Suggested Response - Q6

** Using this example, answer the questions below.

The mashing process has three temperature stands, namely 62°C, 65°C, and 72°C. Furthermore, at the end of the mash profile, the mash is heated up to 76°C before transfer to the lauter tun. A minimum heat up rate of 1 °C min-1 is required.

  • Maximum mash vessel volume is 500 hL. Mash vessel diameter is 4 m.

  • Heat is provided by steam at 130°C. The heat of evaporation of steam at 130°C is 2,200 kJ kg-1.

  • Typical overall heat transfer coefficient for an agitated jacketed vessel using steam the way this one does is 1,100 W m-2 K-1.

  • Specific heat capacity and density of mash are 3,600 J kg-1 °C-1 and 1,060 kg m-3 respectively.

(a)  What is the required heat transfer area of the mash vessel?

(b)  What would be the approximate height of the jackets, assuming they were positioned only on the side walls?

(c)  What is the maximum required mass steam flow rate?

 

Response

Let’s start with part (a) and work through the question step by step.

Firstly, we need to understand that we are dealing with unsteady state heat transfer and we need to calculate the solution for the worst-case scenario (highest rate of heat transfer scenario) which is where the mash temperature is 76°C which has the smallest ΔT.

First, we calculate the mass of mash (Mmash):

begin mathsize 14px style text M end text subscript mash equals straight rho straight V equals 1 comma 060 cross times 500 cross times 0.1 equals 53 comma 000 kg end style

Next, we need to calculate the heat transfer rate:

begin mathsize 14px style text q= end text straight Q over straight t end style

To find out the amount of specific heat required to increase the mash temperature we can use the formula:

begin mathsize 14px style text Q=M end text subscript mash Cp increment straight T end style

Therefore:

begin mathsize 14px style text q=M end text subscript mash fraction numerator Cp increment straight T over denominator straight t end fraction end style

text q=53,000 × 3.6 ×  end text 1 over 60 equals 3 comma 180 space kW

The heat transfer from the surface to the mash can be calculated using:

begin mathsize 14px style text q=UA∆T end text end style

From this we can derive the amount of surface area required to achieve this heat transfer rate as follows:

begin mathsize 14px style text A= end text fraction numerator straight q over denominator straight U increment straight T end fraction equals fraction numerator 3 comma 180 over denominator 1.1 cross times left parenthesis 130 minus 76 right parenthesis end fraction equals 53.54 straight m squared end style

Next let’s tackle part (b) and work through it step by step.

Using the formula for calculating the surface area of a cylinder, we can determine the approximate height of the jackets.

begin mathsize 14px style text A= πdL end text end style

begin mathsize 14px style text L= end text fraction numerator straight A over denominator straight pi straight d end fraction equals fraction numerator 53.54 over denominator straight pi cross times 4 end fraction equals 4.26 space straight m end style

To achieve this heat transfer, the required height of the jackets for this vessel is 4.26 m.

It may be the case that the liquid level of the mash vessel is below 4.26 m. This is not ideal, as part of the heating surface may not be covered and this can cause vessel fouling. Rather than using a single jacket on the shell, we could use two jackets: one covering the bottom cone and one on the sides. Adding a cone jacket means the shell jacket can be smaller, and thus not exposed above the mash.

Finally, let’s cover part (c) and work through it step by step.

As the heating medium is changing state, we will use the latent heat formula for the heat transfer:

begin mathsize 14px style text Q=M end text subscript steam cross times straight H subscript vap end style

We need to work out the mass flowrate of steam and therefore can divide both sides by time:

begin mathsize 14px style text q= end text straight Q over straight t equals fraction numerator straight M subscript steam cross times straight H subscript vap over denominator straight t end fraction end style

And because begin mathsize 14px style text m= end text straight M over straight t end style:

begin mathsize 14px style text q=m end text subscript steam cross times straight H subscript vap end style

We can now rearrange to isolate msteam:

begin mathsize 14px style text m end text subscript steam equals straight q over straight H subscript vap equals fraction numerator 3 comma 180 over denominator 2 comma 200 end fraction 1.445 space kg space straight s to the power of negative 1 end exponent equals 5 comma 203 space kg space straight h to the power of negative 1 end exponent end style

 

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