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adminKeymaster
kamruzzaman said:
if i want to know, whats load will be need in a apertment in where 3*1.5 ton ac,5*100watts tube light,2*40watts celling fan,1 electric heater 1000watts. i mean how much current need?
1. 3*1.5 ton AC cunsumption in current is 16.5Amps when its calculated on 3 phase
2. 5*500 Watts tube light consumption is 0.69 Amps when its calculated on 3 phase
3. and all other loads will consume only 1.14 Amps
Total Requirement of current will be 18.4 Amps when its calculated on 3 phase.kamruzzaman said:
if i want to know, whats load will be need in a apertment in where 3*1.5 ton ac,5*100watts tube light,2*40watts celling fan,1 electric heater 1000watts. i mean how much current need?
kamruzzaman said:
if i want to know, whats load will be need in a apertment in where 3*1.5 ton ac,5*100watts tube light,2*40watts celling fan,1 electric heater 1000watts. i mean how much current need?
adminKeymasterW=VxI.
If the light remains 'on' for one hour, the consumption is 600 watt hour (.6KW)
1000 wats consumed over a period of one hour is one kilowatt hour (also called unit).
adminKeymasterelectrical deals with only generation and trasmission of energy while electronics controls deals with control of flow of energy
adminKeymasterAdd up your total wattage devide the total wattage by 230 or whatever you house voltage is that will give you current used watts/volts=current
adminKeymaster%impedence=impedence voltage/rated voltage *100
%Z=IZ/Vr *100
adminKeymasterI am not the specialist, but as I know well electrical material sales side – there is always the effect : – does the people in jungle not wear shoes because there are not shoes for sale or does nobody sell shoes because nobody buys them ?
Producer of such components should push his innovative product and the price of that product should be in last end still competitive.
For panel builder means the addittional product range higher stock cost and still supply uncertainty, at the same time the supplier has not the lovest price level to offer as the volume is low.
But always where is the problem there is an opportunity.
Be the first who will use such system!
adminKeymasterHi
Isn't the surge protective philosophie concerning T2 arresters described in Wiki-EIG a little bit old -fashioned?
http://www.electrical-installation.org/wiki/Selection_of_a_Type_2_SPD
I understand that it is coming from the time when France had own standard which knew only todays T2 arresters and they were suggested to be choosen concerning by risk of lightning (and by more valuable equipment).
Today, when international standard uses 'german' (my term) approach – T1, T2, T3 arresters – I do not see the reason to split T2 arresters (which are mainly varistors) into so many different levels. The Up of smaller varistors is mainly lower only for the reason that Up is always measured by In, so by the same surge current the actual Ures is the same , or by same cases – more powerful arresters have even lower Ures thah the smaller ones.
There is not significant price difference either (at least on our market) between T2 arresters having Imax 40, 20, 15 , or 5 kA.
T2 arresters having Imax 65 kA are today sold under label T1+2 with Iimp =7kA
Price depends more from how famous is the brand name of the producer or of his activity on the market.
I find that there is enough to have one standard T2 arrester 40/20kA and in the last end to use T3 arrester.
Btw. If I read the definition of the “Lighting Protection Zones” in EN 62 305 , than I understand, that LPZ 3 requires defenitely screening of the zone. So – it means that there is no sense to install T3 arrester into distribution board if the device to be protected is located not in the same DB but situated somewhere in distance and feeded via normal PVC cable without screen?
(of course – installation T3 there makes not the situation worse)
PS
What You as specialists think about so called 'T1+2' arresters, which itself are only a little bit more powerful varistors.
Is their effect really comparable with devices where 2 physically different devices (spark gap and varistor) are mounted beside each other (sometimes on the same base only, sometimes into the same housing)?
adminKeymasterusman said:
Answer????????
Difrencne b/w Electrical and Electronics is simply that Elctronics is a subset of Electrical and moreover in Electrical we deal with currents ranging from ameres to MEGA AMPERS, While in Electronics we deal with only Mili ampers.
adminKeymasterAll you need is to work out the reactive power byfirst finding the cos inverse of the current power factor then the tan of the samethen work out at therequired power factor aand get the difference this will be the value of the capacitor to be installed to improve the power factor.
adminKeymasterWhat is dc machines.
2011/04/10 at 4:10 am in reply to: Reading & understanding of the control and protection drawings of electrical installations e.g electrical Sub-station / Grid Station #11947adminKeymasterinstrumentpedia said:
by knowing internationally accepted standard electrical symbols you can easil read the drawing
thanks friend. It is a very helpfull to learn about electrical drawings
adminKeymasteryou save half of you money because you consumption of power i half the wattage. and half the consumption is half the bill. but with exemption if the utility company has a minimum charges of power bill…
2011/04/09 at 9:49 am in reply to: if a generator is generating 20MW and only 5MW is used where does the other 15MW goes? #11945adminKeymastertibbie said:
if a generator is generating 20MW and only 5MW is used where does the other 15MW goes?
if a generator is generating 20MW and only 5MW is used you have two cases . case one if this generator is connected to a grid ,then the 15 MW will be consumed in the grid. case 2 , if this generator in feeding the load directly it will never reach 20 MW , only the speed and the frequency will increase but the generator power will remain 5 MW.
P=T.w , p is the power , t is torqur, w is the angular speed , w=2.pi .f
=
adminKeymasterYes , take the line voltage 350v in place of 415v
2011/04/07 at 2:19 pm in reply to: if a generator is generating 20MW and only 5MW is used where does the other 15MW goes? #11942adminKeymasterPower Formula :
P= E.V.Sin δ / X
(Generator)
(Load) Ø angle(δ-ø) X
If load runs Ø angle then generator produce load w.r.t (δ-Ø) angle.
Where E = Energy,
V = Voltage,
δ = Power factor angle
X = Reactance
If u are not still clear you can contact me on my H.P (00919350130187)
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