Question 1 |
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 1 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}}
Question 2 |
A venturimeter as shown in the figure (not to scale) is connected to measure the flow of water in a vertical pipe of 20 cm diameter.
Assume g=9.8 \mathrm{~m} / \mathrm{s}^{2}. When the deflection in the mercury manometer is 15 cm, the flow rate (in lps, round off to two decimal places) considering no loss in the venturimeter is ___________
Assume g=9.8 \mathrm{~m} / \mathrm{s}^{2}. When the deflection in the mercury manometer is 15 cm, the flow rate (in lps, round off to two decimal places) considering no loss in the venturimeter is ___________
49.4 | |
23.36 | |
87.64 | |
68.22 |
Question 2 Explanation:
\begin{aligned} \text { Discharge }(Q) &=C_{d} \frac{A_{1} A_{2}}{\sqrt{A_{1}^{2}-A_{2}^{2}}} \sqrt{2 g h} \\ h &=X\left(\frac{\rho_{m}}{\rho}-1\right)=0.15\left(\frac{13.6 \times 10^{3}}{10^{3}}-1\right) \\ &=1.89 \mathrm{~m}\\ Q &=\frac{A_{1} A_{2}}{A_{2} \sqrt{\left(\frac{A_{1}}{A_{2}}\right)^{2}-1}} \times \sqrt{2 \times 9.8 \times 1.89} \\&= \frac{\frac{\pi}{4}(0.2)^{2}}{\sqrt{(2)^{4}-1}} \times \sqrt{2 \times 9.8 \times 1.89} \\&= 49.395 \mathrm{l} / \mathrm{s} \simeq 49.40 \mathrm{l} / \mathrm{s} \end{aligned}
Question 3 |
A fire hose nozzle directs a steady stream of water of velocity 50 m/s at an angle of 45^{\circ} above the horizontal. The stream rises initially but then eventually falls to the ground. Assume water as incompressible and inviscid. Consider the density of air and the air friction as negligible, and assume the acceleration due to gravity as 9.81 \mathrm{~m} / \mathrm{s}^{2}. The maximum height (in m,round off to two decimal places) reached by the stream above the hose nozzle will then be _________
63.71 | |
56.21 | |
98.36 | |
88.24 |
Question 3 Explanation:
As we know that
\begin{aligned} {V}^{2}-u^{2}&=2 a S\\ \text{In vertical direction} (\uparrow)\\ 0^{2}-\left(50 \sin 45^{\circ}\right)^{2} &=2(-9.81) h_{\max } \\ h_{\max } &=\frac{\left(50 \sin 45^{\circ}\right)^{2}}{2(9.81)}=63.71 \mathrm{~m} \end{aligned}
Question 4 |
A fluid flowing steadily in a circular pipe of radius R has a velocity that is everywhere parallel to the axis (centerline) of the pipe. The velocity distribution along the radial direction is V_r=U\left ( 1-\frac{r^2}{R^2} \right ), where r is the radial distance as
measured from the pipe axis and U is the maximum velocity at r=0. The average velocity of the fluid in the pipe is
\frac{U}{2} | |
\frac{U}{3} | |
\frac{U}{4} | |
\frac{5}{6}U |
Question 4 Explanation:
\begin{aligned} u&=U\left ( 1-\frac{r^2}{R^2} \right ) \\ \dot{m}&=\int_{0}^{R} \rho (2 \pi r\; dr)u\\ &= 2 \pi \rho U\int_{0}^{R} \left ( 1-\frac{r^2}{R^2} \right ) r\; dr\\ \rho (\pi R^2)\bar{u} &= 2 \pi \rho U \left ( \frac{R^2}{2}-\frac{R^4}{R^2 \times 4} \right )\\ \bar{u}&= \frac{U}{2}\\ \bar{u} &= \text{Mean velocity} \\ U&=\text{Max velocity} \end{aligned}
Question 5 |
A cast iron pipe of diameter 600 mm and length 400 m carries water from a tank and
discharges freely into air at a point 4.5 m below the water surface in the tank. The friction
factor of the pipe is 0.018. Consider acceleration due to gravity as 9.81 m/s^2. The velocity
of the flow in pipe (in m/s, round off to two decimal places) is __________.
2.56 | |
1.52 | |
4.12 | |
6.82 |
Question 5 Explanation:
Apply energy equation between (1) and (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_f\\ 4.5&=\frac{flV^2}{2gD}+0.5\frac{V^2}{2g}+\frac{V^2}{2g} \\ 4.5&=\frac{(0.018)(400)V^2}{2(9.81)(0.6)}+\frac{1.5V^2}{2g} \\ 4.5&=\frac{12V^2}{2g}+\frac{1.5V^2}{2g} \\ V^2&=6.54 \\ V&=2.557 m/s\simeq 2.56m/s \end{aligned}
There are 5 questions to complete.
In question no 7, you are adding the wrong diagram means not according to the question
Kindly Correct the fig of question no 7. Point Q,R & S are not shown.
Thanks