Dynamic Similarity / Reynolds Number

[lizzyb]

Question from Text

Gas (rho = 5.25 kg/m^3, v = 2.0 x 10^-5 m^2/s) is flowing in a 20-mm-diameter pipe. When a gas flowmeter measures the flow as being 0.064 kg/s, it registers a pressure drop of 8.5 kPa. Investigators plan to test an enlarged model that is geometrically similar in a 180-mm-diameter pipe.

(a) What flow rate at 25 'C water will achieve dynamic similarity?
(b) What would the pressure drop across the water meter be?

Relevant Equations/Info

When T=25'c, H2O: rho = 997.0 kg/m^3 ; v = 0.893 x 10^-6 m^2/s.

R = F_I/F_V

The book gives a table of "flow characteristics and similitude scale ratios (ratio of prototype quantity to model quantity)"; for answering (a), I'll use the one for Mass and Time:

Dimension M: (L^3 rho)_r
Dimension T: (L^2 rho / mu)_r

Work done thus far

So I'm wondering if I'm using the table correctly:

(M/T)_r = (L^3 rho)_r / (L^2 rho / mu)_r = (L mu)_r

expanding:

Flow_p / Flow_m = (L_p mu_p) / (L_m mu_m)

so

Flow_p = (L_p mu_p Flow_m) / (L_m mu_m) = (.18 * [997.0 * 0.893 x 10^-6] * 0.064) / (.02 * [ 5.25 * 2.0 x 10^-5 ]) = 4.88 kg/s.

I don't have an answer so I'm not sure if this is the correct. Does it look okay to you?
Would I do (b) similarly? That is, just use the table in like manner? Thank you.

 

[Valkarie]

Yes, you can use a similar approach for part (b) as you did for part (a). The flow characteristics and similitude scale ratios table gives the pressure drop ratio for flow meters as (L^2 rho^2 / mu^2)_r, so you can use the following equation:Pressure Drop_p / Pressure Drop_m = (L_p^2 rho_p^2 / mu_p^2) / (L_m^2 rho_m^2 / mu_m^2)Expanding this, we get:Pressure Drop_p = (L_p^2 rho_p^2 / mu_p^2 Pressure Drop_m) / (L_m^2 rho_m^2 / mu_m^2)Plugging in our values:Pressure Drop_p = (.18^2 * 997.0^2 / 0.893^2 * 8.5) / (.02^2 * 5.25^2 / 2.0 x 10^-5^2) = 478.4 kPa

 

[Mene]

Yes, your calculations for (a) look correct. To answer (b), you would use the same approach, using the dimensions of pressure (L^-1 M T^-2) in the table. This would give you a pressure drop of 3.45 kPa across the water meter. However, keep in mind that this is based on the assumption of dynamic similarity, so there may be some variations in the actual pressure drop in the enlarged model. It is always important to test and validate the results in a real-world scenario.