This post would be quite long since agitation scale-up involved trials and evaluation of the parameters. This post is based on Handbook of Chemical Engineering Calculations Fourth Edition by Nicholas P. Chopey.

To understand better about how to scale-up nongeometric liquid agitator, I will use an example. We will make one change at a time to understand its effect.

The example is below:

*A process involving water-like liquid must be scaled up from an agitated 18-in diameter, 15-gallon pilot-scale reactor to a 120-in diameter, 7000-gal large scale reactor. *

*The pilot scale has a 18-in straight side and the large-scale reactor will have a 168-in straight side. Both reactors have ASME dished heads on the top and bottom.*

*Successful process performance was obtained in the pilot scale with two 6.0-in diameter pitched-blade turbines operating at 350 rpm. It is proposed that the large-scale reactor use hydrofoil impellers instead of pitched-blade turbines, for improved liquid motion. *

*Each pitched-blade turbine has a turbulent power number of 1.37 and each hydrofoil has a power number of 0.3.*

*Past scale-up experiences with similar processes, but with geometrically similar tanks, were successful when impeller tip speed was held constant.*

**Step 1 – Calculate Liquid Levels**

The very first step is to calculate liquid levels. The purpose is to check if straight-side as mentioned in the above example is sufficient to hold liquid. After liquid level is calculated, then we will check ratio of liquid level to tank diameter for pilot scale and large scale, respectively.

Liquid level is expressed using the following formula:

Figure below illustrates which straight side and depth to straight side are. Contingency should be added because 5% to 10% of calculated tank volume is typically filled by tank internals, such as impeller, shaft, baffles.

Figure below illustrates which straight side and depth to straight side are. Contingency should be added because 5% to 10% of calculated tank volume is typically filled by tank internals, such as impeller, shaft, baffles.

*How to determine depth to straight side?*

The depth to straight side is function of dished head type. In this example the type of head is ASME dished head. There are several resources that tabulates depth to straight side and its volume as function of diameter.

For example, by using table below for ASME F&D, 18-in vessel has depth to straight side 3.000 in and volume 2.0 gallon.

As for 120-in vessel, the depth to straight side is 20.351 in and its volume is 607.3 gallon.

__For Pilot Scale:__

- Volume of straight side = tank volume – volume of depth to straight side = 15 – 2 =
__13 gallon or 3003 in__^{3} - Liquid level at straight side = Volume of straight side x 4 / (pi x tank diameter
^{2}) = 3003 x 4 / (pi x 18^{2}) =__8 in__ - Total liquid level = liquid level at straight side + depth to straight side = 11.8 + 3 =
__8 in__ - If contingency is 10%, then total liquid level after contingency = (1+10%) x 14.8 =
__3 in__ - Total liquid level is less than straight side of the tank (18 in), so that tank straight side is
- Ratio of liquid level to tank diameter =
__3/18 = 0.9__

__Follow the same procedure for large-scale:__

- Volume of straight side = tank volume – volume of depth to straight side = 7000 – 607.3 =
__6393 gallon or 1,476,714 in__^{3} - Liquid level at straight side = Volume of straight side x 4 / (pi x tank diameter
^{2}) = 1,476,714 x 4 / (pi x 120^{2}) =__6 in__ - Total liquid level = liquid level at straight side + depth to straight side = 130.6 + 20.351 =
__9 in__ - If contingency is 10%, then total liquid level after contingency = (1+10%) x 150.9 =
__166 in__ - Total liquid level is less than straight side of the tank (168 in), so that tank straight side is
- Ratio of liquid level to tank diameter =
__166/120 = 1.38__

We can see here that the ratio of liquid level to tank diameter is not the same between pilot scale and large scale. For geometric similarity, all such length ratios should be equal. Since the ratio is not equal in this example, it is necessary to use nongeometric scale-up.

**Step 2 – Direct Scale-up for Non-geometrically Similar Reactor and Evaluate the Issues**

In this step, we will directly scale-up the reactor and evaluate the issues. In the following steps, we will repeatedly calculate several ratios to evaluate if the scale-up is satisfied. Such ratios are power per volume ratio and torque per volume ratio.

