Question

Determine the pressure difference in tube of non-uniform cross sectional area as shown in figure. $\Delta$P = ?
d₁ = 5 cm, V₁ = 4, d₂ = 2 cm, V₂ = ?

• A. 304200 Pa
• B. 304500 Pa
• C. 302500 Pa
• D. 303500 Pa

Answer 1

Answer: (B)

$A_1{V}_1=A_2{V}_2$
$5^2\times 4=2^2\times V_2$
$V_2=25$
$P_1+\frac{1}{2}+V_1^2=P_2+\frac{1}{2}+V_2^2$
$P_1-P_2=\frac{1}{2}+\left(V_2^2-V_1^2\right)$
$=\frac{1}{2}\times 10^3\left(25^2-4^2\right)$
$=500\left(625-16\right)$
$P_1-P_2$ =500 × 609 = 304500 Pa