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Answer:
the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²
Explanation:
Given the data in the question;
To determine the maximum intensity of an electromagnetic wave, we use the formula;
=
ε₀cE
²
where ε₀ is permittivity of free space ( 8.85 × 10⁻¹² C²/N.m² )
c is the speed of light ( 3 × 10⁸ m/s )
E is the maximum magnitude of the electric field
first we calculate the maximum magnitude of the electric field ( E )
E = 350/f kV/m
given that frequency of 60 Hz, we substitute
E = 350/60 kV/m
E = 5.83333 kV/m
E = 5.83333 kV/m × (
)
E = 5833.33 N/C
so we substitute all our values into the formula for intensity of an electromagnetic wave;
=
ε₀cE
²
=
× ( 8.85 × 10⁻¹² C²/N.m² ) × ( 3 × 10⁸ m/s ) × ( 5833.33 N/C )²
= 45 × 10³ W/m²
= 45 × 10³ W/m² × (
)
= 45 kW/m²
Therefore, the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²
Answer:
(1) The maximum air temperature is 1383.002 K
(2) The rate of heat addition is 215.5 kW
Explanation:
T₁ = 17 + 273.15 = 290.15
T₂ = 290.15 × 3.17767 = 922.00139
Therefore,
T₃ = T₂×1.5 = 922.00139 × 1.5 = 1383.002 K
The maximum air temperature = T₃ = 1383.002 K
(2)
Therefore;
Q₁ = 1.005(1383.002 - 922.00139) = 463.306 kJ/jg
Heat rejected per kilogram is given by the following relation;
= 0.718×(511.859 - 290.15) = 159.187 kJ/kg
The efficiency is given by the following relation;
Where:
β = Cut off ratio
Plugging in the values, we get;
Therefore;
Heat supplied =
Therefore, heat supplied = 215491.064 W
Heat supplied ≈ 215.5 kW
The rate of heat addition = 215.5 kW.