__For Pilot Scale:__

- Total power of impellers (hp) = no of impellers x Np x SG x N
^{3}x D^{5}/(1.524 x 10^{13})

Np = Power Number. It is a function of impeller type

SG = Specific Gravity of Liquid

N = Agitator speed (rpm)

D = Impeller diameter (in)

Total power of impellers (hp) = 2 x 1.37 x 1 x 350^{3} x 12^{5} / (1.524 x 10^{13}) = __0.06 hp__

- Power per volume = 0.06 x 1000/15 =
__4 hp/1000 gallon__ - Torque = 63,025 x total power / Agitator speed = 63,025 x 0.06 / 350 =
__79 in-lb__ - Torque per volume = 10.79 x 1000 / 15 =
__720 in-lb/1000 gallon__

__Then, for large scale:__

- Since diameter of impeller is not the example, we will use geometric similarity.

Diameter of impeller at large scale = tank diameter at large scale/tank diameter at pilot scale x diameter of impeller at pilot scale.

Diameter of impeller at large scale = 120 / 18 x 6 = __40 in.__

- Assume number of impellers is the same as in pilot scale, which is two units.
- Agitation speed at large scale = agitation speed at pilot scale x impeller diameter at pilot scale/impeller diameter at large scale

Agitation speed at large scale = 350 rpm x 6 in / 40 in = ** 52.5 rpm**.

- Total power of impellers (hp) = no of impellers x Np x SG x N
^{3}x D^{5}/(1.524 x 10^{13})

Np = Power Number. It is a function of impeller type

SG = Specific Gravity of Liquid

N = Agitator speed (rpm)

D = Impeller diameter (in)

Total power of impellers (hp) = 2 x 0.3 x 1 x 52.5^{3} x 40^{5} / (1.524 x 10^{13}) = __0.58 hp__

Note: power number is 0.3 because we use hydrofoil impeller (see example narrative).

- Power per volume = 0.58 x 1000/7000 =
__08 hp/1000 gallon__ - Torque = 63,025 x total power / Agitator speed = 63,025 x 0.58 / 52.5 =
__700 in-lb__ - Torque per volume = 700 x 1000 / 7000 =
__100 in-lb/1000 gallon__

**Evaluation:**

We will compare power volume ratio between large scale and pilot scale = 0.08 / 4 = 2%.

Torque per volume ratio between large scale and pilot scale = 100 / 720 = 14%.

Drastic change in power per volume and torque per volume indicates that direct scale-up is incorrect. We need to adopt step-by-step nongeometric scale-up. We will identify the problems and find the solutions.

**Step 3 – Conduct Geometry Similarity Scale-up**

A good starting point even for nongeometric scale-up is making geometric similarity one. For geometric similarity, the ratios of all the length dimensions are the same from small to large scale. Therefore:

- Large-scale-impeller diameter = pilot-scale impeller diameter x (large-scale tank diameter / pilot-scale tank diameter)

Large-scale-impeller diameter = 6 in x (120 in/18 in) = __40 in__

For tip speed (Π x D x N) to remain constant, the rotational speed of pilot scale must be adjusted to inverse proportion to the geometric ratio. Thus,

- Large-scale impeller rotational speed = pilot-scale impeller rotational speed x (pilot-scale impeller diameter / large-scale impeller diameter)

Large-scale impeller rotational speed = 350 rpm x (6 in / 40 in) = __52.5 rpm__

Since all the geometric ratios to be the same, the liquid level in large-scale vessel has to be:

- Large-scale liquid level = pilot-scale liquid level x (large-scale tank diameter / pilot-scale tank diameter)

Pilot-scale liquid level is 16.3 in as calculated in Step 1.

Thus, large-scale liquid level = 16.3 x (120/18) = __109 in__

After considering large-scale head dimension and volume calculation, the geometric-similarity volume in a 120-in diameter is ** 4318 gallons**.

To verify the scale-up step, let’s calculate tip speed.

