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 )
/ 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.
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:
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