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Efficiency of Gas Turbine
how to calculate the power in MW that drives the compressor...

I would like to inquire to how I can calculate the power in MW that drives the compressor. As we know the efficiency of gas turbine is about 25%.

I did some calculation which is as follows:
If the efficiency is 100% and GT efficiency 25% that means 25% is equal to 60 MW 75% is equal to 180 MW (loses).
This 180MW is the power that drives the Gas turbine compressor.

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Ayed,

Your questions sure cover the gambit!

The efficiency of most GE-design heavy duty gas turbines *in simple-cycle application* (i.e, without a waste heat recovery boiler (or heat recovery steam generator) in the exhaust is approximately 32-35% at Base Load in new and clean condition at nameplate rated conditions (ambient pressure, ambient temperature, ambient humidity).

New and clean condition implies clean turbine inlet air filters, clean axial compressor bellmouth and IGVs, clean axial compressor without excessive clearances, hot gas path parts equal to new specifications--including combustion liners, transition pieces, turbine nozzles, turbine buckets, and an exhaust duct back pressure within design specifications). It also implies fuel with the expected characteristics as defined in Sect. 05 of the Control Specification drawing. And lastly, it implies the IGV LVDTs have been properly calibrated, the compressor discharge pressure transmitter(s) are properly calibrated, and the exhaust T/Cs and wiring are properly installed and are all working properly with a minimal exhaust temperature spread at Base Load.

So, this means that for roughly ever three horsepower developed by the turbine section while operating at Base Load in a new and clean condition that two horsepower are required to drive the axial compressor. That's a pretty good rule of thumb.

I don't believe I've ever seen formulae outside of computers at GE Engineering that can accurately calculate the amount of energy being consumed by the compressor at a given operation condition. I'm sure there are companies selling software that can perform these calculations based on estimates of compressor efficiencies and such, but without a lot of instrumentation that's not normally installed on a compressor and turbine and without intimate knowledge of the design parameters of the compressor it would be very difficult to come up with an exact figure.

32-35% efficiency is better than many fossil fuel-fired boilers (coal or oil or natural gas) can achieve on their best day during initial operation. And, if a gas turbine is operated in combined cycle mode, using the exhaust heat to produce steam to drive a steam turbine, well, efficiencies of more than 55% and close to 60% are now being achieved for the entire cycle. The rule of thumb for combined cycle power plants is that you will be able to produce approximately 0.5 MW with a steam turbine for every MW produced by the gas turbine, for the same fuel flow-rate (presuming no extractions). And that's much better than any fossil fuel-fired boiler and turbine could ever hope to achieve, with fewer emissions in many cases, as well (gotta get that green aspect in there!).

Again, there are probably some people somewhere who could give you some formulae that could calculate it down to the second decimal place, but it will still likely be an estimate based on assumptions, and would fall in the range described above (+/- 1%).

You need something more than that, you better start talking to some power plant designers and architect/engineering firms.

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Efficiency of the gas turbine

The total energy balance in a gas turbine can be summarized as

Total energy input = Compressor load + Generator power output + Flue gas energy loss + rotational losses

with the exception of the rotational losses all others can be calculated in a easy manner , what you need is the following information.

1. Calorific value of the fuel (kcal/kg)

2. Mass flow rate of the fuel (kg/s or t/hr) (if only the volume flow is available then you need to know the density of the fuel)

3. air flow into the turbine ( this is tricky most of the sites do not have a air flow meter , you have to get it from the characteristic graphs which GE provides , you can also get the value form the site acceptance test or the performance guarantee test done at commissioning)

if you have a HRSG , which you probably will have then you can calculate the efficiency of the HRSG and also the combined efficient of the total co generation power plant, for this you need the additional data

Note - i do not know the layout of your plant or its operating nature , i am assuming that the steam used for deaeration is got from the plant itself and CPH if present is inside the HRSG itself. you need to work out the details yourself , i am giving the calculations for a self sustaining plant. ie it takes only water at room temp and fuel and gives out power and steam. The plant axillary consumption which will be around 2-4% of the plant full load is neglected in the calculation. mainly because i do not have sufficient data.

