Originally Posted by Teus Tuinenburg

Dear kj,

1-

If you do not cool the compressed air and maintain the 200 degrC (Which is only reached at a certain pressure) the the RH at 3.5 bar(o) and 200 degrC becomes 0%, as the water is boiling at hat temperature (149 degrC) and pressure, when the intake conditions are 35 degrC at ambient pressure (1 bara) and 100% RH.

Conclusion, there will be no condensation at all.

2-

See attached file

3-

An aftercooler is not necessary.

Many installations (even in cement) operate perfectly without condensation problems

Because of the high fly ash temperature condensation is not occurring.

If you have 3 ESP’s, operating parallel, then you need 3 separate are flows.

Combining the 3 compressors into one header is then not allowed.

If the 3 compressors serve one conveying line at the same moment it is OK.

However, why 3 compressors parallel? Is the airflow that big?

Have a nice day

Teus

Sir

Still iam not able to understand , how the RH figure of 0.2% arrive , can you explain ■

Dear kj,

From physics, we know that:

RH=100*Watercontent/0.622 * (Pressure-Saturatedvaporpressure)/ Saturatedvaporpressure

At 4.5 bar pressure and 149 degrC, the saturated vapor pressure is equal to the pressure=4.5 bar

In the formula this value leads to RH=0%

All water is converted into gas.

At 4.5 bar pressure and 145 degrC, the saturated vapor pressure is equal to the pressure=4.1325 bar

In the formula this value leads to RH=0.2%

The RH is so low at 4,5 bar and 145 degrC, because the water is almost boiling.

The physics tables and formulas are used in the calculation program.

Have a nice day

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

From physics, we know that:

RH=100*Watercontent/0.622 * (Pressure-Saturatedvaporpressure)/ Saturatedvaporpressure

At 4.5 bar pressure and 149 degrC, the saturated vapor pressure is equal to the pressure=4.5 bar

In the formula this value leads to RH=0%

All water is converted into gas.

At 4.5 bar pressure and 145 degrC, the saturated vapor pressure is equal to the pressure=4.1325 bar

In the formula this value leads to RH=0.2%

The RH is so low at 4,5 bar and 145 degrC, because the water is almost boiling.

The physics tables and formulas are used in the calculation program.

Have a nice day

Teus

Dear sir

Can you tell me how to calculate teh mixture temperature for a given temp of ash and air ■

Dear kj,

Basic thermo dynamics.

The heat-content of air and fly ash is equalized over both, resulting in a mixture-temperature.

T(mix) = [ Cp * Q(air) * T(air) + Cfa * Q(FA) * T(FA)]/[ Cp * Q(air) + Cc * Q(FA)]

Divided by Q(air) and Q(fly ash)/Q(air) = SLR

T(mix) = [ Cp * T(air) + Cfa * SLR * T(FA)]/[ Cp + Cc *SLR]

in which :

Cp = specific heat-content of air at constant pressure

Q(air) = Mass flow of air

T(air) = Temperature of air

Cfa = specific heat of fly ash

Q(FA) = Mass flow of fly ash

T(FA) = Temperature of fly ash

Example for cement:

Cp = 0.24 10^3 cal/kg

Q(air) = 4.54 kg/sec

T(air) = 150 degr.C

Cc = 0.2 10^3 cal/kg

Q(cem) = 111.1 kg/sec

T(cem) = 50 degr.C



T(mix) = [0.24 * 4.54 * 150 + 0.2 * 111.1 * 50]/[ 0.24 * 4.54 + 0.2 * 111.1]= 54.7 degr C



Good day

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

Basic thermo dynamics.

The heat-content of air and fly ash is equalized over both, resulting in a mixture-temperature.

T(mix) = [ Cp * Q(air) * T(air) + Cfa * Q(FA) * T(FA)]/[ Cp * Q(air) + Cc * Q(FA)]

Divided by Q(air) and Q(fly ash)/Q(air) = SLR

T(mix) = [ Cp * T(air) + Cfa * SLR * T(FA)]/[ Cp + Cc *SLR]

in which :

Cp = specific heat-content of air at constant pressure

Q(air) = Mass flow of air

T(air) = Temperature of air

Cfa = specific heat of fly ash

Q(FA) = Mass flow of fly ash

T(FA) = Temperature of fly ash

Example for cement:

Cp = 0.24 10^3 cal/kg

Q(air) = 4.54 kg/sec

T(air) = 150 degr.C

Cc = 0.2 10^3 cal/kg

Q(cem) = 111.1 kg/sec

T(cem) = 50 degr.C



T(mix) = [0.24 * 4.54 * 150 + 0.2 * 111.1 * 50]/[ 0.24 * 4.54 + 0.2 * 111.1]= 54.7 degr C



Good day

Teus

Sir

1) How to calculate teh specific heat of fly ash for given temperarture

2) On of my client is askinf ro comply teh IS8573 requirement for compressed air at inlet of conveying system . Now i am in vague if i have to comply teh class 4 of IS8573 which says the water content in compressed air shall be less than 6g/m3 . Now as i hav eteh two option

a) Either the to take the compressed air before intercooler of arountd temp 176 Deg C where the RH is 0.2 %. But caution to be taken , that the this high temp will have teh bundle of water present , which if at all reaches to soem where may be 80 Deg C will throw a plenty of water

b) Second option is , to consider the air dryer , where the outlet ppm for pressure dew point is 400ppm which does not suit the IS8573 class 4 . Sir what shall i do?

Need your suggetion ■

Dear kj,

1-

“the specific heat of fly ash for given temperature”

This has to be measured by the fly ash producer.

However, the dependence of the specific heat on temperature is as far as I know not much.

2-

The water content of the compressed air should be less than 6 g/m3

As the m3 changes with the pressure, this parameter has no meaning.

A cooler (which can be an air to air cooler) with water separator eliminates the possibility of condensation in the conveying installation. (As discussed before)

If the customer wants to spend money on coolers and dryers ……………………….?

Dewpoint is in degrC as far as I understand and not ppm.

Have a nice day

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

1-

“the specific heat of fly ash for given temperature”

This has to be measured by the fly ash producer.

However, the dependence of the specific heat on temperature is as far as I know not much.

