Physics problems with fluids help?
Alright, there are 2 problems that i am stuck on and really have no clue how to do. Any help/explanation would be extremely beneficial.
1. a 6.0 cm diameter pipe gradually narrows to 4.0cm. water flows though the pipe at a certain rate, the gauge pressures in the 2 sections are 32.0 kpa and 24.0 kpa. What is the volume rate of flow? (I know the equation of continuity and bernoulli's should be used, but don't know how to use them when neither of the velocities are given.
2. a geologist finds that a moon rock whose mass is 9.28 kg has an apparent mass of 6.18 kg when submerged in water. What is the densisty(I think it has something to do with archemides principle and the bouyant force)
- GaryLv 67 years agoFavorite Answer
1. What is the volume rate of flow?
Try this approach:
Using the "simplifed form" of Bernoulli's equation (where the difference in height between the two points is negligible):
(1) static pressure (p) + dynamic pressure (q) = total pressure,
plus the statement of Bernoulli's principle that " total pressure is constant along a streamline", allows us to write:
(2) p1 + q1 = p2 + q2
The dynamic pressure formula is:
(3) q = ½ ρ * V²,
where V = fluid speed.
Substituting (3) into (2):
(4) p1 + ½ ρ * V1² = p2 + ½ ρ * V2²
(5) 2 * (p1 - p2) / ρ = V2² - V1²
We know the values on the left side of (5); let's work on the right side:
The volume rate of flow must be the same in both sections; therefore:
(6) π * r1^2 * V1 = π * r2^2 * V2 ==>
(7) V2 = (r1 / r2)^2 * V1
Substituting (7) into (5) we get:
(8) 2 * (p1 - p2) / ρ = [ (r1 / r2)^2 * V1 ]² - V1² ==>
(9) V1² * [ (r1 / r2)^4 - 1 ] = 2 * (p1 - p2) / ρ ==>
(10) V1² = [ 2 * (p1 - p2) / ρ ] / [ (r1 / r2)^4 - 1 ]
You can solve (10) for V1, since you know all the values on the right side.
The volume flow rate VFR is just Area X Speed, which we used in (6) above, so:
(11) VFR = π * r1^2 * V1 <<<=== Answer for What is the volume rate of flow?
Note: Make sure that you use a consistent set of units when you substitute the values into (10) and (11).
2. What is the densisty(I think it has something to do with archemides principle and the bouyant force)
You're right! It does have to do with Archimedes' principle and buoyancy--see Source 2.
Specifically, if you read that section of the wiki article, you'll find the formula:
(12) (density of object) / (density of water) =
(weight of object) / (weight of object - immersed weight)
So you can solve (12) for the density of object and substitute the values given to get the Answer you need.
.Source(s): 1. http://en.wikipedia.org/wiki/Bernoulli%27s_princip... 2. http://en.wikipedia.org/wiki/Buoyancy#Archimedes.2...
- 4 years ago
SG is a ratio of an merchandise's density to that of water. An merchandise with a SG = a million.00 will glide with one hundred% of its quantity decrease than water. A block of wood with density = 0.seventy 9 floats with seventy 9% of its quantity decrease than water. ANS-a million 550/19.3 = 28.5 cm³ ANS-2 quantity of block of wood = 2500/0.fifty 5 ? 4545 cm³ in water this block floats with fifty 5% of its quantity decrease than water in oil this block floats with 2500/0.80 5 = 2941 cm³ decrease than water ANS-3 10.3 m of clean water = a million.01E5 Pa 750/10.3 = 72.8 atmospheres ANS