Water flows in a horizontal tube (see figure). The pressure of water changes by $$700 \; Nm^{-2}$$ between $$A$$ and $$B$$ where the area of cross section are $$40 \; cm^2$$ and $$20 \; cm^2$$, respectively. Find the rate of flow of water through the tube. (density of water $$= 1000 \; kgm^{-3}$$)

Equation of continuity: $$A_A v_A = A_B v_B \implies 40 v_A = 20 v_B \implies v_B = 2v_A$$

Bernoulli's equation for a horizontal tube:

$$P_A + \frac{1}{2}\rho v_A^2 = P_B + \frac{1}{2}\rho v_B^2$$

$$P_A - P_B = \frac{1}{2}\rho (v_B^2 - v_A^2)$$

$$700 = \frac{1}{2}(1000)\left((2v_A)^2 - v_A^2\right) \implies 700 = 500(3v_A^2)$$

$$1500 v_A^2 = 700 \implies v_A = \sqrt{\frac{7}{15}} \approx 0.68\text{ m/s} = 68.0\text{ cm/s}$$

Volume flow rate $$Q$$: $$Q = A_A v_A = 40 \times 68.0 = 2720\text{ cm}^3\text{/s} $$