WEBVTT
Kind: captions
Language: en
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welcome to massive open online course
on fluidization engineering todays lecture
00:00:34.590 --> 00:00:44.290
is on calculation of gas pumping power consumption
in fluidized bed and what to learn in this
00:00:44.290 --> 00:00:52.860
lecture that we learn here how to calculate
the power consumption to pump the gas in fluidized
00:00:52.860 --> 00:01:05.509
bed the gas is distributed through the
distributor in the fluidized bed so to pass
00:01:05.509 --> 00:01:14.060
this gas to the distributor of course sufficient
power is required and then this power how
00:01:14.060 --> 00:01:23.650
to calculate against the resistance
of the fluidized bed on this gas consumption
00:01:23.650 --> 00:01:35.520
so let us see how to calculate what
should be the actual fiamble of this power
00:01:35.520 --> 00:01:42.659
consumption of the blower you will see that
power consumption is a significant cost factor
00:01:42.659 --> 00:01:51.590
in any process using fluidized beds occasionally
you will see it can be so high that it cancels
00:01:51.590 --> 00:02:02.590
the advantages of fluidization operation therefore
roughly estimation of the power which
00:02:02.590 --> 00:02:10.399
is required for the distribution of gas through
the distributor in the early design stages
00:02:10.399 --> 00:02:24.030
before making a detailed design of the design
of the fluidized bed or or deciding
00:02:24.030 --> 00:02:34.800
to the pilot plant operation of course
this will be adequate now since the distributor
00:02:34.800 --> 00:02:43.730
constitute a considerable function of pressure
drop across a fluidized bed it is always important
00:02:43.730 --> 00:02:53.560
to be noted that the power consumption of
the blower that drives the gas through the
00:02:53.560 --> 00:03:08.840
bed so the power consumption mainly
contributed by this by this by this distributor
00:03:08.840 --> 00:03:14.510
through which this gas is consumed gas is
distributed inside the bed of course other
00:03:14.510 --> 00:03:25.040
different parts of the of the fluidization
design that contribute to the power consumption
00:03:25.040 --> 00:03:31.280
of the fluidization operation in that
case the power consumption through the
00:03:31.280 --> 00:03:37.850
fluidized bed for the distribution gas is
the main contributory part so we will here
00:03:37.850 --> 00:03:48.680
know how to calculate this power consumption
of the compressor or blower by which
00:03:48.680 --> 00:03:56.459
the gas is compressed to distributor through
which gas is distributed
00:03:56.459 --> 00:04:04.440
now power consumption of the blower
let us see here how to calculate in this case
00:04:04.440 --> 00:04:15.879
you will see that a stream of gas which
will be compressed from an initial pressure
00:04:15.879 --> 00:04:23.320
to higher pressure if you consider that
initial pressure is p one and the higher
00:04:23.320 --> 00:04:33.000
pressure is p two the then gas is to be compressed
from this lower pressure of p one to the higher
00:04:33.000 --> 00:04:43.470
pressure p two to pump this gas through
the entire fluid bed in the fluidized bed
00:04:43.470 --> 00:04:53.120
then what should be the difference of this
pressure that can be equal to p two
00:04:53.120 --> 00:05:02.190
minus p one p two minus p one here this equation
so this p two minus p one that is called the
00:05:02.190 --> 00:05:10.440
pressure difference that means the pressure
of higher to lower what should be the difference
00:05:10.440 --> 00:05:18.190
that will be equals to the what should be
the pressure difference of the bed and
00:05:18.190 --> 00:05:25.669
what should be the pressure difference or
pressure drop in the distributor and is there
00:05:25.669 --> 00:05:36.720
any other pressure drop contributed by other
means like cyclones or filters in the fluidized
00:05:36.720 --> 00:05:48.170
bed so summation of this bed pressure distributor
pressure and the pressure drop of different
00:05:48.170 --> 00:05:57.440
mechanical means for the contribution of
this total pressure difference we have
00:05:57.440 --> 00:06:04.970
to then calculate the summation of this three
or if any other pressure contribution to
00:06:04.970 --> 00:06:15.320
equivalence for the pressure from lower
pressure to the higher pressure so as per
00:06:15.320 --> 00:06:22.550
thermodynamics then for adiabatic reversible
operations with negligible kinetic and potential
00:06:22.550 --> 00:06:31.770
energy effects you can calculate this ideal