- Large-scale tip speed = pi x D x N / 12 = pi x 40 in x 52.5 rpm / 12 =
.__550 ft/min__

Assuming type of impeller is the as pilot-scale, which is pitch blade turbine. Therefore, power number is 1.37. Then, the power required for two impellers will be:

- Large-scale impeller total power (hp) = number of impeller x Np x SG x N
^{3}x D^{5}/ (1.524 x 10^{13}) ; where Np = power number, SG = liquid specific gravity, N = rotational speed (rpm), and D = impeller diameter (in)

Large-scale impeller total power = 2 x 1.37 x 1 x 52.5^{3} x 40^{5} / (1.524 x 10^{13}) = ** 2.66 hp**.

- Power per volume is = 2.66 x 1000 / 4318 =
.__62 hp/1000 gal__ - Torque = 63,025 x total power / N = 63,025 x 2.66 / 52.5 =
.__3198 in-lb__ - Torque per volume = 3198 x 1000 / 4318 =
.__741 in-lb/1000 gal__

We can see here, a geometric-similarity scale-up with constant tip speed also keeps the ratio of torque to volume constant. As calculated in Step 2, pilot-scale torque per volume is 720 in-lb/1000 gal, which is similar to calculation result above, 741 in-lb/1000 gal.

In several cases, keeping torque/volume is more important than keeping power/volume constant.

**Step 4 – Apply Volume Adjustment to the Results of Step 3**

Results in Step 3 showed that geometric-similarity scale-up resulted in a volume of 4318 gallon in a 120-in tank diameter. However, the design specification called for 7000 gal. Therefore, higher liquid level is required.

For equal tip speed and impeller with geometric similarity, the rotational speed remains the same in spite of higher liquid level. Therefore, power and torque remain the same for the same impellers, speed, and fluid properties. However, as the volume changes, then power per volume also changes:

- Power per volume for 7000 gal = large-scale impeller total power x 1000 / 7000 = 2.66 x 1000 / 7000 =
__38 hp/1000 gal__ - Torque per volume for 7000 gal = large-scale torque per volume x 1000 / 7000 = 3198 x 1000 / 7000 =
__460 in-lb/1000 gal__

Let us compare power per for volume and torque per volume obtained from calculation above with pilot-scale (see Step 2).

- Comparison of pilot per volume = 0.38/4 =
__5%__ - Comparison of torque per volume = 460/720 =
__5%__

Comparison above shows that there is significant reduction in agitation intensity compared to pilot-scale results.

**Step 5 – Adjust Step 4 Results to Gain Constant Torque per Volume**

Equal torque per unit volume should keep velocity magnitudes similar, thus giving the appearance of similar liquid agitation intensity.

To maintain the torque per volume 720 in-lb/1000 gallon (as in pilot-scale) in 7000 gal vessel, then the torque should be:

- Torque = 720 in-lb/1000 gal x 7000 gal = 5037 in-lb.

Combining power and torque equation:

Then, total power is:

- Total power = 2 x 1.37 x 1 x 65.9
^{3}x 40^{5}/ (1.524 x 10^{13}) =__5.3 hp__ - Power per volume = 5.3 x 1000 / 7000 =
__0.8 hp/1000 gal__ - Tip speed = pi x N x D / 12 = pi x 65.9 x 40 / 12 =
__690 ft/min__

We can see here the tip speed increased compared to calculated in Step 3 (550 ft/min). Whether this increase in tip speed is an advantage or disadvantage depends on the process.

**Step 6 – Adjust Step 5 Results to Gain Constant Tip in Addition to Constant Torque per Unit Volume**

In Step 4, we use equal tip speed for geometric similarity scale-up. On the other hand, in Step 5, we use equal torque/volume criteria. However, with a change in the impeller diameter, both tip speed and torque per unit volume can be kept constant simultaneously.

In the following equation, the subscripts 1 and 2 represent the characteristics for small and large volumes, respectively.

__Constant Tip Speed__

N_{1} D_{1} = N_{2} D_{2}

- Solving total power = 2 x 1.37 x 1 x 47
^{3}x 44.7^{5}/ (1.524 x 10^{13}) =__33 hp__ - Power per volume = 3.33 x 1000 / 7000 =
__5 hp/1000 gal__ - Torque = 63,025 x 3.33 / 44.7 =
__4689 in-lb__ - Torque per volume = 4689 x 1000 / 7000 =
__670 in-lb/1000 gal__ - Tip speed = pi x 44.7 x 47 / 12 = 550 ft/min

With a small adjustment in impeller diameter, both torque per volume and tip speed can be held constant. Since liquid level has changed, geometric similarity no longer applies and impeller diameter adjustment may be acceptable option.

