A horizontal pipe of uniform cross sectional area A empties into a bucket, �filling it at a rate R (unit volume per unit time).
Use dimensional analysis to determine the speed,v, of the fluid in the pipe to within a dimensionless multiplicative constant of order 1.
What happens to v as A decreases (for fi�xed R)?
I understand how dimensional analysis works, however when it says "to within a dimensionless multiplicative constant of order 1" i don't understand what its asking for.
In class we were shown that you change distance=L, Mass=M, Time=T.
Where Rate=L^3/T.
R = v A (look at a drawing of what you have here)
now the way you are supposed to do it
v = L/T
A = L^2
R = L^3/T = (L/T)L^2
so
R = L A/T
but L/T = v
so
R = v A
well in this case the result is exact, but in general there could be a constant of proportionality between the output and the input that we could not get from dimensional analysis
for example Force on a plate of area A in a fluid of density rho in kg/meter^3 and speed v in L/T
F = ma --> M L/T^2
= L^3 ( M/L^3) L /T^2
= ( M/L^3)L^2 L^2/T^2
= density * Area * v^2
but obviously the actual number will depend on the shape f the object
so
F = some shape constant * density * area * v^2
When using dimensional analysis, we can relate variables based on their dimensions. In this case, we want to determine the speed, v, of the fluid in the pipe using dimensional analysis.
Let's start by identifying the relevant variables and their dimensions:
- The cross-sectional area of the pipe is denoted as A and has dimensions of [L^2] (length squared).
- The rate at which the fluid fills the bucket is denoted as R and has dimensions of [L^3/T] (length cubed per unit time).
- The speed of the fluid in the pipe is denoted as v and has dimensions of [L/T] (length per unit time).
To relate these variables, we can use the principle of dimensional homogeneity, which states that any equation must have consistent dimensions on both sides.
First, we can see that the rate of filling the bucket, R, is directly related to the cross-sectional area, A, and the speed, v. Since the rate involves length cubed divided by time, we can write:
R ∝ Av
Here, "∝" denotes proportionality.
To eliminate the proportionality sign and find the exact relationship, we need to introduce a dimensionless constant. In this case, the question asks for a dimensionless multiplicative constant of order 1. This means that the constant is not expected to introduce a large variation in the result and can be approximated as 1.
Therefore, we can write the equation as:
R = kAv
Where "k" is a dimensionless constant of order 1.
Finally, we can isolate the speed, v, by rearranging the equation:
v = R/(kA)
Now let's discuss what happens to v as A decreases (for fixed R).
As we can see from the equation, if the cross-sectional area, A, decreases while R remains constant, the speed, v, will increase. This is because the fluid has to flow through a smaller area, so it will have to travel faster to maintain the same flow rate, R.
In conclusion, when A decreases (for fixed R), the speed, v, of the fluid in the pipe increases.