2-

The water content of the compressed air should be less than 6 g/m3

You

A cooler (which can be an air to air cooler) with water separator eliminates the possibility of condensation in the conveying installation. (As discussed before)

If the customer wants to spend money on coolers and dryers ……………………….?

Dewpoint is in degrC as far as I understand and not ppm.

Have a nice day

Teus

sir

1) Apparantly , if the client has asked to comply the requirement of 6g/m3 , in such case if i avoid using the after cooler (as you told earlier) how iam making sure that the water will be less than the 6g/m3. As i ubderstand when the compressed air temp elivates it has the more capability to hold the water . Indeed if iam maintaining the mixture temp of 145 Deg c in such case there may not be condensation . but how about if by any means the temp comes below , as in such case i understand there will be pleanty of water as the same was not removed in after cooler

2) As per the IS 8573 class 4 the pressure dew point shall be 3 Deg C and water content shall be less than 6g/m3, whereas my supplier of refrigerant air dryer says the carryover ppm from dryer will be 400ppm which is much higher but at the same time he is complying the pressure dew point of 3 deg c

2) You said As the m3 changes with the pressure, this parameter has no meaning.

How? ■

Dear kj,

6 g/m3

1-

Can you give the involved temperatures and pressures again, so that I can calculate the conditions.

Pressures and temperatures at specific locations:

Intake compressor (Ambient)

Outlet compressor (before cooler)

After cooler (condensed water drained)

When mixed with (hot) fly ash.

2-

Atmospheric conditions:

Air pressure = 1 bar(a)

Temperature = 25 degrC

Volume = 1m3

Water content 6 g

Pressurized conditions:

Air pressure = 5 bar(a)

Temperature 50 degrC

Water content 6 g

Volume under pressurized conditions = (1*1*323)/(5*298)=0.2167 m3

Water content per m3 = 6/0.2167 = 27.69 g/m3

Therefore, water content is always given as mass/kg of dry air.

RH is no direct indication for the amount of water vapor in the air.

Condensation starts when the RH has reached 100%.

Good evening

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

6 g/m3

1-

Can you give the involved temperatures and pressures again, so that I can calculate the conditions.

Pressures and temperatures at specific locations:

Intake compressor (Ambient)

Outlet compressor (before cooler)

After cooler (condensed water drained)

When mixed with (hot) fly ash.

2-

Atmospheric conditions:

Air pressure = 1 bar(a)

Temperature = 25 degrC

Volume = 1m3

Water content 6 g

Pressurized conditions:

Air pressure = 5 bar(a)

Temperature 50 degrC

Water content 6 g

Volume under pressurized conditions = (1*1*323)/(5*298)=0.2167 m3

Water content per m3 = 6/0.2167 = 27.69 g/m3

Therefore, water content is always given as mass/kg of dry air.

RH is no direct indication for the amount of water vapor in the air.

Condensation starts when the RH has reached 100%.

Good evening

Teus

Sir

Is there any maor difference between material density and bulk density ■

Dear kj,

Bulk density:

The mass of 1 m3 of bulk material.

Bulk density = Bulkmass/Volume

Inter particle voids are included in the volume, however are normally lighter than the material itself

Material density:

The mass of 1 m3 of solid material.

Material density = material mass/material volume

Inter particle voids are excluded from the volume.

Particle density:

The mass of 1 particle of material.

Particle density = particle mass/particle volume

That is all

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

6 g/m3

1-

Can you give the involved temperatures and pressures again, so that I can calculate the conditions.

Pressures and temperatures at specific locations:

Intake compressor (Ambient)

Outlet compressor (before cooler)

After cooler (condensed water drained)

When mixed with (hot) fly ash.

2-

Atmospheric conditions:

Air pressure = 1 bar(a)

Temperature = 25 degrC

Volume = 1m3

Water content 6 g

Pressurized conditions:

Air pressure = 5 bar(a)

Temperature 50 degrC

Water content 6 g

Volume under pressurized conditions = (1*1*323)/(5*298)=0.2167 m3

Water content per m3 = 6/0.2167 = 27.69 g/m3

Therefore, water content is always given as mass/kg of dry air.

RH is no direct indication for the amount of water vapor in the air.

Condensation starts when the RH has reached 100%.

Good evening

Teus

Dear sir

1) Inlet temp 32 Deg C

2) Inlet pressure to compressor 1 bar

Now from here i have two option

1) Either i take the compressed air before after cooler , where the temp is 175 Deg C, but than iam compramising with the risk not eliminating teh water inside the compressed air

2) Either i take the compresssed air after after cooler where the temp is 50 Deg C .

In botah the case as you have already specified that the RH will be less ie around 0.2 %

But if i ubderstand your previous response 6g/m3 does not plays any role .. Than what shall i consider to comply the IS 8573 class 4.

My worry is client is asking for above IS code for conveying air compressed air , now for given parameter how iam making sure to suit teh requirement, and which option i should use

Fly ash temp 150 Deg C ■

Dear kj,

Can you specify exactly what the code IS 8573 class 4 describes?

Is it a code for compressed plant air for tools and motors?

Or id the code really for pneumatic conveying air?

Thanks

Teus ■

Dear kj,

Inlet:

1 bar(a)

32 degrC

RH=100%

Water vapor content = 0.03061 kg/kg dry air

After compressor:

5 bar(a)

175 degrC

RH=0%

Water is boiling

After cooler:

5 bar(a)

50 degrC

RH=100%

Water vapor content = 0.01552 kg/kg dry air

Condensated water = 0.01509kg/kg dry air

Specific volume of air at 5 bar(a) and 50 degrC = 0.18328 m3/kg

Water vapor content = 0.01552/0.1832 = 0.08471 kg/m3 # 84.7 g/m3

If you want at 5 bar(a) approx 6 g/m3 water vapor content per m3 of compressed air, you need to cool the air to approx. 3 degrC

Again, this not necessary in pneumatic conveying.

Even cement is conveyed with un-cooled air without any problems.

Have a nice day

Teus ■

Dear ansoni,

Your concern is the possible condensation of water vapor in the compressed conveying air.

Condensation takes place when the Relative Humidity (RH) of the convey air reaches 100%.

There are 2 parameters that influence the RH: temperature and pressure.

Assume a 100% RH at the compressor intake.

1)

After the compressor, the RH should be lower, due to the higher temperature and the RH should be higher, due to the higher pressure.

