Question 5.3: Determine the average velocity of the flow of water in the p......
Determine the average velocity of the flow of water in the pipe in Fig. 5–13, and the static and dynamic pressure at point B. The water level in each of the tubes is indicated. Take ρ_w = 1000 kg/m³.
Fluid Description. We have steady flow. Also, we will consider the water to be an ideal fluid.
Bernoulli Equation. At A the total pressure is the static pressure, which is the result of both the weight of fluid and the internal pressure within the pipe. It is found from p_A = \rho_w gh_A. At B the total (or stagnation) pressure is a combination of static and dynamic pressures found from p_B = \rho_w gh_B. Knowing these pressures, we can determine the average velocity of flow V_A using the Bernoulli equation, applied at points A and B, where B is a stagnation point on the streamline. We have
\frac{\rho _w gh_A}{\rho _w g} + \frac{V_A^2}{2g} + 0 = \frac{\rho _w gh_B}{\rho _w g} + 0 + 0
V_A = \sqrt{2g (h_B – h_A)} = \sqrt {2(9.81 m/s^2)(0.150 m – 0.090 m)}
V_A = 1.085 m/s = 1.08 m/s
The static pressure at both A and B is determined from the piezometric head.
(p_A)_{static} = (p_B)_{static} = \rho _w gh_A = (1000 kg/m^3) (9.81 m/s^3) (0.09 m) = 883 PaThe dynamic pressure at B is determined from
\rho _w \frac{V_A^2}{2} = (1000 kg/m^3) \frac{(1.085 m/s)^2}{2} = 589 PaThis value can also be obtained by noticing that
h_{dyn} = 0.15 m – 0.09 m = 0.06 m
so that