shaft works to compress each kilogram of gas
00:06:31.770 --> 00:06:40.170
from this lower pressure to the higher
pressure to pump it the pump is gas through
00:06:40.170 --> 00:06:47.610
the entire fluidized bed system fluid bed
system is given by this this will be
00:06:47.610 --> 00:06:55.100
your notation about this shaft work this will
be of course it will be ideal shaft work which
00:06:55.100 --> 00:07:07.250
will be the integration of this pressure
per unit mass of the gas from this initial
00:07:07.250 --> 00:07:15.500
pressure to this higher pressure p two so
here of course this pressure will be negligible
00:07:15.500 --> 00:07:22.229
with compare to this this kinetic and potential
energy effects will be negligible compare
00:07:22.229 --> 00:07:28.690
to this shaft work here so based on this
will be able to calculate what should be the
00:07:28.690 --> 00:07:38.050
what should be the work done by this compressor
for compressing this gas from pressure p one
00:07:38.050 --> 00:07:39.650
to p two
00:07:39.650 --> 00:07:45.849
now if we consider that there is a ideal gas
which is being compressed from this lower
00:07:45.849 --> 00:07:52.310
pressure to the higher pressure and if it
is considered that ideal gas behavior is there
00:07:52.310 --> 00:08:01.310
then you can say this equation will be followed
at that p v is equal to n r t so the ideal
00:08:01.310 --> 00:08:09.940
pumping requirement then becomes after substitution
of this p value here in this equation and
00:08:09.940 --> 00:08:17.160
then what should be the d p and finally we
can get this equation of this shaft work ideal
00:08:17.160 --> 00:08:26.000
shaft work as this this is a function of this
at that gamma and what is that v and the
00:08:26.000 --> 00:08:36.440
pressure so this v is nothing but the volume
per unit mass of the gases is so here so w
00:08:36.440 --> 00:08:45.040
s ideal that is called shaft work rate shaft
work and ideal condition it will be gamma
00:08:45.040 --> 00:08:52.360
gamma minus one into p one v one into p two
by p one to the power gamma minus one by gamma
00:08:52.360 --> 00:08:58.850
minus one or you can say this will be is equal
to gamma by gamma minus one into p two into
00:08:58.850 --> 00:09:06.351
v two into one minus p one by p two to
the power gamma minus one by gamma so in
00:09:06.351 --> 00:09:12.560
this case v is the volumetric flow rate of
gas in meter cube per second so here in this
00:09:12.560 --> 00:09:24.810
case gamma gamma is nothing but the ratio
of ratio of heat capacity of gas at
00:09:24.810 --> 00:09:33.040
constant pressure and constant volume so this
gamma is defined as this which is denoted
00:09:33.040 --> 00:09:40.300
by this expression this gamma is equal to
c p g by c v g so from this equation you can
00:09:40.300 --> 00:09:47.110
ideally calculate what should be the shaft
work done whenever pumping of this gas through
00:09:47.110 --> 00:09:53.120
this distributor and distributed into the
bed from this lower pressure to the higher
00:09:53.120 --> 00:09:55.130
pressure
00:09:55.130 --> 00:10:01.510
now adiabatic reversible compression of the
gas from this lower pressure p one to the
00:10:01.510 --> 00:10:09.959
higher pressure that may cause an increase
in temperature because of this pressure
00:10:09.959 --> 00:10:18.680
difference now from thermodynamics points
of view of course you one can calculate
00:10:18.680 --> 00:10:26.240
the temperature at p two as what should be
the p two t two as this p two that means if
00:10:26.240 --> 00:10:31.279
we compress this gas from this lower pressure
to higher pressure at that higher pressure
00:10:31.279 --> 00:10:37.670
what should be the temperature change and
then that change of temperature at then that
00:10:37.670 --> 00:10:44.610
final temperature at this p two will be is
equal to t two that can be impulse to t
00:10:44.610 --> 00:10:51.220
one into p two by p one to the power gamma
minus one by gamma so from this equation you
00:10:51.220 --> 00:10:58.029
will be able to calculate what should be the
temperature at this at this pressure of p
00:10:58.029 --> 00:11:06.440
two this gamma is the ratio of specific heats
of gas is equal to one point six seven for
00:11:06.440 --> 00:11:14.790
monatomic gases and one point four zero for
diatomic gas and if it is triatomic gas then
00:11:14.790 --> 00:11:21.320
you have to consider this gamma value as one
point three three however for real operation
00:11:21.320 --> 00:11:26.990
of course you will see that there will
