Use dimensional analysis to determine the pressure, P, in the pipe to within a dimen- sionless multiplicative constant of order 1.(Hint: P may depend on another physical variable besides A and R.) Using your result from part (a), reexpress your answer in a form that involves v. By what factor does P increase when v is doubled?
LOL, again ?
R = A V = L^2 L/T
P = F/A = m L/T^2 / L^2
P = m L/ (T^2L^2)
m = rho L^3
P = rho L^3 /(T^2 L)
P = rho L^2/T^2 = rho v^2
4
by the way this does not give us the sign and rho is M/L^3 but is constant in the usual incompressible fluid problem
To determine the pressure, P, in the pipe using dimensional analysis, we need to consider the variables involved and their dimensions. Let's break down the information given:
- A: The cross-sectional area of the pipe.
- R: The gas constant.
- P: The pressure in the pipe.
- v: The velocity of the fluid flowing through the pipe.
We can start by writing down the dimensions of each variable:
[A] = L^2 (area is measured in square units)
[R] = ML^2T^-2K^-1 (gas constant has units of energy per temperature per mole)
[P] = ML^-1T^-2 (pressure is measured in force per area)
[v] = LT^-1 (velocity is measured in length per time)
Now, let's consider the equation relating these variables. Dimensionally, pressure can be related to a combination of the other variables using powers:
P = A^a * R^b * v^c
To find the exponents (a, b, c) in the equation, we need to equate the dimensions on both sides.
Equating the dimensions of mass, length, and time, we have:
L^-1T^-2 = L^(2a) * (ML^2T^-2K^-1)^b * (LT^-1)^c
This equation gives us three separate equations involving the exponents:
Equating powers of length:
-1 = 2a + 2b + c
Equating powers of mass:
0 = b
Equating powers of time:
-2 = -2b - c
From the second equation, we find b = 0, which means pressure, P, does not depend on the gas constant R. Therefore, we can rewrite the equation as:
P = A^a * v^c
Plugging this into the equation from the first equation, we have:
-1 = 2a + c
Solving these equations simultaneously, we find a = -1/2 and c = 1/2. Therefore, we can write the pressure equation as:
P = k * (A/v^2)^(1/2)
where k is a dimensionless multiplicative constant of order 1.
Now let's reexpress the answer in a form that involves v. We have:
P = k * (A/v^2)^(1/2)
To find how P changes when v is doubled, we can substitute 2v for v in the equation:
P' = k * (A/(2v)^2)^(1/2)
P' = k * (A/4v^2)^(1/2)
P' = (1/2) * k * (A/v^2)^(1/2)
P' = (1/2) * P
Therefore, when v is doubled, P increases by a factor of 1/2 or decreases by half.
In summary, the pressure in the pipe, P, can be expressed as P = k * (A/v^2)^(1/2). This means that P depends on the cross-sectional area, A, and the velocity, v, of the fluid flowing through the pipe. When the velocity, v, is doubled, the pressure, P, decreases by half.