Answer:
L = 475.718
T = 240.89 ft
M = 23.0195
LC = 472.728
R = 1225 ft
Explanation:
See the attached file for the calculation.
Answer:
8.85 Ω
Explanation:
Resistance of a wire is:
R = ρL/A
where ρ is resistivity of the material,
L is the length of the wire,
and A is the cross sectional area.
For a round wire, A = πr² = ¼πd².
For aluminum, ρ is 2.65×10⁻⁸ Ωm, or 8.69×10⁻⁸ Ωft.
Given L = 500 ft and d = 0.03 in = 0.0025 ft:
R = (8.69×10⁻⁸ Ωft) (500 ft) / (¼π (0.0025 ft)²)
R = 8.85 Ω
Answer:
|W|=169.28 KJ/kg
ΔS = -0.544 KJ/Kg.K
Explanation:
Given that
T= 100°F
We know that
1 °F = 255.92 K
100°F = 310 .92 K
We know that work for isothermal process
Lets take mass is 1 kg.
So work per unit mass
We know that for air R=0.287KJ/kg.K
W= - 169.28 KJ/kg
Negative sign indicates compression
|W|=169.28 KJ/kg
We know that change in entropy at constant volume
ΔS = -0.544 KJ/Kg.K
Answer:
We know that all petrol engines are works on Otto cycle.Otto cycle have four process out of four two are constant volume process and others two are isentropic processes.
There are lots of limitations for practical Otto cycle these are as follows
1.In practical cycle heat can not add at constant volume.
2.In practical cycle there is a gap between combustion of two set of fuel.
3.Lots of heat is dissipated by cylinders.
4.Valve opening and closing is not a sudden process it requires some time.
5.There is a limitations of cylinder material ,it means that temperature of cycle can not rise after a specified limit of material.
Due to these above limitations practical cycles have low efficiency as compare to ideal cycle.
