Variable Speed Orbital

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Orbital Welding, Orbital Welding Systems

Orbital welding is Automatic Tungsten inert gas welding. It eliminates chances of manual errors in welding. It produces identical welds for hundred of times hence accuracy in welding. The Physics of the GTAW Process: The orbital welding process uses the Gas Tungsten Arc Welding process (GTAW) as the source of the electric arc that melts the base material and forms the fusion weld. The metal to be welded is melted by the intense heat of the arc and fuses together. Reasons for Using Orbital Welding Equipment There are many reasons for using orbital welding equipment. The ability to make high quality, consistent welds repeatedly at a speed close to the maximum weld speed offer many benefits to the user: Productivity. An orbital welding system will drastically outperform manual welders, producing identical welds hundreds of times as per set program. Quality. The quality of a weld created by an orbital welding system with the correct weld program will be superior to that of manual welding. In applications such as semiconductor or pharmaceutical tube welding, orbital welding is the only means to reach the weld quality requirements. It eliminates chances of manual errors in welding. It produces identical welds for hundred of times hence accuracy in welding. Consistency. Once a weld program has been established an orbital welding system can repeatedly perform the same weld hundreds of times, eliminating the normal variability, inconsistencies, errors and defects of manual welding. Orbital welding systems may be used in applications where a tube or pipe to be welded cannot be rotated or where rotation of the part is not practical. Orbital welding equipments may be used in applications where access space restrictions limit the physical size of the welding device. Weld heads may be used in rows of boiler tubing where it would be difficult for a manual welder to use a welding torch or view the weld joint. Many other reasons exist for the use of orbital equipment over manual welding. Examples are applications where inspection of the internal weld is not practical for each weld created. By making a sample weld coupon that passes certification, the logic holds that if the sample weld is acceptable, that successive welds created by an automatic machine with the same input parameters should also be of same quality. Industries and Applications for Orbital Welding

Pharmaceutical Industry
Semiconductor Industry
Aerospace
Boiler Tube
Food, Dairy and Beverage Industries
Nuclear Piping
Offshore Applications

We undertake turnkey orbital welding jobs with supply of world class fittings and tubing / piping. We also undertake subcontracting jobs for orbital welding. We have our own 5 state of the art Orbital welding machines. More machines are available on requirement. The welding machines are microprocessor based with weld documentation facility. We have 30 well trained orbital welders with vast experience. So far we have installed around 300 Kilometers of tubing with orbital welding. We use only genuine electrodes / consumables as specified by manufacturer so that quality of the weld will be identical. All the support equipments like tube facing machines, flow measurement devices, etc are from world class manufacturers.

About the Author

Nitin Godse - http://www.excelgas.net - Excel Gas & Equipments Pvt. Ltd. - <a href="<a href="http://www.excelgas.net">http://www.excelgas.net" title="orbital welding">Orbital welding</a>

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Physics Question:?

I'm sure it's really simple but I'm terrible at physics. I think I'm doing something wrong. Please correct. Thank you.
Here's the question:

A 100 kg satellite is placed in a low Earth circular orbit with radius 1.2*10^7 m. An identical satellite is to be placed in circular orbit with twice the orbital period.
(a) Find the speed of the first satellite.
(b) Find the radius of the second satellite.

Here's what I have:
(a) v= square root of (Gm/r)
G=6.67*10^--11
Plug in and get v= 2.36 * 10 ^ -8 or 2.36 times 10 to the negative 8
(b) T (period) of the firs satellite = 2 (pie) square root of (r to the third / Gm)
multiply this by two and set that equal to the same equation where r is the variable wanted and then I find that r= 1.9 times 10 the the 7...

Thank you for helping. I think I might have plugged the wrong mass in. Should I plug in the mass of earth or the satellite?

Yes, ‘m’ in your formula is the mass of earth.
You have used 100 kg for m and hence the answer is
2.36 x10^ (-8) m/s

Always use capital M for mass of earth to avoid the mistake.

M = 5.977x 10^24 kg.
G = 6.673 x 10^ (-11) N m^2 /kg^2.
It may useful to remember the value of
G M = 3.99 x 10^14 in the case of earth.

(V)^2 = 3.99 x 10^14 / 1.2 x 10^7

V = 5765 m/s.
0000000000000000000000000000000000000000000

For the second problem,

Since (V) ^2 = GM / r and V = 2pi r / T.

(2pi r / T.)^2 = GM / r

T^2 is proportional to (r) ^3. (Kepler’s law)

(r2 / r1) ^3 = (T2 / T1) ^2

(T2 / T1) = 2 given.

(r2 / r1) ^3 = 4

r2 / r1 = 1. 5874

r2 = 1.5874 x 1.2 x 10^7

r2 = 1.9 x 10^7 m.

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