Cooling the air increases the RH to 100% at convey air pressure.

2)

Convey air mixes with fly ash.

Depending on the fly ash temperature, the RH decreases with a high fly ash temperature and increases when the fly ash temperature is lower than the conveying air temperature.

3)

Further in the pipeline, the pressure decreases and the temperature is normally higher than the ambient temperature. The RH must be lower than 100%, because the air pressure is lower than at the beginning of the pipeline (maximum 100%) and the temperature after mixing with the fly ash is evolving at least to the ambient temperature.

If the parameters of the fly ash pneumatic conveying system are known, the air conditions in the pipeline are also known and condensation can be calculated at any location.

In cement conveying, where possible condensation can bond with the cement, the convey air is never dried, only in some installations cooled and water drained.

These untreated installations work without any condensation problems.

Condensation calculations of cement conveying installations show maximum 0.5% bonding of cement in the most extreme circumstances.

The fly ash is inert and therefore, the bonding problem is not an issue.

If you can give the pneumatic conveying pressures, temperatures, capacity and airflow of tour system, I can calculate the possible condensation of water.

Have a nice day

Teus ■

Originally Posted by Teus Tuinenburg

Dear ansoni,

Your concern is the possible condensation of water vapor in the compressed conveying air.

Condensation takes place when the Relative Humidity (RH) of the convey air reaches 100%.

There are 2 parameters that influence the RH: temperature and pressure.

Assume a 100% RH at the compressor intake.

1)

After the compressor, the RH should be lower, due to the higher temperature and the RH should be higher, due to the higher pressure.

Cooling the air increases the RH to 100% at convey air pressure.

2)

Convey air mixes with fly ash.

Depending on the fly ash temperature, the RH decreases with a high fly ash temperature and increases when the fly ash temperature is lower than the conveying air temperature.

3)

Further in the pipeline, the pressure decreases and the temperature is normally higher than the ambient temperature. The RH must be lower than 100%, because the air pressure is lower than at the beginning of the pipeline (maximum 100%) and the temperature after mixing with the fly ash is evolving at least to the ambient temperature.

If the parameters of the fly ash pneumatic conveying system are known, the air conditions in the pipeline are also known and condensation can be calculated at any location.

In cement conveying, where possible condensation can bond with the cement, the convey air is never dried, only in some installations cooled and water drained.

These untreated installations work without any condensation problems.

Condensation calculations of cement conveying installations show maximum 0.5% bonding of cement in the most extreme circumstances.

The fly ash is inert and therefore, the bonding problem is not an issue.

If you can give the pneumatic conveying pressures, temperatures, capacity and airflow of tour system, I can calculate the possible condensation of water.

Have a nice day

Teus

Sir let me give a parameter here

1) Fly ash temp inlet to vessel 150DEG C

2) Inlet conveying air pressure 3.5 Bar

3) Inlet conveying air temperature 50 Deg

4) Fly ash conveying rate 60 TPH

5) Proposed Compressed air volume 1200SCFM

With this what could be the possible condensation in pipeline and what is its implication on conveying . I am stressing this point , as why we require air dryer . One thing that it is it will avoid having the condensation in pipeline which otherwise will create corrosion and may happily convey teh material without any pipelinechocking.

But can you suggest wheteher really it is require ■

Dear kj,

The calculation goes as follows:

1)

Inlet ambient air conditions:

Ambient pressure = 1 bar(a)

Ambient temperature = 35 degrC

RH = 100%

2)

Compressed air conditions:

Pressure = 4.5 bar(a)

Temperature = 50 degrC

Between the ambient conditions and the compressed conditions, there is a cooler installed, because the compressed air has a RH=100% and 42.48 kg/hr water is condensated and drained from the cooler/condenser.

3)

Then the air (RH = 100% and 50 degrC) mixes with the fly ash (150 degrC).

The mixture temperature becomes approx. 145 degrC

The convey air conditions become then:

Pressure = 4.5 bar(a)

Temperature = 145 degrC

RH = 0.2 % (Very dry)

In case there was no cooler, the moisture content of the fly ash would become:

42.48/60000*100=0.07% moisture content of the fly ash.

This is no problem at all for a pneumatic conveying system

have a nice day

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

The calculation goes as follows:

1)

Inlet ambient air conditions:

Ambient pressure = 1 bar(a)

Ambient temperature = 35 degrC

RH = 100%

2)

Compressed air conditions:

Pressure = 4.5 bar(a)

Temperature = 50 degrC

Between the ambient conditions and the compressed conditions, there is a cooler installed, because the compressed air has a RH=100% and 42.48 kg/hr water is condensated and drained from the cooler/condenser.

3)

Then the air (RH = 100% and 50 degrC) mixes with the fly ash (150 degrC).

The mixture temperature becomes approx. 145 degrC

The convey air conditions become then:

Pressure = 4.5 bar(a)

Temperature = 145 degrC

RH = 0.2 % (Very dry)

In case there was no cooler, the moisture content of the fly ash would become:

42.48/60000*100=0.07% moisture content of the fly ash.

This is no problem at all for a pneumatic conveying system

have a nice day

Teus

Sir

1) Indeed the the air will be dry after mixing with hot matreail. But by the time teh compressed air reaches the user point , teher begins the condessation which erodes the pipi and otehr accessories , and theerby may have some hidden problem in conveyinhg which i dont know . Can you throw a light on this problem

1) Decondaly how you arrived to RH = 0.2 % (Very dry) ■

Dear kj,

The RH calculation went as follows:

1)

Inlet ambient air conditions:

Ambient pressure = 1 bar(a)

Ambient temperature = 35 degrC

RH = 100%

Calculate water vapor content = 0.0365 kg/kg dry air

2)

Compressed air conditions:

Pressure = 4.5 bar(a)

Temperature = 50 degrC

Calculate water vapor content = 0.0173 kg/kg dry air

Condensed water=0.0365-0.0173=0.0192 kg/kg/dry air

Between the ambient conditions and the compressed conditions, there is a cooler installed, because the compressed air has a RH=100% and 42.48 kg/hr water condensates and is drained from the cooler/condenser.

3)

Then the air (RH = 100% and 50 degrC) mixes with the fly ash (150 degrC).

