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Old Apr 26th 2019, 06:52 AM   #1
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Finding new Reynold's Number (Re) with decrease and increases in percentages

This is the OG question:

"A small hose in a machine carries a viscous fluid.

When the machine is first turned on the fluid flows through the hose and the Reynold's number is 1800.

As it warms up, there is a reduction of the density of the fluid of 1.61%, a reduction in the viscosity of the fluid of 6.74%, and an increase in the diameter of the hose by 1.50%.

Assuming the speed of the fluid is kept constant, what is the Reynold's number now?"

I know what I did probably looks like the dumbest thing ever, but how would I go about figuring out the new Reynold's number?

My working for this question is the section at the VERY bottom, with the Reynold's equation, ofc (ignore the work in progress above lolol)

https://imgur.com/a/ecjv0eM

Thanks for any and all help

-I
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Old Apr 29th 2019, 06:06 AM   #2
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Reynolds Number is the ratio between
the inertial forces (which act to keep the fluid moving exactly as it is now) and
the viscous forces (which act to resist the fluids movement and slow it down).

Re=Vflow.Rho.Lref/Visc.

Vflow is the speed of the fluid
Rho is the density
Visc is the viscosity
Lref is "a representative length"
Lref is subject to interpretation depending on the practical application.
(for a wing it is usually the distance between the leading a trailing edge of the wing)
For a pipe, I am guessing that Lref might be the pipe diameter.

If so, your problem becomes:

Re'=Vflow.Rho*{(100-1.61)/100}.Lref*{(100+1.5)/100}/(Visc*{(100-6.74)/100})
Re'=Vflow.Rho*{0.9839}.Lref*{1.015}/(Visc*{0.9326})
Re'=(Vflow.Rho.LRef/Visc) * {0.9839*1.015/0.9326}

but (Vflow.Rho.LRef/Visc) is what we originally started with
which the question tells us is 1800.
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Last edited by Woody; Apr 29th 2019 at 06:11 AM.
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