Coal Consumption Calculation

A fifth-grade arithmetic teacher could help; all that is needed is to multiply/divide the various numbers that are given; plus remembering that a day is 24 hours.

I think the issue here is your asking a question that you yourself answered in the question. I don't see how we can help you or others with these types of questions. I most assuredly would make every effort to see to it that someone needing help here would get the information needed to the best of my ability. As I am sure most everyone else would and have done also.

Do we become harsh at times, you bet. Should we, no maybe not. But, just thumb through some of the sections and count the number of really bad questions and you can see why we are what we are.

It is sad that you seem offended when someone speaks the truth to you.

Your question indicates that you have no grasp of the problem that you want someone else to solve for you.

Your reply to Tornado was rude and uncalled for.

You should apologize to him.

No. Not at all. What I commented was accordingly. Even I tried for the same question of mine. But, was missing to multiply by 24. You can overlook to what he said. If I replied accordingly, what's harsh in that.? We all start criticizing for harsh words, but in this process, forget to what the motive actually was..

If this is so, please teach me the entire power plant operation & calculations. It seems, you are 100% of everything. I don't think, I did any mistake in posting such comments. Rather, I'd like you to look again both of ours' reply once.

Assuming of course the plant is producing full capacity 24 hours per day. Few plants actually perform at that rate that I know of, load varies so output varies.

Thank you Sir. I was committing a mistake not by multiplying by 24 & so was encountering the problem. Else, I was doing it correctly.

I'm done here.

You are beyond help.

Remember, there are 24 hours in every day. Your facility will not be running at 100% all the time.

I'd also apologize to your fifth grade teacher, if you had one.

i believe Ace answered you. however, understanding you are in a power plant 300Mw the size and obvious must be a college graduate or a graduate engineer where all you need to do is to multiply the 300 Mw x 1864300 and divide by 4250 = Coal Tph/300 Mw 100% Hope you know how to make this at 80% boiler efficiency?

With 6 entries you have the nerve to disrespect Tornado. I along with other members regard him as a mentor. The guy who hired you must have brain of a Class 5 because he'd not hire anybody more intelligent than him.I'd have been more harsher than him had i come in before. You do need to apologise.

By the way where did you get the figure 1864.3 kcal/kwh. All my life I knew this as 860 kcal/kw ?

I was just passing by read the question and the reply. Got quite disturbed. Further more words spelt would take no where. Our goal should be to encourage more of people to join and participate. That would really motivate people to hit the button "Tell a Friend about CR4". Level of knowledge a person has is just a comparison with others with lesser, if ever they exists. Correct me if I am wrong. Have a nice day

Agreed.

Lyn: Actually the question appears to be simple, but one really needs to predict coal consumption in a coal-fired plant, then the REAL answers are not simple. One would have to really know the REAL calorific value of each lot of coal, the precise load schedule each hour, and the base and incremental heat rates for the unit to get a realistic answer.

For just playing around with what the so-called maximum coal order would be in a typical day if the unit is behaving typically, then yeah, just do the arithmetic.

Otherwise some basic calculus is in order.

James,

I understand, and agree with, what you say about heat content and usage rates.

My point is that the OP obviously does not.

This question should have been directed toward the person responsible for determining these numbers in his plant. (local knowledge) Unfortunately, that may well be the OP in this case.

I'm all for helping people with legitimate engineering problems if they need help.

I believe that they should come to the table with at least the basic knowledge required to perform the task at hand, or some reasonable (at least a sentence) explanation of why they are requesting help. That might also help the members determine the level of competence of the requester. I know that's asking a lot, for people of many varied and linguistic backgrounds.

Cheers.

So Lyn, when the time comes, how do you want yours, one lump or two? Man or sheep?

Dear Mr.dhanandhan,

I get extremely happy when I see you.

Please correct me then :

300 mw x 1000 = 300000 kw x 860 = 258000000/4250 = 60.7 Tph Coal to be used?

At 88% efficiency about 70.0 Tph Coal?

