Tip:
Highlight text to annotate it
X
This lecture is in continuation of the lecture of additional topics, Industrial Furnaces.
Why I selected this lecture is, simply to illustrate the various concepts which I have
delivered in earlier lecture on Industrial
Furnace. Now, here, I will be illustrating those concepts by solving few examples. Here
is the example number one.
Now, in this particular example, what have I done? Let us consider a fuel fired furnace which is maintained at the temperature
1327 degree Celsius. Now, this furnace is fuel fired, having 70 percent Carbon. This
is solid fuel fired or Coal fired furnace; it is given that, it has 70 percent Carbon.
Here, the calorific value of this fuel that is Net calorific value - NCV of this fuel
is given - 27900 kilojoule per kg. Then, of course, for combustion, air is also
passed along with the fuel so that combustion occurs inside the furnace. The products of
combustion that is POC - they consist of Nitrogen, Oxygen, H 2 O and CO
2. Now, their volume percent is given - the composition is given: say Nitrogen is 76 percent,
Oxygen is 7 percent, H 2 O is 5 percent and CO 2 is 12 percent; they are all in percent.
Now, what we have to find out? We have to find out, for this particular situation, what
is the fuel consumption? Also, it is given that the gross available heat requirement
of the furnace is 1.5 into 10 to the power 3
kilowatt - that is what is given to us. Now, under this condition, we are required
to find out what is the fuel consumption. Now, here, it is also given. If it is not
given, as I have said in my earlier lecture, that when temperature or products of
combustion is not given, then you can assume that the products of combustion will discharge
at least at the temperature of the furnace; It cannot be discharged below the temperature;
of course, not above also,
but we can assume safely that it discharges, let us say in this particular problem, at
the temperature 1327 degree Celsius. This is the temperature of POC. In this particular
problem, air is supplied at 25 degree
Celsius. Of course, fuel is also supplied at 25 degree Celsius. So, we are required
to find out the fuel consumption.
So, let us proceed now. We will select now, first of all, the basis of calculation. As
I said, in all thermo dynamic calculation, consider, it is a basis is at 298 Kelvin.
That is, at this temperature, the heat content
value should be known to us. Now, first of all, in order to calculate the fuel consumption,
we have to know - what is the amount of heat that is carried away by POC. If we know that
amount of heat which is
carried away by POC and if we know the calorific value, then we can calculate gross available
heat per kg of the fuel. We are already given with the gross available heat requirement.
If we divide both with the
gross available heat per kg of the fuel, then we get the fuel consumption. So, just like
that. So, first of all, we have to calculate the
amount of POC. Since I have calculated this on several occasions, you can to do the Carbon
balance. You can assume that as, say Y kg mole of POC and you can do the
Carbon balance. You cannot do anything else except Carbon balance because the Coal is
having 70 percent Carbon and nothing else is given.
So, in this particular problem, the approach would be to go from the POC side and then
to calculate. So, here, first we will calculate amount of POC, and this can be obtained by Carbon balance. So,
if we do
the Carbon balance, straightaway I will write down, the amount of POC for this particular
problem - that will be equal to 0.486 kg mole per kg of Carbon. Now, since we know the amount
of POC, we calculate
the heat to POC. Now, we are in a position to calculate heat to POC.
In order to calculate heat to POC, we have to calculate the heat content of the various
components of the POC - that is Nitrogen, Oxygen, H 2 O and CO 2. Now, remember here,
Net calorific value of the fuel is
given. So, by definition in the calculation of net calorific value, the state of the product
of combustion is chosen as vapor because the H 2 O is in the vapor state. So, while calculating
the heat carried by products
of combustion, you should also select the reference state -- gaseous, and their H 2
O is in the vapor state.
Now heat to POC: first of all, I give you the values of the heat content. The heat content
of various components are H 1600 minus H 298 for CO 2, for O 2, for H 2 O vapor; mind you
- these are at 298 Kelvin
and of Nitrogen. For CO 2 this value is 66350; For O 2 -- 43710; for H 2 O vapor -- 54880;
for Nitrogen -- 41620. These values are given in kilojoule per kg mole. So, now, heat to
POC is straightforward form to
calculate. I will write down first of all CO 2, then H 2 O vapor, then Oxygen and then
Nitrogen. Now, I can calculate kg moles because I know
the amount of POC. So, CO 2 will be 0.05832, H 2 O vapor is 0.0243, O 2 is 0.034 and Nitrogen
is 0.3694. So, I can calculate now heat carried by each
component of the flue gas or products or combustion in kilojoule. So, for CO 2 it is simply multiplied
by CO 2 value, heat content CO 2 per kg mole with the kg mole. So, this will be 3869, I
have rounded off,
13341486 and 15374. So, if you sum it, the heat to POC will be 22063 kilojoule.
