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Lng Uploading In A Road Tanker - Temperature And Bog

lng uploading bog transient analysis

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#1 opel

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Posted 02 April 2017 - 10:08 AM

LNG needs to be uploaded in a road tanker. Following is the relevant information:

LNG:
-Pressure: 3barg
-Temperature: -160 °C
-Flow rate: 50 m3/h

Road tanker:
-Tanker volume: 25 m3
-Temperature: +10 °C
-Pressure: Atmospheric
-Insulation: Perlite


Following two cases needs to be calculated:


Case-I: By fueling LNG in tanker, How long (time) will it take to reach a temperature of -160°C in tanker and how much LNG will be boiled-off (Vaporized) during this time?


Case-II:The tanker is 10% filled (ca. 2.5 m3) and is at -100°C and at 6barg. By fueling LNG in tanker at these conditions, how much time will it take to reach a temperatureof -160°C in tanker and how much LNG will be boiled-off (Vaporized) during this time?


Both cases should be regarded as being adiabatic for simplicity.

Would appreciate your input regarding the calculation procedure and / or relevant equations for calculations.


Many thanks in advance!

#2 Bobby Strain

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Posted 02 April 2017 - 12:45 PM

This sounds like an academic exercise. But, to answer your question: Conduct a heat balance in both cases.

 

Bobby



#3 opel

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Posted 04 April 2017 - 02:29 AM

Bobby, my query is a real world issue, though it might sound an academic exercise by my formulation of the problem.

 

Coming to the Point,  I have performed a heat balance calculation and have enclosed an excel spreadsheet for your review. The comments of other members are, off course, equally welcome.

 

Thanks!

Attached Files


Edited by opel, 04 April 2017 - 02:48 AM.


#4 Bobby Strain

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Posted 04 April 2017 - 09:56 AM

It does not look even close. And nobody wants to hunt through spreadsheets. So show us a simple procedure. You only need to vaporize enough LNG to provide the energy removal equal to that required to cool the metal mass of the tank. So, back to the drawing board. And, you should have a proper procedure before going to a spreadsheet where everything gets lost. Find this quantity, then decide how long you want to cool. You should also contact the tank supplier for cooldown recommendations.

 

Bobby



#5 opel

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Posted 05 April 2017 - 09:59 AM

It does not look even close. And nobody wants to hunt through spreadsheets. So show us a simple procedure. You only need to vaporize enough LNG to provide the energy removal equal to that required to cool the metal mass of the tank. So, back to the drawing board. And, you should have a proper procedure before going to a spreadsheet where everything gets lost. Find this quantity, then decide how long you want to cool. You should also contact the tank supplier for cooldown recommendations.

 

Bobby

 Bobby, I have enclosed a PDF document here to show the approach that I adopted in the excel spreadsheet, uploaded here earlier. Would appreciate the feedback for yours and other valuable members.

 

Would you please name some tank suppliers in your knowledge having expertise in LNG uploading?

 

Thanks!

Attached Files


Edited by opel, 05 April 2017 - 10:12 AM.


#6 Bobby Strain

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Posted 05 April 2017 - 12:16 PM

You need nothing more than the first two equations. Equation 1, m = mass of metal tank Cp = metal heat capacity, temperature  difference is 10 - (-160). Equation 2 is for LNG. Use Google to find information on the road tankers and respective suppliers.

 

Bobby


Edited by Bobby Strain, 05 April 2017 - 09:09 PM.


#7 opel

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Posted 06 April 2017 - 02:48 AM

You need nothing more than the first two equations. Equation 1, m = mass of metal tank Cp = metal heat capacity, temperature  difference is 10 - (-160). Equation 2 is for LNG. Use Google to find information on the road tankers and respective suppliers.

 

Bobby

 

Bobby, thanks for your feedback. Please clarify:

 

(i) since the outer tank is insulated, shouldn't  I make use of Eq. (3) to account for the heat conduction through insulation layer?

 

(ii) can I use Eq.(1a) to calculate the time to cool down the tanker? 

 

Thanks!


Edited by opel, 06 April 2017 - 02:55 AM.


#8 Bobby Strain

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Posted 06 April 2017 - 09:08 AM

Once you find information on the tanks, the answers will be obvious.

 

Bobby



#9 cea

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Posted 07 April 2017 - 12:40 AM

 

It does not look even close. And nobody wants to hunt through spreadsheets. So show us a simple procedure. You only need to vaporize enough LNG to provide the energy removal equal to that required to cool the metal mass of the tank. So, back to the drawing board. And, you should have a proper procedure before going to a spreadsheet where everything gets lost. Find this quantity, then decide how long you want to cool. You should also contact the tank supplier for cooldown recommendations.

 

Bobby

 Bobby, I have enclosed a PDF document here to show the approach that I adopted in the excel spreadsheet, uploaded here earlier. Would appreciate the feedback for yours and other valuable members.

 

Would you please name some tank suppliers in your knowledge having expertise in LNG uploading?

 

Thanks!

 

Opel, mCpT equation for LNG, is not relevant here, since liquid LNG will only vaporize by absorbing latent heat. There will not be any change in vapor temperature as long as tanker is maintained under atmospheric pressure. Hence, your equation 1a shall be only Q=m Lamda.

 

Secondly, you need to consider effect of heat gain through insulation as well, once temperature starts dropping down.

 

Thus modified eq. will be

 

m Lamda = ((kmetal AMetal surface / L) (Integral) (dT)) metal + ((kInsulation AInsulation sur / L) (Integral) (dT)) insulation

 

You do not have to insert term "time" in Eq. 1a as "m" is supposed to be in kg/hr.

 

This is what I can suggest.

 

(Bobby, correct me if I am wrong.)



#10 Bobby Strain

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Posted 07 April 2017 - 09:13 AM

cea,

     Yes, you are wrong. Follow the advice in post #8 and you will understand why.

 

Bobby






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