Concept:
The pressure drop in the duct is given as, \({\rm{\Delta }}P = \frac{{32\mu L{V_{avg}}}}{{{D^2}}}\)
Where D = 4A/P
Calculation:
Given:
μ = 2 × 10-5 kg/m-s, L = 1 m, Vavg = 1 m/s
\(D = \frac{{4 \times 25 \times 15}}{{2\left( {25 + 15} \right)}} = 18.75~mm = 0.01875~m\),
∴ ΔP =\(\frac{{32 \times \left( {2 \times {{10}^{ - 5}}} \right) \times 1 \times 1}}{{{{\left( {0.01875} \right)}^2}}} = 1.82~{\rm}\) Pa