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https://www.Çℎḙḡḡ.com/homework-help...-balanced-3-phase-load-40-kva-v-v-b-q84727916

daghang salamat
Capacity of two transformers connected in open Delta(V V connection) is =√3VL*IL =√3×kVA rating of one transformer

=20√3 kVA=34.641kVA

Actual load on transformers=40kVA

%Overloading=(Actual load -Actual Capacity)*100 /Actual Capacity

=(40-34.641)*100/34.641=15.47%
 
boss pa unlock po quiz po namin :(
https://www.Çℎḙḡḡ.com/homework-help...v-min-determine-required-diameters--q23439879
thank you po
 
boss pa unlock po quiz po namin :(
https://www.Çℎḙḡḡ.com/homework-help...v-min-determine-required-diameters--q23439879
thank you po
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Master pa unlock po huhu

https://www.Çℎḙḡḡ.com/homework-help...system-pump-pump-b-deliver-water-ad-q46711060
Question 1

Annual worth analysis is being used to select the alternative.

Pump A -

Initial cost = 200,000

Operating cost = 30,000

Salvage value = 10,000

Life time = 12 years

Interest rate = 9%

Calculate the annual worth of Alternative A -

AWA = - initial cost (A/P, i, n) - operating cost + salvage value (A/F, i, n)

AWA = -200,000(A/P, 9%, 12) - 30,000 + 10,000(A/F, 9%, 12)

AWA = [-200,000 * 0.13965] - 30,000 + [10,000 * 0.04965]

AWA = -27,930 - 30,000 + 496.5

AWA = -57,433.50

The annual worth of Pump A is $-57,433.50

Pump B -

Initial cost = 300,000

Operating cost = 23,000

Salvage value = 60,000

Life time = 16 years

Interest rate = 9%

Calculate the annual worth of Alternative B -

AWB = - initial cost (A/P, i, n) - operating cost + salvage value (A/F, i, n)

AWB = -300,000(A/P, 9%, 16) - 23,000 + 60,000(A/F, 9%, 16)

AWB = [-300,000 * 0.12030] - 23,000 + [60,000 * 0.03030]

AWB = -36,090 - 23,000 + 1,818

AWB = -57,272

The annual worth of Pump B is $-57,272

The annual worth of Pump B is numerically higher.

So,

Pump B is better.
 
Question 1

Annual worth analysis is being used to select the alternative.

Pump A -

Initial cost = 200,000

Operating cost = 30,000

Salvage value = 10,000

Life time = 12 years

Interest rate = 9%

Calculate the annual worth of Alternative A -

AWA = - initial cost (A/P, i, n) - operating cost + salvage value (A/F, i, n)

AWA = -200,000(A/P, 9%, 12) - 30,000 + 10,000(A/F, 9%, 12)

AWA = [-200,000 * 0.13965] - 30,000 + [10,000 * 0.04965]

AWA = -27,930 - 30,000 + 496.5

AWA = -57,433.50

The annual worth of Pump A is $-57,433.50

Pump B -

Initial cost = 300,000

Operating cost = 23,000

Salvage value = 60,000

Life time = 16 years

Interest rate = 9%

Calculate the annual worth of Alternative B -

AWB = - initial cost (A/P, i, n) - operating cost + salvage value (A/F, i, n)

AWB = -300,000(A/P, 9%, 16) - 23,000 + 60,000(A/F, 9%, 16)

AWB = [-300,000 * 0.12030] - 23,000 + [60,000 * 0.03030]

AWB = -36,090 - 23,000 + 1,818

AWB = -57,272

The annual worth of Pump B is $-57,272

The annual worth of Pump B is numerically higher.

So,

Pump B is better.
tysm po hehe lifesaver sa quiz
 
https://www.Çℎḙḡḡ.com/homework-help...5_MXHz7BqUp90r89DY6D16HG0qwz3jEzp02AmHWIG2CS4

pa unlock po hehe salamt ulit ng marami sir <3
 
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