The basic equations for the calculation are

`For GT               Efficiency = 860*MW output (MwHr) / (fuel flow(kg) * calorific value of fuel(in kcal/kg))this is the base formula  where fuel flow is normally available in M3 which you have to convert to mass with the known density. for HRSG               efficiency (overall) = steam flow rate * enthalpy of steam / (HRSG inlet temp  * 0.25 * air flow rate)              efficiency (heat exchanger) = steam flow rate * enthalpy of steam / ( (HRSG inlet temp - HRSG outlet temp) * 0.25 * air flow rate)the combined efficiency of the co generation is given by                  efficiency =  ( (860*MW output (MwHr)) +  steam flow rate * enthalpy of steam) / (fuel flow(kg) * calorific value of fuel(in kcal/kg))sample calculation I am now in a frame 5 site , so i am taking the daily production readings from here. the values at your site will be different Naptha consumption = 192m3 Power generation = 379 Mwhr Average power generation / hr  = 15.8 MW Calorific value of naptha = 11250 kcal/kg Naptha density = 0.7the HRSG is a twin drum and produces two different steam varities one a VHP steam at 48Kg/cm2 and 435 deg and other MP steam at 18kg/cm2 at 245 deg Enthalpy of VHP steam = 785 kcal/kgEnthalpy of IP steam = 692 kcal/kg VHP steam production = 805 tonnes average steam production / hr = 33.5 t/hr MP steam production = 99 tonnes average MP steam production /hr = 4.12 t/hr HRSG inlet temperature = average GT exhaust temp                        = 490 deg HRSG outlet temperature = average stack temperature                         = 140 deg ambient temperature = 32 degair flow rate - the air flow for a frame 5 machine at site condition (32 deg ambient) is 408 tonnes at 85 deg IGV opening , as the machine was put in cogen cycle and the average IGV opening is 56 deg , from the chara graph the air flow is estimated as 364 tonnes. so GT efficiency =  (860 * 379 *100) / ( 192 * 0.7 * 11250)              =   21.57 % HRSG efficiency (overall) =  ( ( 33.5 * (785-30) ) + (4.12 * (692-30) ) ) / ( 490 * 0.25 * 364) ( here stack losses are taken into account)                          =   62.83 % HRSG efficiency (heat exchanger) =  ( ( 33.5 * (785-30) ) + (4.12 * (692-30) ) ) / ( (490-140) * 0.25 * 364) ( here stack lossses not taken into account)                                = 88 % overall co-generation efficiency                         =  (860 * 379) + ( ( 805 * (785-30) ) + (99 * (692-30) ) ) / ( 192 * 0.7 * 11250)                         =  67 % `

i am still not get how i can calculate the power that drives the compressor.
i need the equation.

note that we have HRSG and the capacity is 118tons

best regards...........

Ayed,

Without being able to measure the air flow through the compressor, the best you're going to be able to do is estimate the amount of work being done by the compressor.

And the size of the HRSG isn't going to affect the compressor load, either.

>Efficiency of the gas turbine
>
>The total energy balance in a gas
>turbine can be summarized as
>
>Total energy input = Compressor load +
>Generator power output + Flue gas energy
>loss + rotational losses
>
>with the exception of the rotational
>losses all others can be calculated in a
>easy manner , what you need is the
>following information.
>
>1. Calorific value of the fuel
>(kcal/kg)
>
>2. Mass flow rate of the fuel (kg/s or
>t/hr) (if only the volume flow is
>available then you need to know the
>density of the fuel)
>
>3. air flow into the turbine ( this is
>tricky most of the sites do not have a
>air flow meter , you have to get it from
>the characteristic graphs which GE
>provides , you can also get the value
>form the site acceptance test or the
>performance guarantee test done at
>commissioning)

And then this just falls apart. In the calculations, nowhere does it mention the compressor load or provide any indication of a sample value for rotational losses. There's no mention of air flow or how the value 860 came into the formula nor what it represents.

Just doing a quick search of wikipedia.org, more than one article suggests the thermal efficiency of simple-cycle gas turbines is approximately 30-40%, and that combined cycle gas turbine efficiencies are as high as 60%. GE used to market one of their aero-derivative packages as a "40-40" machine because it had 40% thermal efficiency for 40 MW, and that's darned high for a simple cycle machine (that had to have inlet cooling to achieve the 40-40 moniker).

