Mixture calculator

Overview
1. Calculation of the ingredients of a mixture
2. Calculation of the mixture temperature of two liquids
1. Calculation of the ingredients of a mixture
=> Concentration of the mixture cT (total)

Mass m1:
kg


Mass m2:
kg


Mass m3:
kg


Concentration c1:
%, Brix, ...


Concentration c2:
%, Brix, ...


Concentration c3:
%, Brix, ...


Result [ %, Brix, ... ]


Mixture formula:

m(total) * c(total) = (m1*c1)+(m2*c2)+(m3*c3)+ ...

where m = mass, c = concentration


The calculation programme on the left only calculates the total concentration for three ingredients of the mixture. If another quantity is sought or the mixture consists of more than three components, the mixture formula must be adjusted accordingly. As an alternative it is also possible to use the accompanying calculation programme iteratively in order to find out a different quantity. If the component sought is only made up of two ingredients, the programme can be used by entering "0" for the mass and concentration of the third component. The mixture formula can be used with both units of mass and units of volume. Of course this cannot occur within one calculation.

Example 1: -Mass information

The total concentration of 5000kg alkaline solution is to be calculated. The alkaline solution is mixed together in a large tank with the following ingredients:

- Water 4500kg, concentration 0% (does not yet contain any alkaline solution)

- Alkaline solution 50kg, concentration 80% (NaOH)

- Alkaline solution 450kg, concentration 2% (NaOH)

Result: The finished solution contains 0.98% sodium hydroxide solution

 

Example 2: -Volume information-

How high is the Brix value (concentration) of a drink which is made up of the following ingredients:

- Orange juice 10,000 litres, 13 Brix

- Apple juice 7,000 litres, 10 Brix

- Sucrose solution (liquid sugar) 500 litres, 65 Brix

Result: The finished drink has a value of 13.29 Brix

 

Example 3: -Iteration-
A 2% acid solution is to be produced from 400 kg of a 1.5% and a 3% acid solution. How many kg of the 3% solution are required?

Result: 200kg of the 3% solution are required


Solution process: m1=400kg, c1=1.5%, c2=3% are entered in the accompanying fields and "0" is entered for m3 and c3. Subsequently a new value is entered for m2 until cT(total)=2% emerges (iteration, multiple repetition to approach a solution)

Commentary: This procedure avoids the necessity of performing a calculation with a Pearson Square.

2. Calculation of the mixture temperature of two liquids
=> Mixing temperature tm

Mass m1:
kg


Mass m2:
kg


Thermal capacity c1:
kj/(kg/°K)


Thermal capacity c2:
kj/(kg/°K)


Temperature t1:
°C/°K


Temperature t2:
°C/°K


Result [ °C ]


 

The accompanying calculation programme is based on Richmann's law of mixtures. The mixture temperature can also be determined with the mixture formula given in 1 (see above field)  if the thermal capacities of the components to be mixed are identical. Both the mixing formula in 1 and Richmann's law of mixtures can be extended with more components of a mixture than shown here.

m1*c1*(t1-tm) = m2*c2*(tm-t2)

tm = ((m1*c1*t1)+(m2*c2*t2)) / ((m1*c1)+(m2*c2))



 

Substance data: Thermal capacity [kJ/(kgK)]
Water, 20°C 4,1819
Water, 100°C 4,216
Ethylene glycol (45%) water 3,33
Beer wort, 12°P (original wort), 10°C 3,875
Drinking milk, 2.5% fat, 0°C 3,83
Raw milk, 4.2% fat, 20°C

3,9



  Warmer liquid Colder liquid
Mass m1 m2
Thermal capacity c1 c2
Temperature T1

T2

 

 



 

 

 

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