be a frictional losses and then frictional
00:11:26.990 --> 00:11:35.779
losses will be converted to this heat energy
now this actual shaft work required always
00:11:35.779 --> 00:11:44.240
will greater than the ideal case so of course
this actual and practical condition you will
00:11:44.240 --> 00:11:51.630
see the actual shaft work will be required
to compress this gas is greater than this
00:11:51.630 --> 00:11:54.950
ideal shaft work
00:11:54.950 --> 00:12:01.910
now for real operation then or practical operations
then its frictional losses we can say the
00:12:01.910 --> 00:12:10.850
actual shaft work can be represented by this
w s actual that will be is equal to w s
00:12:10.850 --> 00:12:17.970
ideal divided by eta eta is the efficiency
of the compressor which is roughly given by
00:12:17.970 --> 00:12:24.910
this eta will be is equal to zero point five
five to zero point seven five that means efficiency
00:12:24.910 --> 00:12:32.329
fifty five percent to seventy five percent
for turbo blower and it will be sixty percent
00:12:32.329 --> 00:12:39.550
to eighty percent if it is rooters blower
and this efficiency will be eighty percent
00:12:39.550 --> 00:12:47.560
to ninety percent if an axial blower or a
two stage reciprocating compressor is used
00:12:47.560 --> 00:12:53.620
to compress the gas from this lower pressure
to the higher pressure through this distributor
00:12:53.620 --> 00:13:00.930
inside the fluidized bed now you can get more
information of course from this books of handbook
00:13:00.930 --> 00:13:10.660
of fluidization and fluid particle system
that is edited by yang so you can get more
00:13:10.660 --> 00:13:13.110
information on this
00:13:13.110 --> 00:13:21.420
and then actual temperature of the gas leaving
a well insulated adiabatic but not hundred
00:13:21.420 --> 00:13:29.690
percent efficient compressor is then calculated
by this t two actual will be is equal to t
00:13:29.690 --> 00:13:39.149
one plus t one by eta into p two by p one
to the power gamma minus one gamma minus one
00:13:39.149 --> 00:13:45.260
so what should be the actual temperature at
its higher temperature if you are considering
00:13:45.260 --> 00:13:50.670
that there will be a fictional losses during
the operation of the compression from this
00:13:50.670 --> 00:13:56.610
lower pressure to the higher pressure now
in this case you will see this t two of course
00:13:56.610 --> 00:14:03.300
you will see the function of this t one of
course the and also in pressure at higher
00:14:03.300 --> 00:14:10.690
pressure if suppose pressure is this higher
then this it will be of course is greater
00:14:10.690 --> 00:14:19.350
than one and this temperature greater than
thant one that is why the temperature is
00:14:19.350 --> 00:14:26.440
higher than the ideal case here so this eta
will give you the efficient of the compressor
00:14:26.440 --> 00:14:31.279
from which you can calculate what the actual
temperature
00:14:31.279 --> 00:14:39.560
now let us see one example that how to calculate
this pressure compressor power to to to
00:14:39.560 --> 00:14:50.769
calculate to calculate the power of
gas distribution into the plenum of the
00:14:50.769 --> 00:15:00.279
fluidized bed system by this compressor by
compressor or any blower also calculate how
00:15:00.279 --> 00:15:08.149
to calculate the temperature rise due to the
heat of compression here now let us see that
00:15:08.149 --> 00:15:18.910
system parameters are given as here if suppose
grid distributor pressure is here this
00:15:18.910 --> 00:15:26.420
six kilopascal that means here some distributor
like grid type distributor is used to distribute
00:15:26.420 --> 00:15:37.130
the gas into the fluidized bed and bed
pressure is fifteen kilopascal whereas
00:15:37.130 --> 00:15:45.240
this cyclone filters and others pressure
drop is given as twelve k p a and and the
00:15:45.240 --> 00:15:53.640
exist of filters it is given as this three
hundred fifty k p a and gas enters the compressor
00:15:53.640 --> 00:16:01.579
at t one is equal to twenty degree centigrade
and initial pressure at one hundred one k
00:16:01.579 --> 00:16:10.220
p a and the volumetric flowrate of the fluid
is ten meter cube per second the efficiency
00:16:10.220 --> 00:16:17.509
of the compressor is considered as eighty
five percent and the ratio of the specific
00:16:17.509 --> 00:16:27.470
heat ratio is one point four so with these
conditions how to calculate the compressor
00:16:27.470 --> 00:16:34.800
power to pass this reactant gas into the plane
amount of fluidized bed system let us see
