Question

Air flows at the rate of $1.5m^3/s$ through a horizontal pipe with a gradually reducing cross-section as shown in the figure. The two cross-sections of the pipe have diameters of 400 mm and 200 mm. Take the air density as 1.2$kg/m^3$ and assume inviscid incompressible flow. The change in pressure $(p_2-p_1$) (in kPa) between sections 1 and 2 is

• A. -1.28
• B. 2.56
• C. -2.13
• D. 1.28

Answer 1

The correct option is A -1.28

$\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+Z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+Z_2$
[where, Z = constant]
$A_1=\frac{\pi }{4}\times {0.4}^2=0.1256m^2$
$A_2=\frac{\pi }{4}\times {0.2}^2=0.0314m^2$
$\frac{p_2-p_1}{\rho g}=\frac{V_1^2-V_2^2}{2g}$
$\frac{p_2-p_1}{1.2}=\frac{1}{2}[\frac{Q^2}{A_1^2}-\frac{Q^2}{A_2^2}]$
$p_2-p_1=0.6\times (1.5{)}^2[\frac{1}{A_1^2}-\frac{1}{A_2^2}]$
$p_2-p_1=1.35[\frac{1}{0.01577}-\frac{1}{0.000985}]$
=1.35[63.41-1015.22]
=-1284.94 Pa=-1.28 kPa