Question

4. Going back to the situation in question 1 (pipe narrows from an area of 2.0 cm2 to an area of 1.0 cm2 ; the speed of the water in the larger part of the pipe is 2.0 m/s. You already calculated the speed in the narrow part in question 1). The pipe is horizontal so the height of the water doesn’t change. Suppose you measure the pressure in the larger part of the pipe to be 8,000 Pa. What is the pressure in the narrow part

Answer 1

Answer:

$P_2 = 2000 Pa$

Explanation:

we know from continuity equation

A1 V1 = A2 V2

$2 \times 2 = 1 \times V2$

V2 = 4 m/s

$P_1 +\frac{1}{2} \rho v_1^2 + mg y_1 = P_2 +\frac{1}{2} \rho v_2^2 + mg y_2$

we know$y_1 = y_2$, so we have

$P_1 +\frac{1}{2} \rho v_1^2 = P_2 +\frac{1}{2} \rho v_2^2$

$P_2 = P_1 + \frac{1}{2} \rho (v_1^2 -v_2^2)$

$P_2 = 8000 + \frac{1}{2} \times 1000 (2^2 - 4^2)$

after solving we get

$P_2 = 2000 Pa$