B1) 20 pts. The thickness of each of the two sheets to be resistance spot welded is 3.5 mm. It is desired to form a weld nugget that is 5.5 mm in diameter and 5.0 mm thick after 0.3 sec welding time. The unit melting energy for a certain sheet metal is 9.5 J/mm3 . The electrical resistance between the surfaces is 140 micro ohms, and only one third of the electrical energy generated will be used to form the weld nugget (the rest being dissipated), determine the minimum current level required.

Answers

Answer 1

Answer:

minimum current level required =  8975.95 amperes

Explanation:

Given data:

diameter = 5.5 mm

length = 5.0 mm

T = 0.3

unit melting energy = 9.5 j/mm^3

electrical resistance = 140 micro ohms

thickness of each of the two sheets = 3.5mm

Determine the minimum current level required

first we calculate the volume of the weld nugget

v = [tex]\frac{\pi }{4} * D^2 * l[/tex] = [tex]\frac{\pi }{4} * 5.5^2 * 5[/tex] = 118.73 mm^3

next calculate the required melting energy

= volume of weld nugget * unit melting energy

= 118.73 * 9.5 = 1127.94 joules

next find the actual required electric energy

= required melting energy / efficiency

= 1127 .94 / ( 1/3 )  = 3383.84 J

TO DETERMINE THE CURRENT LEVEL REQUIRED  use the relation below

electrical energy =  I^2 * R * T

3383.84 / R*T = I^2

3383.84 / (( 140 * 10^-6 ) * 0.3 ) = I^2

therefore  8975.95 = I ( current )


Related Questions

/ Air enters a 20-cm-diameter 12-m-long underwater duct at 50°C and 1 atm at a

mean velocity of 7 m/s, and is cooled by the water outside. If the average heat

transfer coefficient is 85 W/m2

°C and the tube temperature is nearly equal to the

water temperature of 5°C, determine the exit temperature of air and the rate of heat

transfer.​

Answers

Answer:

A) EXIT TEMPERATURE = 14⁰C

b) rate of heat transfer of air = - 13475.78 = - 13.5 kw

Explanation:

Given data :

diameter of duct = 20-cm = 0.2 m

length of duct = 12-m

temperature of air at inlet= 50⁰c

pressure = 1 atm

mean velocity = 7 m/s

average heat transfer coefficient = 85 w/m^2⁰c

water temperature = 5⁰c

surface temperature ( Ts) = 5⁰c

properties of air at 50⁰c and at 1 atm

= 1.092 kg/m^3

Cp = 1007 j/kg⁰c

k = 0.02735 W/m⁰c

Pr = 0.7228

v  = 1.798 * 10^-5 m^2/s

determine the exit temperature of air and the rate of heat transfer

attached below is the detailed solution

Calculate the mass flow rate

= p*Ac*Vmean

= 1.092 * 0.0314 *  7 = 0.24 kg/s

A charge is distributed uniformly along a long straight wire. The electric field 2 cm from the wire is 36 N/C. The electric field 4 cm from the wire is:

Answers

Answer:

New electric field = 18 N/C

Explanation:

Given:

Length (E1) = 2 cm

New length (E2) = 4 cm

Electric field =  36 N/C

Find:

New electric field

Computation:

New electric field = 36 [2 / 4]

New electric field = 36 [1/2]

New electric field = 18 N/C

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