Question

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

• A. $3020c{m}^3/s$
• B. $2420c{m}^3/s$
• C. $2720c{m}^3/s$
• D. $1810c{m}^3/s$

Answer 1

Answer: (C)

Rate of flow of water $=A_A{V}_A=A_B{V}_B$
$(40)V_A=(20)V_B$
$V_B=2V_A......(1)$
Using Bernoulli's theorem
$P+\frac{1}{2}\rho V^2=P_B+\frac{1}{2}\rho V_B^2$
$P_A-P_B=\frac{1}{2}\rho \left(V_B^2-V_A^2\right)$
$700=\frac{1}{2}\times 1000\left(4V_A^2-V_A^2\right)$
$V_A=0.68m/s=68cm/s$
Rate of flow $=A_A{V}_A$
$-\left(40\right)\left(68\right)=2720c{m}^3/s$