00:16:34.800 --> 00:16:41.730
here this solution what is this you have to
calculate first what should be the what should
00:16:41.730 --> 00:16:49.029
be the pressure that is final pressure
p two then final pressure p two will be is
00:16:49.029 --> 00:16:54.149
equal to what should be the p exit pressure
what should be the pressure contribution by
00:16:54.149 --> 00:17:01.610
the cyclones and filters what should be the
bed pressure drop and what should be the distributor
00:17:01.610 --> 00:17:08.360
pressure drop which is as used as grid here
now in this case we exit pressure is given
00:17:08.360 --> 00:17:15.070
three hundred fifty whereas p cyclones and
filters it is given twelve k p a and bed pressure
00:17:15.070 --> 00:17:23.889
it is given is fifteen k p a whereas this
grid pressure drop is six k p a so now
00:17:23.889 --> 00:17:30.590
in this case this they have total exit total
total pressure that is a final pressure
00:17:30.590 --> 00:17:38.970
will be the summation of this one two three
four pressure contribution and it is coming
00:17:38.970 --> 00:17:46.250
three hundred eighty eight kilopascal now
next part that you have to determine then
00:17:46.250 --> 00:17:53.039
what should be the ideal power consumption
which is required to distribute the gas
00:17:53.039 --> 00:18:05.450
through this grid distributor now as per this
formula of this shaft work at ideal condition
00:18:05.450 --> 00:18:14.299
that is gamma by gamma minus one into p one
into q one into p two by p one to the power
00:18:14.299 --> 00:18:21.940
gamma minus one by gamma minus one so in this
case gamma is given as one point four so one
00:18:21.940 --> 00:18:30.100
point four by one point four minus one into
p one is given as one hundred one kilopascal
00:18:30.100 --> 00:18:36.799
and then q one is given ten meter q per second
and after the substitution of this p two as
00:18:36.799 --> 00:18:43.130
three eighty three and p one as one zero one
to the power this gamma value if you substitute
00:18:43.130 --> 00:18:50.460
and finally if you calculate it you will get
this ideal shaft work as one six three eight
00:18:50.460 --> 00:18:52.100
kilo watt
00:18:52.100 --> 00:19:02.250
then what should be then the actual power
consumption if the if the compressor or blower
00:19:02.250 --> 00:19:07.710
efficiency is eighty five percent then shaft
work actual shaft work will be is equal to
00:19:07.710 --> 00:19:14.039
w s ideal by eta it has deficiency of the
compressor then it will be one six eight three
00:19:14.039 --> 00:19:19.990
divided by zero point eight five then it is
final is coming one nine two seven kilo watt
00:19:19.990 --> 00:19:26.940
or you expressed as two five eight seven horse
power now what should be the then temperature
00:19:26.940 --> 00:19:36.940
rise the t two because of this compression
of gas from this p one p one that is one
00:19:36.940 --> 00:19:43.760
zero one kilopascal two this p two that
is three hundred eighty eight kilopascal then
00:19:43.760 --> 00:19:50.280
you will get this t two from this equation
here so this t two will be is equal to t one
00:19:50.280 --> 00:20:01.710
plus t one by eta one into p two by
p one to the power gamma minus one by gamma
00:20:01.710 --> 00:20:09.110
minus one so from this you substitute this
t one value t one is given here that initial
00:20:09.110 --> 00:20:14.530
temperature is two ninety three and this two
ninety three divided by zero point eight five
00:20:14.530 --> 00:20:20.289
that is efficiency of the compressor and then
p two is three eighty three and p one is one
00:20:20.289 --> 00:20:28.070
zero one and substitute this gamma value finally
you will get it is coming four fifty three
00:20:28.070 --> 00:20:34.220
or one hundred eighty degree centigrade
so in this way you can calculate what should
00:20:34.220 --> 00:20:43.090
be the temperature rise during this
compressor ok of this gas from this lower
00:20:43.090 --> 00:20:48.429
pressure to the higher pressure so we are
getting here t two as four fifty three whereas
00:20:48.429 --> 00:20:59.299
t one is whereas t one whereas t one is is
given you two ninety three k and t two
00:20:59.299 --> 00:21:08.620
now we are getting as four fifty three
k so temperature increase increase delta t
00:21:08.620 --> 00:21:16.780
increased four fifty three minus two ninety
three that will be is equal to that will
00:21:16.780 --> 00:21:29.340
be is equal to that is here you can say here
one sixty k by this increase of by this
00:21:29.340 --> 00:21:36.600
