:

Given:

pressure ratio, $r_{p}$ = 6.75

Formula used:

$\eta = \frac{(r_{p})^{\frac{\gamma -1}{\gamma}}-1}{r_{p}^{\frac{\gamma-1}{\gamma}}}\\$

$\eta = 1- \frac{1}{r_{p}^{\frac{\gamma-1}{\gamma }}}$ (1)

where,

$r_{p}$ = pressure ratio

γ = specific heat ratio of a gas( here, helium gas it is 1.667)

Now,

Eqn (1 ) is for thermal efficiency of an ideal gas, using eqn (1), we get

$\eta = 1- \frac{1}{6.75^{\frac{\ 1.667-1}{\ 1.667 }}}$

\eta = 1- \frac{1}{2.1469} = 0.5342

percentage thermal efficiency, \eta =53.42%

4 0

2 years ago

Explicar el funcionamiento de un multímetro analógico.

Whitepunk [10]

Answer:

Un multímetro analógico funciona como un medidor de bobina móvil de imán permanente (PMMC) para tomar mediciones eléctricas

Explanation:

El multímetro analógico es un medidor o galvanómetro D'Arsonval que funciona según el principio de los medidores de bobina móvil de imán permanente (PMMC)

Un multímetro analógico está formado por un puntero de aguja unido a una bobina móvil colocada entre el polo norte y sur de un imán permanente dispuesto de tal manera que, cuando una corriente eléctrica fluye a través de la bobina, genera una fuerza de campo magnético que interactúa con el imán fuerza de campo de los imanes permanentes que hace que la bobina se mueva junto con el puntero de la aguja sobre un dial graduado

Para controlar el movimiento del puntero de la aguja, de modo que el par requerido para producir una cantidad de movimiento por corriente detectada por el multímetro, se colocan dos resortes a través de la bobina para proporcionar resistencia al movimiento en ambas direcciones y para permitir la calibración del multímetro analógico.

4 0

3 years ago

A carbon resistor has a resistance of 976 ohms at 0 degrees C. Determine its resistance at 89 degrees C

nignag [31]

Answer:

1028.1184 Ohms

Explanation:

<u>Given the following data;</u>

Initial resistance, Ro = 976 Ohms
Initial temperature, T1 = 0°C
Final temperature, T2 = 89°C

Assuming the temperature coefficient of resistance for carbon at 0°C is equal to 0.0006 per degree Celsius.

To find determine its new resistance, we would use the mathematical expression for linear resistivity;

$R_{89} = R_{0} + R_{0}(\alpha T)$

Substituting into the equation, we have;

$R_{89} = 976 + 976*(0.0006*89)$

$R_{89} = 976 + 976*(0.0534)$

$R_{89} = 976 + 52.1184$

$R_{89} = 1028.1184 \ Ohms$

5 0

3 years ago

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the str

Murljashka [212]

The question is incomplete. The complete question is :

The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?

Solution :

Given data :

Diameter of the rod : 46 mm

Torque, T = 85 Nm

The polar moment of inertia of the shaft is given by :

$J=\frac{\pi}{32}d^4$

$J=\frac{\pi}{32}\times (46)^4$

J = 207.6 $mm^4$

So the shear stress at point A is :

$\tau_A =\frac{Tc_A}{J}$

$\tau_A =\frac{85 \times 10^3\times 12 }{207.6}$

$\tau_A = 4913.29 \ MPa$

Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.

3 0

2 years ago

The following electrical characteristics have been determined for both intrinsic and p-type extrinsic gallium antimonide (GaSb)

xxTIMURxx [149]

Answer:

0.5m^2/Vs and 0.14m^2/Vs

Explanation:

To calculate the mobility of electron and mobility of hole for gallium antimonide we have,

$\sigma = n|e|\mu_e+p|e|\mu_h$ (S)

Where

e= charge of electron

n= number of electrons

p= number of holes

$\mu_e=$ mobility of electron

$\mu_h=$ mobility of holes

$\sigma =$ electrical conductivity

Making the substitution in (S)

Mobility of electron

$8.9*10^4=(8.7*10^{23}*(-1.602*10^{-19})*\mu_e)+(8.7*10^{23}*(-1.602*10^{-19})*\mu_h)$

$0.639=\mu_e+\mu_h$

Mobility of hole in (S)

$2.3*10^5 = (7.6*10^{22}*(-1.602*10^{-19})*\mu_e)+(1*10^{25}*(-1.602*10^{-19}*\mu_h))$

$0.1436 = 7.6*10^{-3}\mu_e+\mu_h$

Then, solving the equation:

$0.639=\mu_e+\mu_h$ (1)

$0.1436 = 7.6*10^{-3}\mu_e+\mu_h$ (2)

We have,

Mobility of electron $\mu_e = 0.5m^2/V.s$

Mobility of hole is $\mu_h = 0.14m^2/V.s$

6 0

3 years ago