How to calculate energy in kWh for DC energy measurment in MikroC

[swethamenon]

In MY PROGRAM FOR DC ENRERY MEASUMENT,have some doubt
voltage in V& current inA
so power in Watts.
each second sensing v& i.
1watt=joules/sec
so to convert into kW/hr(will be a small value).say if joules/sec reaches --(how much value) convert to kWhr.
How to code for this??

adcval=Adc_Read(0)*.0048;
adcval1=Adc_Read(1)*.0048;
Is=adcval/330;
I=Is/2.5;
V1=I*5000;
floattostr(V1,val);
val[5]='\0';
lcd_out(1,3,val);
Is1=adcval1/330;
I1=Is1*1000;
floattostr(I1,val);
val[5]='\0';
lcd_out(1,11,val);
power = V1 * I1;
//floattostr(power,p_val);
//p_val[5]='\0';
//lcd_out(2,3,p_val);
energy =(energy + power);
floattostr(energy,val);
//val[5]='\0';
lcd_out(2,3,val);
delay_ms(1000);
}

 

[jayanth.devarayanadurga]

P = V * I = I^2 * R

P (in Watts) = V (in volts) * I (in amperes)

E (in joules) = P (in watts) * T (in sec)

Eg: 40 watt lamp is ON for 1 hour

E = 40 watt * 3600 sec

= 14,400 joules

Power in Kilowatts

6Kw + 4Kw + 3 * 100 watt

= 6 Kw + 4 Kw + (100 * 3) W

= (6 + 4) Kw + 300 W

= (6 + 4) Kw + (0.300 * 1000) W

= 10 Kw + 0.3 Kw

= 10.3 Kw

Energy in Kilowatt hours = Power in kilo watts * time in hours

= 10.3 Kw * 10 hours

= 103 kilowatt hours (kWh)


You have to calculate E every 1 sec using

E (in joules) = E + (P * T)

1 joule = 2.77777778 * 10^(-7) kilowatt hours



eg: if E (in joules) is 10000000 then

E (in kWh) = E (in joules) * 2.77777778 * 10^(-7)


= 10000000 * 2.77777778 * 10^(-7)

= 10^(7) * 2.77777778 * 10^(-7)

= 2.7777778 kWh