**Step 7 – Investigate Changing the Number of Impellers**

In Step 3, we got volume of 4318 gallon if geometric-similarity calculation is used. If using 7000 gal, as per our desired design, then volume is increase ** 7000/4318 = 1.6 times**.

Therefore, the number of impellers would be 1.6 times x 2 impellers = ** 3 impellers**.

For three impellers, then:

- Total Power = 3 x 1.37 x 1 x 52.5
^{3}x 40^{5}/ (1.524 x 10^{13}) =__4 hp__ - Power per Volume = 4 x 1000 / 7000 =
__57 hp/gal__ - Torque = 63,025 x 4 / 52.5 =
__4797 in-lb__ - Torque per volume = 4797 x 1000 / 7000 =
__685 in-lb/1000 gal__ - Tip speed = pi x 52.5 x 40 / 12 =
__550 ft/min__

By adding number of impellers, power per volume is similar (Step 6: 0.5 hp/1000 gal, Step 7: 0.57 hp/gal), torque per volume is also similar (Step 6: 670 in-lb/1000 gal, Step 7: 685 in-lb/1000 gal), and the same tip speed.

**What if we change the impeller type?**

**Step 8 – Investigate Changing Impeller Type While Keeping Tip Speed Constant**

By changing impeller type, power number will also change. The typical hydrofoil impeller has a power number of 0.3. If number of impellers is 3 as calculated in Step 7, by changing impeller type to hydrofoil type, then:

- Total Power = 3 x 0.3 x 1 x 52.5
^{3}x 40^{5}/ (1.524 x 10^{13}) =__88 hp__ - Power per Volume = 0.88 x 1000 / 7000 =
__125 hp/gal__ - Torque = 63,025 x 0.88 / 52.5 =
__1050 in-lb__ - Torque per volume = 1050 x 1000 / 7000 =
__150 in-lb/1000 gal__ - Comparison torque per volume = 150/720 =
__21%__

Torque per volume for pilot scale is 720 in-lb/gal (see Step 2)

Based on the calculation, changing impeller type gave one-fifth torque per volume compared to original torque per unit volume as calculated in Step 2. Therefore, agitation intensity is not as high as in original pilot scale.

**Step 9 – Change Impeller Type with Equal Tip Speed and Adjusted Torque per Volume**

We can use calculation similar with those in Step 6 to keep both tip speed and torque per volume constant when impeller type is changed.

Because type of impeller change, power number must be included in the calculation. A hydrofoil impeller is more efficient at creating liquid motion than a pitched-blade turbine. Therefore, the torque required by the hydrofoil impeller will be assumed to be half that of the pitched-blade turbine.

In the following equation, subscript 1 will be used for pitched-blade turbine and subscript 2 will be used for the hydrofoil impeller.

__Constant Tip Speed__

N_{1} D_{1} = N_{2} D_{2}

For impeller diameter 52.7 in and impeller rotational speed 39.9 rpm:

- Total power = 3 x 0.3 x 1 x 39.9
^{3}x 52.7^{5}/ (1.524 x 10^{13}) =__52 hp__ - Power per volume = 1.52 x 1000 / 7000 =
__22 hp/1000 gal__ - Torque = 63,025 x 1.52 / 39.9 =
__2396 in-lb__ - Torque per volume = 2396 x 1000 / 7000 =
__343 in-lb/1000 gal__ - Tip speed = pi x 39.9 x 52.7 / 12 = 550 ft/min
- Comparison torque per volume = 343/720 =
__48%__

Torque per volume for pilot scale is 720 in-lb/gal (see Step 2)

Based on above calculations, some factors unavoidably must change, while the important ones are held constant. In different cases, power per volume might be held constant.

No specific scale-up method can be applied to all problems. However, certain applications, such as solids suspensions, which is geometry dependent, should follow geometric similarity scale-up.

**Tired of reading?**

I already made above calculation in this spreadsheet. I hope you find this post useful.