Suggested Response - Q1

*/**High gravity (HG) beer is diluted inline, within a 100 mm internal diameter pipe, to sales gravity (SG) through the addition of dilution water. If the HG beer and dilution water volumetric flowrates are the same at 500 hL/h each, what would be the volumetric flowrate of the SG beer? The densities of the HG, SG beers and dilution water are respectively 1,007, 1,004 and 1,000 kg/m3.

Response:

A simple sketch will help us visualise the situation.

Q1

Before we solve this problem, which is a ‘continuity of mass’ problem, let us state what the answer is not.

We cannot simply add the volumetric flowrates of HG beer and dilution water (500 + 500 = 1,000 hL/h) to get the flowrate of SG beer. Why? Well, we know that mass is conserved, but volume is not strictly conserved. As the densities of the three streams are different, we need to account for their impact on the mass and volume flowrates of each stream, however small this effect might be.

 

Conservation of mass at the dilution point can be expressed as:

G subscript H G end subscript space plus space G subscript D W end subscript space equals space G subscript S G end subscript

G is the mass flowrate of the three streams respectively: HG (high gravity), DW (dilution water) and SG (sales gravity).

We can determine the mass flowrates of the HG and DW streams using the respective values of density:

G subscript H G end subscript space equals space rho subscript H G end subscript space Q subscript H G end subscript space equals space 1 comma 007 space cross times space open parentheses fraction numerator 500 space cross times space 100 over denominator 1 comma 000 end fraction close parentheses space equals space 50 comma 350 space k g divided by h

G subscript D W end subscript space equals space rho subscript D w end subscript space Q subscript D W end subscript space equals space 1 comma 000 space cross times space open parentheses fraction numerator 500 space cross times space 100 over denominator 1 comma 000 end fraction close parentheses space equals space 50 comma 000 space k g divided by h

Hence, the mass flowrate of SG beer is:

G subscript S G end subscript space equals space G subscript H G end subscript space plus space G subscript D W end subscript space equals space 50 comma 350 space plus space 50 comma 000 space equals space 100 comma 350 space k g divided by h

The volumetric flowrate of SG beer can be determined from:

Q subscript S G end subscript space equals space G subscript S G end subscript over rho subscript S G end subscript space equals space fraction numerator 100 comma 350 over denominator 1 comma 004 end fraction space equals space 99.95 space m cubed divided by h

Q subscript S G end subscript space equals space 999.5 space h L divided by h

So, the volumetric flowrate of the SG beer stream is 999.5 hL/h, which is slightly less than the speculated value of 1,000 hL/h. One might argue that a 0.05% difference would be acceptable in such a practical situation. What do you think?

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