A non-inductive load takes a current of 15A at 125V. An inductor is then connected in series in order that the same current shal
1 answer:
Answer:
The inductance of the inductor is 0.051H
Explanation:
From Ohm's law;
V = IR .................. 1
The inductor has its internal resistance referred to as the inductive reactance, X, which is the resistance to the flow of current through the inductor.
From equation 1;
V = IX
X =
................ 2
Given that; V = 240V, f = 50Hz, =
, I = 15A, so that;
From equation 2,
X=
= 16Ω
To determine the inductance of the inductor,
X = 2
fL
L =
=
= 0.05091
The inductance of the inductor is 0.051H.
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Answer:
The final specific internal energy of the system is 1509.91 kJ/kg
Explanation:
The parameters given are;
Mass of steam = 1 kg
Initial pressure of saturated steam p₁ = 1000 kPa
Initial volume of steam, = V₁
Final volume of steam = 5 × V₁
Where condition of steam = saturated at 1000 kPa
Initial temperature, T₁ = 179.866 °C = 453.016 K
External pressure = Atmospheric = 60 kPa
Thermodynamic process = Adiabatic expansion
The specific heat ratio for steam = 1.33
Therefore, we have;
Adding the effect of the atmospheric pressure, we have;
p = 1000 + 60 = 1060
We therefore have;
T₁/T₂ = 1.70083
T₁ = 1.70083·T₂
T₂ - T₁ = T₂ - 1.70083·T₂
Whereby the temperature of saturation T₁ = 179.866 °C = 453.016 K, we have;
T₂ = 453.016/1.70083 = 266.35 K
ΔU = 3××(T₂ - T₁)
= cv for steam at 453.016 K = 1.926 + (453.016 -450)/(500-450)*(1.954-1.926) = 1.93 kJ/(kg·K)
cv for steam at 266.35 K = 1.86 kJ/(kg·K)
We use cv given by (1.93 + 1.86)/2 = 1.895 kJ/(kg·K)
ΔU = 3××(T₂ - T₁) = 3*1.895 *(266.35 -453.016) = -1061.2 kJ/kg
The internal energy for steam =
= 2777.12 kJ/kg
= 0.194349 m³/kg
p = 1000 kPa
= 2777.12 - 0.194349 * 1060 = 2571.11 kJ/kg
The final specific internal energy of the system is therefore, + ΔU = 2571.11 - 1061.2 = 1509.91 kJ/kg.