But, ProcessValue needs to patent the plant at his site, and quick. Because a thermal efficiency of 67% is exceptional.

And unrealistic. Particularly if the Frame 5 is only putting out an average of 15.8 MW per hour, which is very low if the unit is operating at Base Load, which is when the efficiency would be highest and when most performance guarantee tests are run. And even more unrealistic if the GT efficiency is only 21.57%.

please let me know the factor 0.25 which has been taken to calculate HRSG efficiency and also why the enthalpies have been substracted by 30?

Ashwin

By Brian Freeman on 12 April, 2013 - 9:10 pm

from what I understand...
MW = (J/s)*3600(s/hr)=3600(J/hr)=>MWhr=3600J

Lets say MWhr/kg value of naptha = 10,000; then your equation should read (there was a parentheses misplaced in your equation, what you wrote was actually equal to 325940):

overall co-generation efficiency

`>                        =  ((860 * 379)>+  ( 805 * (785-30) ) + (99 * (692-30)>) ) / ( 192 * 0.7 * 10,000) >                        =  74 % `

what do you think?

hi dear "Process Value"..

how u got these values?
>Enthalpy of VHP steam = 785 kcal/kg
>Enthalpy of IP steam = 692 kcal/kg

any table?

best regards

I kinda forgot to add this small bit in the calculation above

`Energy balance in Gasturbine inlet ambient air = 25 deg cdp = 6.8kg/cm2 ctd = 302 deg exhaust = 507 deg fuel input = 1.7 kg/s calorific value of naptha = 11250 kcal/kg density of naptha = 0.71 power = 16 MW air flow into the turbine = 360 t/hr specific heat capacity of air = 0.25 kcal/kg deg input energy into the turbine = fuel input + air input  fuel input = 1.7 * 3.6 * 11250 *(1000) kcal             = 68850 Mcal  air input = 360 * 0.25 * 302            = 27180 Mcal total energy input to the turbine = 96030 Mcal  power output from the generator = 16*860                                = 13760 Mcal  flue gas losses = (360+1.6*3.6)*0.25*507                 = 46390 Mcal               compressor load and rotational losses = 96030 - (46390 + 13760)                                        = 96030 - 60150                                       = 35880 Mcal this equated in terms of power = 35880/860                                = 41 MW`

the compressor load does not change much with the loading of the machine , you will see that for a 20 MW gasturbine , the compressor load is about 40MW.

And CSA , are you sure about 30-35% efficiencies ?? to the best of my knowledge GT simple cycle efficiencies does not exceed 25 +/- 2 %.

Let's see, if the total power produced by the gas turbine in your example is 60 MW (20 MW generator output + 40 MW compressor load), then isn't 20 MW (the output rating of the gas turbine) 1/3 (33%) of 60 MW? So, roughly two of every three horsepower produced by the gas turbine are used to drive the compressor, and only one of every three horsepower is available to drive the generator?

Lastly, compressor load varies with IGV angle.

It's kinda ironic that the fortune at the bottom of the control.com page when I read this post was:

`Speed is a substtitute for accurancy.`

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Quiery 1

" And then this just falls apart. In the calculations, nowhere does it mention the compressor load or provide any indication of a sample value for rotational losses. There's no mention of air flow or how the value 860 came into the formula nor what it represents."

ok , i kinda assumed that people will be familiar with the joules constant , power energy equivalency now i have to derive the formulas as well i guess. well here goes