Now, since we know heat to POC, we can calculate the gross available heat per kg of the fuel.
GAH - gross available heat per kg of fuel, as I have done in a previous problem that
is equal to 27900 is the calorific value of fuel minus 22063. Heat to POC - that value
is 5837; this is kilojoule. Now, you see
that much amount of heat is available. If you just do a simple calculation, then you
will note that, out of the calorific value of the fuel that is 27900, the heat carried
by POC, if I just calculate in percent, heat
carried by POC is equal to 22063 upon 27900 into 100. A simple calculation will give you
- approximately 79 percent of the calorific value of the fuel is carried away by flue
gases. That means in this particular
problem, the 21 percent of the calorific value of the fuel is utilized by the useful part
of the function, which is the heating air; 79 percent is carried away by the flue gases.
So, higher is the temperature, more percentage of the calorific value will be available to
you in the products of combustion; of course, lower is the temperature, lower amount will
be available. Now, this particular
thing will constitute a different dimension of the capturing of the energy. You are seeing
that a large portion of the energy is still available. 79 percent of the calorific value
of the fuel which is still available to you,
depends on what you do with that available heat in the products of combustion.
Now, before this, we can calculate fuel consumption. We can now calculate fuel consumption which
is gross available heat required per unit of time divided by gross available heat per
kg of the fuel; here, of
course, per unit of time. So, we know, the 1.5 into 10 to the power
3 is what it is joule per second; so, into 3600 divided by gross available heat per kg
of the fuel. We have calculated 5837. So, you note that, the fuel consumption is 925
kilogram per hour. So, this is the fuel consumption when the furnace is operated by 70 percent
Carbon of the fuel, and POC or the heat to POC is at 1327 degree Celsius. Now, the question
before us is - can we
cut down the fuel requirement? The answer is yes.
In the same process, we can do that. How will we do that if we analyze this problem? You
note that the heat carried by the POC is of considerable amount that is of the order of
79 percent. What we can do with
that? We can let it go waste because it is joule;
the 79 percent of the calorific value is in fact the fuel, where fuel is the calorific
value. So, that means it is going to waste. So, the first thing that should come into
your
mind after this analysis is - you should think of the ways how can we use the sensible heat
of products of combustion in order to cut down the fuel requirement? Now, by cutting
down the fuel requirement, you
are doing two functions simultaneously. One - you are cutting down the fuel requirement
as well as you are reducing the Carbon emission to the environment. This is what I am going
to illustrate by the same problem. I am continuing and putting as part
b.
What I am telling is when combustion air is preheated. Now, important is, with heat recovered from flue gases, there is no point
if you preheat the air by combusting extra amount of fuel; then you have not done
a very intelligent job. The intelligent job that you have done is that, you are preheating
air by recovering the heat which is going with the POC; that is the important thing,
my dear friends. So, it is preheated with heat recovered from
flue gases, flue gases or products of combustion - one and the same, in a heat exchanger which
is having 50 percent relative efficiency. This concept I already
introduced to you in my earlier lecture on industrial furnace.
So, this is an intelligent way. You are capturing the heat and trying to reuse it. So, with
that, you are reducing the fuel requirement; bound to reduce. Because of the amount of
heat you are supplying now, fuel
consumption will be reduced; so, fuel saving, contributing to the energy saving and also
contributing to the environment cleanliness because then you will be discharging less
Carbon into the atmosphere. That is
what I am going to illustrate now.
So, let me represent this particular problem. What have I done? I have done the same thing.
This is my furnace, as usual operating at 1327 degree Celsius. Coal - 70 percent Carbon,
NCV - 27900 kilojoule per
kg. Now, this is a flue gas or products of combustion - one and the same thing. The composition
is same as it was earlier; so, there is no need to write here. The temperature discharge
is at again 1327 degree
Celsius. Now, what do I do? I teach now, a heat exchanger;
so, I put a heat exchanger; this is heat exchanger. Now, the objective of heat exchanger is to
extract the heat of POC and transfer it to the air, incoming air. So,
here somewhere I put my air which I require for combustion. In (a) part, it was not preheated;
now, the same air - I am preheating it. So, air and here is the outlet for air and here
is the flue gas. Of course, its temperature, you have to find
out from heat balance. Presently, it is not our objective. So, what we are doing now?
