Steam Control Valve Specification Problem

[Pavan Kumar]

Hi All,

I need guidance to know if the way I am calculating the pressure upstream and downstream of a steam control valve for three different flow rates for its specification is correct or not.

The system sketch is attached.

I have to specify a Steam Control valve that controls steam flow to a Butane Vaporizer, through a steam bundle, that vaporizes liquid butane to vapor butane. The butane vaporizer has a Steam bundle immersed in liquid butane supplied from the bottom nozzle. The Steam control valve supplies steam to the Steam bundle at the desired opening which is controlled by the butane vaporizer pressure. The desired vaporizer pressure is 55 psig. The min, normal and max liquid butane flow rates are 10, 36 and 40 gpm respectively.

Taking the normal flow rate case, I determined that the required steam flow rate is 2659.38 lb/hr of superheated steam at 50 psig. As you can see in the sketch the steam is supplied to the control valve from the 265 psig Saturated (HP) steam header through a pressure let down regulator that is set to reduce the steam pressure from 265 to 70 psig pressure. Then there is a 80 ft long 3" Sch 40 [pipe between the let down regulator and the steam control valve. There is 2 feet of 3" sch 40 pipe between the downstream of the control valve and the steam bundle.

A) To calculate the steam pressure and temperature at the control valve inlet, I calculated the following :

1. I calculated the Steam super heat temperature for isenthalpic expansion from 265 psig to 70 psig to be 175.79 Deg C

2. I calculated the heat loss from the 3" sch 40 pipe with 2" mineral wool insulation and it is 7472.6 Btu/hr after considering the resistance of ambient air at 21.1 Deg C ( 70 Deg F).

3. I calculated the steam pressure prop using isothermal flow equation and calculated it as
1.5 psi. This means the steam pressure at the Control valve inlet is 70-1.5 =68.5 psig.

4. The specific enthalpy of Superheated steam at 70 psig is 1202.49 Btu/lb. Substracting the heat loss of 7472.6 from the heat content of steam at 70 psig ( 1202.49 Btu/lb)flowing at 2659.38 lb/hr, I get the specific enthalpy at the control valve inlet as 1199.68 Btu/lb

Specific enthalpy at Control valve inlet= Heat content of 70 psig superheated steam - Heat loss
= (2659.38*1202.49-7472.6)/ 23659.38 = 1199.68 Btu/lb

Knowing the steam pressure = 68.5 psig and specific enthalpy = 1199.68 Btu/lb at control valve inlet I get the steam condition to be superheated with superheat temperature to be 172.68 Deg C.

So at the control valve inlet:

Flow Rate = 2659.38 lb/hr
Pressure = 68.5 psig
Temperature = 172.68 Deg C
Specific Enthalpy = 1199.68 Btu/lb
Condition = Superheated.


B)To Calculate the Steam pressure and temperature at control valve outlet:

1. I fixed the pressure at the steam bundle as 50 psig and then back calculated the pressure drop for the flow of 26549.38 lb/hr to be 0.78 psi. So the pressure at the control valve outlet is 50.78 psig. With the steam expansion across the control valve being isenthalpic, the specific enthalpy at the control valve outlet is also 1199.68 Btu/lb

2. Calculated the heat loss in 3 feet of 3" sch 40 outlet pipe for 2659.38 lb/hr as 267.9 Btu/hr.

3. This gives the condition of steam at the bundle inlet to be 1199.579 Btu/lb. At 50 psig and 1199.579 Btu/lb the steam is superheated with temp = 168.547 Deg C.

At the bundle inlet the condition is

Pressure = 50 psig
Temp = 168.547 Deg C

Using this information for the required heat load of 2.481 MM Btu/hr required to maintain 55 psig Vaporizer pressure at 40 gpm liquid butane ( at 0 Deg C), I recalculate the required steam flow rate. Then after a couple of iterations I arrive at the following conditions at teh control valve inlet and outlet.

Case1: Butane Flow rate = 40 gpm

At Control Valve Inlet:
1. Steam flow rate = 2659.38 lb/hr.
2. Pressure = 68.5 psig
3. Temperature = 172.628 Deg C
4. Specific Enthalpy = 1199.68 Btu/lb
5. Condition = Superheated

At Control Valve Outlet :
1. Steam flow rate = 2659.38 lb/hr.
2. Pressure = 50.78 psig
3. Temperature = 168.547 Deg C
4. Specific Enthalpy = 1199.68 Btu/lb
5. Condition = Superheated

If this way of calculation is correct I will run the calcs for Cases 2 and 3 also and send the specification to the vendor.