`Power energy equivalents 1 kilowatt (KW) = 1 kilo joule/ sec 1 kilowatt sec ( Kw s) = 1 kilo joule converting sec to hours , as Kw Hr is the standard for electrical energy measurement 1 kilowatt hour (Kw Hr) = 3600 Kilo joule now we know 1 calorie = 4.187 joules , this is the joules constant so 1 Kilo calorie = 4.187 Kilo joules 1 kilo joules = 0.23883 Kilo calories so 1 Kilowatt hour (Kw Hr) = 3600 * 0.23883 Kilo calories                         = 859.80416 Kilo calories                         = 860 kilo calories ( this is a reasonable approximation) this is the electrical energy and heat energy equivalence . This is how the 860 in the formula came from. Efficiency of the gas turbine efficiency of the turbine =  energy equivalent of generator Generator power output / energy input into the turbine energy equivalent of generator Generator power output = 860 * Kw-Hr energy input into the turbine = Calorific value of the fuel (Kcal /kg) * fuel flow ( Kg/hr)                               = 860 * Kw-hr /  Calorific value of the fuel (Kcal /kg) * fuel flow ( Kg/hr) now multiplying both the numerator and denominator by thousand                              = 860 * Kw-hr  * 1000 /  Calorific value of the fuel (Kcal /kg) * fuel flow ( Kg/hr) * 1000 this converts the Kw-Hr to Mw-Hr and kg/hr to t/hr                             = 860 * Mw-hr / calorific value of the fuel (Kcal/kg) * fuel flow ( t/hr) so this is how the energy efficiency of the gas turbine is derived. i am also deriving a formula for heat rate of the turbine to the efficiency heat rate of the turbine is defined as the The ratio of fuel energy input as heat per unit of net work output. It is expressed mostly in Btu/Kwhr or in kj /Kwhr . I am not a fan of Btu , but i am a ardent fan of SI units :) so i will derive the equation here in Si units. Heat rate = Kj/ Kwhr Efficiency = KwHr * 3600 / Kj Efficiency = 3600 / (kj/kwhr) efficiency = 3600 / Heat rate as far as the rotational and compressor loads are concerned i have given the calculation in the next post. Please go through it.`

...........................................................................................................................................

Quiery 2

" Just doing a quick search of wikipedia.org, more than one article suggests the thermal efficiency of simple-cycle gas turbines is approximately 30-40%, and that combined cycle gas turbine efficiencies are as high as 60%. GE used to market one of their aero-derivative packages as a "40-40" machine because it had 40% thermal efficiency for 40 MW, and that's darned high for a simple cycle machine (that had to have inlet cooling to achieve the 40-40 moniker).

But, ProcessValue needs to patent the plant at his site, and quick. Because a thermal efficiency of 67% is exceptional.

And unrealistic. Particularly if the Frame 5 is only putting out an average of 15.8 MW per hour, which is very low if the unit is operating at Base Load, which is when the efficiency would be highest and when most performance guarantee tests are run. And even more unrealistic if the GT efficiency is only 21.57%."

I did a quick search through my GE manuals and i came up with this. I am uploading a small document , a GE document which has the heat rate for its line of gas turbines.

http://www.2shared.com/document/VUjnzbcw/GEgasTurbine.html

heat rate for a distillate fired frame 5 machine is given as 12847 kj/kw-hr . from the efficiency to heat rate equation derived above it calculates to an efficiency of 28% max. this is the full load efficiency of the machine at iso conditions , at 26 MW , air flow of 450 t/hr pressure ratio of 10.6 etc .... please refer the doc. this is seldom achieved in real life. Base load of the machine at site conditions is 21.5 MW , and the supplier BHEL doc gives a assured heat rate of 3390 kcal/kwhr as the heat rate , ie the base load efficiency of the machine is 25 %. thus for a machine running at a part load of 15 MW an efficiency of 22% is not unrealistic. its just the way the machine is.

wikipedia provides good answers , most of the time it is quite right, but 40% efficiency is way higher and i will bet on the GE manual than wiki.

now that we mention it , frame 9 machine seems to have the lowest heat rate on a gas fired machine , 9930. this equates to a efficiency of 34%. i have contacted my friend in NTPC , he will give the data on the fame 9 machine they are running , but he assured me that they get close to 30% efficiency at base load operation. it seems that the smaller machines have a higher heat rate and thus a lower efficiency. machines in refineries are mostly frame 6. they also do not fare well. not above 25-26%.

and about the cycle efficiencies. I calculated the efficiency of a co-generation power plant , not a combined cycle power plant. in a co-generation power plant the plant output is both power and steam. in a combined cycle power plant the plant output is power only. the steam from the HRSG is routed to a steam turbine to produce power. combined cycle power plants have a lower efficiency than co generation power plants. in co generation power plants the heat equivalent of the steam generated is taken as the output. this gives them a higher efficiency than a combined cycle power plant.

in a combined cycle power plant , the steam is used to run a steam turbine at the downstream which has a 30% efficiency due to the condenser losses.

in a cogeneration cycle power plant as the steam equivalent is directly taken , this 70% reduction in efficiency does not come into play. only the stack losses are accounted for.

this is the reason why the efficiencies of a co generation plant is higher than a combined cycle power plant.

combined cycle gives efficiencies in the range of 50 - 60% while co- generation system give efficiency of 70-80 %. the plant efficiency shown in the calculation is quite low as it was operating in part load conditions.