We are taking air from here and we supplying this air into this furnace. So, here
we have a preheated air. Now, in the preheated air..., so, that is what now our loop is.
Now, what we have done in the furnace? We installed a heat exchanger; it extracted the
heat, and heat the combustion air. So, now we have to calculate back, the fuel
consumption. Of course, we have installed a heat exchanger; we have invested certain
amount of money; we want see what profit it brings; how much it can save? So,
again, we have to calculate the so called heat to POC, then heat recovered, and again,
we have to do the heat balance. So, let us do that.
Now, first of all, since we are preheating the air, we have to find out the amount of
air. So, the first thing is that.
You have to find amount of air because we are preheating the air. Only heat is being
put into the furnace through preheated air and nothing else. So, find amount of air - how
can you find out amount of air? In
several occasions, how have you done? You will do the Nitrogen balance; you will perform
the Nitrogen balance. So, if you perform the Nitrogen balance, the Nitrogen from air - that is equal to 0.3694 kg mole
because all Nitrogen of air goes into the products of combustion, provided the Coal
does not contain any amount of Nitrogen because there is no composition given over here. So,
you can do nothing from the
Coal side; all that you can do from the products of combustion side.
Similarly, we can then find out Oxygen from air; say, 79 21 percent. So, that will be
0.0982 kg mole. Now, mind you, this air which you have calculated 0.092 - it consist of
theoretical Oxygen as well as excess
Oxygen as well as both; just for your information. So, now, we know the amount of air. So, next
thing we can calculate now - the relative efficiency.
We can calculate relative efficiency. As already you know, that is equal to sensible heat in
preheated air upon sensible heat in air at hot flue gas temperature; already, I have illustrated how
to calculate this. Now, here, relative efficiency is given. So,
what we can calculate? Relative efficiency is given. So, we can calculate sensible heat
in air at hot flue gas temperature. Hot flue gas temperature is 1327 degree Celsius.
The heat content values are given. So, you can calculate both things. So relative efficiency
let us take as 50; of course, it is multiplied by 100. So, when I do percent, that is fifty
by hundred that is equal to
preheated air is delta H 1. So, this will be equal to 0.3694 into 41620 plus 0.0982
into 43710 - that is what we will calculate. So, this calculation comes delta H 1 that
is equal to 9833 kilojoule. That means that
much amount of kilojoule you are able to recover from the heat which was taken away by products
of combustion by installing a heat exchanger. So, now the gross available heat would be
- calorific value of this fuel plus sensible heat of air minus heat to POC, or let us put
as delta H 1, heat to POC. So, everything is known. Now, this value will be equal to
15670 kilojoule. Now, I can calculate. Fuel consumption heat
requirement is the same -- 1.5 into 10 to the power 3 kilowatt. I have illustrated through
the earlier formula. So, this fuel consumption now will be coming 344 kilogram
per hour. What do you think now, my dear friends?
By installing a preheater, now you are consuming for the same heat requirement 344 kg per hour.
So, what will be the fuel saving? Fuel saving by installing a preheater or exchanger - that
will be equal to 580
kilogram per hour.
Now, extending this thing per , let us say, if now furnace operates for 20 hours per day, then saving per day
will be equal to 11140. So, this will be in kg. Now, the Carbon saving is the fuel. Carbon
saving would
be 70 percent of this; 8127 kilogram per day you are saving Carbon. Now, this is equivalent
to reduction in CO 2; reduction in CO 2 - that will be equal to 29800 kg. How is it? C plus
O 2 is CO 2. Same you
have convert to kg mole; multiply by 44. So, you are reducing that much amount of Carbon
dioxide. Now, here comes the concept of motivation.