The mixture temperature becomes approx. 145 degrC

The convey air conditions become then:

Pressure = 4.5 bar(a)

Temperature = 145 degrC

Already calculated water vapor content = 0.0173 kg/kg dry air

RH = 0.2 % (Very dry)

4)

Air conditions at end of pipeline:

Pressure = 1 bar(a)

Temperature = 35 degrC

Already calculated water vapor content = 0.0173 kg/kg dry air

RH = 47.3 % (dry)

Conclusion: In the pipe line there is NO CONDENSATION.

Calculations are based on the Psychrometric Chart.

Calculating physical phenomena and applying this knowledge on technical installations is why engineers do exist.

Best regards

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

The RH calculation went as follows:

1)

Inlet ambient air conditions:

Ambient pressure = 1 bar(a)

Ambient temperature = 35 degrC

RH = 100%

Calculate water vapor content = 0.0365 kg/kg dry air

2)

Compressed air conditions:

Pressure = 4.5 bar(a)

Temperature = 50 degrC

Calculate water vapor content = 0.0173 kg/kg dry air

Condensed water=0.0365-0.0173=0.0192 kg/kg/dry air

Between the ambient conditions and the compressed conditions, there is a cooler installed, because the compressed air has a RH=100% and 42.48 kg/hr water condensates and is drained from the cooler/condenser.

3)

Then the air (RH = 100% and 50 degrC) mixes with the fly ash (150 degrC).

The mixture temperature becomes approx. 145 degrC

The convey air conditions become then:

Pressure = 4.5 bar(a)

Temperature = 145 degrC

Already calculated water vapor content = 0.0173 kg/kg dry air

RH = 0.2 % (Very dry)

4)

Air conditions at end of pipeline:

Pressure = 1 bar(a)

Temperature = 35 degrC

Already calculated water vapor content = 0.0173 kg/kg dry air

RH = 47.3 % (dry)

Conclusion: In the pipe line there is NO CONDENSATION.

Calculations are based on the Psychrometric Chart.

Calculating physical phenomena and applying this knowledge on technical installations is why engineers do exist.

Best regards

Teus

Sir

1) Apparantly , my concern is teh compressed air after machine when reaches upto user point(ie inlet of pressure vessel in densphase) , generates soem condensate as because of teh condensation nature of compressed air as velocity changes in pipeline . I am figuring out the compressed air deispite of mixture of compressed air with ash . In such a case teher may be teh possiibility of getting condensed water with ash and result in fudion.

2) How dis you arrive to RH of 0.2% can you throw a more light

3) In our application we are using teh oil free screw compressor for conveying . Now i have two option either using aftercooler to reduce teh compressed air temp from 200Deg to 50Deg or bypassing and taking teh elevated temp of 200 drg compressed air directly to ash conveying . Which otion you recomend . Please note that we are using three compressor combining in main header and sending the same directly to 3 field receiver ( as teher is 3ESP ) ■

Dear kj,

1-

If you do not cool the compressed air and maintain the 200 degrC (Which is only reached at a certain pressure) the the RH at 3.5 bar(o) and 200 degrC becomes 0%, as the water is boiling at hat temperature (149 degrC) and pressure, when the intake conditions are 35 degrC at ambient pressure (1 bara) and 100% RH.

Conclusion, there will be no condensation at all.

2-

See attached file

3-

An aftercooler is not necessary.

Many installations (even in cement) operate perfectly without condensation problems

Because of the high fly ash temperature condensation is not occurring.

If you have 3 ESP’s, operating parallel, then you need 3 separate are flows.

Combining the 3 compressors into one header is then not allowed.

If the 3 compressors serve one conveying line at the same moment it is OK.

However, why 3 compressors parallel? Is the airflow that big?

Have a nice day

Teus

Attachments

temperaruresrh (PDF)

■

Originally Posted by Teus Tuinenburg

Dear kj,

1-

If you do not cool the compressed air and maintain the 200 degrC (Which is only reached at a certain pressure) the the RH at 3.5 bar(o) and 200 degrC becomes 0%, as the water is boiling at hat temperature (149 degrC) and pressure, when the intake conditions are 35 degrC at ambient pressure (1 bara) and 100% RH.

Conclusion, there will be no condensation at all.

2-

See attached file

3-

An aftercooler is not necessary.

Many installations (even in cement) operate perfectly without condensation problems

Because of the high fly ash temperature condensation is not occurring.

If you have 3 ESP’s, operating parallel, then you need 3 separate are flows.

Combining the 3 compressors into one header is then not allowed.

If the 3 compressors serve one conveying line at the same moment it is OK.

However, why 3 compressors parallel? Is the airflow that big?

Have a nice day

Teus

Sir

What shall be teh arrangement if i have teh two compressor to feed teh air at 3 ESPS ■

Dear kj,

One conveying system requires its own air supply.

This supply can consist of multiple compressors in parallel.

Two conveying systems, operating at the same time do need two dedicated air supply systems.

If two pneumatic conveying systems share one air supply, then only one of the conveying systems can operate.

If one air supply has to serve two conveying systems simultaneously, then the air molecules cannot remember for which system they are intended.

Success

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

Inlet:

1 bar(a)

32 degrC

RH=100%

Water vapor content = 0.03061 kg/kg dry air

After compressor:

5 bar(a)

175 degrC

RH=0%

Water is boiling

After cooler:

5 bar(a)

50 degrC

RH=100%

Water vapor content = 0.01552 kg/kg dry air

Condensated water = 0.01509kg/kg dry air

Specific volume of air at 5 bar(a) and 50 degrC = 0.18328 m3/kg

Water vapor content = 0.01552/0.1832 = 0.08471 kg/m3 # 84.7 g/m3

If you want at 5 bar(a) approx 6 g/m3 water vapor content per m3 of compressed air, you need to cool the air to approx. 3 degrC

Again, this not necessary in pneumatic conveying.

Even cement is conveyed with un-cooled air without any problems.

Have a nice day

Teus

Dear sir

You have mentioned , to achive 6g/m3 . teh temperature of compressed air has to be quencehed to 3 deg C which is teh dew point . Now how do i quantify and in formula which wil result in 6g/m3 for a given temperature of 3 deg ■

Dear kj,

Using the table :

-Saturated water vapor pressure = function(Temperature)

and the

-definition of Relative Humidity as actual water vapor pressure/saturated water vapor pressure at considered temperature

and the

-gas laws for gas mixes and partial pressures

and the

-psychrometric table (or chart)

it is possible to calculate the all the parameters, related to water content and water vapor condensation.

