Q2GATE 2019NAT

Consider the reactor shown in the figure. The flow rate through the reactor is $Q \; m^3/h$ . The concentrations (in mg/L) of a compound in the influent and effluent are $C_0$ and C, respectively. The compound is degraded in the reactor following the first order reaction. The mixing condition of the reactor can be varied such that the reactor becomes either a completely mixed flow reactor (CMFR) or a plug-flow reactor (PFR). The length of the reactor can be adjusted in these two mixing conditions to $L_{CMFR}$ and $L_{PFR}$ while keeping the cross-section of the reactor constant. Assuming steady stateand for $C/C_0 =0.8$ , the value of $L_{CMFR}/L_{PFR}$ (round off to 2 decimal places) is _______

🚀 Solve in Practice Mode

📖 Explanation

Hydrocarbons and nitrogen oxides are considered primary air pollutants. $c=\frac{c_0}{1+kt}$ For (PFR) plug flow reactor

\begin{aligned} c&=c_0e^{-kt} \\ \text{As}\;\; c/c_0 &=0.8 \\ \text{For CMFR} \;\; 0.8&=\frac{1}{1+kt_{CMFR}} \\ \Rightarrow \;\;t_{CMFR} &=\frac{0.25}{k} \\ \text{For PFR} \;\; 0.8&=e^{-kt_{PFR}} \\ \Rightarrow \;\; t_{PFR}&=\frac{0.22314}{k} \end{aligned}

Now for steady state

\begin{aligned} v&=\text{constant}\\ L&=vt\\ So, \;\;\frac{L_{CMFR}}{L_{PFR}}&=\frac{vt_{CMFR}}{vt_{PFR}}\\ &=\frac{0.25}{0.22314}=1.12 \end{aligned}

Q3GATE 2016NAT

A square plate is suspended vertically from one of its edges using a hinge support as shown in figure. A water jet of 20 mm diameter having a velocity of 10 m/s strikes the plate at its mid-point, at an angle of $30^{\circ}$ with the vertical. Consider g as $9.81\: m/s^{2}$ and neglect the self-weight of the plate. The force F (expressed in N) required to keep the plate in its vertical position is _______________

🚀 Solve in Practice Mode

📖 Explanation

Force exerted by jet in x direction

\begin{array}{l} F_{x}=m[V \sin \theta-0 \mid \\ =\rho Q \times V \sin \theta \\ =\rho A V \times V \sin \theta \\ =1000 \times \frac{\pi}{4}(0.02)^{2} \times (10)^{2} \sin 30^{\circ} \\ =157079 \mathrm{N} \end{array}

Taking mornent about hinge

\begin{array}{r} F_{x} \times \frac{0.2}{2}=F \times 02 \\ F=7.85 \mathrm{N} \end{array}

Q4GATE 2016NAT

A penstock of 1 m diameter and 5 km length is used to supply water from a reservoir to an impulse turbine. A nozzle of 15 cm diameter is fixed at the end of the penstock. The elevation difference between the turbine and water level in the reservoir is 500 m. Consider the head loss due to friction as 5% of the velocity head available at the jet. Assume unit weight of water = $10\: kN/m^{3}$ and acceleration due to gravity (g)= $10\: m/s^{2}$ . If the overall efficiency 80%, power generated (expressed in kW and rounded to nearest integer) is ___________

🚀 Solve in Practice Mode

📖 Explanation

Apply energy equaiton at the free surlace of reservior and exit of nozzle

\begin{aligned} 500&=\text{Head loss } \frac{v_{1}^{2}}{2 g}\\ 500&=0.05 \frac{v_{1}^{2}}{2 g}+\frac{v_{1}^{2}}{2 g}\\ \sqrt{\frac{2 \times 10 \times 500}{1.05}}&=v_{1} \\ V_{1}&=97.59 \mathrm{m} / \mathrm{sec}\\ \text{Water Power WP }\\ &=\frac{1}{2} m v_{1}^{2} \\ &=\frac{1}{2}\left(10^{2}\right) \times \frac{\pi}{4}(0.15)^{2}(9759) \\ &=8212178 \mathrm{kW} \\ \text{Now,}\quad \eta_{0}&=\frac{\text { Shaft power }(\$ P)}{W P} \\ 08&=\frac{S P}{8212178} \\ S P&=6569.74 \mathrm{kW} \\ &\simeq 6570 \mathrm{kW} \end{aligned}

