When they carry lots of current high voltage wires get hot d

When they carry lots of current, high voltage wires get hot due to electrical tance in the wire. When they get hot, they thermally expand and sag more than al. potentially coming in contact with trees. The largest blackout in US history caused by an event like this, with over 50 million people losing power, some for several days. Below is a picture of some lines. The transmission line is made of copper and has heated up from 20 C (ambient temperature) to 250 C when a large surge of current passes through. If the wind blowing over the wire is 8 m/s, how much heat is being carried away by the wind for a 1 km long section of wire that is 15 mm in diameter? What is the drag force on the wire due to the wind?

Solution

a) Here we can find the heat being carried with formula of heat loss by air flow.

Q = m_a. c_p.density of air. (t_h- t_a)

m_a= Air flow rate (m^3/ s)

c_{p= Specific heat of air =1.005 (kJ/ Kg C)}

Density of air= 1.2 kg/m³

t_h = temperature of the wire (max) = 250 C

t_a = ambient air temperature = 20 C

Speed of air = 8 m/s

Caluate air flow rate = Area . speed of air

= ( curved surface area of wire air faces ) . speed of air

Curved surface area of wire = ( .Radius . lenght of the wire taken i.e 1 km ) ( since only semi surface of the wire faces the air) only half of the surface is taken

Rasdius of wire= Diameter /2= 15/2=7.5mm= 0.0075m

Curved surface are = 3.14*.0075*1000=23.5619 m² (1km=1000m)

Air flow rate (m_a)= 23.5619 * 8= 188.4956 m³ /s

Q=188.4956 *1.005*1.2* (250-20)

Q ( heat carried out by air for 1 km) =52284.9095 kJ/s= 522.84.91 kW

b)

FD (Drag Force)

F_D= ½*Density of air * (velocity of air)²* C_D* Area of the curved surface area

C_D= Drag coefficient of wire( Assume 1 for real object)

F_D = .5 *1.2 * 8*8 * 1* 23.5619

=45.2388 kg.m/s²

F_d   = 45.24 N