The motivation is - you can earn this Carbon, that is so called Carbon oxide or Carbon credit,
and that is available to you. Whereas, to define a Carbon offset which is
a financial instrument which is aimed at a reduction in greenhouse gas emissions, typically,
it has come from the concept of Kyoto protocol and so on, to encourage
the high temperature users based on the fossil fuel that well you earn Carbon credit or you
Carbon oxide. Now, how can we earn? Carbon oxides are measured
in terms of metric tons of CO 2 equivalent. That means how much amount of CO 2 we are
saving; that is your Carbon oxide. So, for this say one Carbon
offset that is equal to reduction in 1000 kg of CO 2 or its equivalent; equivalent means, suppose
we are reducing other NOX of water then it is 1000 kg CO 2.
So, in this case, Carbon offset or Carbon - they are same thing. We can call Carbon
offset or Carbon credits are same thing. That will be approximately, 30 Carbon credits you
are storing per day. So, we can
imagine now per month or per year and so on. So, this is the illustration of this.
Further if you think, this is one way of reduction of the fuel.
Now, suppose, if you do not have a heat exchanger, but Coal; Part (c) - there is no heat exchanger.
Now, Coal is burnt with a mixture of cold air which is 25 degree Celsius and Oxygen;
no heat exchanger - that
is the condition. Now, excess of Oxygen the condition is excess
of Oxygen that is in air plus pure O 2 that you will be supplying over theoretical for
combustion will be kept same as with preheater. That means whatever amount
of heat that you added through the preheater that will not be added now, but then, that
you have to compensate it somewhere by adding Oxygen.
Now, what we have to calculate? Calculate amount of Oxygen in kg per hour to obtain same fuel consumption
as in part (b). So, now what is required to be done? That is, you have to obtain this
same fuel
consumption, but without preheater, as you got in part (b). Now, this objective can be
attained because excess of Oxygen, you have to keep the same. How can you attain this
particular objective? You have
noted that, every last percentage of heat is carried away by Nitrogen. So, if I reduce
the amount of Nitrogen, then I can do my job. So, let us solve this particular problem.
We note that preheater brings 9833 kilojoule of heat. When preheater is not used, then
Nitrogen of the flue gas must be reduced corresponding to the amount of 9833
kilojoule; then my job is over. So let us consider let x kg mole. O 2 is now
required. Then 3.76 x will be the moles of N 2. So, that means 3.76 x into 41620 - that
will be equal to 9833. So, the amount of x, that I can calculate is 0.0628 kg
mole. That much amount of Oxygen if I add, then I will be removing Nitrogen and hence
the less amount of heat will be carried away by Nitrogen. Hence, I equal into 9833.
So, whatever amount of heat which preheater was bringing, now, I have cut down that heat
by reducing the amount of Nitrogen; that is the only trick.
So, now Oxygen, I can calculate. The Oxygen will come 691 kilogram per hour because I
know the fuel consumption already given to you. So, from that, you can get the amount
of Oxygen; that is 691 kg per
hour. So, what my objective of illustration of this particular problem is that, there
are several ways in which one can cut down the fuel requirement and hence the concept
of Carbon credit. Now, I will give you another problem. So,
you can think and you can do that. .
Now, another problem is like this: the problem number 2. Let us consider a batch reverberatory
furnace which is melting. I mind you, which is melting 305 tons of Copper and heating it to 1327 degree Celsius.
Now, it is estimated that about one half of that required for heating and melting for heating
and melting is lost. That means, the heat which is required for melting and heating
- half of that is being lost. Fuel oil is used whose composition is 85 percent
Carbon, 12 percent Hydrogen and 3 percent Oxygen. Its NCV - that is equal to 9446 kilo
calorie per kg, burnt with one 120 percent theoretical dry air. So, flue
gases enter into a waste heat boiler. So, here instead of a heat exchanger, what we
have done? We have installed a waste heat boiler.
Now, waste boiler is a typical appliance for heat recovery and converting water into a
steam. It is very important heat recovery device. Where, in the heat boiler it is said
that heat lost is estimated to be 15 percent
of the sensible heat of entering flue gases. That is, only 85 percent of the heat is recovered
and 15 percent is lost which is of only Stack gases. That means the gases from the boiler
leave at 320 degree Celsius.
