Question 1 |

Two discrete spherical particles (P and Q) of equal mass density are
independently released in water. Particle P and particle Q have diameters of
0.5 mm and 1.0 mm, respectively. Assume Stokes' law is valid.

The drag force on particle Q will be________ times the drag force on particle P. (round off to the nearest integer)

The drag force on particle Q will be________ times the drag force on particle P. (round off to the nearest integer)

4 | |

8 | |

10 | |

12 |

Question 1 Explanation:

In case of discrete particle settling and Stoke's
law valid, at terminal velocity, since there is no
change in velocity, the net force on the body
is zero. Hence,

F_D=\left ( \frac{\pi}{6}D^3\rho _sg-\frac{\pi}{6}D^3\rho _sg \right )=\frac{\pi}{6}gD^3(\rho _s-\rho _f)

For density of medium (\rho _f) and mass density of sphere (\rho _s) constant,

Drag force (F_D)\propto D^3

For particle P, diameter (D_P)=0.5 mm

For particle Q, diameter (D_Q)=1 mm

\Rightarrow \frac{(F_D)_Q}{(F_D)_P}=\frac{(D_Q)^3}{(D_P)^3}=\left ( \frac{1}{0.5} \right )^3=8

\Rightarrow (F_D)_Q=8 \times (F_D)P

F_D=\left ( \frac{\pi}{6}D^3\rho _sg-\frac{\pi}{6}D^3\rho _sg \right )=\frac{\pi}{6}gD^3(\rho _s-\rho _f)

For density of medium (\rho _f) and mass density of sphere (\rho _s) constant,

Drag force (F_D)\propto D^3

For particle P, diameter (D_P)=0.5 mm

For particle Q, diameter (D_Q)=1 mm

\Rightarrow \frac{(F_D)_Q}{(F_D)_P}=\frac{(D_Q)^3}{(D_P)^3}=\left ( \frac{1}{0.5} \right )^3=8

\Rightarrow (F_D)_Q=8 \times (F_D)P

Question 2 |

A pump with an efficiency of 80% is used to draw groundwater from a well for
irrigating a flat field of area 108 hectares. The base period and delta for paddy
crop on this field are 120 days and 144 cm, respectively. Water application
efficiency in the field is 80%. The lowest level of water in the well is 10 m
below the ground. The minimum required horse power (h.p.) of the pump is
________. (round off to two decimal places)

(Consider 1 h.p. = 746 W; unit weight of water = 9810 N/m^3)

(Consider 1 h.p. = 746 W; unit weight of water = 9810 N/m^3)

25.64 | |

36.25 | |

30.82 | |

48.32 |

Question 2 Explanation:

Base period (B) = 120 days

\begin{aligned} Delta (\Delta )&=144 cm=1.44m\\ Duty(D)&=\frac{8.64B}{\Delta }=\frac{8.64 \times 120}{1.44}\\ D&=720\frac{hec}{cumec}\\ Q&=\frac{Area}{D}=\frac{108}{720}=\frac{3}{20}m^3/s\\ Q_{applied}&=\frac{Q}{\eta _a}=\frac{3}{20 \times 0.8}=\frac{3}{16}\\ \text{Power required}&=\frac{mgh}{t}\\ &=\rho _w \times \left ( \frac{volume}{t} \right )g \times h\\ &=\rho _w Qgh=\gamma Qh\\ &=9810 \times \frac{3}{16} \times 10\\ &=\frac{73575}{4}watt \end{aligned}

Horse power of pump = \frac{\text{Power required}}{746 \times \text{efficiency}}=\frac{73575}{4 \times 746 \times 0.8}=30.82hp

\begin{aligned} Delta (\Delta )&=144 cm=1.44m\\ Duty(D)&=\frac{8.64B}{\Delta }=\frac{8.64 \times 120}{1.44}\\ D&=720\frac{hec}{cumec}\\ Q&=\frac{Area}{D}=\frac{108}{720}=\frac{3}{20}m^3/s\\ Q_{applied}&=\frac{Q}{\eta _a}=\frac{3}{20 \times 0.8}=\frac{3}{16}\\ \text{Power required}&=\frac{mgh}{t}\\ &=\rho _w \times \left ( \frac{volume}{t} \right )g \times h\\ &=\rho _w Qgh=\gamma Qh\\ &=9810 \times \frac{3}{16} \times 10\\ &=\frac{73575}{4}watt \end{aligned}

Horse power of pump = \frac{\text{Power required}}{746 \times \text{efficiency}}=\frac{73575}{4 \times 746 \times 0.8}=30.82hp

Question 3 |

A hydraulic jump takes place in a 6 m wide rectangular channel at a point where
the upstream depth is 0.5 m (just before the jump). If the discharge in the
channel is 30 m^3/s and the energy loss in the jump is 1.6 m, then the Froude
number computed at the end of the jump is ___________. (round off to two
decimal places)
(Consider the acceleration due to gravity as 10 m/s^2.)