The OP is probably looking at the "advertised" heat rate, and not the true operational heat rate you and I would be accustomed to seeing.

3409.52 BTU/kWh, MEH value, here in Texas (we still use at least some English units, just for the sake of confusion). Favorite saying in Texas: "Confusion to the enemy!" Usually done while raising your glass in a toast.

1864.3 kcal/kWh => 7398.4 BTU/kWh (a smashing good plant if true) Some of the "cadillac" coal-fired plants here in the U.S. can do this, and as I understand it, the Japanese even have Supercritical Steam plants that can easily beat this number.

2250 kcal/kWh => 10135.6 BTU/kWh (typical for an incremental heat rate in a smaller 40 year old natural gas fired boiler generating plant).

3600KJ==1kWHr=856Kcal

The OP's Project Report declares THIS POWER STATION's Projected energy conversion at 1864.3 KCal to 0ne KWHr

This means Thermal efficiency= 856 divided by 1864.3=0.459=45.9% .This looks like low but may be OK.

Mukulmahant Thanks for the enhancement to my knowledge. Check and verify my calculation please.

Ducon, at your post #8, you gave the equivalent of a kWh correctly as 860 kCal. But heat rate is different, it indicate the heat needed to convert in to 1 kWh power. It is lower for more efficient plants.

By the way, I was talking of plant heat rate & not the CV of coal. When I mentioned, the plant heat rate is 1864.3 kcal/kWh, it implies, its not CV. So, even though you didn't see this figure your whole life, I am giving you a fair chance to look upon.

Thank you so very very much.

Ducon: I think you are referring to the mechanical equivalent of heat (859.184 kcal/kwh). The OP was referring to the <advertised> nameplate heat rate (efficiency) of the entire plant.

This includes the entire cycle, including auxilliaries as best I understand it, but in some instances, this does not include some of the plant auxilliaries (which IMHO is simply cheating). There is no need to break down the efficiency of the boiler (heat output/heat input), feedwater circuit pumping and/or regnerative heating efficiency, turbine efficiency, generator efficiency, transformer efficiency, ....ad infititum.

What is being looked at in a power plant from an operations point of view is (1) Fuel consumption (in terms of heat value, usually the LHV), and (2) Power output at the point of sale to the grid (meter). Obviously, this varies depending on % maximum load, weather conditions (affects cooling efficiency) if there is open-system evaporative cooling tower, and general state of repair of the equipment.

Aplly this formula:

300x1000x24x3600/4.18/0.45/4250 = 3,242,330 kg / day of coal max.

Where: 4.18 is kj/kcal and 0.45 is the boiler convertion efficiency (which seems to be suggested by your 1864.3 kcal/kWh).

I (we ) think that you should use an efficiency of 40% at best.

Also, the value you will get from the above should be multiplied by the estimated (or real) average output % from the 300MW maximum.

LAA lucke , thank you a lot. It was a great help.

Using data given

=300 x 1000 x 24 x1864.3 / 4250 = 3,158,343.5 kg of coal

the 1864.3 value is due to either using 4.2 instead of 4.18 kj /kcal, and 45.9% efficiency ... or just 46% ~

It is better to use the formula and replace the efficiency figure as necessary. This will depend on the process and/or on what the OP is looking at in term of efficiency.

I will make the following assumption: when you say plant heat rate, I can only assume you mean the unit heat rate to which the particular boiler is attached. Perhaps this is the only boiler in the plant, which clears things up a lot.

300 MW in 1 hour = 300,000 kWh

The heat required in your "plant" for one hour = 300,000 kWh x 1864.3 kcal/kWh

and this is 559,290,000 kcal/hr (the per hour unit is important not to miss).

The coal needed per hour: 559,290,000 kcal/hr / 4250 kcal/kg = 131,598 kg/hr

Coal per day = 131,598 kg/hr x 24 hour/day = 3158344 kg (3158.3 metric ton).

Thank you very much Sir. I got it