compression of this gas from lower pressure
of that is one zero one kilopascal to three
00:21:36.600 --> 00:21:39.110
hundred eighty three kilopascal
00:21:39.110 --> 00:21:48.350
now let us see another example if if we
use the if we use the distributor in such
00:21:48.350 --> 00:21:57.860
way that its the pressure drop is some
value and if we if we distribute the gas
00:21:57.860 --> 00:22:05.720
in two locations or it will bypass this gas
in such way different condition then what
00:22:05.720 --> 00:22:14.370
should be the power consumption is there any
benefit of this bypassing of gas to distribute
00:22:14.370 --> 00:22:20.240
the gas into the fluidized bed whether it
will be beneficial or not let us see some
00:22:20.240 --> 00:22:28.390
one example whether this bypassing of gas
is useful or not or is there any reduction
00:22:28.390 --> 00:22:36.480
of power consumption or not let us see
this things with this example now calculate
00:22:36.480 --> 00:22:41.990
the compressor power requirement to run and
atmospheric fluidized bed coal combustor under
00:22:41.990 --> 00:22:49.330
the following condition condition one is that
distributor pressure will be three k p a another
00:22:49.330 --> 00:22:56.120
condition distributor pressure will be ten
k p a and third condition that if it is
00:22:56.120 --> 00:23:03.070
the condition like that fifty percent of the
required air that will be bypass bypassed
00:23:03.070 --> 00:23:16.260
to the bed and is introduced into the freeboard
of the fluidized bed to run it and also
00:23:16.260 --> 00:23:23.570
because of this you will see some volatile
gases will be burned in the fluidized bed
00:23:23.570 --> 00:23:29.690
and released in the bed by the coal now at
this condition you have to take distributor
00:23:29.690 --> 00:23:36.580
pressure as this initial condition whatever
it is the three three k p a now you have to
00:23:36.580 --> 00:23:43.640
consider that that gas is entered initial
at one hundred one kilopascal and initial
00:23:43.640 --> 00:23:50.940
temperature is twenty degree centigrade
and the gamma value that is ratio of specific
00:23:50.940 --> 00:24:01.960
heat capacity it is constant pressure constant
volume as one point four and across the bed
00:24:01.960 --> 00:24:11.020
the bed pressure it will be is equal to
ten k p a and at the bed exit that one
00:24:11.020 --> 00:24:17.470
hundred three k p a it will considered now
coal is supplied into the bed at a rate
00:24:17.470 --> 00:24:27.190
of eight tons per hour and gross heating value
is twenty five mega joule per k g air at
00:24:27.190 --> 00:24:34.600
saturated conditions needed at ten normal
meter cube per k g at fifteen percent excess
00:24:34.600 --> 00:24:40.919
and efficiency of the compressor is given
seventy five percent of power plant it
00:24:40.919 --> 00:24:42.470
is thirty six percent
00:24:42.470 --> 00:24:52.409
let us see the solution here it is given that
initial pressure is one hundred one kilopascal
00:24:52.409 --> 00:24:59.400
and this p three on the bed it is given this
p three here in this figure you will see p
00:24:59.400 --> 00:25:07.260
three here at this condition this one zero
three and p one it is not there but shaft
00:25:07.260 --> 00:25:13.120
work you have to calculate and p zero is one
zero one kilopascal what is the v zero what
00:25:13.120 --> 00:25:18.790
is the t zero is not given let us see here
what should be the v zero that is volumetric
00:25:18.790 --> 00:25:30.529
flowrate of the gas here it is seen that
that it is given that eight hundred k g
00:25:30.529 --> 00:25:39.230
or eight ton per hour eight ton per hour
feed rate is coal feed rate is eight tons
00:25:39.230 --> 00:25:48.179
per hour and air at saturated condition needed
this ten normal meter cube per k g of coal
00:25:48.179 --> 00:25:55.299
so in this case air is required ten meter
cube per k g of coal but whereas this k g
00:25:55.299 --> 00:26:01.980
is supplied as eight tons that is eight thousand
k g per hour and then if you multiply it
00:26:01.980 --> 00:26:10.170
by ten then you will get this how what
would be the volume of gas per hour to be
00:26:10.170 --> 00:26:18.960
supplied and then at this condition of this
that is here at this t zero condition that
00:26:18.960 --> 00:26:23.380
is at the initial condition what should be
the volume of the gas at this temperature
00:26:23.380 --> 00:26:29.330
of twenty degree centigrade then you will
get this then it will this meter cube per
00:26:29.330 --> 00:26:35.150
hour and again if you divide by this thirty
