Dimension and Pipe Diameter
The drag, FD, on a sphere located in a pipe through which a fluid is flowing is to be determined experimentally. Assume that the drag is a function of the sphere diameter, d; the pipe diameter, D; the fluid velocity, v; and the fluid density, ρ. (a) What dimensionless parameters would you use for this problem? (b) Some experiments using water indicate that for d = 0.005 m, D = 0.0125 m, and v = 0.6 m/s, the drag is 6.5×10-3 N. If possible, estimate the drag on a sphere located in a 0.6 m diameter pipe through which water is flowing with a velocity of 1.85 m/s. The sphere diameter is such that geometric similarity is maintained.
Solution: This problem has to be solved by dimensional analysis. First, we have to set up the dimensional matrix,
FD
d ρ v
D
M
1
0
1
0
0
L
1
1
-3
1
1
t
-2
0
0
-1
0
As there are 5 parameters, and the rank is 3, so there will be two dimensionless groups or two isolated variables. The drag force is taken to be one of the isolated variable. The other will simply be the pipe diameter, D. This is due to the fact that D and d are having the same dimension, so they can not be both taken as core variables. For the 3 core variables, check to ensure rank is 3. π1 = da ρb vc D = Mb La-3b+c+1 t-c = M0 L0 t0
Upon solving the above, we have a = -1, b = 0, c = 0. So we have
Similarly, we can find the second group π2 = da ρb vc FD = Mb+1 La-3b+c+1 t-c-2 = M0 L0 t0
Upon solving the above, we have a = -2, b = -1, c = -2. So we have
To find the sphere diameter, it is known that we have geometric similarity, which is also supported by our π1.
To find the drag force, we can make use of π2,
7
In the above application, since π2 = f( π1 ) and π1 remains the same, so we have constant π2.
Solution: This problem has to be solved by dimensional analysis. First, we have to set up the dimensional matrix,
FD
d ρ v
D
M
1
0
1
0
0
L
1
1
-3
1
1
t
-2
0
0
-1
0
As there are 5 parameters, and the rank is 3, so there will be two dimensionless groups or two isolated variables. The drag force is taken to be one of the isolated variable. The other will simply be the pipe diameter, D. This is due to the fact that D and d are having the same dimension, so they can not be both taken as core variables. For the 3 core variables, check to ensure rank is 3. π1 = da ρb vc D = Mb La-3b+c+1 t-c = M0 L0 t0
Upon solving the above, we have a = -1, b = 0, c = 0. So we have
Similarly, we can find the second group π2 = da ρb vc FD = Mb+1 La-3b+c+1 t-c-2 = M0 L0 t0
Upon solving the above, we have a = -2, b = -1, c = -2. So we have
To find the sphere diameter, it is known that we have geometric similarity, which is also supported by our π1.
To find the drag force, we can make use of π2,
7
In the above application, since π2 = f( π1 ) and π1 remains the same, so we have constant π2.