Question 1 |
The figure shows two fluids held by a hinged gate. The atmospheric pressure is P_a = 100 kPa. The moment per unit width about the base of the hinge is
________ kNm/m. (Rounded off to one decimal place)
Take the acceleration due to gravity to be g = 9.8 m/s^2 .
Take the acceleration due to gravity to be g = 9.8 m/s^2 .
45.8 | |
57.2 | |
68.2 | |
87.6 |
Question 1 Explanation:
Forces from Pressure diagram
\mathrm{F}_{1}= Area of \triangle \mathrm{ADE} \times width
F_{1}=\frac{1}{2} \times A D \times D E \times width
=\frac{1}{2} \times 1 \times 9810 \times 1
\mathrm{F}_{1}=4905 \mathrm{~N}
\mathrm{F}_{2}= Area of rectangle \mathrm{DEBF} \times width
F_{2}=2 \times 9810 \times 1
\mathrm{F}_{2}=19620 \mathrm{~N}
\mathrm{F}_{3}= Area of triangle \Delta \mathrm{EFC}
=\frac{1}{2} \times \mathrm{EF} \times \mathrm{FC} \times width
=\frac{1}{2} \times 2 \times 39240 \times 1
\mathrm{F}_{3}=39240 \mathrm{~N}
Location of forces from top
\begin{aligned} & \mathrm{F}_{1} \rightarrow \frac{2}{3} \times 1=\frac{2}{3} \mathrm{~m} \\ & \mathrm{~F}_{2} \rightarrow\left(1 \mathrm{~m}+\frac{2 \mathrm{~m}}{2}\right)=2 \mathrm{~m} \\ & \mathrm{~F}_{3} \rightarrow 1 \mathrm{~m}+2 \mathrm{~m} \times \frac{2}{3}=\frac{7}{3} \mathrm{~m} \end{aligned}
Pressure at depth 1 \mathrm{~m}=\rho_{1} \mathrm{gh}_{1}
1000 \times 9.81 \times 1=9810 \frac{N}{m^{2}}
\mathrm{DE}=9810 \frac{\mathrm{N}}{\mathrm{m}^{2}}
BF=9810 \frac{N}{m^{2}}
Pressure at depth 3 m
\begin{aligned} & \rho_{1} g h_{1}+\rho_{2} g h_{2} \\ & 9810+2000 \times 9.81 \times 2 \\ & 9810+39240=49050 \frac{\mathrm{N}}{\mathrm{m}^{2}} \end{aligned}
(B F)+(F C)=B C
Location of forces from bottom hinge
\begin{aligned} & \mathrm{F}_{1} \rightarrow 3 \mathrm{~m}-\frac{2}{3} \mathrm{~m}=\frac{7}{3} \mathrm{~m} \\ & \mathrm{~F}_{2} \rightarrow 3 m-2 m=1 \mathrm{~m} \\ & \mathrm{~F}_{3} \rightarrow 3 m-\frac{7}{3} \mathrm{~m}=\frac{2}{3} \mathrm{~m} \end{aligned}
Moment about hinge \mathrm{F}_{1} \times \frac{7}{3}+\mathrm{F}_{2} \times 1+\mathrm{F}_{3} \times \frac{2}{3}
4905 \times \frac{7}{3}+19620+39240 \times \frac{2}{3}
\Rightarrow \quad 57225 \mathrm{Nm}=57.22 \mathrm{kNm} / \mathrm{m}
Question 2 |
A uniform wooden rod (specific gravity = 0.6,
diameter = 4 cm and length = 8 m) is immersed
in the water and is hinged without friction at point
A on the waterline as shown in the figure. A solid
spherical ball made of lead (specific gravity = 11.4)
is attached to the free end of the rod to keep the
assembly in static equilibrium inside the water.
For simplicity, assume that the radius of the ball is
much smaller than the length of the rod.
Assume density of water = 10^3 kg/m^3 and \pi = 3.14.
Radius of the ball is _______ cm (round off to 2 decimal places).
Assume density of water = 10^3 kg/m^3 and \pi = 3.14.
Radius of the ball is _______ cm (round off to 2 decimal places).
2.42 | |
3.62 | |
8.26 | |
3.59 |
Question 2 Explanation:
As the system is in equilibrium,
\Sigma M_o=0
(F_{B1}-W_1) \times \frac{L}{2} \cos \theta +F_{B2}-W_2) \times L \times \cos \theta =0
(\rho gV_1-\rho _1gV_1) \times \frac{1}{2}+(\rho gV_2-\rho _2gV_2) =0
(\rho -\rho _1) \times g \times \frac{\pi d^2L}{4} \times \frac{1}{2}+(\rho -\rho _2) \times \frac{4}{3} \times \\pi R^3 \times g=0
(100-600) \times \frac{0.04^2 \times 8}{8} +(1000-11400) \times \frac{4}{3} \times R^3=0
\Rightarrow R=0.03587m=3.59cm
Question 3 |
Consider a steady flow through a horizontal
divergent channel, as shown in the figure, with
supersonic flow at the inlet. The direction of flow is
from left to right.
