Oil Fired Boiler Steam exam revision question

[BenC1994]

Homework Statement

See image attached, have completed the parts a and b using steam tables and interpolation and got the answers

a) Wet steam h at 10.5 bar: h = 2378 kj/kg

b) Superheated steam at 10.5 bar @ 285 Degrees Celsius: h = 3018.12 kj/kg

Which I believe to be correct now, parts c,d and e I have not seen before on any revision papers and it would be greatly appreciated if someone could explain this to me?

Homework Equations

Q = Mc(Delta)T ?
1 bar is 1x10^5 pa or n/m^2

The Attempt at a Solution

42000 kj/kg I know I can use the units for this to calculate the mass of oil but unsure of process?

 

[SteamKing]

View attachment 83246

Homework Statement

See image attached, have completed the parts a and b using steam tables and interpolation and got the answers

a) Wet steam h at 10.5 bar: h = 2378 kj/kg

b) Superheated steam at 10.5 bar @ 285 Degrees Celsius: h = 3018.12 kj/kg

Which I believe to be correct now, parts c,d and e I have not seen before on any revision papers and it would be greatly appreciated if someone could explain this to me?

Homework Equations

Q = Mc(Delta)T ?
1 bar is 1x10^5 pa or n/m^2

The Attempt at a Solution

42000 kj/kg I know I can use the units for this to calculate the mass of oil but unsure of process?

You didn't need to start a new thread.

There, that's much better. It's hard to read these images when they're rotated 90 degrees.

 

[BenC1994]

New to this haha will bear it in mind next time

 

[SteamKing]

View attachment 83246

Homework Statement

See image attached, have completed the parts a and b using steam tables and interpolation and got the answers

a) Wet steam h at 10.5 bar: h = 2378 kj/kg

I agree with this value.

b) Superheated steam at 10.5 bar @ 285 Degrees Celsius: h = 3018.12 kj/kg

I agree with this value.

Which I believe to be correct now, parts c,d and e I have not seen before on any revision papers and it would be greatly appreciated if someone could explain this to me?

Homework Equations

Q = Mc(Delta)T ?
1 bar is 1x10^5 pa or n/m^2

The Attempt at a Solution

42000 kj/kg I know I can use the units for this to calculate the mass of oil but unsure of process?

For part c) You've calculated the enthalpy of the steam coming out of the superheater at P = 10.5 bar and T = 285 °C. You are given the fact that the feed water supplied to the boiler has a temperature of 50 °C and a specific enthalpy h = 209.3 kJ/kg.

How much heat would the boiler have to add to 1 kg of feedwater at the condition above to turn it into 1 kg of superheated steam at P = 10.5 bar and T = 285 °C?
[Hint: this is why you had to complete part b) above. Since we are using steam tables, Q = mc ΔT is no longer valid.]

Once you calculate how much heat is required to make 1 kg of superheated steam, how much heat is required to make 1.5 tonnes of superheated steam per hour?
Note: 1 tonne = 1000 kg.

Chew on answering part c) and see what you come up with. Please show your calculations in your reply.

 

[BenC1994]

I have attempted something but not being the greatest at steam calculations I have been confused at a probably easy question.

Up to now I have

Steam h = 3018.12 kJ/kg

Fw h = 230 kJ/kg

For a 235 c temperature rise it it
obvious that
3018.12 - 230 = 2788.12 kJ/kg

Then I thought 2788.12 kJ per 1KG
So per 1500kg it must be
2788.12 x 1500 = 4182180 kJ/kg

Confused and bewildered then divided this figure by 60 just because. And got...
69703 kJ/hr

This is either a stroke of genius or just an emphasis of my lack of ability to work with steam...

Thanks,

 

[SteamKing]

I have attempted something but not being the greatest at steam calculations I have been confused at a probably easy question.

Up to now I have

Steam h = 3018.12 kJ/kg

Fw h = 230 kJ/kg

It's not clear why you are using the enthalpy of the feedwater as h = 230 kJ/kg. The problem clearly stated that h(fw) = 209.3 kJ/kg

For a 235 c temperature rise it it
obvious that
3018.12 - 230 = 2788.12 kJ/kg

See comment above about enthalpy of the feed water.

