Lesson 12 - Heat Transfer End of Unit SAQ

*** You have been asked to design a yeast collection vessel for a small scale brewery. It is required to hold 10 hL (1,110 kg) of yeast collected from a fermentation vessel. The specific heat capacity of yeast is 3,900 J kg^-1 K^-1.

a)  The yeast is cropped at 7°C but must be stored at 3°C. What is the heat transfer surface area required to achieve this cooldown in 30 minutes?

• Cooling is done using glycol supplied at 0°C and returned at 2°C.

• The overall heat transfer coefficient is 350 Wm^-2 K^-1.

• Assume negligible heat gain from the environment.

b)  Comment on the area calculated in a). Discuss what changes you would make to the vessel design to address this. The vessel was originally designed with the following characteristics:

• Straight conventional jackets  

• Diameter = 0.9 m

• Height = 2.5 m

• Surface area = 6.9 m²

• Total volume = 1.3 m³ (30% headspace)

c)  What is the actual heat gain during the 30 min cooldown?

• Vessel surface temperature = 5°C

• Ambient temperature = 20°C

• Convective heat transfer coefficient = 5 Wm^-2 K^-1

• Absorptivity of stainless steel = 0.4

• The Stefan-Boltzmann constant = 5.672 × 10^-8 Wm^-2 K^-4

d)  In question (a), we assumed negligible heat gain from the environment. Is this a valid assumption?

e)  Describe how insulation would assist in reducing the heat gain from the environment. How much is the heat gain reduced by when the vessel is insulated?

• The surface temperature of the insulated vessel is 18°C

Response:

a)  Because the temperature difference between the product and utility fluid is changing over time, we can use the equation for unsteady state heat transfer in order to determine the required heat transfer surface area.

This equation assumes that the temperature of the utility fluid (glycol) is not changing. In this example, the glycol is changing temperature. The temperature change is relatively small, so we can determine the average glycol temperature as follows:

And so,

 

b)  To be realistic, the required heat transfer area has to be less than the available surface area of the vessel itself. In this example, the available surface area is only 6.9 m², and so there will not be sufficient space on the vessel walls for the required jackets.

The heat transfer area is determined using the two equations below. Let’s review this equation and see what changes can be made to reduce  to a more reasonable number that will fit the vessel.

M, Cp, t, T₁ and T₂  are all fixed parameters, as defined by the process requirements of the vessel.

The glycol supply temperature could be decreased, but this will have only a very small impact on the heat transfer area and could cause other issues such as freezing.

So, there are three variables that can be manipulated: U, d and h.

Let’s look first at d and h. The original vessel design had a height:diameter ratio of 2.8:1. Brewing vessels have a typical height:diameter ratio of between 2:1 and 4:1.

Increasing the height:diameter ratio of the vessel is a viable option as this will increase the available surface area, while still maintaining the same volume.

U is the other variable that can be manipulated in the design of this jacketed vessel. The following design changes could be considered in order to increase the overall heat transfer coefficient.

• Using a dimple jacket will improve the overall heat transfer coefficient (between 550-700 Wm^-2 K^-1). Repeating the above calculation with a heat transfer coefficient of 650 Wm^-2 K^-1 gives a required heat transfer area of 4.1 m². This is a far more realistic transfer area and will constitute 54% of the available surface area.

Dimple jackets are a popular choice for brewing vessels. This is because they are flexible and relatively low in cost. With the increased overall heat transfer coefficient offered by a dimple jacket, there would be sufficient vessel surface area for a dimple jacket.

• Placing baffles inside the conventional jacket will increase the heat transfer coefficient to between 650-850 . Repeating the above calculation with a heat transfer coefficient of 750  gives a required heat transfer area of 3.5 m², which is 47% of the available surface area. This is a simple and affordable modification that can be made to the vessel design. It also, will have no impact on the internals of the vessel. With the increased overall heat transfer coefficient offered by the baffles, there would be sufficient vessel surface area for the jacket.
• Agitating the vessel will deliver a similar improvement in overall heat transfer coefficient as baffles. Agitation has the added benefit of keeping the yeast in suspension. However, it will also increase the operating costs, and incorrect or excessive agitation could damage the yeast.
• Changing the jacket type to a dimple jacket will also improve the overall heat transfer coefficient (between 550-700 Repeating the above calculation with a heat transfer coefficient of 650  gives a required heat transfer area of 4.1 m². This is a far more realistic transfer area and will constitute 54% of the available surface area. Dimple jackets are a popular choice for larger vessels because they are flexible and relatively low in cost. However, for small vessels like in this example they may be too expensive.
• Using a half-pipe jacket will give an even greater heat transfer coefficient (between 750-950 Wm^-2 K^-1). Repeating the above calculation with a heat transfer coefficient of 850 Wm^-2 K^-1 gives a required heat transfer area of 3.1 m², which is 41% of the available surface area. This is not recommended for this example, as the vessel will be significantly more expensive, and the smaller heat transfer area is not required.
• Another possible solution is to change the design so that cooling is achieved using an internal coil rather than a jacket. Internal coils have an even higher heat transfer coefficient (between 800-1,200 Wm^-2 K^-1) and the heat transfer area is not limited by the available surface area of the vessel.

Vessels containing internal coils must be cleaned effectively. Typically, this would be done by removing the coil and cleaning manually. The vessel in this example, vessel is likely to be too large for this practice so a cooling coil is not a viable option. Using an external heat exchanger is also not a viable option because it is difficult and very expensive to pump yeast slurry hygienically and without damaging the yeast. This option would not be considered.

 

c)  We now wish to calculate the total heat loss rate from the vessel using the equation:

Note that in this equation, because temperature is absolute (and not a temperature difference), the temperature used must be in K, so:

The overall heat loss from the vessel during the 30 min of cooldown can be calculated as follows:

 

d)  To determine if the heat gain can be considered negligible or not, we can compare it to the heat transfer required to cool down the vessel content.

First, let’s calculate the total heat that we must remove in order to decrease the temperature of the yeast. The sensible heat is given using:

It makes sense that heat transfer is negative as heat is being lost by the product (the temperature is decreasing)

The heat loss as a percentage of total heat transfer is shown as follows:

This means that the assumption of negligible heat gain from the environment is not valid. 

e)  To reduce heat gain from the environment the tank can be insulated. Insulation material is made of a low thermal conductivity material. Therefore, a layer of insulation adds a layer of resistance (r) to multi-layer heat transfer system. The insulation material should also have a low water vapour permeability to prevent the movement of water vapour from the warm ambient air onto the cool vessel surface, carrying heat with it. Surrounding the insulation with a layer of cladding and vapour barrier would also prevent the ingress of water vapour. 

Let’s now calculate the heat loss for an insulated tank. We can use the same equations from part c), with a surface temperature of 18°C. This increases the temperature difference but it will also reduce the radiative heat loss by the 4^th power.

Remember that

Again, be sure to use temperature in K, so:

 

So, the heat loss with insulation is:

Let’s recalculate the % loss with insulation:

To compensate for heat gain from the environment, the overall heat that the utilities stream (glycol) must remove has to increase. By insulating the vessel, the heat loss is reduced significantly from 7.7% to 1.0%. This means that rather than having to remove an additional 7.7% heat, the utility fluid only needs to remove an additional 1.0%. This will reduce the demand loading on the refrigeration plant.