Mathcad 07-18 - rpm 1 min kVA kW Problem 7-18 Solution kVAR kW HP 40 = V L 200 V ⋅ = I L 115 A ⋅ = 93 ⋅ = RPM 1745 = f 60 Hz ⋅ =

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
I LR Start kva 3 V L = I LR 726.307A = b. Rated torque can be found from rated horsepower and speed T rated 5252 HP RPM lb ft = T rated 120.39 lb ft = c. The minimum expected starting torque can be found from table 7-5: Factor TLR 140 % = T LR Factor TLR T rated = T LR 168.546 lb ft = d. To find the rated power factor we need the input real and apparent power S in 3 V L I L = S in 3.984 10 4 × W = P in P out η = P in 32.073kW = F p P in S in = F p 0.805 = of course the power factor is lagging for an induction motor
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: rpm 1 min ≡ kVA kW ≡ Problem 7-18: Solution kVAR kW ≡ HP 40 = V L 200 V ⋅ = I L 115 A ⋅ = η 93 % ⋅ = RPM 1745 = f 60 Hz ⋅ = NEMA Code letter is G a. For NEMA Code G, the maximum kVA/HP (from table 7-3) at starting is 6.29 Factor LR 6.29 kVA hp ⋅ = P out HP hp ⋅ = P out 40hp = Start kva Factor LR P out ⋅ = Start kva 251.6kW =...
View Full Document

This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue University-West Lafayette.

Ask a homework question - tutors are online