GATE Papers >> Mechanical >> 2018 >> Question No 144

Question No. 144

Air flows at the rate of 1.5 m3/s through a horizontal pipe with a gradually reducing crosssection 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/m3 and assume inviscid incompressible flow. The change in pressure $\left(p_2-p_1\right)$ (in kPa) between sections 1 and 2 is

Solution of Question No 144 of GATE 2018 Mechanical Paper

Air flow rate $\mathrm Q=1.5\;\mathrm m^3/\sec$

Velocity at sec $1:\;{\mathrm V}_1=\frac{1.5}{\displaystyle\frac{\mathrm\pi}4(0.4)^2}\;\mathrm m/\sec\;=11.937\;\frac{\mathrm m}\sec$

Velocity at sec $2:\;{\mathrm V}_2=\frac{1.5}{\displaystyle\frac{\mathrm\pi}4(0.2)^2}\;\frac{\mathrm m}\sec\;=47.746\;\frac{\mathrm m}\sec$

Applying Bemoulli's theorem at section 1 and 2

$\begin{array}{l}\frac{{\mathrm P}_1}{\mathrm{ρg}}+\frac{\mathrm V_1^2}{2\mathrm g}+{\mathrm Z}_1=\frac{\displaystyle{\mathrm P}_2}{\displaystyle\mathrm{ρg}}+\frac{\displaystyle\mathrm V_2^2}{\displaystyle2\mathrm g}+{\mathrm Z}_2\\\frac{{\mathrm P}_1-{\mathrm P}_2}{\mathrm{ρg}}=\frac{\mathrm V_2^2-\mathrm V_1^2}{2\mathrm g}\;\;\lbrack\mathrm{as}\;{\mathrm Z}_1={\mathrm Z}_2\rbrack\\({\mathrm P}_1-{\mathrm P}_2)=\frac{(\mathrm V_2^2-\mathrm V_1^2)\mathrm\rho}2\;=\frac{(47.746^2-11.937^2)\times1.2}2\\({\mathrm P}_1-{\mathrm P}_2)=1282.34\;\mathrm{pa}=1.282\;\mathrm{kpa}\\\therefore\;\mathrm{The}\;\mathrm{change}\;\mathrm{in}\;\mathrm{pressure}\;({\mathrm P}_1-{\mathrm P}_2)=-1.282\;\mathrm{kpa}\end{array}$