My Brother's Wedding Invitation

Dear Friends,

It’s my great pleasure to invite

You and your family on the auspicious occasion of

My Brother's Wedding and Reception

PRABHUGOUDA

Weds

PURVIKA

SANGANAGOUDA

Weds

KAVITA

On 7th Jun 2015,

Muhurtham:12:14pm

@

Buddha , Basava, Ambedkar Bhavan

#Koppal Road, Yelburga, Koppal Dist

Karnataka,  

Please treat this as my personal invitation and kindly grace this occasion with your presence and blessings.

Reception
on 7^th Jun 2015

Between 1:30pm to 4pm

@
Buddha , Basava, Ambedkar Bhavan

#Koppal Road, Yelburga, Koppal Dist

Karnataka

EAGERLY EXPECTING YOUR PRESENCE

Cheers,

BASANAGOUDA. S.MALIPATIL

ಶ್ರೀಬಸವೇಶ್ವರರ ವಚನಗಳು

ದಯವಿಲ್ಲದ ಧರ್ಮವಾವುದಯ್ಯಾ ?
ದಯವೇ ಬೇಕು ಸಕಲ ಪ್ರಾಣಿಗಳಲ್ಲಿಯೂ
ದಯವೇ ಧರ್ಮದ ಮೂಲವಯ್ಯೂ
ಕೂಡಲಸಂಗಯ್ಯನಂತಲ್ಲದೊಲ್ಲನಯ್ಯೂ !!

ತನಗೆ ಮುನಿವವರಿಗೆ ತಾ ಮುನಿಯಲೇಕಯ್ಯಾ?
ತನಗಾದ ಆಗೇನು? ಅವರಿಗಾದ ಚೇಗೇನು?
ತನುವಿನ ಕೋಪ ತನ್ನ ಹಿರಿಯತನದ ಕೇಡು?
ಮನದಾ ಕೋಪ ತನ್ನರಿವಿನಾ ಕೇಡು
ಮನೆಯೊಳಗಣ ಕಿಚ್ಚು ಮನೆಯ ಸುಟ್ಟಲ್ಲದೆ
ನೆರೆಮನೆಯ ಸುಡದು ಕೂಡಲಸಂಗಮದೇವಾ !!

ಜನಿತಕ್ಕೆ ತಾಯಾಗಿ ಹೆತ್ತಳು ಮಾಯೆ!
ಮೋಹಕ್ಕೆ ಮಗಳಾಗಿ ಹುಟ್ಟಿದಳು ಮಾಯೆ!
ಕೂಟಕ್ಕೆ ಸ್ರೀಯಾಗಿ ಕೂಡಿದಳು ಮಾಯೆ!
ಇದಾವ ಪರಿಯಲ್ಲಿ ಕಾಡಿತ್ತು ಮಾಯೆ?
ಈ ಮಾಯೆಯ ಕಳವೊಡೆ ಎನ್ನಳವಲ್ಲ;
ನೀವೇ ಬಲ್ಲಿರಿ ಕೂಡಲಸಂಗಮದೇವಾ!!

ನೆಲವೊಂದೆ ಹೊಲೆಗೇರಿ ಶಿವಾಲಯಕ್ಕೆ
ಜಲವೊಂದೆ ಶೌಚಾಚಮನಕ್ಕೆ
ಕುಲವೊಂದೆ ತನ್ನ ತಾನರಿದವಂಗೆ
ಫಲವೊಂದೆ ಷಡುದರುಶನ ಮುಕ್ತಿಗೆ
ನಿಲವೊಂದೆ ಕೂಡಲಸಂಗಮದೇವಾ ನಿಮ್ಮನರಿದವಂಗೆ !!

ತನು ನಿಮ್ಮದೆಂದ ಬಳಿಕ ಎನಗೆ ಬೇರೆ ತನುವಿಲ್ಲ
ಮನ ನಿಮ್ಮದೆಂದ ಬಳಿಕ ಬೇರೆ ಮನವಿಲ್ಲ
ಧನ ನಿಮ್ಮದೆಂದ ಬಳಿಕ ಎನಗೆ ಬೇರೆ ಧನವಿಲ್ಲ
ಇಂತೀ ತ್ರಿವಿಧವೂ ನಿಮ್ಮದೆಂದರಿದ ಬಳಿಕ ಎನಗೆ ಬೇರೆ
ವಿಚಾರವುಂಟೆ ಕೂಡಲಸಂಗಮದೇವಾ!!