0.40 | |

0.85 | |

0.65 | |

0.75 |

Question 3 Explanation:

Q=30m^3/sec

B=6m

y_1=0.5m

E_L=1.6m

q= \frac{Q}{B} =\frac{30}{6}=5m^3/sec/m

We know that,

\begin{aligned} E_L=\frac{(y_2-y_1)^3}{4y_1y_2}&=1.6m\\ \frac{(y_2-0.5)^3}{4 \times 0.5 \times y_2}&=1.6\\ y_2^3-1.5y_2^2+0.75y_2-0.125&=3.2y_2\\ y_2=2.5m,-0.0527m,&-0.0947m \end{aligned}

Hence, y_2=2.5m

Post jump Froude's No.

(F_2)=\frac{V_2}{\sqrt{\sqrt{gy_2}}}=\frac{\left ( \frac{30}{6 \times 2.5} \right )}{\sqrt{10 \times 2.5}}=0.4

Question 4 |

Water is flowing in a horizontal, frictionless, rectangular channel. A smooth
hump is built on the channel floor at a section and its height is gradually increased
to reach choked condition in the channel. The depth of water at this section is y_2 and that at its upstream section is y_1. The correct statement(s) for the choked and
unchoked conditions in the channel is/are

In choked condition, y_1 decreases if the flow is supercritical and increases if the flow is subcritical. | |

In choked condition, y_2 is equal to the critical depth if the flow is supercritical or subcritical. | |

In unchoked condition, y_1 remains unaffected when the flow is supercritical or subcritical. | |

In choked condition, y_1 increases if the flow is supercritical and decreases if the flow is subcritical. |

Question 4 Explanation:

Question 5 |

The dimension of dynamic viscosity is:

ML^{-1}T^{-1} | |

ML^{-1}T^{-2} | |

ML^{-2}T^{-2} | |

ML^{0}T^{-1} |

Question 5 Explanation:

Unit of dynamic viscosity =\frac{kg}{m.s} \; or \; \frac{Ns}{m^2}=ML^{-1}T^{-1}

Question 6 |

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?

\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?

(P)-(IV), (Q)-(III), (R)-(V), (S)-(I), (T)-(II) | |

(P)-(III), (Q)-(IV), (R)-(V), (S)-(I), (T)-(II) | |

(P)-(IV), (Q)-(III), (R)-(II), (S)-(I), (T)-(V) | |

(P)-(III), (Q)-(IV), (R)-(I), (S)-(V), (T)-(II) |

Question 7 |

Depth of water flowing in a 3 m wide rectangular channel is 2 m. The channel carries a discharge of 12 m^3/s . Take g = 9.8 m/s ^2.

The bed width (in m) at contraction, which just causes the critical flow, is _________ without changing the upstream water level. (round off to two decimal places)

The bed width (in m) at contraction, which just causes the critical flow, is _________ without changing the upstream water level. (round off to two decimal places)

2.85 | |

4.25 | |

2.15 | |

1.55 |

Question 7 Explanation:

Given: B = 3m, y = 2m, Q = 12 m^3
/sec

Velocity(v)=\frac{Q}{A} =\frac{12}{2 \times 3}=2m/sec

Specific energy at section (1-1)

E=y+\frac{v^2}{2g}=2+\frac{2^2}{2 \times 9.8}=\frac{108}{49}m

When channel section is contracted to minimum width and for constant discharge Q, the flow over contracted section will be critical flow and under the assumption that no energy loss has taken place.

