Water (η = 1.00E-3 Pa·s) is flowing through a horizontal pipe with a volume flow rate of 0.0153 m3/s. As the drawing below shows, there are two vertical tubes that project from the pipe. Assume that H = 0.0477 m and L = 0.670 m
I know how to plug in everything but, my main problem is solving for r^4. Please help me!!
To solve for r^4 in this problem, we can use the formula for the volume flow rate of a fluid through a pipe.
The volume flow rate (Q) of a fluid through a pipe is given by the equation:
Q = (π * r^2 * ΔP) / (8 * η * L)
where:
- Q is the volume flow rate (0.0153 m^3/s in this case)
- r is the radius of the pipe
- ΔP is the pressure difference across the pipe (which is equal to the difference in height of the water columns in the vertical tubes, H)
- η is the viscosity of the fluid (1.00E-3 Pa·s for water)
- L is the length of the pipe (0.670 m in this case)
To solve for r^4, we need to rearrange the formula:
r^4 = (8 * η * Q * L) / (π * ΔP)
Now we can substitute the known values into the formula:
r^4 = (8 * 1.00E-3 * 0.0153 * 0.670) / (π * 0.0477)
Let's calculate the value:
r^4 = 0.020355 m^4
Therefore, the value of r^4 is approximately 0.020355 m^4.