Is it advisable to use polytropic process to calculate the gas pressure as opposed to isothermal / adiabatic and exclude all the pipe heat loss calculations?. There is a paper by William M. Kirkland ( attached in next message) which shows how mass flow can be calculated using polytropic approximation of compressible flow.

Your suggestions will be very helpful to me.

Thanks and Regards,
Pavan Kumar

 

[Pavan Kumar]

Hi georgeverghese,

For flashing flow, break up the pipe into 3-4segments, use an initial set of trial values for mean pressure for each segment and use the Lockhart- Martinelli procedure on pages 6-26,6-27. Change trial values to suit till they converge.

This procedure is only good for "Non-flashing" Liquid /gas flow. It is not suitable for Flashing flow. I think I will have re-read JC Leung's paper ( attached) and read some more papers which I need to search.

You can figure some way of making liquid butane at -10degC to 0degC to mix with the bulk liquid (so it gets prewarmed above 0degC) before it contacts the HX bundle. During normal operation the Liquid butane at 0 Deg C mixes with 50 Deg C butane in the vessel. The problem is during start-up which we need to brainstorm.

During shutdown, as long as kettle vapor space pressure is kept high, you can avoid steam condensate from freezing up inside the heating bundle. OK.

Ok, you've got a SS heating bundle, which is good, but not bulletproof. If you have a tube rupture during shutdown and butane rushes into the flash tank, condensate may freeze in the flash tank, block the vapor exit resulting in overpressure. Also, if C4 does not freeze, you've got a large flammable vapor cloud of C4 around the flash tank vent. Yes this is a serious life threatening hazard. How can we prevent the tube rupture from happening?. The tube side design pressure is 200 psig at 400 Deg F while the shell side maximum pressure is only 70 psig (limited by the PSV on the vessel set at 70 psig).

Thanks and Regards,
Pavan Kumar

 

[georgeverghese]

The Lockhart Martinelli procedure can be used given the assumption that in each segment, a mean vapor fraction can be applied. Its approximate, but it should give you a ballpark estimate.
During startup, may be you could
a)Pressurise the kettle vapor space with 50psig N2, then introduce liquid C4.
b)Crack open a steam trap bypass and keep the bundle warm while C4 is being introduced
c)Vent out the C4 / N2 vapor mix to LP flare for a while until the vapor space is expected to be all C4.

To me, it looks like the condensate flash drum upper design pressure should be 200psig, and its lower design temp should be the min ambient temp Tamb - 5degC.
The vent location should be re examined and a flame arrestor installed at the end. The entire HX and bundle min design temp should also be Tamb-5degC.

 

[Pavan Kumar]

Hi georgeverghese ,

I will follow your advise and work with Lockhart Martinelli method as you suggested.

I still have the following questions from what you said below.

georgeverghese

a)Pressurise the kettle vapor space with 50psig N2, then introduce liquid C4. How will pressurizing the vapor space of the vessel ( shell side) prevent freezing of the steam inside the tubes?. Increasing the vapor space pressure will raise the boiling point of butane and would require either more steam or more sub-cooling of condensate to provide the higher latent heat for butane. Wouldn't this be counter productive?
b)Crack open a steam trap bypass and keep the bundle warm while C4 is being introduced
c)Vent out the C4 / N2 vapor mix to LP flare for a while until the vapor space is expected to be all C4.

To me, it looks like the condensate flash drum upper design pressure should be 200psig, and its lower design temp should be the min ambient temp Tamb - 5degC. I think you are saying that the steam side and downstream is all designed for 200 psig ( the design pressure of the steam bundle). But tube side is the higher pressure side and the shell side ( butane side ) is the low pressure side. The reason why Butane is entering the steam condensate side is due to condensation of steam which is creating a vacuum with butane passing through the pinhole. The original question of how to prevent tube rupture still remains unanswered.

The vent location should be re examined and a flame arrestor installed at the end. The entire HX and bundle min design temp should also be Tamb-5degC.

 

[georgeverghese]

a) By pressuring the shellside to say 5-10psig will prevent C4 liquid from boiling at atm pressure = -1degC, which will otherwise freeze the steam in the bundle.

But if you do (b), it may be possible to avoid step (a). Having to vent C4 vapor - N2 vapor mix then wont be necessary on startup.

There is no such HX where tube or plate rupture cannot occur - in all cases, tube rupture remains an operational and/or safety risk.