Applying the ideal gas equation:
PV = nRT
PV = (mass/Mr)RT
mass = PVMr/RT
mass = (101325 x 4672.2 x 4) / (8.314 x 297)
= 766887.3 kg
= 7.7 x 10⁵ kg
The answer is increased. A power factor of one or "unity power factor" is the aim of any electric utility business from the time when if the power factor is less than one, they have to give more current to the user for an assumed amount of power use. In so doing, they gain more line harms.
Air flowing from areas of high pressure to low pressure creates wind.
Answer:
a) 90 kJ
b) 230.26 kJ
Explanation:
The pressure at the first point 
= 10 bar —> 10 x 102 = 1020 kPa
The volume at the first point 
= 0.1 m^3
The pressure at the second point 
= 1 bar —> 1 x 102 = 102 kPa
The volume at the second point 
= 1 m^3
Process A.
constant volume V = C from point (1) to P = 10 bar.
Constant pressure P = C to the point (2).
Process B.
The relation of the process is PV = C
Required
For process A & B
(a) Sketch the process on P-V coordinates
(b) Evaluate the work W in kJ.
Assumption
Quasi-equilibrium process
Kinetic and potential effect can be ignored.
Solution
For process A.
V=C
There is no change in volume then
The work is defined by
║
V║limit 1--0.1
90 kJ
Process B
PV=C
By substituting with point (1) C = 10^2 x 1= 10^2
The work is defined by
║ ln(V)║limit 1--0.1
=230.26 kJ
Answer:
31.8 × 10⁻⁴ J = 3.18 mJ
Explanation:
We know the intensity I of a wave is I = P/A where P = power and A = area = 0.500 m²
The intensity of an electromagnetic wave is also equal to I = E₀²/μ₀c
where E₀ = maximum electric field strength = √2E where E = rms value of electric field = 0.0200 N/C, μ₀ = 4π × 10⁻⁷ H/m ,c = 3 × 10⁸ m/s
P/A = E₀²/μ₀c = 2E²/μ₀c
P = 2E²A/μ₀c = 2 × (0.02 N/C)² × 0.5 m²/(4π × 10⁻⁷ H/m × 3 × 10⁸ m/s)
= 1.06 × 10⁻⁴ W = 0.106 mW
Since P = E/t where E = Energy and t = time
E = Pt with t = 30 s
E = 1.06 × 10⁻⁴ W × 30 s = 31.8 × 10⁻⁴ J = 3.18 mJ
So the wave carries 3.18 mJ of energy through the window in 30 s
