# GATE ME 2011

 Question 1
A streamline and an equipotential line in a flow field
 A Are parallel to each other B Are perpendicular to each other C Intersect at an acute angle D Are identical
Fluid Mechanics   Fluid Kinematics
Question 1 Explanation:
Stream line and equipotential line in a flow field are perpendicular to each other
 Question 2
If a mass of moist air in an airtight vessel is heated to a higher temperature, then
 A Specific humidity of the air increases B Specific humidity of the air decreases C Relative humidity of the air increases D Relative humidity of the air decreases
Refrigeration and Air-conditioning   Properties of Moist Air
Question 2 Explanation:
The moist air in an airtight vessel is heated to a higher temperature. This process is treated as sensible heating. In sensible heating process, the relative humidity of the air decreases with increase in dry bulb temperature at constnat specific humidity
This process is shown on psychrometic chart in Fig. Question 3
In a condenser of a power plant, the steam condenses at a temperature of $60^{\circ}$C. The cooling water enters at $30^{\circ}$C and leaves at $45^{\circ}$C. The logarithmic mean temperature difference (LMTD) of the condenser is
 A $16.2^{\circ}C$ B $21.6^{\circ}C$ C $30^{\circ}C$ D $37.5^{\circ}C$
Heat Transfer   Heat Exchanger
Question 3 Explanation: \begin{aligned} \therefore \quad \mathrm{LMTD} &=\frac{\Delta T_{1}-\Delta T_{2}}{\ln \left(\frac{\Delta T_{1}}{\Delta T_{2}}\right)} \\ &=\frac{30-15}{\ln (2)}=21.64^{\circ} \mathrm{C} \end{aligned}
 Question 4
A simply supported beam PQ is loaded by a moment of 1kN-m at the mid-span of the beam as shown in the figure. The reaction forces $R_{P}$ and $R_{Q}$ at supports P and Q respectively are A 1kN downward, 1kN upward B 0.5kN upward, 0.5kN downward C 0.5kN downward, 0.5kNupward D 1kN upward, 1kN upward
Strength of Materials   Bending of Beams
Question 4 Explanation: Let us assume the direction of $R_{P}$ and $R_{Q}$ as ,
upward. Now,
$\therefore \quad R_{P}+R_{Q}=0$
$\therefore$Taking moment about point Q we get:
\begin{aligned} R_{P} \times 1+1 &=0 \\ R_{P} &=-1 \mathrm{kN}\\ \therefore R_{Q}&=+1 \mathrm{kN} \end{aligned}
since our assume direction of $R_{p}$ is wrong therefore
$R_{P}=1 \mathrm{kN}$ which act in the down was direction
and $R_{Q}=1 \mathrm{kN}$ acting in upward direction.
 Question 5
A double - parallelogram mechanism is shown in the figure. Note that PQ is a single link. The mobility of the mechanism is A -1 B 0 C 1 D 2
Theory of Machine   Planar Mechanisms
Question 5 Explanation:
This is a special case because the link between & parallel to P and Q is redundant as its presence does not effect the mechanism. Hence the DOF is 1.

There are 5 questions to complete.