Lesson 5 - Gas Transfer SAQ

**/*** A yeast propagation vessel – rated as a pressure vessel – is to be provided with either compressed air or pure oxygen for yeast growth. With reference to Henry’s law, amongst others, describe which gas, if any, would allow greater quantities of oxygen and/or at a faster dissolution rate to be dissolved in the propagation broth.

Response:

Perhaps intuitively you feel that pure oxygen is the gas which would result in greater quantities of dissolved oxygen and at a faster dissolution rate? Well, yes … and no! You see, one has to be careful about jumping to the conclusion that because air contains less oxygen than pure oxygen its use will ‘naturally’ lead to smaller quantities of dissolved oxygen (DO) and at a lower rate. We need to use our (recently acquired!) understanding of gas laws and gas transfer to make a compelling argument. So, let’s start with stating a few assumptions:

1. Besides the option of supplying air or oxygen to the propagation vessel, all other parameters are the same. E.g. the flowrate and pressure of the gases is identical; the vessel pressure is identical; and the gas transfer device is identical.

2. Yeast metabolism and physiological considerations will be ignored.

3. Practical issues, including relative gas costs, will be ignored.

Having stated our assumptions, let’s first consider what drives the process of gas transfer. This is defined as:

And briefly the terms are:

• C_t – the dissolved gas concentration at any time t

• dC_t/dt – the rate (or speed) of change of the dissolved gas concentration

• C_E – the equilibrium (or saturated) dissolved gas concentration

• A – the area between the gas and liquid phase

• V – the liquid volume

• k_L – the ‘mass (gas) transfer coefficient’

As we have assumed that the vessel and its contents are identical, then we can infer that V, the liquid volume, is the same.

We also stated that our gas flowrates of oxygen and air are identical, so we can infer that A, the area between the gas and liquid phases, is the same.

Likewise, the values of k_L, the mass gas transfer coefficient, would be the same. I.e. the gas flowrates are the same, so the disturbance they create in the propagation broth (the turbulence) would be the same and lead to the same mass transfer coefficient.

That leaves us to consider C_E, the equilibrium dissolved oxygen concentration, and C_t, the dissolved gas concentration at any time t. Let’s start with C_E.

C_E, the equilibrium dissolved oxygen concentration, is defined by Henry’s law:

Where p is the partial pressure of the gas above the solution, H is the Henry’s law constant, and x is the molar fraction of the gas dissolved in the liquid. At equilibrium, C_E = x. So let’s rearrange for x:

As we are dealing with the dissolution of oxygen in a propagation broth, the value of H for air or oxygen will be the same. So, C_E is directly proportional to the partial pressure of oxygen in the gas supplied to the propagation vessel. As air contains approximately 21% v/v of oxygen, and pure oxygen is 100% v/v oxygen, we know that the partial pressure of pure oxygen is much higher than that of oxygen in air. E.g. if the gas supply pressure was 1 bar a, then the partial pressures of oxygen in pure oxygen and air respectively would be 1.0 and 0.21 bar a. So, the partial pressure of pure oxygen is 1/0.21 = 4.76 times that of oxygen in air. By this reasoning, we would expect C_E for pure oxygen to be 4.76 times that of air.

This supports our ‘intuition’ that the use of pure oxygen, under our assumptions, will lead to greater quantities of dissolved oxygen, as the equilibrium dissolved oxygen concentration is greater for pure oxygen than air.

So how about the impact, if any, on the dissolution rate? From the above expression for the rate of gas transfer, we can simplify it to:

And:

Comparison of the two gas transfer rate expressions suggests that the rate of oxygen transfer using pure oxygen will be greater than when using air.

So, our ‘intuition’ was correct … you can achieve greater quantities of dissolved oxygen at a faster dissolution rate using pure oxygen than air … under the assumptions we stated.

If we adjusted our assumptions, could air ever achieve similar or even better quantities of dissolved oxygen at a faster dissolution rate than that of pure oxygen? In brief, yes! E.g. if we were willing to supply air to the propagation vessel at a pressure 4.76 times the supply pressure of pure oxygen, then the partial pressure of oxygen in air would be equal to that for pure oxygen and the equilibrium dissolved oxygen concentrations would be the same. This probably wouldn’t be very beneficial for optimal yeast growth, and would have implications on the capital and operating costs of the propagator, but it highlights that assumptions have to be clearly considered in such comparative problems.