Correct Answer - Option 3 : 20.0
Given that,
Pressure drop (ΔP1) = 5 kg/m2
Flow rate (Q) is doubled.
= Q2 = 2Q1
The relation between pressure drop and flow rate is,
\(Q \propto \sqrt {{\rm{\Delta }}P}\)
\(\Rightarrow \frac{{{Q_1}}}{{{Q_2}}} = \;\sqrt {\frac{{{\rm{\Delta }}{{\rm{P}}_1}}}{{{\rm{\Delta }}{{\rm{P}}_2}}}}\)
\(\Rightarrow \frac{{{Q_1}}}{{2{Q_1}}} = \;\sqrt {\frac{5}{{{\rm{\Delta }}{{\rm{P}}_2}}}}\)
⇒ ΔP
2 = 5 × 4 = 20 kg/m
2