Grain weights and volumes for hand feeding of sheep
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When feed budgeting, or deciding on supplementary feeding rates, it is often necessary to convert the feeding rate (in kilograms or tonnes) to a volume for the feed-out bin. Therefore, it is important to know the density of the grain and also the volume of your feed bin so that accurate amounts of feed are used. Just as important is the feed out rate if only a proportion of the bin is to be fed to a particular mob.
Grains have variable bulk densities (usually given in kg/hectolitre or tonnes/cubic metre), and can often be lighter than expected. This affects the amounts actually fed out.
Average bulk densities of grains:
(1 cubic metre (1 m3) = 1000 litres or 10 hectolitres)
Actual values depend on moisture content, variety, quality and contamination of the grains.
A quick and easy method of determining the bulk density of a grain is to weigh a 20 l container of the grain. Divide the weight of the grain by the 20 l to give a bulk density in t/m3 (note it is the same as kg/l) or to get the answer in kg/hectolitres multiply the answer by 100. For example, a 20 l container of barley which weighs 10.8 kg has a bulk density of 0.54 t/m3 or 54 kg/hectolitre.
Converting Grain Weight to Volume
If the feed requirement for a mob of sheep is given in kilograms, it is a simple calculation to convert it into a volume for use with a feed bin if you know the grain's bulk density.
A mob of sheep needs to be fed 500 kg of oats (bulk density = 0.50 t/m3 ) and the feed bin holds 3 m3.
Volume needed = 500 kg oats / 0.5 t/m3 = 1 m3
Volume needed = 500 kg /50 kg/hectolitres = 10 hectolitres
This means 1/3 of the bin capacity should be filled.
When feeding out two grains at a time, eg. lupins and oats, both volumes need to be calculated so that accurate amounts are fed.
A mob of sheep needs to be fed 500 kg of a 25/75 lupin: oats mix. Lupins at 0.78 t/m3 and oats at 0.50 t/m3 bulk densities and the feed bin holds 3 m3.
Volume of lupins needed = 25% of 500 kg = 0.125 t (125 kg) / 0.78 t/m3 = 0.16 m3
Volume of oats needed = 75% of 500 kg = 0.375 t (375 kg) / 0.5 t/m3 = 0.75 m3
= 0.16 + 0.75 = 0.91 m3 total
The volume of lupins is about 1/3 less than an equivalent amount, in weight, of oats.
Page amended : 4 May 2007
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