Q5GATE 2014NAT

A horizontal nozzle of 30 mm diameter discharges a steady jet of water into the atmosphere at a rate of 15 liters per second. The diameter of inlet to the nozzle is 100 mm. The jet Impinges normal to a flat stationary plate held close to the nozzle end. Neglecting air friction and considering the density of water as 1000 kg/ $m^3$ , force exerted by the jet (in N) on the plate is _______

🚀 Solve in Practice Mode

📖 Explanation

$\text{Force }=\rho a V^{2}$ $\mathrm{Q}=\mathrm{a} \mathrm{V}=15 \; litre/sec (given)$ $\text{Force }=\rho \frac{\mathrm{Q}^{2}}{\mathrm{a}}$ $=1000 \times \frac{15^{2}}{\pi} \times 30^{2} \mathrm{m}^{3} \times 10^{-6} \frac{\mathrm{m}^{6}}{\mathrm{s}^{2}} \times \frac{1}{10^{-6}} \mathrm{m}^{2}$ $=1000 \times \frac{1}{\pi} \mathrm{N}=318.3 \mathrm{N}$

Q6GATE 2014MCQ

A horizontal jet of water with its cross-sectional area of 0.0028 $m^2$ hits a fixed vertical plate with a velocity of 5 m/s. After impact the jet splits symmetrically in a plane parallel to the plane of the plate. The force of impact (in N) of the jet on the plate is

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \text { Force on plate } &=\left(\rho_{w} a V\right) \mathrm{V} \\ &=\rho_{w} a V^{2} \\ &=1000 \times 0.0028 \times (5)^{2} \\ &=70 \mathrm{N} \end{aligned}

Q7GATE 2007MCQ

A horizontal water jet with a velocity of 10 m/s and cross sectional area of 10 $mm^{2}$ strikes a flat plate held normal to the flow direction. The density of water is 1000 kg/ $m^{3}$ . The total force on the plate due to the jet is

🚀 Solve in Practice Mode

📖 Explanation

The force is given by

\begin{aligned} F &=\rho a V^{2} \\ \therefore F &=1000 \times 10 \times 10^{-6} \times (10)^{2} \\ &=1 \mathrm{N} \end{aligned}

Q8GATE 2005MCQ

A tank and a deflector are placed on a frictionless trolley. The tank issues water jet (mass density of water = 1000 kg/ $m^{3}$ ), which strikes the deflector and turns by 45 $^{\circ}$ . If the velocity of jet leaving the deflector is 4 m/s and discharge is 0.1 $m^{3}$ /s, the force recorded by the spring will be

🚀 Solve in Practice Mode

📖 Explanation

Force in spring will be the force in horizonta direction.

\begin{aligned} \therefore \quad F_{H} &=\rho Q V \cos \theta \\ &=1000 \times 0.1 \times 4 \times \cos 45^{\circ} \\ &=\frac{400}{\sqrt{2}}=200 \sqrt{2} \mathrm{N} \end{aligned}

Q9GATE 2003MCQ

A horizontal jet strikes a frictionless vertical plate (the plan view is shown in the figure). It is then divided into two parts, as shown in the figure. If the impact loss is neglected, what is the value of $\theta$ ?

🚀 Solve in Practice Mode

📖 Explanation

From continuity equation, we get,

\begin{aligned} & & Q_{0}=Q_{1}+Q_{2} \\ \Rightarrow & & Q_{0}=0.25 Q_{0}+Q_{2} \\ \Rightarrow & & Q_{2}=0.75 Q_{0} \end{aligned}

since, the impact losses are neglected, the velocity will remain unchanged in the direction of $Q_{1}$ and $Q_{2}$ $\text{i.e.},\quad V_{0}=V_{1}=V_{2}$ Applying impulse momentum equation, we get,

\begin{aligned} \rho Q_{0} V_{0} \sin \theta&=\rho Q_{2} V_{2}-\rho Q_{1} V_{1} \\ \Rightarrow \quad Q_{0} \sin \theta&=0.75 Q_{0}-0.25 Q_{0} \\ \Rightarrow \quad \sin \theta&=1 / 2 \\ \Rightarrow \quad \theta&=30^{\circ} \end{aligned}

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