So i do not think i need to go and patent my site , unfortunately it is running at kinda low efficiencies. all refineries and process plants which operate the gas turbine at part loads have the same situation.

So my concluding remarks are , my calculations and math are quite right. there are no deviations in the equations or in the values provided.

and one last thing , i just saw CSA's post now

" Let's see, if the total power produced by the gas turbine in your example is 60 MW (20 MW generator output + 40 MW compressor load), then isn't 20 MW (the output rating of the gas turbine) 1/3 (33%) of 60 MW? So, roughly two of every three horsepower produced by the gas turbine are used to drive the compressor, and only one of every three horsepower is available to drive the generator? "

well CSA you are talking about only the useful power output from the turbine. ( ok the compressor load is not got as useful power , but i am taking it as the work done by the turbine) . but every heat engine needs to reject out certain heat , in gas turbines it is in the form of flue gases , that should also be accounted as a loss is it not ??

Speed is a substtitute for accurancy. LOL , as long as you cover up / rectify your mistakes as soon as you make them , no one is going to notice right ;) . just trying to lighten up the serious discussion :)

in a combined cycle power plant, the steam is used to run a steam turbine at the downstream which has a 30% efficiency due to the condenser losses.

Sir, please explain where you got this number for steam turbine efficiency. Steam turbines are pretty good at extracting the energy from the steam--I have no reference off hand, but I think 75%+ is not an unheard of number

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Efficiency of the steam turbine

Kurush , steam turbine power plants have only 30% efficiency , i do not know from where you got 75% efficiency and all . i am giving a sample calculation again.

if you look at the above equations you will see that 1 MWhr is equivalent to 860*1000 kcal of energy.

let us suppose that i am having a fully condensing turbine , with input 90Kg and 500 deg steam and condenser operating at vacuum of -0.9 kg/cm2 and a dynuss fraction of 0.85

`the enthalpy of the input steam is given by = 812 kcal/kg enthalpy of steam going into the condenser is given by = 533 kcal/kg the heat energy available for the turbine to convert to work is = 812 - 533                                                                 = 280 kcal/kg`

the heat energy which goes to the condenser is lost , this is what i was referring to the condenser losses.

assuming that turbine has no rotational and radiation losses , they will be small , so they can be neglected you can see that for producing 1 MW output you will need 860/280 = 3.1 tonnes of steam at 90 kg pressure and 500 deg. thus the efficiency can be calculated as

`efficiency of the steam turbine = 860/(3.1*812)                                  = 34.12 % `

well , now you see that neglecting all the losses i get a ideal steam turbine efficiency as 34% only. i have not taken into account that the boiler will be only 88% efficient , there will be extraction steam for deaeration etc . if you have all the data , and if you calculate you will see that the efficiency of a conventional steam power plant is only 30% +- 2% .

>Efficiency of the steam turbine
>
>Kurush , steam turbine power plants
>have only 30% efficiency , i do not know
>from where you got 75% efficiency and
>all . i am giving a sample calculation
>again.

I did not say efficiency of "steam plant"; I said efficiency of steam TURBINE.

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I calculated the efficiency of the steam turbine only. which part of it did you not understand. you have said the you have a steam turbine running right, with the equations above calculate the efficiency yourself you will get to know.

efficiency of a fully condensing turbine is given by

= 860 * power output of the turbine / ( inlet steam flow (t/hr) * enthalpy of steam kcal/kg)

once again i say steam turbines have 30-35% efficiency only.

if you need any second reference , i suggest you look into power plant engineering by black & veatch, or PK Nag or steam plant operation by lammers and woodruff. it will give you good idea about how to calculate steam turbine efficiencies and also a overall idea about design and operation of steam turbines and power plants in general.