Then I thought 2788.12 kJ per 1KG
So per 1500kg it must be
2788.12 x 1500 = 4182180 kJ/kg

Couple things here:
1. The boiler is making steam at the rate of 1500 kg/hr, so time is already included in the units.
2. If the change in enthalpy is measured in kJ/kg, then multiplying the change in enthalpy by the rate of steam production = the rate of heat transfer
Working out the units, (kJ/kg) × (kg / hr) = kJ / hr

Always check the units of your calculation (separately if you need to).

Confused and bewildered then divided this figure by 60 just because. And got...
69703 kJ/hr

This is either a stroke of genius or just an emphasis of my lack of ability to work with steam...

Thanks,

If you are confused by a step, think carefully before plunging ahead.

Correct your calculations for the enthalpy of the feed water, and correct the rate of heat furnished by the boiler.

 

[BenC1994]

So you're saying that,

Steam h = 3018.12 kJ/kg
Fw h = 209.3 kJ/kg

Therefore 3018.12 - 209.3 = 2808.82 kJ/kg

Therefore 2808.82 x 1500
= 4213230 kJ/hr

This seems a very odd figure...

I then assumed for part d,

Boiler efficiency = heat absorbed by steam/heat added by steam

So 0.7 = 4213230/m x 42000 = 100.315

So m = 100.315/0.7
m = 143.3 kg/hr

Is this correct method?

Part e, have attempted but unsure of how the density of the oil ties in with the calc? Or of any process to do it.

Cheers,

 

[SteamKing]

So you're saying that,

Steam h = 3018.12 kJ/kg
Fw h = 209.3 kJ/kg

Therefore 3018.12 - 209.3 = 2808.82 kJ/kg

Therefore 2808.82 x 1500
= 4213230 kJ/hr

This seems a very odd figure...

It doesn't matter if it seems an odd figure ... it's correct.

I then assumed for part d,

Boiler efficiency = heat absorbed by steam/heat added by steam

So 0.7 = 4213230/m x 42000 = 100.315

So m = 100.315/0.7
m = 143.3 kg/hr

Is this correct method?

Yes, it's correct.

Part e, have attempted but unsure of how the density of the oil ties in with the calc? Or of any process to do it.

Cheers,

Well, fuel oil is sold by the liter, not the kilogram.
In part d), you calculated how many kg of fuel oil you needed to burn for each hour of operating this boiler to make steam.

For part e), you are asked to calculate the cost of purchasing fuel for this boiler, assuming it is in operation for 24 hours per day for 350 days per year.

The oil costs 1.85 pounds sterling per liter.

How to figure out how many kg in each liter of oil?

The fuel has a relative density of 0.85, which means that a given volume of fuel oil is only 85% as heavy as an equal volume of fresh water.

How much does a liter of fresh water weigh?

 

[BenC1994]

That's great, with a,b,c,d, complete,
So onto part e) thanks for the information provided,

I need 143.307 kg of oil per hour (kg/hr)

1 litre of fresh water is 1kg
So 1 litre of this oil is 0.85kg

So,
143.307/0.85 = 168.596 (what units would this be?)

168.596 x £1.85 = 311.90 £ per hour

311.90 x 24 Hours = 7485.68 £/day!

7485.68 x 350 days =
£2,619,989 per 350 day year!

That's $4,045,918.01 per a 350 day year! I think.

Not your average household boiler price, if so then I shall increase the amount of board I pay to my parents significantly...

Is this the correct method for calculating e? If so then I will formalize the method and revise it for the exam,

Thanks,

 

[SteamKing]

That's great, with a,b,c,d, complete,
So onto part e) thanks for the information provided,

I need 143.307 kg of oil per hour (kg/hr)

1 litre of fresh water is 1kg
So 1 litre of this oil is 0.85kg

So,
143.307/0.85 = 168.596 (what units would this be?)

If 1 liter of oil weighs 0.85 kg, and you are burning oil a a rate of 143.307 kg/hr, then the units are ...

168.596 x £1.85 = 311.90 £ per hour

311.90 x 24 Hours = 7485.68 £/day!

7485.68 x 350 days =
£2,619,989 per 350 day year!

That's $4,045,918.01 per a 350 day year! I think.

Not your average household boiler price, if so then I shall increase the amount of board I pay to my parents significantly...

Is this the correct method for calculating e? If so then I will formalize the method and revise it for the exam,

Thanks,

Yes, everything looks good.