# Fluid Dynamics and Flow Measurements

 Question 1
Match Column X with Column Y:
$\begin{array}{|l|l|}\hline \text{Column X}&\text{Column Y} \\ \hline \text{(P) Horton equation} & \text{((I) Design of alluvial channel} \\ \hline \text{(Q) Penman method} & \text{(II) Maximum flood discharge}\\ \hline \text{(R) Chezys formula}& \text{(III) Evapotranspiration}\\ \hline \text{(S) Lacey's theory}& \text{(IV) Infiltration}\\ \hline \text{(T) Dicken's formula}& \text{(V) Flow velocity}\\ \hline \end{array}$
Which one of the following combinations is correct?
 A (P)-(IV), (Q)-(III), (R)-(V), (S)-(I), (T)-(II) B (P)-(III), (Q)-(IV), (R)-(V), (S)-(I), (T)-(II) C (P)-(IV), (Q)-(III), (R)-(II), (S)-(I), (T)-(V) D (P)-(III), (Q)-(IV), (R)-(I), (S)-(V), (T)-(II)
GATE CE 2022 SET-2   Fluid Mechanics and Hydraulics
 Question 2
Two reservoirs are connected by two parallel pipes of equal length and of diameters 20 cm and 10 cm, as shown in the figure (not drawn to scale). When the difference in the water levels of the reservoirs is 5 m, the ratio of discharge in the larger diameter pipe to the discharge in the smaller diameter pipe is ____________. (round off to two decimal places)
(Consider only loss due to friction and neglect all other losses. Assume the friction factor to be the same for both the pipes) A 2.25 B 6.32 C 4.22 D 5.66
GATE CE 2022 SET-1   Fluid Mechanics and Hydraulics
Question 2 Explanation:
\begin{aligned} h_{f1} &=h_{f2} \\ \frac{8Q_1^2}{\pi ^2 g} \times \frac{fl}{D_1^5}&= \frac{8Q_2^2}{\pi ^2 g} \times \frac{fl}{D_2^5} \\ \left ( \frac{Q_1}{Q_2} \right )^2&= \left ( \frac{D_1}{D_2} \right )^5\\ \frac{Q_1}{Q_2} &= \left ( \frac{D_1}{D_2} \right )^{5/2}\\ &= \left ( \frac{0.2}{0.1} \right )^{5/2}\\&=5.66 \end{aligned}

 Question 3
The ratio of the momentum correction factor to the energy correction factor for a laminar flow in a pipe is
 A $\frac{1}{2}$ B $\frac{2}{3}$ C 1 D $\frac{3}{2}$
GATE CE 2021 SET-2   Fluid Mechanics and Hydraulics
Question 3 Explanation:
For laminar flow through pipe.
$\frac{\text { Momentum correction factor }}{\text { Kinetic energy correction factor }}=\frac{4 / 3}{2}=\frac{4}{6}=\frac{2}{3}$
 Question 4
Two water reservoirs are connected by a siphon (running full) of total length 5000 m and diameter of 0.10m, as shown below The inlet leg length of the siphon to its summit is 2000m. The difference in the water surface levels of the two reservoirs is 5 m. Assume the permissible minimum absolute pressure at the summit of siphon to be 2.5 m of water when running full. Given:
friction factor f=0.02 throughout,
atmospheric pressure= 10.3 m of water, and
acceleration due to gravity g= 9.81 $m/s^2$.

Considering only major loss using Darcy-Weisbach equation, the maximum height of the summit of siphon from the water level of upper reservoir,h (in m, round off to 1 decimal place)is _____
 A 2.8 B 6.9 C 7.8 D 5.8
GATE CE 2019 SET-1   Fluid Mechanics and Hydraulics
Question 4 Explanation: Applying Bernoulli's equation between 1 & 3
\begin{aligned} \frac{P_1}{\rho g}+\frac{v_1^2}{2g}+z_1&= \frac{P_3}{\rho g}+\frac{v_3^2}{2g}+z_3+h_{f13}\\ z_1-z_3&=h_{f13}\;\;\;(\because \; P_1=P_3=P_{atm}\; and \; v_1=v_3=0) \\ 5&=\frac{fLv^2}{2gd} \\ v^2&=\frac{5 \times 2\times 9.81\times 0.1}{0.02\times 5000} \\ v&=0.313m/s \end{aligned}
Applying Bernoulli's equation between 1 & 2
\begin{aligned} \frac{P_1}{\rho g}+\frac{v_1^2}{2g}+z_1&= \frac{P_2}{\rho g}+\frac{v_2^2}{2g}+z_2+h_{f12}\\ 10.3+0+0&=2.5+\frac{0.313^2}{2 \times 9.81}+h+\frac{0.02 \times 2000 \times 0.313^2}{2 \times 9.81 \times 0.1}\\ h&=5.798m \end{aligned}
 Question 5
Water flows through a 90$^{\circ}$ bend in a horizontal plane as depicted in the figure. A pressure of 140 kPa is measured at section 1-1. The inlet diameter marked at section 1-1 is $\frac{27}{\sqrt{\pi }}cm$, while the nozzle diameter marked at section 2-2 is $\frac{14}{\sqrt{\pi }}cm$. Assume the following:

(i) Acceleration due to gravity = 10 $m/s^{2}$
(ii) Weights of both the bent pipe segment as well as water are negligible.
(iii) Friction across the bend is negligible.

The magnitude of the force (in kN, up to two decimal places) that would be required to hold the pipe section is ______
 A 1.85 B 2.6 C 4.45 D 6.58
GATE CE 2017 SET-1   Fluid Mechanics and Hydraulics
Question 5 Explanation:
\begin{aligned} P_{1} A_{1} &=F_{x} \\ F_{x} &=\left(140 \times 10^{3}\right)\left[\frac{\pi}{4}\left(\frac{0.27}{\sqrt{\pi}}\right)^{2}\right] \\ &=2551.5 \mathrm{N}\\ F_{y} &=P_{2} A_{2} \\ &=\left(1.013 \times 10^{5}\right)\left[\frac{\pi}{4}\left(\frac{0.14}{\sqrt{\pi}}\right)^{2}\right] \\ &=49637 \mathrm{N} \\ \text { Resultant force } &=\sqrt{F_{x}^{2}+F_{y}^{2}} \\ &=\sqrt{(2551.5)^{2}+(49637)^{2}} \\ &=259933 \mathrm{N} \\ &=2.60 \mathrm{kN} \end{aligned}

There are 5 questions to complete.