By Namatimangan08 on 1 February, 2011 - 8:41 am

>in a combined cycle power plant, the
>steam is used to run a steam turbine at
>the downstream which has a 30%
>efficiency due to the condenser losses.
>
>Sir, please explain where you got this
>number for steam turbine efficiency.
>Steam turbines are pretty good at
>extracting the energy from the steam--I
>have no reference off hand, but I think
>75%+ is not an unheard of number

Probably you are referring to isentropic or cylinder efficiency. In this case 75% is the least you can expect. It can be as high as 90%. Isentropic efficiency is the ratio between the actual expansion power to the theoretical expansion power assuming the expansion takes place at constant entropy.

As far as thermal efficiency is concern you can find somewhere around 45% for a big ST (1000MW), 3-pressure type. Thermal efficiency for the steam cycle (exclude boiler) is the ratio between the total expansion power to the total energy received by the steam cycle.

>
>Probably you are referring to
>isentropic or cylinder efficiency. In
>this case 75% is the least you can
>expect. It can be as high as 90%.
>Isentropic efficiency is the ratio
>between the actual expansion power to
>the theoretical expansion power assuming
>the expansion takes place at constant
>entropy.

*ding* *ding* Guess we should define our control volumes and "efficiencies" better.

Perhaps I'm ignorant, but whenever referring to turbine efficiency, we are talking about isentropic efficiency of the turbine. By convention, thermal efficiency (in my experience) is used for the entire plant.

Is one or us right and the other wrong? No. We just were ambiguous on our definitions.

By Namatimangan08 on 2 February, 2011 - 1:38 pm

You are right. Turbine efficiency by convention is an isentropic efficiency.

I take note that you define thermal efficiency for the entire plant. Nothing wrong with it.

That will give overall thermal efficiency. There are many efficiencies entangle in it. They are isentropic efficiency, ST cycle thermal efficiency, boiler efficiency, generator efficiency, mechanical efficiency and combustion efficiency. Sometimes we need to look specifically on each of them. It is very helpful if we can configure our energy balance boundary to evaluate each of those efficiencies.

Please i am working on the filtration system for gas turbines. help me with information on

conditions of the inlet air (into the filter)

conditions of the outlet air (from the filter into the compressor)

Filtration efficiency of the filters

Differential pressure in the filter

By Namatimangan08 on 3 February, 2011 - 2:36 am
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In practice probably we use net change in enthalpy between two terminals of your compressor, namely enthalpy of air at compressor discharge and enthalpy of ambient air. In this case you are assuming intake filter is a part of your compressor. You can do correction to normalize work done to overcome filter differential separately.

In order for you to know enthalpies at both terminals you have to measure mainly:

Ambient air: Temp & pressure

Compressor discharge : Temp & pressure

These 2 sets of data in fact defined enthalpies of ambient air and at comp discharge. You can use standard table for properties of air to get the desired enthalpies.

Enthalpy diff between these two terminals in defined work done to compress each kg or air. Obviously this is done by the compressor. The engineering unit shall be in Joule/kg.

Finally you have to know air mass flow rate (kg/s). Here is tricky part since it is rare you have measurement for that. The best & also one of the standard methods used is via index pressure. Basically this method utilizes index pressure changes to predict deviation of air mass flow when compared to the known reference (e.g. commissioning data).

In brief your calculation should look like this one.

`Enthalpy change across comp = h2(P2,T2)- h1(P1,T1)                            = Delta h (J/kg)Compressor input power = (Delta h)*dm_air/dt (J/s)`

The most critical parameter to get good result is to get accurate value for dm_air/dt.

What I mentioned above is very simplified task. Depends on your applications more often you have to do a lot more fine tuning/correction/normalization to ensure what you are measuring serves the intended purpose.

How can i get HRSG outlet temp if i know exhaust temp going in hrsg and water at saturated condition and process heat given at 5 bar

By Namatimangan08 on 1 February, 2012 - 6:52 am

> How can i get HRSG outlet temp if i know exhaust temp going in hrsg and
> water at saturated condition and process heat given at 5 bar

You have to measure it. I'm sure your plant has measurement equipment to measure stack temperature. For most of practical purposes stack temperature can assume HRSG outlet temperature.

