Q = 3 Vph Iph sin(phase angle) = 31/2 Vline Iline sin(phase angle)
mw is the unit of real power and mvar is unit of reactive power. You should now the current and power factor angle to calculate the voltage.p=vi cos piq=vi sin piAnswerI think you mean MW, not mw -capital 'M' mega; lower-case 'm' milli!!! And the symbol for watt is a capital 'W', not a lower case 'w'. Also, I think you mean 'Mvar'(mega, not milli!).
Mega - Volt - Ampere - Reactive
By definition, MVA is equivalent to the vector sum of MW and MVAR: MVA^2 = MW^2 + MVAR^2 = 2500 MVA = 50
As you have written it (mvar), it stands for 'reactive millivolt amperes', but I think you probably mean 'Mvar', which stands for 'reactive megavolt amperes'. These are units for reactive power of an alternating-current load.A lower-case 'm' represents 'milli', whereas an upper-case 'M' represents 'mega'. 'V' represents 'volts', 'A' represents 'amperes', and 'r' represents 'reactive'.SI does not specify a symbol for reactive volt amperes (it only recognises 'watts'), so it is seen in written in various ways, including: var, VAR, VAr, VA(r), and VAr -in each case, it is also common to see a 'raised period' between the V and the A -e.g. V.Ar .
1mva = 1000kva so you simply divide by 1000. 10000KVA = 10MVA K = kilo = 1000 M = mega = 1000000
mw/mva=power factor reactive power(Q)=I2XL or E2/XL where XL= REACTANCE apparent power = square root of (MW2 + MVAR2 )
mw is the unit of real power and mvar is unit of reactive power. You should now the current and power factor angle to calculate the voltage.p=vi cos piq=vi sin piAnswerI think you mean MW, not mw -capital 'M' mega; lower-case 'm' milli!!! And the symbol for watt is a capital 'W', not a lower case 'w'. Also, I think you mean 'Mvar'(mega, not milli!).
MVA is the apparent power. MVA=( MW+ MVAr)1/2
Mega - Volt - Ampere - Reactive
By definition, MVA is equivalent to the vector sum of MW and MVAR: MVA^2 = MW^2 + MVAR^2 = 2500 MVA = 50
For 230 kv application, Its around 75k/MVar
There are two concerns here regarding loading on transformers of this size. First is the difference between MVA and MW. MW is just real power -- watts. MVA is total power which includes real power (MW) and reactive power (MVAR).--- http://en.allexperts.com/q/Electric-Power-Utilities-2405/operation-limit-oof-power.htm
As you have written it (mvar), it stands for 'reactive millivolt amperes', but I think you probably mean 'Mvar', which stands for 'reactive megavolt amperes'. These are units for reactive power of an alternating-current load.A lower-case 'm' represents 'milli', whereas an upper-case 'M' represents 'mega'. 'V' represents 'volts', 'A' represents 'amperes', and 'r' represents 'reactive'.SI does not specify a symbol for reactive volt amperes (it only recognises 'watts'), so it is seen in written in various ways, including: var, VAR, VAr, VA(r), and VAr -in each case, it is also common to see a 'raised period' between the V and the A -e.g. V.Ar .
Potentially. This would depend on the size of the cogeneration, and on the location in the electrical grid. Generally, the larger the cogeneration, the more requirements will be placed on it to act like a typical power plant (specific MVAR/MW output within specified power factor ranges, must stay online for faults, etc.).
MVA= square root of (MW2 + MVAR2 )
The SIL=(KV LL / Zo), where the V(LL) is the receiving end voltage in kV and Zo is the surge impedance in ohms. when the line is loaded over its SIL, it behaves like a shunt reactor - absorbing Mvar from the system, and when is loaded less its SIL it behaves like a shunt capacitor, supply Mvar back to the system. So to increase the Surge Impedance Loading (SIL), we need to decrease the the surge impedance of the line, and that can be done by introducing series capacitors (capacitors in series with the transmission line) or shunt capacitors (capacitors in parallel with transmission lines), which means providing Mvar to the system and reducing the Mw. hopefully that helps
Type your answer here... I t saves lot of power for variable load drives,Like FD ,ID fans and Feed pumps .The centrifugal pump consumes almost constant power at the rated speed VFD helps to reduce the speed and Flow NAD HENCE POWER REDUCTION. It enables smooth start up of the pump and hence reduced maintenance cost of the Pipe lines. It improves the power factor and hence the GRID to supply less MVAR to the induction Motors. Less penalty to the power Boards.