Question

The velocity of a liquid flowing in a circular pipe of radius \(R\) varies from zero at the wall to a maximum at the pipe center. The velocity distribution in the pipe can be represented as \(V(r),\) where \(r\) is the radial distance from the pipe center. Based on the definition of mass flow rate \(\dot{m}\) obtain a relation for the average velocity in terms of \(V(r)\) \(R,\) and \(r\).

Short answer

Answer: The average velocity of the liquid is given by the expression, \(\overline{V} = \frac{2 \int_{0}^{R} r V(r) dr}{R^2}\).

Step-by-step solution

Find the mass flow rate \(\dot{m}\)

To find the mass flow rate, we will use the following equation: \(\dot{m} = \int_{0}^{R} \rho A(r) V(r) dr\) where: - \(\rho\) is the density of the fluid (a constant), - \(A(r)\) is the cross-sectional area of a circular pipe at a given radial distance \(r\), - \(V(r)\) is the velocity of the fluid at the radial distance \(r\), and - \(R\) is the radius of the circular pipe. The cross-sectional area \(A(r)\) can be obtained by considering small circular rings (called cylindrical shells) of radius \(r\), and thickness \(dr\). The area of such a ring is: \(A(r) = 2\pi r dr\) Substitute this into the mass flow rate equation to get: \(\dot{m} = \int_{0}^{R} \rho 2\pi r dr V(r)\)

Solve the integral

To find the mass flow rate, we must now solve the integral we obtained in step 1: \(\dot{m} = \int_{0}^{R} \rho 2\pi r dr V(r)\) Since \(\rho\) and \(2\pi\) are constants, they can be taken out of the integral: \(\dot{m} = \rho 2\pi \int_{0}^{R} r V(r) dr\) We cannot simplify this expression further, as we do not have an explicit form for \(V(r)\). So, we will leave it as is: \(\dot{m} = \rho 2\pi \int_{0}^{R} r V(r) dr\)

Write the relation between mass flow rate and average velocity

The average velocity, denoted as \(\overline{V}\), is the mass flow rate divided by the product of density and total cross-sectional area of the pipe. Therefore, we can write the relationship as follows: \(\overline{V} = \frac{\dot{m}}{\rho A_{total}}\) The total cross-sectional area of the pipe, \(A_{total}\), is given by the area of a circle with radius \(R\): \(A_{total} = \pi R^2\) Substitute this expression into the relation above: \(\overline{V} = \frac{\dot{m}}{\rho \pi R^2}\)

Find the average velocity in terms of \(V(r)\), \(R\), and \(r\)

To find the average velocity, we will substitute the mass flow rate we derived in step 2 into the relation we derived in step 3: \(\overline{V} = \frac{\rho 2\pi \int_{0}^{R} r V(r) dr}{\rho \pi R^2}\) Notice that the density, \(\rho\), and \(\pi\) cancel out, leaving us with: \(\overline{V} = \frac{2 \int_{0}^{R} r V(r) dr}{R^2}\) This is the final expression for the average velocity of the fluid in terms of \(V(r)\), \(R\), and \(r\).