Chapter 12 Problems
1. Mixing of fat in chopped meat
Fat added at beginning = 0.15%
Fractional composition of samples: 0.234, 0.104, 0.164, 0.196, 0.304, 0.076
Samples are 5g
Weight of fat particle 0.01g
Mean composition, ( x )
= 1.078/6
=
0.180
Mean overall = 0.15 (fat added) so it is the mean that has to be used in calculations for
mixing indices and so this means the mathematical relationship using values rather than
differences will not be correct.
So deviations are (0.234 – 0.15), (0.0104 – 0.15), etc. and working out
the sum of the squares = 4.067 x 10-2 = Ns2 = 6s2
thence mean squared deviation ,
s2
=
6.778 x 10-3 ( s = 0.0823)
=
0.68 x 10-2
Overall fat = 15% = 0.15 = p
so2
=
p(1 - p)
=
0.15 x 0.85
=
0.1275
=
12.75 x 10-2
The samples are 5g and the particles 0.1g, so there are 5/0.1 = 50 particles in each sample
so
sr2
=
so2/N.
=
12.8 x 10-2 /50
=
0.26 x 10-2
2. Mixing Index for fat/meat mix
After10mins. mixing
(M)10
=
=
=
=
e-10K
-10K
K
After 15mins mixing
Therefore
-15K
e-5.05
(M)15
Therefore
So
3. Liquid mixer in water
D (propellor)
= 0.3m
D (tank)
= 0.6m
Speed = 300revsmin-1
Water  = 1000kgm-3
Reynolds Number =
=
(so2 - s2)/( so2 - sr2)
(12.75 x 10-2 – 0.68 x 10-2)/(12.75 x 10-2 – 0.26 x 10-2)
(12.07 x 10-2)/ (12.49 x 10-2)
0.966 (or 96.6%) after 10 minutes
= 1-0.966
= -3.37
= 0.337
= 0.0344
= -15 x 0.337 = -5.05
= 0.0641
= 1 – M15
= 0.994 (or 99.4%)
=
5 revs-1
 = 1.16 x 10-3 Nsm-2 (kg m-1 s-1)
(D2 N /)
(0.3)2 x 5 x1000/ 1.16 x 10-3 m2 s-1 kgm-3 /kg m-1s-1
=
=
0.45 x105/1.16 x 10-3
3.88 x 105
(Po)
=
K(Re)n
Assuming K = 41 and n = -1
Dick should this be ??
(Po)
=
41 x (3.88 x 105)-1
=
159.08 x 10-5
(P/D5 N3 )
=
1.59 x 10-3
m5 (rev s-1)3 kgm-3
-3
5
P
=
1.59 x 10 x (0.3) x (5)3 x 1000
=
1.59 x 10-3x 2.16 x 10-4 x 1.25 x102 x 103
=
4.3 x 10-2Js-1
And since 1 horsepower = 746 Js-1
???
Required motor??
4. Stirring of olive oil
D (propellor)
= 0.3m
D (tank)
= 0.6m
Speed = 300revsmin-1
=
5 revs-1
Olive oil  = 910kgm-3
 = 84 x 10-3 Nsm-2 (kg m-1 s-1)
Reynolds Number =
(D2 N /)
=
(0.3)2 x 5 x 910/ 84 x 10-3
m2 s-1 kgm-3 /kg m-1s-1
3
-3
=
0.41 x10 / 84 x 10
=
4.88 x 103
Assuming K = 41 and n = -1
(Po)
= 41 x ( 4.88 x 103)-1 =
8.40 x 10-3
5 3
-3
(P/D N )
=
8.40 x 10
m5 (rev s-1)3 kgm-3
P
=
8.40 x 10-3
x (0.3)5 x (5)3 x 910
-3
=
2322 x 10
=
2.322 Js
And since 1 hp = 746 Js-1
5. Nutrient in baby food
Out put per day is 50 tonnes(40x103 kg) per day, nutrient is 15ppm, total nutrient = 0.75kg
The large mixer can do 3 tonnes at a time.
This is not usefully divisive, giving 17 mixes of 2.94 tonnes
Therefore preferable to use 20 mixes, each of 2.5 tonnes. This decision depends on the
time of mixing and the total time in the day, and also the cost of 3 extra mixes. But
difficult to weigh 2.94 tonnes.
The mix contains 15ppm, therefore each 2.5 tonnes (2,500 kg) must contain 15/106 x 2500
= 0.0375kg nutrient
It would be preferable if the second mixer were also doing 20 mixes per day.
The second mixer can mix 0.5 tonnes, maximum 20 x 0.5 = 10 tonnes
Each mix would need to contain 0.0375kg nutrient i.e. 0.0375kg in 500kg
First mixer, 0.05 tonnes = 50kg , minimum 8.3kg.
0.0375kg in 50kg i.e. 37.5 g per mix which is easy to weigh.
Start with 37.5 g in 50kg, with 20 batches, total nutrient is 0.75kg
In second mixer, 50kg mixed with 450kg, i.e. 500kg per mix
In third mixer 0.5 tonnes (500kg) mixed with 2 tonnes (2,000kg) of other material, giving
2.5 tonnes.