Suggested Response - Q5

**/*** A bright beer tank has a beer depth of 2.5 m and a headspace pressure of 1 bar g. A 15 mm diameter sample valve is located 0.5 m above the base. It is left partially open (with a loss coefficient of 200). By considering the valve loss coefficient only, what would be the initial flowrate of beer through the valve? (Beer ρ = 1,005 kg/m3).

Response:

This is a slightly unusual problem in that we are only given the loss coefficient of the partially closed valve. It also does not appear to be like previous pressure drop complications because it is a combination of both a static and a dynamic problem. Let us consider this in a little more detail.

The important things to recognise include:

  • The pressure on the upstream side of the valve consists of a hydrostatic head of (2.5 – 0.5) = 2 m of beer and a 1 bar g headspace pressure.

  • The beer discharges from the sample valve into atmospheric pressure, i.e. 0 bar g. so all the supply pressure will be dropped across the valve.

  • As beer passes through the valve, the level of beer in the bright beer tank will drop, and the supply pressure will also drop. However, we are only concerned with the initial flowrate through the valve.

Hopefully, you will see that the problem has become one of equating velocity heads lost across the valve with the supply pressure. This being the sum of headspace (H/S) pressure and hydrostatic pressure which in the case of this example is 2 m.

P subscript H divided by S end subscript plus P subscript 2 m end subscript equals left parenthesis k u squared over 2 right parenthesis rho

Rearranging for the flow velocity u:

u equals square root of fraction numerator 2 over denominator rho k end fraction left parenthesis P subscript H divided by S end subscript plus P subscript 2 m end subscript right parenthesis end root

The hydrostatic pressure due to the 2 m head of beer is given by:

P subscript 2 m end subscript equals rho g h equals 1 comma 005 cross times 9.81 cross times 2 equals 19.72 k P a

Substituting this into the equation for u:

u equals square root of fraction numerator 2 over denominator 1 comma 005 cross times 200 end fraction space left parenthesis 100 comma 000 plus 19 comma 720 right parenthesis space end root

u equals space square root of 1.19 end root equals 1.1 space m divided by s

The diameter of the valve is 15 mm, so the flowrate would be:

Q equals A u equals pi over 4 cross times 0.015 squared cross times 1.1 equals 0.00019 m cubed divided by s

Q equals 0.19 space l divided by s

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