A series ac circuit consists of an inductor having a reactance of 80 Ω and an inductance of 190 mH, a 40-Ω resistor, a capacitor
1 answer:
V = IZ
V = rms voltage, I = rms current, Z = total circuit impedance
The impedance Z of a device is sometimes expressed as the sum of a real and imaginary component:
Z = R + jX
R = resistance, j = , X = reactance
The impedance of a capacitor is -j/(ωC)
ω = ac source frequency, C = capacitance
We can see that the impedance of a capacitor has no real component and has a negative imaginary component, that is to say:
Z = -jX, where we're given X = 100Ω, so Z = -j100Ω
The impedance of an inductor is jωL, ω = source frequency, L = inductance
We can see that the impedance of an inductor has no real component and has a positive imaginary component, that is to say:
Z = jX, where we're given X = 80Ω, so Z = j80Ω
The impedance of a resistor is simply R, R = resistance
WE can see that the impedance of a resistor has a positive real component and no imaginary component, that is to say:
Z = R, where we're given R = 40Ω, so Z = 40Ω
Add up the impedances to get the total impedance:
Z = 40Ω + j80Ω - j100Ω = (40-j80)Ω
To get the rms source voltage, we multiply the rms current I and the impedance Z. These two quantities will be complex numbers, and the math behind multiplying two complex numbers involves <em>multiplying their magnitudes</em>. We already have the magnitude of the rms current 2.2A, so let's calculate the magnitude of the impedance:
|Z| = √(40²+(-80)²)Ω = 89.44Ω
Now let's calculate the rms source voltage V = IZ:
V = 2.2A(89.44Ω)
V = 197V
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Answer:
W = 0 J
Explanation:
The amount of work done by gas at constant pressure is given by the following formula:
where,
W = Work done by the gas
P = Pressure of the gas
ΔV = Change in the volume of the gas
Since the volume of the gas is constant. Therefore, there is no change in the volume of the gas:
<u>W = 0 J</u>
Answer:
The answers to the question are
(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W
(ii) The temperature of the outer surface of the insulation is 49.89 °C
Explanation:
To solve the question, we note that the heat transferred is given by
Where
= Temperature at the inside of the pipe = 300 °C
= Temperature at the outside of the pipe = 20 °C
r₁ =internal radius of pipe = 4.0 cm
r₂ = Outer radius of pipe = 4.5 cm
r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm
= 15 W/m·K
= 0.038 W/m·K
= 75 W/m²·K
= 10 W/m²·K
Plugging in the values in the above equation where for a unit length L = 1 m, we have
Q = 131.32 W
From which we have, for the film of air at the pipe outer boundary layer
Where
for the air film on the pipe outer surface is given by
where A =area of the outside of the pipe
= = 0.227 K/W
Therefore
131.32 W = which gives
= 49.89 °C
Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)
Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)
T₁ = Surface temperature of the pipe = 49.89 °C and
T₂ = Temperature of the surrounding = 20.00 °C
Plugging in the values gives, q' = 0.307 W per m²
Total heat lost per unit length = 131.32 + 0.307 =131.62 W