The calculation algorithm requires some efforts in the field of thermodynamics.

I did that many years ago for a tapioca pelletizing project in Thailand, whereby the warm pellets were air cooled.

During that cooling process, the tapioca pellets were also dried and that was measured with the wet bulb method as increase in water content in the out coming cooling air.

Use:

http://en.wikipedia.org/wiki/Relativehumidity

and

http://en.wikipedia.org/wiki/Psychrometrics

Success

Teus

PS.

At:

5 bar

50 degrC

The dew point temperature is 63.86 degrC (not 3 degrC) ■

Originally Posted by Teus Tuinenburg

Dear kj,

Using the table :

-Saturated water vapor pressure = function(Temperature)

and the

-definition of Relative Humidity as actual water vapor pressure/saturated water vapor pressure at considered temperature

and the

-gas laws for gas mixes and partial pressures

and the

-psychrometric table (or chart)

it is possible to calculate the all the parameters, related to water content and water vapor condensation.

The calculation algorithm requires some efforts in the field of thermodynamics.

I did that many years ago for a tapioca pelletizing project in Thailand, whereby the warm pellets were air cooled.

During that cooling process, the tapioca pellets were also dried and that was measured with the wet bulb method as increase in water content in the out coming cooling air.

Use:

http://en.wikipedia.org/wiki/Relativehumidity

and

http://en.wikipedia.org/wiki/Psychrometrics

Success

Teus

PS.

At:

5 bar

50 degrC

The dew point temperature is 63.86 degrC (not 3 degrC)

Dear sir

In your earlier calculation sheet and technical formula , you taught me to calculate the condensation between two condition . Ie

Condion1 :- Inlet compressed air at 1 Bar and 35 Deg

Condion2:- Outlet compressed air 50 deg c at 4.5 Bar

Now condenstaion stated is 0.0192kg/kg of air . But as in compressor the temperature is elevated from 30 deg c to 170 deg c and then being quenched to 35 or 40 degc by means of inter cooler & after cooler . Now how to quantify the condensation between this three phase pulsation of compressed air. Indeed the condensation primarily depends on after cooler eficiency and overall heat transfer coeff of after cooler , but assuming 100% efficiency how do we calculate the water condensed from elevated temp of 140 (highly armed with water vapour) to 50 deg c (outlet of compressed air) . I will be geatful if you send me the claculated example of above case

2) How to claculate dew point ■

Dear kj,

Attached the calculations.

Condition 1:

1 bar(absolute)

30 degrC

100 % RH

Condition 2:

5.5 bar(absolute)

170 degrC

As the boiling temperature of water at 5.5 bar(absolute) is 156 degrC (>170 degrC), the gas is now a mix of 2 gases [Dry air(RH=0) and overheated steam] of 170 degrC.

Condition 3 (condition 2 in the attachment):

5.5 bar(absolute)

40 degrC

The calculation for 140 degrC and 50 degrC cannot be executed, because the initial RH and the pressures are not given.

Have a nice day

Teus

Attachments

aircondensation (PDF)

■

Originally Posted by Teus Tuinenburg

href="showthread.php?p=64600#post64600" rel="nofollow">

Dear kj,

Attached the calculations.

Condition 1:

1 bar(absolute)

30 degrC

100 % RH

Condition 2:

5.5 bar(absolute)

170 degrC

As the boiling temperature of water at 5.5 bar(absolute) is 156 degrC (>170 degrC), the gas is now a mix of 2 gases [Dry air(RH=0) and overheated steam] of 170 degrC.

Condition 3 (condition 2 in the attachment):

5.5 bar(absolute)

40 degrC

The calculation for 140 degrC and 50 degrC cannot be executed, because the initial RH and the pressures are not given.

Have a nice day

Teus

Sir

I could not able to get your atention on my specific quiry

Here is the statement

Condition I

Ambient air obsolute pressure 1BAR

Temp 30 Degc

Condition 2

Ambient air cmpressed to 5.5 Bar

obsolute pressure 5.5BAR

Temp 170 Degc (before inter cooler)

Consition 3

Compressed air quenched to 50 deg c from 170 deg c

Temp 50 Degc

Pressure 5.5 bar

Now for above three condition there will be different range of water vapor content . Now what is apparant from there specified consition is

More the temperature , more the capacity of water acumulation . Hence water vapor at 170 Deg c could be more than water vapou at 50 Deg C

Now by means of intercooler the vaporised quantity of water vapour will condensed out depending on the intercooler efficiency

Now how do we qualtify that how much the vapor being condensed from 170 deg c to 50 deg c

I have attached one file from ATLAS COPCO & exercising to correlate your calc with th attached one

My objective laiddown for me is to find out the amount of water condensed at stage wise of compressed air

Attachments

doc1 (DOC)

■

Dear kj,

The missing information is the RH in the starting condition 1.

please verify

BR

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

The missing information is the RH in the starting condition 1.

please verify

BR

Teus

Sir sorry for incomplete information

RH WILL BE 100% ■

Dear kj,

The calculation results are attached.

Condition 1:

1 bar (absolute)

30 degrC

100% RH

Watervapor content : 0.02717 kgH2O/kg dry air

Condition 2:

5.5 bar(absolute)

170 degrC

Mixture of dry air and 0.02717 kg overheated steam/kg dry air

No condensation

Condition 3:

5.5 bar(absolute)

50 degrC

Watervapor content : 0.01408 kgH2O/kg dry air

Condensed water : 0.01309 kgH2O/kg dry air

Watervapor content + Condensed water = 0.01408 + 0.01309 = 0.02717 kgH2O/kg dry air, equaling the original amount of H2O from condition 1.

Success

Teus

Attachments

aircondensation2 (PDF)

■

Originally Posted by Teus Tuinenburg

Dear kj,

The calculation results are attached.

Condition 1:

1 bar (absolute)

30 degrC

100% RH

Watervapor content : 0.02717 kgH2O/kg dry air

Condition 2:

5.5 bar(absolute)

170 degrC

Mixture of dry air and 0.02717 kg overheated steam/kg dry air

No condensation

Condition 3:

5.5 bar(absolute)

50 degrC

Watervapor content : 0.01408 kgH2O/kg dry air

Condensed water : 0.01309 kgH2O/kg dry air

Watervapor content + Condensed water = 0.01408 + 0.01309 = 0.02717 kgH2O/kg dry air, equaling the original amount of H2O from condition 1.

