Suggested Response - Q4

** Bright beer is flowing down a pipe with an internal diameter of 125 mm. If the density and viscosity are 1,005 kg/m3 and 2 cP respectively, and the flow velocity is 2 m/s, confirm that the flow is turbulent. What internal pipe diameter would be required to ensure the flow was laminar, whilst achieving the same volumetric flowrate?

Response:

We calculate the Reynolds number to confirm if the flow is turbulent:

R e space equals space fraction numerator rho u d over denominator mu end fraction space equals space fraction numerator 1 comma 005 space cross times space 2 space cross times space 0.125 over denominator 2 space cross times space 10 to the power of negative 3 end exponent end fraction space equals space 1.26 space cross times space 10 to the power of 5

Hence, Re > 4,000, which confirms the flow is turbulent.

Laminar flow is characterised by a Reynolds number below 2100. To ensure our flow is laminar, we would require Re < 2,100, however, we cannot select a lower pipe diameter to reduce the Re as doing so would result in a lower volumetric flowrate (for the same velocity), and we are required to achieve the same flowrate. We therefore need to use a higher diameter pipe, with lower velocity, to achieve Re < 2,000 for the same flowrate. So first, let us work out what the flowrate was in the original 125 mm pipe:

Q space equals space A u space equals space fraction numerator pi d squared over denominator 4 end fraction space cross times space u space equals space fraction numerator pi space cross times space 0.125 squared over denominator 4 end fraction space cross times space 2 space equals space 0.024 space m cubed divided by s

To set a Re = 2,100 requires the following:

R e space equals space fraction numerator rho u d over denominator mu end fraction space equals space 2 comma 100

Rearranging for u:

u space equals space 2 comma 100 space cross times space fraction numerator mu over denominator rho d end fraction space equals space 2 comma 100 space cross times space fraction numerator 2 space cross times space 10 to the power of negative 3 end exponent over denominator 1 comma 005 space cross times space d end fraction space equals space fraction numerator 0.004 over denominator d end fraction

u space equals space fraction numerator 0.004 over denominator d end fraction

Consequently, we cannot solve directly for u or d. To solve this for these two parameters, we need another equation involving u and d.  For this, we can turn to the requirement for the new pipe diameter to not only have laminar flow but achieve the same volumetric flowrate as previously. So expressing the flowrate as:

Q space equals space 0.024 space equals space A u space equals space fraction numerator pi d squared space cross times space u over denominator 4 end fraction

d squared u space equals space 4 over pi space cross times space 0.024 space equals space 0.031

d squared u space equals space 0.031

u space equals space fraction numerator 0.031 over denominator d squared end fraction

Equating this expression with the previous one for u gives:

u space equals space fraction numerator 0.031 over denominator d squared end fraction space equals space fraction numerator 0.004 over denominator d end fraction

fraction numerator 0.031 over denominator d squared end fraction space equals space fraction numerator 0.004 over denominator d end fraction

fraction numerator 0.031 over denominator 0.004 end fraction space equals space d squared over d

d space equals space 7.75 space m

And hence u:

u space equals space fraction numerator 0.004 over denominator d end fraction space equals space fraction numerator 0.004 over denominator 7.75 end fraction space equals space 0.0005 space m divided by s

Let us just double-check that this combination of d and u gives Re = 2,000 and Q = 0.024 m3/s:

R e space equals space fraction numerator rho u d over denominator mu end fraction space equals space fraction numerator 1 comma 005 space cross times space 0.0005 space cross times space 7.75 over denominator 2 space cross times space 10 to the power of negative 3 end exponent end fraction space equals space 1947

(There is some rounding error in the calculations.)

And Q:

Q space equals space A u space equals space fraction numerator pi d squared over denominator 4 end fraction space cross times space u

Q space equals space fraction numerator pi space cross times space 7.75 squared over denominator 4 end fraction space cross times space 0.0005 space equals space 0.024 m cubed divided by s

What should be obvious in this problem, in this case if we wish to transport bright beer, is that under laminar flow conditions, at matched flowrates, very large diameter pipes are required.

You have completed 0% of the lesson
0%