Answer to Question #109857 in Physics for Jess

Question

Fluid of density 995 kg m−3 flows through a horizontal pipe that curves down and then curves back to horizontal. (the height between point 1 and 2 is 16cm as the horizontal pipe curves down then curves back to horizontal)

What is the pressure difference ΔP between points 1 and 2 where ΔP = P2 − P1? (to 2 s.f and in kPa)

A=0.032m2

Answer 1

Using Bernoulli's equation:

$p_1 + \rho g h_1 + \frac{\rho v_1^2}{2} = p_2 + \rho g h_2 + \frac{\rho v_2^2}{2}$.

Since the flux is continuous, and the the cross-section is the same at both heights, the amount of water, that goes though the pipe at both heights in time $\Delta t$ is: $A v_1 \Delta t = A v_2 \Delta t$, from where $v_1 = v_2$.

Hence, from Bernoulli's equation: $\Delta p = p_2 - p_1 = \rho g (h_1 - h_2) = 995 kg \cdot m^{-3} \cdot 9.81 \frac{m}{s^2} \cdot 16 \cdot 10^{-2} m \approx 1.56 kPa$.