Explanation:
because it helps in our daily life
Answer:
Mechanic: a person who <u>repairs and maintains machinery</u>
Mechanical engineers: design <u>power-producing machines</u>
Explanation:
Answer:
t=5.5 mm
Heat dissipation per unit length = 90.477 W/m
Explanation:
Given that
Diameter d = 5 mm ⇒r = 2.5 mm
Conductivity of insulated material K = 0.16 W/mK
Heat transfer coefficient = 20
When thickness reaches up to critical radius of insulation then heat dissipation will be maximum
We know that critical radius of insulation of wire is given as follow
Now by putting the values
So the thickness of insulation
t=8-2.5 mm
t=5.5 mm
As we know that heat transfer due to convection given as follows
Q = hAΔ T
Q=20 x 2 x π x 0.008 x (120-30)
Q = 90.477 W/m
So heat dissipation per unit length = 90.477 W/m
Answer:
The required cross-sectional area of the copper wires used to connect the source to the mine lights is 0.029mm²
Explanation:
Given.
Copper Resistivity = 1.69 *10^-8Ωm
The mine lights use a total of 5 kW and operate at 120 V dc.
So,
Power = 5kW
Power Required = 5% of 5kW
Power Required = 0.05 of 5000W
Power Required = 250W
Calculating the Resistance
The power lost in the wires is given by 120² / R, where R is the resistance.
250 = 120²/R
R = 120²/250
R = 57.6 Ω
This is small amount over 100 m.
Calculating the Cross-sectional area.
The resistance of the wires is given by:
R = 1.69 * 10^(-8) *100 / A
R = 1.69 * 10^-6/A, where A is the cross-sectional area.
1.69 * 10^(-6)/A = 57.6
A = 1.69 * 10^-6/57.6
A = 2.9340277777777E−8m²
A = 0.029mm²