SAQ Suggested Response - Q4

 ** You are brewing a beer using isothermal mashing, and you need the mash to be 66°C. From the data in the table below, what should the strike temperature be, and what is the percentage fermentability of the wort produced?

Grist mass

5,500 kg

Water volume

137.50 hL

Slaking heat

2.8°C

OG

11.2°P

PG

0.8°P

Grist temperature

19.5°C

Response:

Our equation for calculating mash temperature from the temperatures and amounts of mashing water and grist is as follows:

begin mathsize 14px style text Mash temperature=  end text open square brackets fraction numerator left parenthesis straight T subscript straight w space cross times space straight M subscript straight w right parenthesis space plus space left parenthesis straight T subscript straight G space cross times space straight M subscript straight G right parenthesis over denominator straight M subscript straight m end fraction close square brackets space plus space straight H subscript straight s end style

Where:

begin mathsize 14px style text T end text subscript straight w space equals space Water space temperature space space space
text M end text subscript straight w space end subscript equals space Water space mass space space
straight T subscript straight G space equals space Grist space temperature space
straight M subscript straight G space end subscript equals space Grist space mass space space
straight M subscript straight m space end subscript equals space Mash space mass space space
straight H subscript straight S equals space Slaking space heat end style

We can rearrange this to make the mashing water the subject:

begin mathsize 14px style text T end text subscript straight w space equals space fraction numerator left square bracket left parenthesis Mash space temperature space minus space straight H subscript straight s right parenthesis space cross times space straight M subscript straight m right square bracket space minus space left parenthesis straight T subscript straight G space cross times space straight M subscript straight G right parenthesis over denominator straight M subscript straight w end fraction end style

Inputting the figures from the table gives us:

begin mathsize 14px style text T end text subscript straight w space equals space fraction numerator left square bracket left parenthesis 66 space minus space 2.8 right parenthesis space cross times space 19 comma 250 right square bracket space minus space left parenthesis 19.5 space cross times space 5 comma 500 right parenthesis over denominator 13 comma 750 end fraction end style

And calculating out the brackets gives us:

begin mathsize 14px style text T end text subscript straight w space equals space fraction numerator 1 comma 216 comma 600 space minus space 107 comma 250 over denominator 13 comma 750 end fraction equals space 80.68 degree straight C end style

We can check we are correct by entering our water temperature into the original equation:

begin mathsize 14px style text Mash temperature=  end text open square brackets fraction numerator left parenthesis straight T subscript straight w space cross times space straight M subscript straight w right parenthesis space plus space left parenthesis straight T subscript straight G space cross times space straight M subscript straight G right parenthesis over denominator straight M subscript straight m end fraction close square brackets space plus space straight H subscript straight s end style

 begin mathsize 14px style text end text open square brackets fraction numerator left parenthesis 80.68 space cross times space 13 comma 750 right parenthesis space plus space left parenthesis 19.5 space cross times space 5 comma 500 right parenthesis over denominator 19 comma 250 end fraction close square brackets space plus space 2.8 space equals space 66 end style

To work out the fermentability, we need to calculate the apparent attenuation using this equation:

begin mathsize 14px style text % Apparent Attenuation = end text fraction numerator left parenthesis text O end text straight G space minus PG right parenthesis over denominator 0 straight G end fraction space cross times space 100 end style

So if we enter our numbers from the table, we get:

begin mathsize 14px style text % Apparent Attenuation = end text fraction numerator left parenthesis 11.2 space degree straight P minus space 0.8 space degree straight P right parenthesis over denominator 11.2 space degree straight P end fraction space cross times space 100 space equals space 92.9 space percent sign end style

And to convert the apparent attenuation to the real attenuation, we multiply by a factor that accounts for the effect of ethanol on the gravity:

begin mathsize 14px style text % Real Attenuation (fermentability) = 92.9 × 0.819 = 76.05% end text end style

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