# Is it ok to say this?

#### IntelInside

10+ Year Member
7+ Year Member
Is it safe to say:

"A non-ideal fluid does not behave in an opposite manner to an ideal fluid. Narrowing a pipe increases the velocity of an ideal fluid. It will probably increase the velocity of a non-ideal fluid as well. However, with a non-ideal fluid you must also consider drag, which impedes flow. Thus, if you narrow the pipe in a non-ideal fluid, velocity will probably increase, but not as much as if there were no drag."

#### RogueUnicorn

##### rawr.
7+ Year Member
erm this doesn't seem very mcat?

#### PhilIvey

10+ Year Member
5+ Year Member
Is it safe to say:

"A non-ideal fluid does not behave in an opposite manner to an ideal fluid. Narrowing a pipe increases the velocity of an ideal fluid. It will probably increase the velocity of a non-ideal fluid as well. However, with a non-ideal fluid you must also consider drag, which impedes flow. Thus, if you narrow the pipe in a non-ideal fluid, velocity will probably increase, but not as much as if there were no drag."
This is likely EK. They stressed this in their Fluids section. Yes, a non-ideal fluid would increase its velocity but not as much as an ideal fluid. Also, realize that drag is greatest at the edge and least in the middle. So, in a non-ideal fluid, the velocity is not equal throughout a specific section of the pipe. In ideal you assume that it flows equally at the edge as the middle which doesn't apply in a real fluid. What you said is correct.

#### kentavr

10+ Year Member
It depend what other parameters are constant. If delta P is const, the of course, the energy will be spent on drug. However, if Throughput(Q) is constant, then EK may be wrong. It is not obvious and need some math. It will be great if somebody can check the calcs below:

1. Let's assume that the speed at the wall is zero. (Standard assumption for non ideal liquids)
2. At radius Ro at center of tube speed(Vo) is maximum.
3. The speed profile is linear or, v(r) = Vo (1-r/Ro)
4. The throughput via the ring of size dR at radius r is:
dQ = v(r) * d(Pi*R^2) = 2*Pi*v(r)*r dR
5. Total integral from zero to Ro is:
Q=Integral(dQ)=Integral(2*Pi*v(r)*r ) dR
6. substitution v(r) from (3) to (5) gives:
Q=Integral( 2*Pi*Vo*(1-r/Ro) * r)dR
7. Taking the integral:
Q=2*Pi *Vo *(r^2/2 - r^3/3Ro)
8. Substituting limits (Ro,0) =>
Q = 2*Pi* Vo*(Ro^2/2 - Ro^2/3) = 2*Pi*Vo*Ro^2/6=Pi*Vo*Ro^2/3
9. From here Vo = 3*Q/(Pi*Ro^2)
10. Cool. the Ro^2 is in denumerator. It means that: Decreasing the radius will increase the speed of the liquid in the center quadratically.

For ideal liquid the speed increase is also quadratic. Indeed Q=v*Pi Ro^2 => V = Q/(Pi *Ro^2).
However, for non-ideal liquid the coefficient is higher.

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