six zero zero second you will get ultimately
00:26:35.150 --> 00:26:43.289
meter cube per second it is coming twenty
three point eight five meter cube per second
00:26:43.289 --> 00:26:52.830
now ok this is your initial gas supplied
at the initial condition to this fluidized
00:26:52.830 --> 00:27:00.170
bed now what should be the shaft work for
this now if you substitute all the values
00:27:00.170 --> 00:27:05.980
in this equation of the shaft work in ideal
condition then you will see that it will be
00:27:05.980 --> 00:27:14.660
your gamma by gamma minus one into this p
one into this volume v one that is twenty
00:27:14.660 --> 00:27:19.309
three point eight five meter cube per second
into p two by p one p two here it is coming
00:27:19.309 --> 00:27:26.970
one one six one one six how it is calculated
here p two is nothing but here this p one
00:27:26.970 --> 00:27:32.500
zero three plus ten one zero three what one
zero three this is nothing but here what should
00:27:32.500 --> 00:27:38.720
be the p three and this now according to this
here see this will be your p three this
00:27:38.720 --> 00:27:45.020
is p three then what should be the p two and
p one p two and p one p two is nothing a one
00:27:45.020 --> 00:27:50.980
zero three plus ten ten is what this is the
your distributor pressure ten then p two plus
00:27:50.980 --> 00:27:57.100
then p two is equal to p three plus ten
that will be is equal to one hundred thirteen
00:27:57.100 --> 00:28:02.010
and then p one is equal to one hundred thirteen
plus three that will be is equal to one hundred
00:28:02.010 --> 00:28:08.080
sixteen how it is coming three here see three
is nothing but equivalent distributor pressure
00:28:08.080 --> 00:28:15.809
as three so distributor pressure as three
so p one p one plus three that will be p one
00:28:15.809 --> 00:28:21.000
will be is equal to then this p one this p
one how it will be this p one p one how to
00:28:21.000 --> 00:28:30.390
calculate this p one p one is nothing but
this one hundred thirteen that means p two
00:28:30.390 --> 00:28:40.110
plus three p two plus three why because this
p two minus p one p two minus p one that will
00:28:40.110 --> 00:28:46.840
be is equal to delta p d distributor pressure
drop then p two will be is equal to then p
00:28:46.840 --> 00:28:56.840
two is equal to p one plus delta p d so p
one is given to you that pone is p one is
00:28:56.840 --> 00:29:05.600
given to you p one is nothing but p one
what is the p one minus p two is equal to
00:29:05.600 --> 00:29:13.420
p one minus p two then it will be is equal
to it will be p one p one minus p two is
00:29:13.420 --> 00:29:18.789
equal to delta p d and then here p two is
given to you that is one hundred thirteen
00:29:18.789 --> 00:29:25.039
plus ten then it will be is equal to one
hundred thirteen this is one one minus one
00:29:25.039 --> 00:29:33.940
hundred thirteen minus one hundred thirteen
that will be is equal to that will be is
00:29:33.940 --> 00:29:46.240
equal to minus minus p one plus delta p
d three so ultimate p one will be is equal
00:29:46.240 --> 00:29:54.520
to p one will be is equal to one hundred sixteen
k p a so in this way we know that what should
00:29:54.520 --> 00:29:58.820
be the p zero what should be the p three p
three is nothing but what here in this case
00:29:58.820 --> 00:30:04.990
it is the atmospheric pressure atmospheric
pressure p three that is one zero one plus
00:30:04.990 --> 00:30:12.460
two here and this is one zero three p three
and then p three will be is equal to one zero
00:30:12.460 --> 00:30:19.610
one plus two that will be is equal to one
zero three one zero three why it is coming
00:30:19.610 --> 00:30:30.380
like that this p three is given to you actually
this p three is this nothing but p zero
00:30:30.380 --> 00:30:37.890
plus p zero p zero minus p two minus p
two here so this will be given as one zero
00:30:37.890 --> 00:30:42.919
three so p two will be is equal to what one
zero three plus ten ten is what ten is nothing
00:30:42.919 --> 00:30:48.700
but the p d distributor pressure and p
one is equal to this so finally we are getting
00:30:48.700 --> 00:30:52.870
the pressure as four fifty five kilowatt
00:30:52.870 --> 00:30:59.470
now if we consider that v condition where
delta p d is equal to ten k p a the this
00:30:59.470 --> 00:31:05.190
represents a distributor plate with excessive
pressure drop since delta p d is equal to
00:31:05.190 --> 00:31:13.730
delta p b here then evaluating pressures gives
