Wort Seperation

Question 1

Wort seperation

Answer 1

Wort is filtered using the husk as the filter bed.

The flow of wort is optimised to give us the right wort clarity (minimising the amount of solids) and to maximise filtration efficiency.

The strong (first) worts are run off into the kettle first.

The grain bed is then sparged (washed) with hot water to remove the maximum amount of soluble extract as more dilute worts.

The process continues until we have the right amount of wort, at the right original gravity, in the kettle.

On completion of filtration, the spent grains (waste husk and embryo) are removed and can be used as co-products (e.g. for animal feed or energy generation).

Question 2

Key process parameters

Answer 2

Work gravity, wort collection flow rate, turbidity/haze, differential pressure, sparge volume, pH

Question 3

Trubidity / EBC

Answer 3

Highly effective wort quality
High hazes in wort are bad news. They can lead to the following problems.
- excessive volumes of trub
- flavour problems
-fermentation problems
-problems with yeast
- poor filterability
-potential shortening of shelf life due to haze formation.

Question 4

Process: mash tun

Answer 4

Both mashing and seperation
-run off
-sparge
-rate to match the flow of discharge

Question 5

Process: lauter tun

Answer 5

Wider than a mash tun, receives mash from MCV. Insulated vessel with false floor.
-filling, recirc, run off, raking, end run off, grain out.

Question 6

Process: mash filter

Answer 6

Filter cloth and membrane. Filter bed is distributed across filter plates; high flow rate.
-filling, filtering,pre compression, sparking, final compression.
- final compression: increase etraxt yield, drier spent grains, reduced cost of effluent disposal.

Question 7

Grist: Mash tun

Answer 7

35% coarse grist
35% fine grist’s
20% husk
10% flour

Question 8

Grist: lauter tun

Answer 8

25% coarse grits
30% fine grits
15% hurl
30% flour

Question 9

Grist: mash filter

Answer 9

15% coarse grits
40% fine grits
5% husk
40% flour

Question 10

Extract calculation

Answer 10

We express the amount of extract in the malt using % weight. Wort extract is expressed as °Plato (°P).

100g of wort at 9°P contains 9 g of extract.

Let us look at an example calculation:

We are making a wort by using 1,000 kg of malt with a lab extract of 80.25%. We have collected 5,500 litres of wort at 12.4°P, which has a density of 1.05 kg/L. What is the extract efficiency?

First, we find the amount of extract in our kettle. We find the weight of our wort by multiplying its volume by its density:

5,500 litres x 1.05 kg/L = 5,775 kg

As °P is the percentage weight of extract, 12.4% of our wort’s weight is extract. So, to find the extract, we multiply the weight of the wort by this percentage:

5,775 x 12.4 / 100 = 716.1 kg of extract

We can now calculate how many kg of extract are in the grist according to the lab. We do this by multiplying the weight by the percentage weight of extract.

1,000 kg x 80.25 / 100 = 802.5 kg

To calculate our efficiency, divide the amount of extract by the amount we would have if we achieved the same extraction as the lab method.

Then multiply this by 100 to give the figure as a percentage:

716.1 / 802.5 x 100 = 89.23%

Question 11

Brewhouse cycle time

Answer 11

The brewhouse cycle time refers to the time taken in one brew from the start of mashing, to the time the mashing vessel is clean and ready for the next.