Steam Out Scenario

I recently sized a vacuum vent for a steam-out scenario and API 2000 did not provide guidance to calculate required relief rate. I thought I should share.

Tank can be cleaned with high pressure steam. After the cleaning, as the tank cools back down to ambient temperature, steam condenses and can create a vacuum inside the tank if the tank is blocked-in and/or rain accelerates the condensation rate. Tank vacuum implosion isn’t pretty.

A vacuum vents or pressure/vacuum vents (e.g. Groth vaccum breaker or Varec P/V vent) can be used to provide relief for an over-vacuum scenario (i.e. letting air entering the tank when a vacuum exists inside the tank). The amount of air entering the tank during relief must at least equal to the volumetric change caused by steam condensation.

Math!

The equations needed for calculations are simple. We assume steady state and start from the basics:

1. Heat loss to the enviroment is the convective heat transfer:

   $$\dot{Q}=UA(T-T_w)$$

2. Heat loss is provided by the condensation of steam, which can be related to the volumetric change ($\dot{V}$) by:

   $$\dot{Q}=\dot{m}\lambda=\rho\dot{V}\lambda$$

3. Combining two two expressions, we have the volumetric change:

   $$\dot{V}=\frac{UA(T-T_w)}{\lambda\rho}$$

   $T$: Steam saturation temperature

   $λ$: Steam latent heat at saturation

   $ρ$: Steam density at saturation

   $U$: Overall heat transfer coefficient

   $T_w$: Tank wall temperature

   $A$: Surface area

   $\dot{V}$: Volumetric change

Defining Relieving Condition

Regardless of the pressure of the steam used to clean the tank, the steam condition must be evaluated at relieving condition. For a steam-out scenario, relieving condition is saturated steam at relieving vacuum.

Often, relieving vacuum is very close to vacuum (within 0.5 psig or so), it is often sufficient to evaluate steam at saturated condition at 0 psig(vacuum).

Do not use the steam operating pressure to evaluate the volumetric change. For example, if 150# steam was used to clean the tank, the steam does not actually cause any vacuum until it condenses to ~0 psig. Thus, evaluating this scenario with 150# steam is incorrect. Think about it, why would steam at 150 psig cause a vacuum? It’s at 150 psig! That’s no vacuum.

I made that mistake once.

FYI, steam at 0 psig saturation has

$$T = 212^{\circ}\: F$$ $$λ = 970.189\: btu/lb$$ $$ρ = 0.0373018\: lb/ft^3$$

Heat Transfer Coefficient

Let’s assume heat transfer coefficient $U$ is constant throughout the tank and constant during the relief. Picking which value to use is a topic of debate. I myself contemplate about this.

However, at this point, pick a side and stick with it.

I have been using $3\; btu/hr\cdot ft^2\cdot F$ value for my evaluations since I have seen another engineering analysis used it before. Or we can be conservative and treat the condensation as if it’s occuring in a exchanger. According to engineering toolbox, heat transfer coefficient for an exchanger with combination of fluid and material as steam-cast-iron-water is $160\; btu/hr\cdot ft^2\cdot F$. That should be safe.

It could be confusing to know which heat transfer coefficient to pick and I unfortunately cannot provide a better answer. I myself is very uncertain of what to do. Hopefully there will be research and data to answer this.

So what do I say when I don’t know the answer to your question? Use your engineering judgement. Or if you know the answer, please let me know! Cheers.

Wall Temperature

Let’s assume we have a cold day in Texas $60^{\circ}F$ and wall temperature is uniform and the same as ambient temperature during relief. You ought to consider your weather when picking this $T_w$ value. Folks in Alaska will say $60^{\circ}F$ is pretty toasty.

Surface Area

Only take into accound the area that is exposed to ambient air. That is, exclude the bottom area and include the roof area.

Putting It All Together

Once you have made a decision on which wall temperature to use for your location, and stick with 0 psig saturated steam, the only variable left is the area ($A$). You can be wise and combine all the constants into one and vary $U\cdot A$ for different tanks:

$$\dot{V}=\frac{hA(T-T_w)}{\lambda\rho}=\frac{U\cdot A(212-60)}{973\cdot 0.0373018}=4.18794\cdot h\cdot A$$

I do this when I have 16 tanks to work with, nothing special if you have one or two at at time.