Calculation of pressure drop in high-pressure argon capillary tubes?
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This post was last edited by A Chao syc62100 on 2017-2-27 20:56. Given conditions: high-pressure argon at 15 MPa, inner diameter of the capillary tube is 0.6 mm, length of the tube is 0.5 m, flow rate is 35 L/min. Question: What is the pressure drop in this tube? Using the formula from the book: q=πd^4△P/128μL, which means △P=q*128μL/πd^4. When these values are substituted, the pressure drop turns out to be 2.5 MPa – that’s an astonishing result. Something must be wrong here Dear sea friends, is it possible that this formula is not suitable for use in high-pressure thin pipelines? If it doesn’t apply, which formula should be used? Thank you. Notes: 1. The dynamic viscosity μ here is taken from Table, at 15 MPa and 20°C; μ = 27.675*10^-6 Pa·s, which is correct. 2. During the calculation, it’s mainly d^4 = (0.6*10^-3)^4 = 0.6^4*10^-12; as a result of this calculation, the value increases significantly right away:):lol
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