Question

The voltage in an EBW operation is 45 kV. The beam current is 50 milliamp. The electron beam is focused on a circular area that is 0.50 mm in diameter. The heat transfer factor is 0.87. Calculate the average power density in the area in watt/mm2.

Answer 1

`P_d=6203.223062W/mm^2`

Step-by-step explanation:

From the question we are told that:

Voltage
`V=45kV`

Current
`I=50mAmp`

Diameter
`d=0.50mm`

Heat transfer factor
`\mu= 0.87.`

Generally the equation for Power developed is mathematically given by

`P=VI\\\\P=45*10^3*50*10^(-3)`

`P=2.250`

Therefore

Power in area

`P_a=1400*0.87`

`P_a=1218watt`

Power Density

`P_d=(P_a)/(Area)`

`P_d=(1218)/(\pi(0.5^2/4))`

`P_d=6203.223062W/mm^2`