Now, you are required to calculate average fuel consumption, air supplied, and heat recovered
in boiler. If you wish, you can also express that heat recovered in terms of calorific
value or whichever ways you
like do it. So, I will just give you the hint to solve
this particular problem. I will omit few steps so that you can fill in that blanks and you
can solve the problem. Now, first of all, I will illustrate. I will
say just draw a line diagram. So, what this problem tells is that, we have say 305 tons
of copper which is melting and its temperature is raised to 1327 degree Celsius. Now,
this is an oil fired furnace; that is a fuel oil, and its composition - you have Carbon,
you have Hydrogen, you have Oxygen; Carbon is 85 percent, hydrogen is 12 percent and
Oxygen is 3 percent. It is burnt with
plus 120 percent theoretical air. Now, its NCV is also given; that is 9446;
it is given in kilo calorie per kg. Now, the POC which is being discharged from here is
taken to a waste heat boiler. Somewhere here, we have a waste heat boiler and
from the waste heat boiler, the gases are discharged at a particular temperature.
Now, here say heat loss is estimated to be 15 percent from the waste heat boiler. Heat
loss to the surrounding is given; say, heat loss from the furnace part. Heat loss to surrounding
- that is equal to 50 percent,
that required for heating and melting, heating plus melting; that is all given to us. So,
now we have to calculate the fuel consumption. Now, as such, first of all, we have to calculate
the material balance and then we have to do the heat balance.
So, here, now let me give you some values that will be requiring say it is given say
for example, the heat content that is H 600 minus H 298 for Copper. It is given 11980
kilo calorie per kg mole - that is given to
you. Now, say H 1600 minus H 298 - this is for
CO 2, for H 2 O, for O 2 and for Nitrogen - that is given. 15850, 13110, 10442, and
9943 - this is given in kilo calorie per kg mole.
Now, the flue gases after the waste heat boiler have been discharging at 320 degree Celsius.
So, these values are given, say H 320 plus 273 that is 593 minus H 298 - these values
are given. Again, for CO 2 that is
equal to O 2 -- 965, for H 2 O, this value is 2409, for say O 2 this value is given 2121,
and for Nitrogen, this value is 2043; again, all their units in kilo calorie per kg mole.
This value will be needed while doing other parts of the problem. So, first of all, you
can calculate heat content in copper because you have to find out the losses. So, heat
content in copper all the values are
given; you can convert kg to kg mole, and so on. So, heat content in copper - that will
be equal to 9515 kilo calorie per hour. Now, again, I can find out the amount of flue
gases. Take one kg of the fuel; compositions are given. So, we can find out the flue gases
or products of combustion. So, we have CO 2, H 2 O, O 2 and Nitrogen.
So, you can calculate their respective amount - kilo calorie per kg mole at the respective
temperature of the discharge. The products of combustion are discharging at 1327 degree
Celsius. So, you can calculate
all and the heat carried by POC should come of the order of 6047.5 degree Celsius.
So, I can calculate now, the fuel consumption, that is, the heat required per unit of time.
You know the heat you will require for melting and heating of the Copper that 9.515 into
10 to the power 6 and plus half is being lost; so, plus 4.5 a sorry 4.7575 into 10 to the
power 6. This is the heat required
for melting and rising the temperature to 1327 or 1600 Kelvin and then half of it is
being lost. So, that much amount of heat is required per unit of time. Available heat
is only 7446 minus heat to POC 6048. So,
fuel consumption is coming 4.2 tons per hour. That is a equal to 4.2 tons per hour; that
is what is the answer for part (a). We can now easily find out the air consumption.
Of course, you have to do the Nitrogen balance and then you know the fuel consumption, and
so on. So, if air consumption per hour we have to find out, that
will come 8048 meter cube per minute. I hope you can calculate the tonnage is give and
air amount is given. So, you can find out that is at one atmosphere and 25degree Celsius;
this is the part (b). Now, the next thing that comes is the heat
recovered in boiler. We have to calculate heat recovered in boiler. What will be the
heat recovered in boiler? Very simple; that is, heat input to boiler minus heat output
from boiler. That is the heat that is stored in the boiler. Heat is being input at 1327
Kelvin of which 15 percent is lost. So, only 85 percent is recovered. The heat output from
boiler - that is heat taken by POC
after exiting from the boiler, that is at 327 degree Celsius. The amount remains the
same. The amount is not being utilized anywhere; only heat being extracted.
So, amount is the same and the temperature is different. You can calculate heat output
from the boiler from the sensible heat value which I have given to you; heat input to the
boiler, we already calculated, and
then 85 percent you take it. So, this value, heat recovered in the boiler - that will come
3928 kilo calorie per kg of fuel.
Right now, just for information, fraction of CV recovered, because we are always interested.