E=E_c=\frac{108}{49}

We have that, E_c=\frac{3}{2}y_c (for rectangular crosssection)

y_c=\frac{2}{3}E_c=\frac{2}{3} \times \frac{108}{49}=\frac{72}{49}

For critical flow condition,

\begin{aligned} \frac{Q^2T}{gA^3}&=1\\ \frac{Q^2B_{min}}{g(B_{min} \times y_c)^3}&=1\\ (B_{min})^2&=\frac{Q^2}{gy_c^3}\\ B_{min}&=\left ( \frac{(12)^2}{9.8 \times \left ( \frac{72}{49} \right )^3} \right )^{1/2}\\ &=2.152m \end{aligned}

Velocity(v)=\frac{Q}{A} =\frac{12}{2 \times 3}=2m/sec

Specific energy at section (1-1)

E=y+\frac{v^2}{2g}=2+\frac{2^2}{2 \times 9.8}=\frac{108}{49}m

When channel section is contracted to minimum width and for constant discharge Q, the flow over contracted section will be critical flow and under the assumption that no energy loss has taken place.

E=E_c=\frac{108}{49}

We have that, E_c=\frac{3}{2}y_c (for rectangular crosssection)

y_c=\frac{2}{3}E_c=\frac{2}{3} \times \frac{108}{49}=\frac{72}{49}

For critical flow condition,

\begin{aligned} \frac{Q^2T}{gA^3}&=1\\ \frac{Q^2B_{min}}{g(B_{min} \times y_c)^3}&=1\\ (B_{min})^2&=\frac{Q^2}{gy_c^3}\\ B_{min}&=\left ( \frac{(12)^2}{9.8 \times \left ( \frac{72}{49} \right )^3} \right )^{1/2}\\ &=2.152m \end{aligned}

Question 8 |

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)

(Consider only loss due to friction and neglect all other losses. Assume the friction factor to be the same for both the pipes)

2.25 | |

6.32 | |

4.22 | |

5.66 |

Question 8 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 9 |

A rectangular channel with Gradually Varied Flow (GVF) has a changing bed
slope. If the change is from a steeper slope to a steep slope, the resulting GVF
profile is

S_3 | |

S_1 | |

S_2 | |

either S_1 or S_2 , depending on the magnitude of the slopes |

Question 9 Explanation:

Question 10 |

With respect to fluid flow, match the following in Column X with Column Y:

\begin{array}{|l|l|}\hline \text{Column X}& \text{Column Y}\\ \hline \text{(P) Viscosity} & \text{(I) Mach number}\\ \hline \text{(Q) Gravity}&\text{(II) Reynolds number}\\ \hline \text{(R) Compressibility}&\text{(III) Euler number}\\ \hline \text{(S) Pressure} &\text{(IV) Froude number}\\ \hline \end{array}

Which one of the following combinations is correct?

\begin{array}{|l|l|}\hline \text{Column X}& \text{Column Y}\\ \hline \text{(P) Viscosity} & \text{(I) Mach number}\\ \hline \text{(Q) Gravity}&\text{(II) Reynolds number}\\ \hline \text{(R) Compressibility}&\text{(III) Euler number}\\ \hline \text{(S) Pressure} &\text{(IV) Froude number}\\ \hline \end{array}

Which one of the following combinations is correct?

(P) - (II), (Q) - (IV), (R) - (I), (S) - (III) | |

(P) - (III), (Q) - (IV), (R) - (I), (S) - (II) | |

(P) - (IV), (Q) - (II), (R) - (I), (S) - (III) | |

(P) - (II), (Q) - (IV), (R) - (III), (S) - (I) |

Question 10 Explanation:

Reynold's number (R_e) is defined when apart from inertial force, viscous forces are dominant.

R_e=\frac{\text{Inertial force}}{\text{Viscous force}}

Froude?s number (F_e): It is used when in addition to inertial force, gravity forces are important.

F_e=\frac{\text{Inertial force}}{\text{Gravity force}}

Euler number (E_u): It is used when apart from inertial force, only pressure forces are dominant.

E_u=\frac{\text{Inertial force}}{\text{Pressure force}}

Mach number (M): It is used when in addition to inertial force, compressibility forces are dominant

M=\frac{\text{Inertial force}}{\text{Elastic force}}

R_e=\frac{\text{Inertial force}}{\text{Viscous force}}

Froude?s number (F_e): It is used when in addition to inertial force, gravity forces are important.

F_e=\frac{\text{Inertial force}}{\text{Gravity force}}

Euler number (E_u): It is used when apart from inertial force, only pressure forces are dominant.

E_u=\frac{\text{Inertial force}}{\text{Pressure force}}

Mach number (M): It is used when in addition to inertial force, compressibility forces are dominant

M=\frac{\text{Inertial force}}{\text{Elastic force}}

There are 10 questions to complete.