Enqiry about an electrical motor's power

[HHJ]

Any clue to to show that it is 10kW? Please tell me where i went wrong and thank you.

I tried and i just couldn't figure it out.

 

[HHJ]

Only Q5b)

 

[CWatters]

I would calculate the required flow rate first (125% of pumping requirement - from part a?), then the power required to move the gas. Then double that to account for the 50% loss in the fan. Then divide by two to get power of each fan.

 

[HHJ]

I would calculate the required flow rate first (125% of pumping requirement - from part a?), then the power required to move the gas. Then double that to account for the 50% loss in the fan. Then divide by two to get power of each fan.

Thanks i will try out get back to you soon..

 

[CWatters]

For info: I notice you wrote "243kW" on the question sheet next to part a). If you multiply 234 * 100/65 you get 374kW.

 

[CWatters]

I've just worked the numbers for Q5b and get a different answer to the book (eg not 10kW). I'll let you have a go before giving you my answer.

 

[HHJ]

Tried i can get the answer, but i do not understand the working. Please Guide me

 

[CWatters]

I looked at solving Q5b using three methods...

Method 1

This uses the same method as you used for Q5a...

h = 1.2*10⁵/(1.225 * 9.81) = 9,986m

Flow rate Q is 125% of 2500 = 3125m³/h = 0.868 m³/S

P = ρ Q g h
= 1.225(kg/m³) * 0.868 (m³/S) * 9.81(m/S²) * 9,986(m)
= 104,163W
= 104kW

The fan and motor efficiency is 50% so the total power required is...
104 * 100/50 = 208kW

However there are two fans and motors and the question asks for the power of each fan so
208/2 = 104kW

So I think the answer is 104kW. (eg Not 10kW).

Method 2

From method 1...
P = ρ Q g h
but
ρgh = pressure
so..
P = Pressure * Volumetric flow rate (Q)

So this method does not use the density ρ and you do not need to calculate the equivalent head h...

They give you the pressure in the question (1.2bar = 120,000Pa).
Flow rate Q is same as method 1 = 0.868 m³/S

Power = 1.2*10⁵ * 0.868 m³/S
= 104,160W
= 104kW

Then the same *2 and /2 gives answer 104kW for each motor. Same answer as method 1.

Method 3

Just as a check... Compare the data in parts a and b...

In part b the flow rate is 1.25x greater than part a
In part b the pressure is about 1/3 of that in part a (3.5bar/1.2bar is about 1/3)

So I would expect the total power in part b to be about 1/3 of that calculated in part a.
In part a the power was about 370kW so a figure of 104kW for part b seems reasonable (right order of magnitude).

So I think there is an error in the answer given in the question.

 

[HHJ]

I looked at solving Q5b using three methods...

Method 1

This uses the same method as you used for Q5a...

h = 1.2*10⁵/(1.225 * 9.81) = 9,986m

Flow rate Q is 125% of 2500 = 3125m³/h = 0.868 m³/S

P = ρ Q g h
= 1.225(kg/m³) * 0.868 (m³/S) * 9.81(m/S²) * 9,986(m)
= 104,163W
= 104kW

The fan and motor efficiency is 50% so the total power required is...
104 * 100/50 = 208kW

However there are two fans and motors and the question asks for the power of each fan so
208/2 = 104kW

So I think the answer is 104kW. (eg Not 10kW).

Method 2

From method 1...
P = ρ Q g h
but
ρgh = pressure
so..
P = Pressure * Volumetric flow rate (Q)

So this method does not use the density ρ and you do not need to calculate the equivalent head h...

They give you the pressure in the question (1.2bar = 120,000Pa).
Flow rate Q is same as method 1 = 0.868 m³/S

Power = 1.2*10⁵ * 0.868 m³/S
= 104,160W
= 104kW

Then the same *2 and /2 gives answer 104kW for each motor. Same answer as method 1.

Method 3

Just as a check... Compare the data in parts a and b...

In part b the flow rate is 1.25x greater than part a
In part b the pressure is about 1/3 of that in part a (3.5bar/1.2bar is about 1/3)

So I would expect the total power in part b to be about 1/3 of that calculated in part a.
In part a the power was about 370kW so a figure of 104kW for part b seems reasonable (right order of magnitude).

So I think there is an error in the answer given in the question.

Sir, you’re right. I consulted my lecturer he told me some answers are intentionally set “wrongly” allowing student to find mistakes, understand and solve. Since the answer for 5b) has been concluded 5c) to solve the additional power required :

No. Cargo oil pumps, Nc = 3 ( from part a.)
cargo oil pump power, Pc = 374 kW ( from part a.)
Load factor = 85%
Diversity factor = 90%

Additional power of cargo pump for the factors,
P_EL = Nc * Pc * load factor * diversity factor
P_EL = 3* 374* 85/100 * 90/100
Pc_EL = 853 kW

No. Inert gas fans,N_f = 2 ( from part b.)
Inert gas fan power, P_f = 104 kW ( from part b.)
Load factor = 85%
Diversity factor = 90%

Additional power of cargo pump for the factors,
Pf_EL = N_f * P_f * load factor * diversity factor
Pf_EL = 2* 104* 85/100 * 90/100
Pf_EL = 318 kW

Sum of total power required for the 24 hours discharge operation, P_EL.

∴P_EL = Pf_EL* Pc_EL
P_EL = 853 + 318 = 1171kW


In this case again, the given answer(889kW) is also incorrect from what i can find

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[CWatters]

I think the question is badly written. It asks you to calculate the "additional electrical power". I can think of three possible answers..

1) This is the average power over the discharge period.
This is the method you used but check your working (PfEL = 2* 104* 85/100 * 90/100 = 159.12 not 318 kW)

I got the following answer..
P_ave = Pump Power * 85/100 * 90/100
= (3*374kW + 2*104kW) * 85/100 * 90/100
= 1017kW

2) Peak power during the discharge period. This is the power when both oil and inert gas pumps are running at the same time...

P_peak = Pump Power * 85/100
= (3*374kW + 2*104kW) * 85/100
= 1131kW

3) Total energy used
E_Tot = Pump Power * 85/100 * 90/100 * 24
= (3*374kW + 2*104kW) * 85/100 * 90/100 * 24
= 24,419 kWH

None of my answers match the book answer.

I'm not sure why they use the word "additional"? Perhaps that should read "average"?