as p zero is equal to one zero p zero is equal
00:31:13.730 --> 00:31:25.120
to one zero one and p is equal to one zero
three and here this p three and p two that
00:31:25.120 --> 00:31:31.070
would be is equal to one one three and
p is equal to one two and twenty three one
00:31:31.070 --> 00:31:37.951
twenty three k p a and following the same
procedure as in part a then we find this actual
00:31:37.951 --> 00:31:45.330
shaft work is equal to six five one kilowatt
or eight seventy three h h p thus the power
00:31:45.330 --> 00:31:52.190
requirement increases by almost two hundred
kilowatt if the distributor pressure is increased
00:31:52.190 --> 00:31:58.940
from this three k p a to the ten k p a now
in the third condition for five bypass into
00:31:58.940 --> 00:32:07.270
the freeboard of fifty percent of the air
and distributor pressure is three k p a then
00:32:07.270 --> 00:32:13.409
we have no change in pressure drops from part
a so in that case p zero is equal to one zero
00:32:13.409 --> 00:32:19.210
one and p three is equal to one zero three
p two is equal to one one three and p one
00:32:19.210 --> 00:32:26.390
is equal to one one six and then v s will
be is equal to what since here very interesting
00:32:26.390 --> 00:32:37.080
that this this whole amount of gas is not
supplied from this compressor and whereas
00:32:37.080 --> 00:32:45.560
this fifty percent of this gas is bypassed
and it is it is first this compressor here
00:32:45.560 --> 00:32:56.090
in this freeboard region so what should
be then v s for this compressor here it should
00:32:56.090 --> 00:33:02.390
be of course half of this total amount
of gas which is being compressed so for the
00:33:02.390 --> 00:33:10.300
primary air from part we can say that this
actual shaft work will be is equal to half
00:33:10.300 --> 00:33:16.950
of that actual shaft work which is obtained
as per that equation in initial condition
00:33:16.950 --> 00:33:24.200
that is this is in part a then that is
four fifty five divided by two this will be
00:33:24.200 --> 00:33:29.779
is equal to two hundred twenty seven point
five kilowatt so by this equation you will
00:33:29.779 --> 00:33:34.860
be able to calculate what should be then what
should be the what should be the power consumption
00:33:34.860 --> 00:33:42.730
if we if we by pass this only then fifty
percent of this air will pass through this
00:33:42.730 --> 00:33:47.940
compressor remaining fifty percent again it
will be compressed through this compressor
00:33:47.940 --> 00:33:53.430
another compressor is this now if we use this
compressor to compress this fifty percent
00:33:53.430 --> 00:33:59.659
of air within this operating condition
of initial pressure of one zero one and this
00:33:59.659 --> 00:34:06.320
volumetric flow rate of this gas as half of
this initial volumetric flowrate of the
00:34:06.320 --> 00:34:10.179
gas or air then what should be that
00:34:10.179 --> 00:34:14.520
so for the air bypassed into the freeboard
we need another blower of course or compressor
00:34:14.520 --> 00:34:20.270
its blower requirement is again given by this
again the same equation to be used to calculate
00:34:20.270 --> 00:34:27.250
the actual shaft work as this so it is coming
here of course this is function of pressure
00:34:27.250 --> 00:34:33.159
this is initial pressure this is this pressure
here it will be one zero three because this
00:34:33.159 --> 00:34:38.570
at this region the atmospheric pressure
is one zero three kilopascal here so based
00:34:38.570 --> 00:34:45.159
on this this actual shaft work will be is
equal to thirty one point six kilowatt
00:34:45.159 --> 00:34:53.171
so that total power requirement for the two
blowers will be the summation of this two
00:34:53.171 --> 00:34:59.890
work like this here initially it was two
hundred twenty seven point five and for the
00:34:59.890 --> 00:35:07.560
bypass gas it is thirty one point six so total
two hundred fifty nine kilowatt so this design
00:35:07.560 --> 00:35:13.260
gives at least see forty three percent almost
forty three percent savings in pumping power
00:35:13.260 --> 00:35:19.810
over the design of part a so what we observe
that if we bypass some amount of gas here
00:35:19.810 --> 00:35:26.030
we can get the benefit of the power consumption
if we if we if we bypass then of course
00:35:26.030 --> 00:35:30.670
the lowering the power consumption as per
this example of course some other operating
00:35:30.670 --> 00:35:36.200
condition to be designed or to be kept in
such way that this power consumption will