ಕಳಬೇಡ ಕೊಲಬೇಡ
ಹುಸಿಯ ನುಡಿಯಲು ಬೇಡ
ಮುನಿಯಬೇಡ ಅನ್ಯರಿಗೆ ಅಸಹ್ಯ ಪಡಬೇಡ
ತನ್ನ ಬಣ್ಣಿಸಬೇಡ ಇದಿರ ಹಳಿಯಲು ಬೇಡ
ಇದೇ ಅಂತರಂಗ ಶುದ್ಧಿ ಇದೇ ಬಹಿರಂಗ ಶುದ್ಧಿ
ಇದೇ ನಮ್ಮ ಕೂಡಲಸಂಗಮನೊಲಿಸುವ ಪರಿ!!

ತಂದೆ ನೀನು ತಾಯಿ ನೀನು
ಬಂಧು ನೀನು ಬಳಗ ನೀನು
ನೀನಲ್ಲದೆ ಮತ್ತಾರೂ ಇಲ್ಲವಯ್ಯಾ
ಕೂಡಲಸಂಗಮದೇವಾ
ಹಾಲಲ್ಲದ್ದು ನೀರಲ್ಲದ್ದು ನಿಮ್ಮ ಧರ್ಮ!

On Line Capacitor bank Calculation

On Line Capacitor bank Calculation
http://ambercaps.com/technicalinfo/capkvarcalculator.htm

Capacitor bank selecting formula

Power Factor is defined as the ratio of Real power and
Apparent power.

for calculating Capacitor bank value the formula is :

kVAr = KW {tan(cos^-1 Phi e) - tan(Cos^-1 Phi d)}

Where -  Phi e = existing Power Factor.
         Phi d = Desired Power Factor.

 

Electrical Eng

http://engineering.electrical-equipment.org/electrical-distribution/how-choose-surge-protection-device.html

"ಶಿವರಾತ್ರಿ ಹಬ್ಬದ ಶುಭಾಶಯಗಳು"

"ಶಿವರಾತ್ರಿ ಹಬ್ಬದ ಶುಭಾಶಯಗಳು"

Heat Conversion

Hello All,
This is an edited re-posting. I am looking for verification on a formula. As part of a personal design project (not for school or homework) I’m designing coils of magnet wire (i.e. coils in a generator) and I have all of the computational resources I need, with the exception of a formula to calculate the conversion of a power flow in watts to a temperature (°C or °F). I’ve been doing a good bit of searching and researching on the web and I’ve found a formula that I would like feedback on.

Now, to utilize a very useful example I found on the web; let’s say that silly me is goofing around with a 15 foot piece of 8 gauge (AWG) copper wire, which has .01 ohms of resistance and that the wire is connected directly to the positive terminal of a 12V automotive battery (no fuse, silly me) and I clumsily allow the other end to contact the negative terminal of the same battery. Dummy me is now holding a wire with 1200 amps running through it. (ouch)

I = E/R
I = 12/0.01
I = 1200 amps

I take this current and plug it into the formula P=I^2*R (or more simply E*I=P) and we get:

P = I2*R
P = (1200*1200)*0.01
P = 14,400 Watts
Or
E*I=P
12V * 1200A = 14,400W

This shows that the wire would dissipate 14,400 watts of heat which would melt the wire's insulation and more than likely ignite everything that comes into contact with the wire (fuel lines, other wires, carpet, plastic, insulation, my skin…).

I can figure out all the same information on the wire coils I'm dealing with and making for my project; however, because the manufacturers all primarily site the thermal limits of their wire insulation in terms of °C or °F etc., that's where I need to be able to figure the conversion from the above example's 14,400 Watts to °C or °F.

On a very educational site called Aus-e-tute at http://www.ausetute.com.au/. They list a specific heat formula for calculating the gain or loss of heat energy in a substance. The formula is based as: The amount of heat energy (q) in Joules gained or lost by a substance is equal to the mass of the substance (m) in grams multiplied by its specific heat capacity (Cg) multiplied by the change in temperature (final temperature - initial temperature) and is stated as: q = m x Cg x (Tf - Ti).

Now, knowing that 1 watt = 1 Joule/second; and using the conversion of 1 lb. = 453.59g, 15 feet of 8 AWG copper wire is 338.83 grams; the specific heat capacity (Cg) for copper is listed (on average) as being 0.385 and lets say the wire was laying in the shade during one of our balmy Florida summer days and it’s starting temperature (Ti) is a cool 30°C (86°F) and finally, we know from my silliness above that the wire has 14,400 watts running through it. So, from the basic laws of mathematics, I believe I can restate the formula as:
q / [(m x Cg) + Ti] = Tf

which, when the numbers are plugged in would give me:
14,400 (J/s) / {[338.83173g * 0.385 (Copper’s Cg)] + 30(°C)} = 89.75(°C) or 193.55(°F)

Preliminary calculations seem to work out well for my needs, but I would like opinions on my restatement of the formula. Basically, what I need to know is, will this calculation actually work as I think it will; even if it only gets me an accuracy of +/- a couple of °F; or am I only wishfully hallucinating.