Question 5.3: A venturimeter is fitted in a horizontal pipe to measure the......

Area A_{1}=\frac{\pi }{4}×(0.4 )^{2}

= 0.1256 m²

Area A_{2}=\frac{\pi }{4}×(0.2 )^{2}

= 0.031415 m²

The difference of pressure head in manometer is

h=x(\frac{s_{m} }{s_{p} }-1 )

=0.08(\frac{13.6}{1}-1 )=1 m of water

Now, using the discharge equation of venturimeter,

Q=C_{d}\frac{A_{1}A_{2} }{\sqrt{A^{2}_{1}-A^{2}_{2} } }\sqrt{2gh}

Substituting values in Eq. (i), we have

Q=0.98\frac{0.1256 \times 0.31415}{\sqrt{0.12562^{2} – 0.031415^{2} } }\sqrt{2×9.81×1}

Q =1.409 m³/s