Success

Teus

Dear sir

Thanksyou verymuch for your extenmded support and guidance. This time , apparantly i happen to quantify the structural weight for mild steel silo. Now for a given volulume storage of bulk material , i have calculated the geometrical spec ir dia and height of cylinder and conical portion . Now what dont no is how to calculate teh structural supprt weight require to support the dead and live load ■

Dear kj,

This last question should be replaced to another section of this forum.

(Silos,Bins,Hoppers&Domes).

That will keep the subjects organized in a structural way.

Structural silo calculation is not my field of experience, others have to take up this one.

Have a nice day

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

From physics, we know that:

RH=100*Watercontent/0.622 * (Pressure-Saturatedvaporpressure)/ Saturatedvaporpressure

At 4.5 bar pressure and 149 degrC, the saturated vapor pressure is equal to the pressure=4.5 bar

In the formula this value leads to RH=0%

All water is converted into gas.

At 4.5 bar pressure and 145 degrC, the saturated vapor pressure is equal to the pressure=4.1325 bar

In the formula this value leads to RH=0.2%

The RH is so low at 4,5 bar and 145 degrC, because the water is almost boiling.

The physics tables and formulas are used in the calculation program.

Have a nice day

Teus

Sir

1) What is difference between vapor pressure and saturated vapour pressuer of water . The reason iam asking , as if i increase the temp keeping the pressure constant the vapour pressure and saturated vapour pressure of water is same . But at 4.5 bar and 145 degc the vapour pressure is 0.02 bar and saturated vapour pressure is 4.13 bar ■

Dear kj,

The dry air and the water vapor are considered to be 2 mixed gases, each delivering a part of the total pressure.

If the dry air contains no water vapor, the vapor pressure is 0.

If the dry air contains some water vapor, the vapor pressure is pvapor.

If the dry air contains the maximum amount of water vapor, the vapor pressure is pvaporsaturated.

The RH = pvapor / pvaporsaturated * 100

If pvapor = pvaporsaturated, then RH = 100%

If you start with water with a certain water content, then:

If you pressurize the dry air-water vapor mixture, the RH increases or the pvaporsaturated decreases.

If you heat the dry air-water vapor mixture, the RH decreases or the pvaporsaturated increases to a maximum until the dry air contains no water vapor anymore (RH=0), which occurs at the boiling temperature.

At that state, there are 2 gases; dry air and dry steam.

see:

http://en.wikipedia.org/wiki/Relativehumidity

have a nice day

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

The dry air and the water vapor are considered to be 2 mixed gases, each delivering a part of the total pressure.

If the dry air contains no water vapor, the vapor pressure is 0.

If the dry air contains some water vapor, the vapor pressure is pvapor.

If the dry air contains the maximum amount of water vapor, the vapor pressure is pvaporsaturated.

The RH = pvapor / pvaporsaturated * 100

If pvapor = pvaporsaturated, then RH = 100%

If you start with water with a certain water content, then:

If you pressurize the dry air-water vapor mixture, the RH increases or the pvaporsaturated decreases.

If you heat the dry air-water vapor mixture, the RH decreases or the pvaporsaturated increases to a maximum until the dry air contains no water vapor anymore (RH=0), which occurs at the boiling temperature.

At that state, there are 2 gases; dry air and dry steam.

see:

http://en.wikipedia.org/wiki/Relativehumidity

have a nice day

Teus

Dear sir

Tonnage of thanks for your kind attention on opur certain clariifcation

Taking this subject forward , i have below clarification

I tried to calculate the temperature changes of mixture throughout the pipeline in attempt to quantify the temperature of mixture at terminal point to see the possibility of condensation and calculation goes as follows

A) Calculate temp changes due to heat convection

T2= (T1-Tamb) exp(-2pi R2 U L / mcp) + Tamp

Above equation derived from equating following below eqn

1)Q=2 Pi L R2 U LMTD Eq 1

2)Q= mcp (T1-T2) Eq 2

U= Overall heat rransfer

R2= Outer dia of pipe

T1= Inlet temperature of pipe

Tamb:- Ambient temp

L= Length of pipe

Cp= Specific heat of mixuture

T2= Temperature of mixture at terminal point

B) Calculate heat generated by particle collision

* m * vel^2 = h(specific heat) * m * dT (got ref from one of your article)

Then dT = * vel^2 / h(spec)

Now Add B in above Value of T2 in A , we will get final T2 temperature at terminal point

QUIRY

a) In above expression , How do I determine the velocity ? , is it the velocity of material being conveyed or the velocity of particle after collision ? . If it is later then how do I determine the same (velocity of particle after collision) ?

b) Secondly , Please correct me with the above stated calculation process & comment on the necessary modification and addition to be done ■

Dear kj,

This part looks OK

The formula should be:

* m * (vel2^2 – vel1^2) = h(specific heat air) * mair * dT + h(specific heat material) * mmaterial * dT

vel2 = particle velocity after collision

vel1 = particle velocity before collision

This is done by using a mathematical trick on physics.

The above calculation is executed every 0.001th of a second in the time domain.

Extra heat loss through the pipe wall is radiation (can be neglected below 400 degrC)

Extra heat generation is friction in the bends and the losses in airflow resistance (turbulence is converted into heat)

Then the heat balance must be divided over the material and the air.

Then the RH humidity calculation based on the found pressure and temperature results in the maximum water vapor content at that location.

If the maximum water vapor content (in kg water/kg dry air) at that location is lower than the water vapor content at the previous location, there is condensation.

This condensation generates condensation heat.

In addition the moisture content influences the gas density and thereby the velocities.

By incorporating all these (and more) effects every 0.001th second, the real situation is approached.

Actually, this is a (very) simple version of DEM.

Calculating the (in this case average) position and velocity of a particle based on the acting forces in the time domain and within space boundaries (the pipe wall).

The smaller the time domain, the more accurate the calculation, but the longer the calculation takes.

