16 replies to this topic

[LeoLeo]

Hi all,

 

In the API 521 (fifth edition, para 5.14.4.1), there is a formula to calculate the final pressure of trapped liquid for thermal expansion case. I have two questions;

 

I want to use this formula and justify that for my case, athermal relief valve is not required due to huge difference between maximum operating pressure (3.5 barg) and design pressure (100 barg) of pipe.

 

I tried to use the procedure (seems to be trial and error) but I do not get a logical answer. the fluid I used for the calculation is cooling water (in the tube side of heat exchanger).

 

I know that most of times we do not bother to calculate the final pressure and/or load of the relief valve for thermal expansion and we just consider 3/4" to 1" valve.

 

Please let me know if you used these formula for such a mentioned case.

 

Thanks in advance.

 

ARM

Edited by ARM, 15 September 2014 - 08:49 AM.

[PingPong]

the fluid I used for the calculation is cooling water (in the tube side of heat exchanger).

That API rule is for piping only, not for equipment that happens to have tubes inside, like a heat exchanger.

[Marc-Andre Leblanc]

Hello,

 

As PingPong said, the intent of API was to rule for piping only, heat exhanger is an other case. 

 

----

 

That being said:

 

A quick and easy comparison will show that you a thermal relief valve is required in almost all practical situations.

 

Let me use an exemple to demonstrate: (for reference I will use Spirax sarco Sub-saturated water tool to get relevant data, it is free and easy to access)

 

In your case, if you have water at 3.5 barg as the cooling medium in a heat exchanger : lets put and hypothesis and assume initial temperature of the water is 30°C

 

The mass and volume need to remain constant since you are blocked in and no liquid is allowed to leak. (no mass or volume change!)

 

This mean your density stay is constant. 

 

At 30°C and 3.5 barg the density of water is 995.810 kg/m3

 

The density need to remain constant at 995.810 kg/m3, while temperature and pressure increase.

 

The maximum pressure is known to be 100 barg

 

The temperature resulting in the this increase in pressure is : 41.88 °C

 

In a thermal relief event for a heat exchanger, the temperature of the blocked in liquid in the cold side should ultimatly reach the temperature of the inlet on the hot side if its not allowed to releive.

 

So, in this case, unless the inlet temperature of the hot side is less than 41.88 °C, you need a pressure relieving device 

 

For exemple if the temperature reach 50°C ... the pressure in the vessel would reach 182 barg ( or leak before it reach this pressure)

 

So the answer to the question ... do you need a thermal relief : YES you do!

 

Regards

 

Marc-Andre

Edited by Marc-Andre Leblanc, 15 September 2014 - 01:34 PM.

[fallah]

ARM,

 

Huge difference between maximum operating pressure and design pressure of a piping segment including a blocked-in liquid (here inside the tube side of a heat exchanger) which is subject to thermal expansion due to heat absorbing from hot side cannot be a credit based on which you avoid installing a TSV to relieve the relevant relief load; but because the blocked-in liquid volume in your case isn't so considerable, it's adequate using a 3/4" * 1" TSV to protect the tube side and associated piping from over pressure...

[fallah]

@PingPong

@Marc-Andre

 

Even though the API 521 para. 5.14.4.1 is submitted under a general title of "Piping", but the formula included in that para. to calculate the final pressure of blocked-in liquid for thermal expansion case, as OP's exact query, can be either applied to piping or equipment...See the attached...

Attached Files

•  Pages from API_STD_5211.pdf   122.55KB   344 downloads

[LeoLeo]

Thanks all for the replies !

 

@ Marc: I believe that we can't assume the volume remains constant. Due to temperature increase piping system will also have expansion. Even the formula in API 521 is considering the linear expansion of metal wall. In addition the density is also is changing for this case is deceasing. That's why we can not assume volume and density of water are both constant.

 

@ Fallah: The huge difference between operating and design condition of cooling water is because of 77% rule to eliminate the installation of PSV for tube failure (of heat exchanger) scenario. I agree with you that we shouldn't consider any credit for high design pressure and eliminate the installation of TRV. My problem is that I am trying to use the formula and calculate what will be the final pressure if we increase the temperature to 85, but I don't get a logical answer.

[fallah]

ARM,

 

Would you please submit the answer you got after final pressure calculation using equation (3) of API 521?

[LeoLeo]

Fallah,

 

Below, you will find the calculation sheet.

Attached Files

•  Thermal Expansion.xls   20KB   526 downloads

[PingPong]

.... but the formula included in that para. to calculate the final pressure of blocked-in liquid for thermal expansion case, as OP's exact query, can be either applied to piping or equipment.

If API RP521 is specified to be used on a project, then 15.14.4.1 is only to be used for piping, simply because it specifically states that it is not for pressure vessels or heat exchangers.

 

Moreover I expect that the formulas given are only valid for a cylindrical shape, like a piece of pipe.

Any other piece of equipment, with a different shape than a simple pipe, will have a different relation between stress in the metal, and volume of the metal shape.

