# GATE Papers >> Mechanical >> 2017 >> Question No 31

Question No. 31

Consider steady flow of an incomressible fluid through two long and straight pipes of diameters $d_1$ and $d_2$ arranged in series. Both pipes are of equal length and the flow is turbulent in both pipes. The friction factor for turbulent flow though pipes is of the form, $f=K(Re)^{-n}$, where $K$ and $n$ are known positive constants and Re is the Reynolds number. Neglecting minir losses, the ratio of the frictional pressure drop in pipe 1 to that  in pipe 2, $\style{font-family:'Times New Roman'}{\left(\frac{\triangle P_1}{\triangle P_2}\right)}$, is given by

##### Answer : (A) $\style{font-family:'Times New Roman'}{\left(\frac{d_2}{d_1}\right)^{\left(5-n\right)}}$

Solution of Question No 31 of GATE 2017 Mechanical Paper

$\begin{array}{l}{\mathrm h}_\mathrm f=\frac{\mathrm{flv}^2}{2\mathrm{gd}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm Q=\mathrm{flow}\;\mathrm{rate}\\\mathrm{Where}\;\mathrm f=\mathrm k=\left({\mathrm R}_\mathrm e\right)^{-\mathrm n}\;\;\;:\;{\mathrm R}_\mathrm e=\frac{\mathrm{ρvd}}{\mathrm\mu}\;\;\;:\;\mathrm v=\frac{4\mathrm Q}{\mathrm{πd}^2}\\{\mathrm h}_\mathrm f=\frac{\mathrm{kμ}^\mathrm n\;4^{2-\mathrm n}\;\mathrm l\;\mathrm Q^{2-\mathrm n}}{\mathrm\rho^\mathrm n\;\mathrm d^{5-\mathrm n}}\\\mathrm{thus}\frac{\triangle{\mathrm P}_1}{\triangle{\mathrm P}_2}=\frac{{\mathrm\rho}_\mathrm g\;{\mathrm h}_{\mathrm f1}}{{\mathrm\rho}_\mathrm g\;{\mathrm h}_{\mathrm f2}}=\frac{{\mathrm h}_{\mathrm f1}}{{\mathrm h}_{\mathrm f2}}\\\frac{\triangle{\mathrm P}_1}{\triangle{\mathrm P}_2}=\left(\frac{{\mathrm d}_2}{{\mathrm d}_1}\right)^{5-\mathrm n}\end{array}$