Have a nice day

Teus ■

Sir below is the clarification

This part looks OK

The formula should be:

Sir

It is very difficult to quantify the velocity of material after collision for every 0.001th because of ample times of calculation required . Would you suggest any alternate method to quantify the same

OK

How do i determine the same

Noted & thanks

Noted & thanks

Have a nice day

Teus[/QUOTE] ■

Dear kj,

That is why mathematics and computers were invented.

Success

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

That is why mathematics and computers were invented.

Success

Teus

Sir if you were to calculate the tempearture drop of mixture (fly ash and air) for a given atmospheric condition adn pipe length as mentioned below , what could have been the possible result?

1) Fly ash @ 120 TPH .

2) Temperature of fly ash at picku[p point:- 150 Deg C

3) Pipe length :- 1.2km

4) Sp heat of ash :- 0.25btu/lb

5) Compressed air FAD :- 6900m3/hr

6) Operating presure :- 2.75

7) Pipe dia :- 12 inch

8) Possible mixture temperature at pickup point :- 135 Deg C

9) Atmospheric condition :- 35 Deg C @ 80% RH

10) What shall be the mixture temperature at terminal point ?

Awaiting for your positive response ■

Dear kj,

Attached the requested calculations.

Success

Teus

Attachments

flyash conveying 1200m (PDF)

■

Originally Posted by Teus Tuinenburg

Dear kj,

Attached the requested calculations.

Success

Teus

Dear sir

Thanks you very much.

Apparantly i posed to evaluate the temperature profile of mixture throughout the conveying pipe as similar you did. To start with i need to determine the temperaure reduction for each node of pipe , accounting following mode of heat transfer .

1) Convective heat transfer from the fly ash to the convey air

2) Convective heat transfer from fly ash to the wall

3) convective heat transfer from conveying air to the pipe wall

4) Convective heat transfer from the pipe wall to surrounding atmosphere

All above set of heat transfer modes involves part calculation of certain fluid dyanamics parameter and assimilating the same for underlying objective (ie determining the temperature) .

Now i know certain algorithm calculation of heat transfer from pipe wall to atmosphere and etc , but without concrete practical knowledge and involved influential parameter . This been a reason for me to once again trigerred the same subject across you and understand set of onvolved calculation to determine the temperature at terminal point

Would you please help me enlightning the formulas for all above heat transfer mode for determining teh mixture (air & ash) temp at terminal point ? . This is my engineering curosity perhabs . ■

Dear kj,

Air:

-cools down with expansion.

-heats up because of friction resistance

-heats up because of pressure drop caused by keeping the product in suspension

Product:

-heats up because of collision- and friction losses.

The involved energies are calculated, using the calculated pressure drops.

The energy is assumed to be equalized instantaneously resulting in a mixture temperature.

The mixture temperature is calculated by dividing the energy over the air and the material based on their specific heat content, resulting in the same temperature for the air and the product. (mixture temperature)

The mixture cools down because of heat convection and radiation to the environment.

These heat changes are calculated in the program at each time interval (0.001) and the corresponding d(Length)=vproduct*dt.

All formulas are standard thermo dynamic formulas.

Arranging them into an algorithm is quite some work (I know now), because there is a relation between the pneumatic conveying calculations and the temperatures and the temperatures have their influence on the air conditions s.a. pressure.

And when temperature and pressure is changing, then that could be a cause for condensation.

And condensation has influence on density.

This can only be approximated in very small calculation steps (In this case 0.001 sec or sometimes shorter)

Sometimes, I get lost in the many relationships to be considered in pneumatic conveying, but, up till now, I think, I found a way out.

Success

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

Air:

-cools down with expansion.

-heats up because of friction resistance

-heats up because of pressure drop caused by keeping the product in suspension

Product:

-heats up because of collision- and friction losses.

The involved energies are calculated, using the calculated pressure drops.

The energy is assumed to be equalized instantaneously resulting in a mixture temperature.

The mixture temperature is calculated by dividing the energy over the air and the material based on their specific heat content, resulting in the same temperature for the air and the product. (mixture temperature)

The mixture cools down because of heat convection and radiation to the environment.

These heat changes are calculated in the program at each time interval (0.001) and the corresponding d(Length)=vproduct*dt.

All formulas are standard thermo dynamic formulas.

Arranging them into an algorithm is quite some work (I know now), because there is a relation between the pneumatic conveying calculations and the temperatures and the temperatures have their influence on the air conditions s.a. pressure.

And when temperature and pressure is changing, then that could be a cause for condensation.

And condensation has influence on density.

This can only be approximated in very small calculation steps (In this case 0.001 sec or sometimes shorter)

Sometimes, I get lost in the many relationships to be considered in pneumatic conveying, but, up till now, I think, I found a way out.

Success

Teus

Sir you already taught me the set of involved parameter influence mixture temp

But , would you please help me furnishing the set of formulas involved to determine the heat gain or loss for air and product ?

Sir i once again make my humble request you to please help me determining the mixture temperature profile throughout conveying pipe line , which will further help to indentify the potential threat of condensation ■

Originally Posted by Teus Tuinenburg

href="showthread.php?p=68557#post68557" rel="nofollow">

Dear kj,

The relevant formulas are:

SpecHeatContMix = (cpGas + MatSpecHeatCont * 4.1875 * MuLocal) / (1 + MuLocal)

dHeatMat = (Abs(Suspensiondp) + Productdp + Gasdp) * GasVolume * 10 * dt

dTempAirHeatMat = dHeatMat / (MatSpecHeatCont * 4.1875 * (dPipeMass - dSedimentMass) * 1000)

dTair = (Tair + 273) / ((AbsolutePressure / (AbsolutePressure - dpLength / 10000)) ^ ((kGas - 1) / kGas)) - (Tair + 273)

Tair = Tair + dTair

TempMix = (cpGas * Tair + MatSpecHeatCont * 4.1875 * Mu * (TempMix + dTempAirHeatMat)) / (cpGas + MatSpecHeatCont * 4.1875 * Mu)

Tair = TempMix

TempDifference = TempMix – AmbientTemperature

dTempdiff = -(HeatRestFactPipeWall * 4.1875 * TempDifference * 3.141596 * Diameter(t) * dLength * dt) / (SpecHeatContMix * (dPipeMass - dSedimentMass))

TempMix = TempMix + dTempdiff

Tair = TempMix

These formulas have to be embedded in the iteration algorithm over each length dL calculated with the local particle velocity and d(t).