In a heat exchanger tubeside there are not only the (cylindrical) tubes, but also the tubesheets, the channel, the channel cover, and in the shellside there is not only the (cylindrical) shell, but also the tubes, however pressurized from the outside, the tubesheet, and the shell cover.

 

As a design engineer I never used 15.14.4.1 myself, nor would I allow it to be used on my project. Even if it would save the investment of a small TSV it is simply not worth the endless discussions with client engineers, and with auditors during the HAZOP. At the end you would decide to add the TSV anyway, simply to end the silly discussions, and finally get a good night's sleep.

Edited by PingPong, 16 September 2014 - 07:21 AM.

[LeoLeo]

PingPong:

 

I understand your point regarding the application of formula.  In addition, I agree that easy way is to install the thermal relief valve and finish the discussion.  But, still I can't get a good night's sleep    since I can't get a logical answer from formula to calculate the final pressure due to temperature increase.  I am looking to this problem only from theoretical point of view and not practical.

[serra]

it is not a problem easy to solve,

assuming a container which expands due to heating of fluid,
you may evaluate dP solving

dV/dT (fluid) - (dV/dT+dV/dP) (container)
-------------------------------------------
       (1/V)*dV/dp (fluid)

dV/dT and (1/V)*dV/dp or isothermal compressibility can be evaluated accurately for water with a model such as IAPWS 1995,

you may also evaluate dV/dT and dV/dP for piping
while, as said by others, for a heat exchanger things seems much more difficult....

[Marc-Andre Leblanc]

Hello, 

 

I agree with PIngpong, API specificaly says this equation is to be used for piping system only and not for pressure vessel or heat exchanger. 

 

@ ARM

 

The method I proposed was in general more conservative, and a simplification of the method in API. 

 

If you tried your calculation at 85°C, considering an initial temperature of 30°C for water the result would have such a high pressure that results would be very approximative and cannot be relied upon.

 

So Instead I will calculate the temperature at resulting into a pressure of 100 barg using the API 521 method:

 

Without pipe dimension, I will assume a SCH80 1” pipe for demonstration

 

P1 = 451.3 kpa

 

P2 = 10101.3 kpa

 

T2 = ? (Initial try at 45°C) (Final value of 42.75°C)

 

T1 = 30°C

 

alphav = 0.000404 (1/°C) ( since we search final temperature, I initially used value at 45°C 0.000420)

 

alphai = 0.0000121 (1/°C)

 

t=0 (no leak allowed) therefore qII and V are negligeable

 

X = 0.000000436 1/kpa

 

d = 0.0243078 m

 

E = 207000000 kpa

 

dw = 0.0045466 m

 

mu = 0.3

 

Resulting T2 using API 521 method = 42.75°C

 

The temperature I calculated with the simplification and more conservative approach was 41.88 °C

 

If you want the resulting pressure at a final temperature of 85°C, they can be approximated but I would not use these value since they are completely irrelevant and impractical because of the extremly high pressure, table cannot be relied upon to give precise value of X.

 

The resulting pressure from API method is around 701 barg and with the method I propose around 705 barg.

 

So in conclusion, the thermal PRV is required.

 

Regards

 

Marc-Andre

[fallah]

PingPong,

 

Yes, the equation (3) is derived such that by which only final pressure of a blocked-in liquid in a cylindrical shape such as a stright piece of pipe with constant diameter and the same material, while is subject to a heat source, can be estimated but do the all blocked-in liquids in piping systems confine in such pieces of pipe? Obviously no; actually there are many situations in which the blocked-in liquid has been happened inside a part of a variable diameter line with several elbows and sometimes with different materials. Is that equations easily applicable for such piping system? Then there is no difference having piping system with elbows and reducers or having an equipment such as tube side of a heat exchanger with different shapes and materials in each part; for both cases if you are going to use such equation, you have to model their confined volumes as a cylindrical shape, in turn, will lead to much deviation from the actual values.

 

Of course, i believe mentioned equation should be used only for estimation of final pressure if one insisted to know it or have a doubt if final pressure is higher than the design pressure, otherwise when the design pressure of the piping or equipment is known it's adequate using a TSV with the set pressure equal to the design pressure of the system and i, same as you, had never used that equation in any relief system design in which i had been involved...

 

As can be seen in the page from API 521 that i attached to one of my previous posts, in introduction of the parameters of the equation, it is sensed that API hadn't focused on piping as only field for which that equation can be applied may be due to limited piping cases for application of the equation as i described at the beggining of current post..

Edited by fallah, 16 September 2014 - 02:15 PM.

[LeoLeo]

Hello, 

 

I agree with PIngpong, API specificaly says this equation is to be used for piping system only and not for pressure vessel or heat exchanger. 

 

@ ARM

 

The method I proposed was in general more conservative, and a simplification of the method in API. 

 

If you tried your calculation at 85°C, considering an initial temperature of 30°C for water the result would have such a high pressure that results would be very approximative and cannot be relied upon.

