Answer to Question #275953 in Physics for Femicrown

Question

One section of a pipe is 2.5cm in diameter, horizontal, and 1m above ground level. This section is connected to a second horizontal section at ground level which is 5cm in diameter. In the elevated section, the velocity of flow of water is 2m/s and the pressure is 1.4×10sN/m². Calculate (a) the velocity of flow in the lower section (b) the pressure in the lower section,and (c) the rate of flow of water through the pipe in kilograms per seconds.

Answer 1

(a) $A_1v_1=A_2v_2\to(\pi d_1^2/4)v_1=(\pi d_2^2/4)v_2\to$

$d_1^2v_1=d_2^2v_2\to v_2=d_1^2v_1/d_2^2=0.025^2\cdot2/0.05^2=0.5\ (m/s)$

(b) $p_1+\rho gh_1+\rho v_1^2/2=p_2+\rho gh_2+\rho v_2^2/2\to$

$p_2=p_1+\rho gh_1+\rho v_1^2/2-\rho gh_2-\rho v_2^2/2=$

$=14+1000\cdot9.8\cdot 1+1000\cdot2^2/2-0-1000\cdot 0.5^2/2=11689\ (Pa)$

(c) $Q=A_2\cdot \rho\cdot v_2=(\pi d_2^2/4)\cdot \rho \cdot v_2=$

$=(3.14\cdot0.05^2/4)\cdot1000\cdot0.5=0.981\ (kg/s)$ or

$Q=A_1\cdot \rho\cdot v_1=(\pi d_1^2/4)\cdot \rho \cdot v_1=$

$=(3.14\cdot0.025^2/4)\cdot1000\cdot2=0.981\ (kg/s)$