00:35:36.200 --> 00:35:41.079
be less compare to this so according to this
problem we are getting at least forty three
00:35:41.079 --> 00:35:49.359
percent power saving by bypassing fifty
percent of the gas through the another blower
00:35:49.359 --> 00:35:55.050
and supply to the freeboard of this fluidized
bed operation
00:35:55.050 --> 00:36:01.101
now if we do another exercise problem here
you can get same way now if pressure drop
00:36:01.101 --> 00:36:06.300
across the gas distributor is six k p a and
what should be the then power consumption
00:36:06.300 --> 00:36:11.250
if pressure drop across the gas distributor
doubled without changing delta p which
00:36:11.250 --> 00:36:17.280
show that power consumption will increase
seventeen point nine eight percent if thirty
00:36:17.280 --> 00:36:23.280
percent of the required air bypass the bed
and air introduced to the freeboard to burn
00:36:23.280 --> 00:36:30.710
the volatile gas is released in the bed by
the biomass in this case delta p d and delta
00:36:30.710 --> 00:36:37.750
p b are same as a so at this operating condition
of this initial pressure of this and t
00:36:37.750 --> 00:36:44.430
zero is this then what should be the power
consumption if the compressor power compressor
00:36:44.430 --> 00:36:49.570
efficiency is seventy five percent and it
is seen that at the same way for the calculation
00:36:49.570 --> 00:36:55.849
it is seen that the power consumption will
be the initial case eight hundred eighty
00:36:55.849 --> 00:37:01.380
two point zero four whereas this power saving
will be seventeen point nine eight percent
00:37:01.380 --> 00:37:09.250
if the pressure drop across the gas distributor
is doubled without changing bed pressure there
00:37:09.250 --> 00:37:17.490
and of course if thirty percent of the
gas if it is bypass the bed and introduced
00:37:17.490 --> 00:37:23.330
into the freeboard then we will see that the
power consumption will be six hundred fifty
00:37:23.330 --> 00:37:29.000
five point seven two here so almost seventeen
point nine eight percent will be power saving
00:37:29.000 --> 00:37:30.440
here
00:37:30.440 --> 00:37:38.589
so from this lecture what we what we
observe that we can calculate the power
00:37:38.589 --> 00:37:48.119
consumption from the thermodynamic equation
and what should be the actual shaft work
00:37:48.119 --> 00:37:53.460
and what should be the ideal shaft work to
compress the gas from this lower pressure
00:37:53.460 --> 00:38:01.070
to the higher pressure and also we can see
if the gas is bypassed and compressed to the
00:38:01.070 --> 00:38:10.960
distributor then we can of course save
some power during the operation and for this
00:38:10.960 --> 00:38:19.109
it is to be noted that of course you have
to consider other pressure drop is there
00:38:19.109 --> 00:38:27.840
any mechanical devices used to compress
or any other operation if it is attached like
00:38:27.840 --> 00:38:39.230
any other heat energy or any other energy
is required to the operation of the
00:38:39.230 --> 00:38:43.700
fluidized bed and what should be the total
energy required for that that you have to
00:38:43.700 --> 00:38:49.120
calculate here in this case only we are considering
that what should be the compressing power
00:38:49.120 --> 00:38:55.520
is required distribute the gas inside the
bed because this part is actually main contributory
00:38:55.520 --> 00:39:04.859
part for the fluidization operation so
this is very important to know and we have
00:39:04.859 --> 00:39:10.119
seen that several other parts also will be
contributing this distributor pressure
00:39:10.119 --> 00:39:17.530
and which will make you the consumption
estimation for this fluidization operation
00:39:17.530 --> 00:39:25.650
in the next lecture we will try to discuss
something about other type of distributor
00:39:25.650 --> 00:39:32.930
is spatial type of distributor like type of
fluidized bed like a bubbling fluidized bed
00:39:32.930 --> 00:39:40.040
what should be the other component of the
bubbling fluidized bed and also what
00:39:40.040 --> 00:39:45.490
should be the bubble characteristics in
the bubbling fluidized bed will be discussed
00:39:45.490 --> 00:39:51.750
in the next lecture and for this this gas
distributor you please follow the following
00:39:51.750 --> 00:39:58.349
books for more reading here kunii and levenspiel
fluidization engineering and yang handbook
00:39:58.349 --> 00:40:04.040
of fluidization and fluid particle system
so thats all todays lecture
00:40:04.040 --> 00:40:04.470
thank you