Chapter 14, Problem 88P

(a) Using Bernoulli's equation, show that be measured fluid speed v for a pitot tube, like the one in figure 14.32(b), is given by $v={\left(\frac{2\rho \text{'}gh}{\rho }\right)}^{1/h}$ , where h is be height of be manometer fluid, p' is the density of the manometer fluid, p is the density of the moving fluid, and g is be acceleration due to gravity. (Note that v is indeed proportional to the square root of h, as stated in text) (b) Calculate v for moving air if a mercury manometer's h is 0.200 m.

Figure 14.32 Measurement of fluid speed on Bernoulli’s principle. (a) A manometer is connected to two tubes close together and small enough not to disturb the flow. Tube 1 is open at the end facing the flow. A dead spot having zero speed is created there. Tube 2 has an opening on the side, so the fluid has a speed v across; thus, pressure there drops. The difference in pressure at the manometer is $\frac{1}{2}\rho v_2^2$ , so h is proportional to . $\frac{1}{2}\rho v_2^2$ (b) This type of velocity measuring device is a Prandtl tube, also known as a pitot tube.