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electrical engineering transformer problem?
A 500 KVA transformer has a 0.6 lagging power factor load of 280 KVA. (a) If additional 0.8 power factor lagging loads are to be added, how many KVA of such loads can be added without overloading the transformer? (b) A 0.8 lagging power factor load of 300 KVA is to be added. Will the transformer be overloaded? (c) If the answer to (b) is yes, determine the KVAR of capacitors needed to reduce the load on the transformer to its rated value.
In (b), please answer this portion providing a solution that will support your answer. Thanks!
4 Answers
- texasmaverickLv 79 years agoFavorite Answer
A) 280 KVA at 0.6 power factor is 168 KW The KVAR for this load is Sqrt(280^2 - 168 ^2) = 224 KVAR
B) 300 KVA at 0.8 power factor is 240 KW.
The KVAR for this load is Sqrt(300 ^ 2 - 240 ^ 20) = 180 KVAR.
The KW of the original load plust the KW of the 300 KVA load is 168 + 240 = 408 KW.
The KVAR of both these loads is 224 + 180 = 404.
This means the KVA of the total load is Sqrt(KW ^2 + KVAR ^ 2) = Sqrt(408 ^2 + 404 ^2) = 572.77 KVA
The transformer will be overloaded.
C) The KVAR required to limit a the KVA of a transformer to 500 KVA, with the KW load of the transformer being 408 KW is KVAR = Sqrt(500 ^2 - 408 ^2) = 289 KVAR. The KVAR of the required capacitor is 404 - 289 = 115 KVAR
TexMav
- JohnB.Lv 59 years ago
Power factor is defined as the ratio of real power to apparent power.
(a) Apparent power must not exceed 500 KVA. Apparent power of the additional load must not exceed 500-280KVA.
(b) Since the apparent power will be 280+300 KVA the total apparent power will be 580 KVA. That sounds overloaded to me.
(c) Since "inductors are said to consume reactive power and capacitors are said to supply it" the inductors of the example are consuming 80 KVAR then the capacitors must supply supply 80 KVAR to reduce the load to rated value.
Source(s): Wikipedia - Anonymous9 years ago
texasmav is right, here is the situation
before the capacitor installed
kva ------------ kw ----------- kvar
m|_x -----------m*cos(x)--- m*sin(x)
280|_53.13---168 ---------- 224----------load#1
300|_36.87--- 240---------- 180----------load#2
574|_44.72--- 408---------- 404---------- total
pf = cos(x) = 0.71
574 > 500kva transformer overloaded
kvar = sqrt(500^2 - 408^2) = 289
capacitor kvar = 289 - 404 = - 115
after the capacitor installed
kva ------------ kw ----------- kvar
m|_x -----------m*cos(x)--- m*sin(x)
280|_53.13---168 ---------- 224----------load#1
300|_36.87--- 240---------- 180----------load#2
capacitor load ----------- (- 115)
500|_35.31--- 408---------- 289---------- total
pf = cos(x) = 0.82
load 500 = transf 500kva
the pf is improve from 0.71 to 0.82 after the capacitor is installed
- ?Lv 44 years ago
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