Water flows in the horizontal pipe shown in the figure.At A the diameter is 5.00 cm and at B the diameter is 4.00 cm The fluid in the manometer is mercury,which has a density of .The manometer reading h is 4.40 cm.We can treat water as an ideal fluid having a density of .What volume of water is flowing through the pipe per second?
A) 0.0206 /s
B) 0.0426 /s
C) 0.00560 /s
D) 0.372 /s
E) 0.186 /s

Question

Water flows in the horizontal pipe shown in the figure.At A the diameter is 5.00 cm and at B the diameter is 4.00 cm The fluid in the manometer is mercury,which has a density of  .The manometer reading h is 4.40 cm.We can treat water as an ideal fluid having a density of  .What volume of water is flowing through the pipe per second? 
A) 0.0206 /s 
B) 0.0426 /s 
C) 0.00560 /s 
D) 0.372 /s 
E) 0.186 /s

Quizplus · Accepted Answer

To find the volume flow rate we will use the equation of continuity which states that the mass flow rate will be constant throughout the pipe. Hence, we have:
$ho_1 A_1 v_1 =ho_2 A_2 v_2$
where $ho_1 = 1000\ kg/m^3$, $ho_2 = 13600\ kg/m^3$, $A_1 =\pi(0.05m)^2/4$, and  $A_2 =\pi(0.04m)^2/4$. The velocity $v_1$ at point A can be calculated using Bernoulli's principle which states that the sum of pressure energy, kinetic energy and potential energy per unit volume in a fluid is constant. Hence, we have:
$P_A + \frac{1}{2} ho_1 v_1^2 + ho_1 gh_A = P_B + \frac{1}{2} ho_1 v_2^2 + ho_1 gh_B$
where $P_A=P_B$; $h_A=0$; $h_B=4.4 cm$; $g=9.8m/s^2$; and $v_2$ can be obtained from $v_2 = \frac{A_1 v_1}{A_2}$. Solving these equations, we get $v_1 = 4.44\ m/s$ and $v_2 = 6.25\ m/s$. Now the volume flow rate can be calculated as $Q= A_1 v_1=\pi(0.05m)^2/4 	imes 4.44 \approx 5.60 	imes 10^{-3} m^3/s \approx 0.0056\ m^3/s$. Therefore, the answer is choice C.

Water Flows in the Horizontal Pipe Shown in the Figure