Hi...You have mentioned a method to measure the air mass flow rate. Can you explain me the method a bit more detail?

By Namatimangan08 on 1 February, 2012 - 8:30 am

I have explained in the other related thread yesterday. Feel free to find it out. Tq.

Hi,

I am working on a project for process integration of a sponge iron plant using gas turbines. I have the inlet conditions for the flue gas and would like to determine the power generation possible. Can anyone help me with this?

By Namatimangan08 on 1 February, 2012 - 8:09 am

Tell us its temperature and mass flow rate. We will tell you how much power you can produce within 3% accuracy.

The temperature is 1050 deg C and the mass flow rate is 33 kg/sec.120.54 t/hr.

By Namatimangan08 on 3 February, 2012 - 12:27 am

> The temperature is 1050 deg C and the
> mass flow rate is 33 kg/sec.120.54 t/hr.

Ok. Let us calculate how much you can get within 3% accuracy.

High temperature reservoir of flu gas = 1050 Cel

Lower temperature reservoir of flu gas = 100 Cel (The current practice based on 30 Cel ambient air-My country)

Flu gas temp gradient = 1050-100= 950 Cel

Average temperature gradient = [(1050+273)+(100+273])*0.5= 846 Kelvin

Cp_air @ 848 K= 1.11kJ/kgK http://www.engineeringtoolbox.com/dry-air-properties-d_973.html

Heat available from flu gas

`Q_flugas =Cp_air (T_inlet-T outle)* dM/dt         = 1.11 X 10^3 X( 1323-373)X 33 Joule/s         = 3.48 X 10^7J/s                 = 34.8MW`

Boiler efficiency can be approximated to be around 0.88 (66%)

Therefore, heat energy that can be absorbed by steam cycle = 0.88 X 34.8 = 30.62MW

I think at most ST will be two pressure type. Its thermal efficiency is estimated to be around 0.33 (33%)- I might be wrong by the order of +/-2% here.

Thus, the gross output you can get will be

P_gross =0.33* 30.62 = 10.1MW

Thanks a lot! :) i am also looking for the mechanical aspects of the design. not much into d details though.. any help on that?

By Namatimangan08 on 15 February, 2012 - 8:17 pm

> i am also looking for the mechanical aspects of the design.

You have to mention more detail about what aspects that you want to know. Tq.

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Hi Namatimangan08,

Not to judge, but I am trying to understand your way of calculating exhaust energy.

Q=m/dt*Cp*(T_inlet-T_outlet)

As Cp, why you take air? Isn't that must be flue gas specific heat?

An example:
Based on gas turbines technical data:
Output Power: 15.290 kW
Heat Rate: 9940 kJ/kWh
Exhaust Flow: 180.050 kg/h
Exhaust Temp: 505 °C
Engine Efficiency: 36,2 %
LHV Natural gas: 9,722 kWh/m³

Input Energy: 15.290/0,362=42.237,5 kW

Energy Loss: 1-0,362=0,628 (62,8%)
According to your assumption, Energy Loss should be bigger than the exhaust energy. Lets use your way of calculating exhaust energy:

Exhaust flue gas temp:505°C
High temperature reservoir of flu gas = 505°C
Lower temperature reservoir of flu gas =100°C

Average temperature gradient=(505+100)/2=302,5°C = 575,6 K
Cp_air=1,04 kJ/kg-K

The usable energy can be calculated as:
Q_usable=50*1,04*(505-100)
Q_usable=21.060 kW

If considered the total extent of exhaust energy, then the output reference temperature must be equal to ambient temperature, lets take 30°C.

Q_exhaust=50*1,04*(505-30)
Q_exhaust=24.700 kW

But;

Input energy was 42237 kW (100%)
Output power is 15290 kW (36,2%)
Exhaust energy is 24700 kW (58,5%)

From this calculation, the total energy losses should be 62,8%. If exhaust energy is 58,5%, this means the rest is only 4,3% which must be the combination of energy losses through the turbine cooling, friction, heat losses etc. 4,3% doesn't seem realistic to me.