In a horizontal tube the water pressure changes by 1500 Nm ⁻² between A and B as shown in figure below. The cross-sectional areas at A and B of the tube are 40 ~cm ^2 and 20 ~cm ^2 , respectively. Find the rate of flow of water through the tube.

Step-by-step solution

Given that, pressure difference By using Bernoulli's equation, p_A+ 1 2 v_A^2=p_B+ 1 2 v_B^2 Density of water, =10^3 ~kg ~m ⁻³ Cross-section areas at points A and B , aligned & a_A=40 ~cm ^2=40 10⁻⁴ ~m ^2 & a_B=20 ~cm ^2=20 10⁻⁴ ~m ^2 aligned By equation of continuity, rate of flow of water through the tube V t =a_A v_A=a_B v_B v_A v_B = a_B a_A = 1 2 Substituting the values from Eqs. (i) and (iii), in Eq. (ii), we get aligned 1500 & = 1 2 10^3 [ (2 v_A )^2-v_A^2 ] 1500 & =500 (4 v_A^2-v_A^2 ) v_A^2 & =1 v_A=1 ~m /