Pressure at location B is observed to be higher than that at an upstream location A. Which among the following options can be the reason?
MSQ
Pressure at location B is observed to be higher than that at an upstream location A. Which among the following options can be the reason?
MSQ
Since volume flow rate is constant, velocity at
B is lower than velocity at A | |
Normal shock | |
Viscous effect | |
Boundary layer separation |
Question 3 Explanation:
If the supersonic flow enters to the diverging duct,
it will act as nozzle the pressure of the flow is
decreases in the nozzle. But it is given the pressure
of the flow at B is more than that A
It will happen only with the development of normal shoot in the diverging flow (the normal shock wave is the characteristic of only supersonic flow) Because of normal shock wave in the diverging nozzle pressure increases and velocity decreases. But mass flow rate remains unchanged.
It will happen only with the development of normal shoot in the diverging flow (the normal shock wave is the characteristic of only supersonic flow) Because of normal shock wave in the diverging nozzle pressure increases and velocity decreases. But mass flow rate remains unchanged.
Question 4 |
A solid spherical bead of lead (uniform density =11000\; kg/m^3 ) of diameter d = 0.1 mm sinks with
a constant velocity V in a large stagnant pool of a
liquid (dynamic viscosity =1.1 \times 10^{-3} kg.m^{-1}.s^{-1} ).
The coefficient of drag is given by C_D=\frac{24}{Re}, where
the Reynolds number (Re) is defined on the basis of
the diameter of the bead. The drag force acting on
the bead is expressed as D=(C_D)(0.5 \rho V^2)\left ( \frac{\pi d^2}{4} \right ), where \rho is the density of the liquid. Neglect the
buoyancy force. Using g = 10 m/s^2, the velocity V
is __________ m/s.
\frac{1}{24} | |
\frac{1}{6} | |
\frac{1}{18} | |
\frac{1}{12} |
Question 4 Explanation:
As buoyancy force is neglected we can neglect
density of fluid ( ) in the following equation.
\begin{aligned} V&=\frac{1}{18\mu _f}(\rho _b-\rho _f)gd^2\\ &=\frac{11000 \times 10 \times (10^{-4})^2}{18 \times 1.1 \times 10^{-3}}\\ &=\frac{1}{18}m/s \end{aligned}
\begin{aligned} V&=\frac{1}{18\mu _f}(\rho _b-\rho _f)gd^2\\ &=\frac{11000 \times 10 \times (10^{-4})^2}{18 \times 1.1 \times 10^{-3}}\\ &=\frac{1}{18}m/s \end{aligned}
Question 5 |
The figure shows a purely convergent nozzle with
a steady, inviscid compressible flow of an ideal gas
with constant thermophysical properties operating
under choked condition. The exit plane shown in
the figure is located within the nozzle. If the inlet
pressure (P_0) is increased while keeping the back
pressure (P_{back}) unchanged, which of the following
statements is/are true?
MSQ
MSQ
Mass flow rate through the nozzle will remain unchanged. | |
Mach number at the exit plane of the nozzle will remain unchanged at unity | |
Mass flow rate through the nozzle will increase. | |
Mach number at the exit plane of the nozzle will become more than unity. |
Question 5 Explanation:
\begin{aligned}
\frac{\dot{m}}{A}&=\sqrt{\frac{\gamma }{R}}\frac{P_0}{\sqrt{T_0}}\frac{1}{\left ( \frac{\gamma +1}{2} \right )^{\frac{\gamma +1}{2(\gamma -1)}}}\\
\frac{\dot{m}}{A}&=0.685 \times \sqrt{P_0\rho _0} \;\;\;...\text{ for Air}\\
\frac{T_c}{T_0}&=\frac{T^*}{T_0}=\frac{2}{\gamma +1}=0.8333\\
&(\text{critical temp ratio})\\
\frac{P_c}{P_0}&=\frac{P^*}{P_0}=\left ( \frac{2}{\gamma +1} \right )^{\frac{\gamma }{\gamma -1}}=0.5282\\
&(\text{critical pressure ratio})
\end{aligned}
where \gamma= Adiabatic index = 1.4
If P_0 increases , mass flow rate increases. Mach number at the exit plane of the nozzle will remain unchanged at unity.
Since the maximum expected velocity from the converging nozzle is sound velocity.
where \gamma= Adiabatic index = 1.4
If P_0 increases , mass flow rate increases. Mach number at the exit plane of the nozzle will remain unchanged at unity.
Since the maximum expected velocity from the converging nozzle is sound velocity.
There are 5 questions to complete.
IN Q.1. ITS BOUSINESSQ THEORY NOT REYNOLD’S…PLS CONFIRM AND RECTIFY!
I have also the same doubt , but everywhere the answer given is Reynolds forces ,
Yes bro u are ✅️ right
in question 19 diameter is 100 mm not 10 mm please correct it
We have updated question in GATE Mechanical 2017 Set-2.