 

So Instead I will calculate the temperature at resulting into a pressure of 100 barg using the API 521 method:

 

Without pipe dimension, I will assume a SCH80 1” pipe for demonstration

 

P1 = 451.3 kpa

 

P2 = 10101.3 kpa

 

T2 = ? (Initial try at 45°C) (Final value of 42.75°C)

 

T1 = 30°C

 

alphav = 0.000404 (1/°C) ( since we search final temperature, I initially used value at 45°C 0.000420)

 

alphai = 0.0000121 (1/°C)

 

t=0 (no leak allowed) therefore qII and V are negligeable

 

X = 0.000000436 1/kpa

 

d = 0.0243078 m

 

E = 207000000 kpa

 

dw = 0.0045466 m

 

mu = 0.3

 

Resulting T2 using API 521 method = 42.75°C

 

The temperature I calculated with the simplification and more conservative approach was 41.88 °C

 

If you want the resulting pressure at a final temperature of 85°C, they can be approximated but I would not use these value since they are completely irrelevant and impractical because of the extremly high pressure, table cannot be relied upon to give precise value of X.

 

The resulting pressure from API method is around 701 barg and with the method I propose around 705 barg.

 

So in conclusion, the thermal PRV is required.

 

Regards

 

Marc-Andre

Marc-Andre,

 

Thanks for the explanation !

There is a formula for isothermal compressibility coefficient which depends on specific volume at initial and final pressure. But, API does not mention that the ν2 (specific volume) should be at P2 and T2 (or P2 and T1).

 

I believe we should consider P2 and T2 for v2 (specific volume) calculation. Do you agree?

[Marc-Andre Leblanc]

Hello,

 

Its is indeed not specified at what temperature to estimate the isothermal compressibility coefficient.

 

For my calculation I calculated both v1 and v2 at T2.

 

The value I estimated was 0.000000436 (1/kpa) at T2

 

If I had used  T1 to get v1 and V2, the value for isothermal compressibility would changed slightly to 0.000000439 (1/kpa)

 

I prefer to use T2 because what I think is of interrest is how compressible the fuild is at the final temperature.

 

But with a difference of less than 1 percent, either value would provide a good estimate. 

 

Regards.

 

Marc-Andre

[coolkashif]

Hello,

I need help on similar problem you all are discussing.

 

I need to know how many minutes it would take to 'pop' open the relief valve if the Economizer BFW isolation valves (10") are closed by mistake while the flue gas is still passing thorough the Economizer duct by heating the trapped water inside Economizer tubes? The BFW supply pressure is around 670 psig (312 F), however when the Economizer is isolated (without draining and venting) then the pressure of BFW (trapped inside the Economizer) would raise from 670 psig to 1119 psig (set pressure of relief valve) and will eventually trigger the relief valve to open.

 

My initial calulation is below. But after going through your discussions above I am wrong in assuming the final temperature T2. Kindly help me on this.

 

Total Volume = 6.159 m³ (10" pipe)

 

Mass of water = 6153 Kg = 13,565 lb

 

Initial pressure = 688 psig

Initial temperature = 312 F

Final pressure = 1119 psig

Final temperature = 560 F (From Steam Tables Sat. Liquid @ 1119 psig)

 

Specific heat @ 560 F Cp = 1.3319 Btu/lb F

Flue gas Heat transfer rate q= 86.8 x 10⁶ Btu/hr

 

Time required t = m Cp dT

q

 

t = 13565 x 1.3319 x (560-312)

86.8 x 10⁶

 

t = 0.051 hr = 3 min

[LeoLeo]

Hello,

I need help on similar problem you all are discussing.

 

I need to know how many minutes it would take to 'pop' open the relief valve if the Economizer BFW isolation valves (10") are closed by mistake while the flue gas is still passing thorough the Economizer duct by heating the trapped water inside Economizer tubes? The BFW supply pressure is around 670 psig (312 F), however when the Economizer is isolated (without draining and venting) then the pressure of BFW (trapped inside the Economizer) would raise from 670 psig to 1119 psig (set pressure of relief valve) and will eventually trigger the relief valve to open.

 

My initial calulation is below. But after going through your discussions above I am wrong in assuming the final temperature T2. Kindly help me on this.

 

Total Volume = 6.159 m³ (10" pipe)

 

Mass of water = 6153 Kg = 13,565 lb

 

Initial pressure = 688 psig

Initial temperature = 312 F

Final pressure = 1119 psig

Final temperature = 560 F (From Steam Tables Sat. Liquid @ 1119 psig)

 

Specific heat @ 560 F Cp = 1.3319 Btu/lb F

Flue gas Heat transfer rate q= 86.8 x 10⁶ Btu/hr

 

Time required t = m Cp dT

q

 

t = 13565 x 1.3319 x (560-312)

86.8 x 10⁶

 

t = 0.051 hr = 3 min

Coolkashif,

 

The case you described is different from what I started here. In your case, water is saturation condition and the reason pressure is increasing is vapor generation in the equipment  (which you need to consider it in your calculation).

 

The case I described is about sub-cooled system containing water which remains sub-cooled at reliving condition. The reason of pressure increase in my case is expansion of water.

 

Regards,

 

ARM