All these - heat to POC, heat recovered, they are all a fraction of the calorific value
of the fuel because the source of
thermal energy is the calorific value of the fuel. So, it is always i mean to know what
fraction you are able to recover that is the important thing.
So, fraction of calorific value of Coal that is recovered is equal to 0.42. Simply divide
3928 by the calorific value of fuel and you get this value. So, this is in short, for
problem number 2.
Now, let me see another problem. Now, this problem I have given. In case you are working
on a shop floor for a long time on the furnaces, and all of a sudden your management has come,
and asks you that --
look, this is the heat balance for which you are operating for very very long years. I
have got an idea that I can save the fuel because I have heard that, if I recovered
the heat from the flue gases, I can save the
fuel. That means I can reduce the fuel consumption. So, my dear engineer, you are being here;
a very simple problem is there; can you convince me, whether by installing a preheater, what
advantage I am going get in my company? So, here is the problem. You are simply given
to heat balance of a continuous furnace. He has told you that heat input, slowly from
combustion of the fuel that is 100 percent of the fuel. That is all. 100 percent of
the total heat is being supplied by combustion of the fuel.
Now, heat output -- again, percentage of total is given. Remember, the process requirement
is 25 percent of the total, sensible heat in fuel gases is 50 percent and heat lost
25 percent. Now, he tells you that, I am considering installation
of a preheater. What is given to you is that. Preheater would
be able to recover 50 percent of the sensible heat of the flue gas and all that 50 percent
you recovered - it would be available to you in the form of combustion air.
That means 50 percent of the heat preheater will recover and the same 50 percent will
be given to the air for preheating. Under this condition, you let me know, what
will be the percent saving in fuel through preheater installation, when the process requirement
and daily heat lost are kept same. I think you understood the problem.
He is very clear in his mind. I want to know from you, if I install a preheater because
that involves a huge amount of investment, what percent saving in fuel I will get, if
the process requirement and daily heat loss
are same? This is his query number one.
Now, his query number 2 is if the daily fuel consumption and heat loss are same.
Now, here, heats are different, he is telling. If the daily fuel consumption and heat loss
are same, what percent increase could be made in heat furnished due to installation of preheater.
There are two things: I install a preheater, naturally I will cut down the fuel requirement;
I install a preheater, I do not cut down the fuel requirement. Then, there will be excess
heat in the furnace. What percent
increase in the heat I will be getting? So, take out your pen and pencil, and notebook.
Convince your management that - look sir, if you install a preheater, that much amount
of fuel you are going to save; if you do not want to save fuel, then you can
increase the efficiency of the furnace. Here, since you are using for melting, you can tell
him -- sir, if you do not cut down your fuel, you will be increasing the efficiency of the
furnace, that is more output will be
there and more money will come into your pocket. So, that is what we have to tell him now.
I will just give you the hint. If suppose you are having what (a) is telling, the process
requirement in daily heat loss is the same and what percent saving in fuel will be achieved.
That means, now, the fuel will not
be 100 percent; fuel will be some x percent because we are saving the fuel. So, if x is
my percentage fuel through which I am supplying the heat of x, x by 2 will go in the flue
gas because in the beginning, 100
percent, 50 percent was going into flue gas. Now, x is the renewed fuel consumption; x
by 2 will enter into the flue gas; out of x by 2, half will be recovered. So, I will
be recovering x by 4.
Now, my heat balance would be input x due fuel plus x by 4. That is coming by the preheater
installation and that will be equal to 50 because I have to keep the process requirement,
heat loss, are same. 50 plus x
by 2 is the heat taken by the flue gas. Now, look now, what I have done? x is the
percentage of fuel. Now, you are required to burn of x. x by 2 will be carried by the
flue gases from x by 2; half you will recover; so, x by 4 will be recovered. So, heat
balance - x plus x by 4. 25 for the process requirement, 25 for the heat loss - that you
to kept same plus x by 2 - that is a flue gas. So, I get the value of x. Very simple.
66.66 percent; so, the fuel saving now
would be 33.34 percent. Look at how convincingly you can tell him that, you can save that much
amount of fuel. Now, the part (b), I leave it to you to think
it over and to do it, and the answer part (b) will come that you can increase the efficiency.
The answer for this is that, you can increase the efficiency of preheating if
you do not cut down the fuel requirement 200 percent.
So, try to get this answer and feel happy that you have got the answer and you understood
the problem.