Success

Teus

Sir

Could not able to interpret the result to meet the system sizing as depcited by you in early case ( please see attched file)

Beffore getting to main point , i pose certain clarification related to formula

1) What is Mu

Main clarification

1) I dont know which formula is for which mode of heat change

You have stated [*]Air cools down with expansion. [*]Air heats up because of friction resistance [*]Air heats up because of pressure drop caused by keeping the product in suspension [*]Product: heats up because of collision- and friction losses. [*]The mixture cools down because of heat convection and radiation to the environment. [*]Mixture is exchanging heat with surrounding [*]heat transfer from the conveying air to the pipe wall

Now as formula stated by you , which formula os for which heat exhanges is something iam not able to conceive

Its my humble request if you may please explain me comprehensively with formula taking simple example of temperature reduction from 155 Deg C to 66 Deg C in attached sheet . I know it will teeth some time for you , bu trust me will really patronage me to eradicate some misconception of possible condensation in pneumatic conveying

Thanks in anticipation as always

Attachments

flyash conveying 1200m (PDF)

■

Dear kj,

The main formula is :

d(pressure)*Volume = specific heat * mass * d(temperature)

Temperature drop of air is calculated according adiabatic expansion without external energy.

Temperature drop through conduction through the pipe wall is:

Energy loss = k * d(temperature) * Area * d(t)

Calculating the temperature curve along the pipeline requires the time increment method and some mathematical manipulation.

Success

Teus ■

Dear kj,

Mu represents the variable SLR

dTair

dTempAirHeatMat

dTempdiff

These formulas (amongst a lot of other formulas, calculating the pressure drops, acceleration, velocities, etc.) are used to calculate the temperature.

For a pneumatic pressure conveying system, the starting temperature is the outlet temperature of the compressor or, in case of an after cooler, the cooled temperature.

It took me some months to make a simple program setup that worked more or less and it took me years to complete the setup into a “reliable” and consistent program.

And after all those years, I find omissions in flexibility and bugs in the program syntax.

The programming work, also made me realize how complex the interactions between the parameters in pneumatic conveying are.

F.i. the heat release, when water vapor in the conveying air condenses, is not yet accounted for in the program. The volume decrease and density aspects of water vapor and condensation are accounted for a short while ago.

Last week, I refined the calculation for air only pressure drop and the possibility to display the air only pressures and temperatures along the pipeline.

Success

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

The main formula is :

d(pressure)*Volume = specific heat * mass * d(temperature)

Temperature drop of air is calculated according adiabatic expansion without external energy.

Temperature drop through conduction through the pipe wall is:

Energy loss = k * d(temperature) * Area * d(t)

Calculating the temperature curve along the pipeline requires the time increment method and some mathematical manipulation.

Success

Teus

Sir my major objective is to determine the potential possibility of condensation

Hence pelase help me in determining the temperature drop . I know the calculation involves for time increment , but the calculation do involve the set of formula and modes of heat transfer . As said by you there are set of heat exchange ie

1) Convective heat transfer from the fly ash to the convey air

2) Convective heat transfer from fly ash to the wall

3) convective heat transfer from conveying air to the pipe wall

4) Convective heat transfer from the pipe wall to surrounding atmosphere

Would you please help me in identifying the temperature reduction my mentioning the formula involved ? Just to take the case , how did you arive the temperature reduction from 155 Deg C to 66 Deg C for first node of pipe in your calculation? . Please help me ■

Dear kj,

This is the formula to calculate the mixturetemperature, when hot air and cold material are mixed.

Cp * Q(air) * T(air) + Cc * Q(cem) * T(cem)

T(mix) = --------------------------------------------------------------

Cp * Q(air) + Cc * Q(cem)

in which :

Cp = specific heat-content of air at constant pressure

Q(air) = Mass flow of air

T(air) = Temperature of air

Cc = specific heat of cement

Q(cem) = Mass flow of cement

T(cem) = Temperature of cement

Example :

Cp = 0.24 10^3 cal/kg

Q(air) = 4.54 kg/sec

T(air) = 150 degr.C

Cc = 0.2 10^3 cal/kg

Q(cem) = 111.1 kg/sec

T(cem) = 50 degr.C

0.24 * 4.54 * 150 + 0.2 * 111.1 * 50

T(mix) = ------------------------------------------------ = 54.7 degr C

0.24 * 4.54 + 0.2 * 111.1

Take care

Teus ■

Originally Posted by Teus Tuinenburg

href="showthread.php?p=68222#post68222" rel="nofollow">

Dear kj,

This is the formula to calculate the mixturetemperature, when hot air and cold material are mixed.

Cp * Q(air) * T(air) + Cc * Q(cem) * T(cem)

T(mix) = --------------------------------------------------------------

Cp * Q(air) + Cc * Q(cem)

in which :

Cp = specific heat-content of air at constant pressure

Q(air) = Mass flow of air

T(air) = Temperature of air

Cc = specific heat of cement

Q(cem) = Mass flow of cement

T(cem) = Temperature of cement

Example :

Cp = 0.24 10^3 cal/kg

Q(air) = 4.54 kg/sec

T(air) = 150 degr.C

Cc = 0.2 10^3 cal/kg

Q(cem) = 111.1 kg/sec

T(cem) = 50 degr.C

0.24 * 4.54 * 150 + 0.2 * 111.1 * 50

T(mix) = ------------------------------------------------ = 54.7 degr C

0.24 * 4.54 + 0.2 * 111.1

Take care

Teus

I apology , i could not able to explain you my concern

My point is how did you arrive the temperature reduction of 66 Deg C from 155 Degc for first node in your calculation attached above for 1200m conveying line? . Sir pleaser explain me the set of formula of heat transfer involved?

Attachments

flyash conveying 1200m (PDF)

■

Dear kj,

I calculated the temperature changes as described in the replies #43 and #45 of this thread.

The mentioned general formulas in #45 are applied on the pneumatic conveying calculations for each time increment dt.

That is all.

In the example, the temperature of 155 degrC is calculated from the flyash of 150 degrC and the compressed aur temperature of 265 degrC.

Because of the high mixture temperature and the low ambient temperature, the cooling is significant and in the intake section, the temperature drops down to 66 degrC.

Success

Teus ■