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Mechanical Engineering Job Interview Questions & Answers by Funda of Mechanical Engineering.
Mechanical Engineering Job Interview Questions & Answers by Funda of Mechanical Engineering.
Description of Questions Remarks
Q1.0: Convert 1 TR to BTU/Hr & Kcal/Hr - MCQ Type
Ans1.0
Since we know 1 TR = 3.51 KW …………………..(1)
We also Know 1 J/S = 1 Watt & 1 Kcal = 4182 Joule
So the 1 Joule = 1/4182 Kcal ………………………(2)
So the 1 TR = 3.51 KW = 3.51x1000 Watt = 3.51x 1000x(1/4182)x3600 Kcal/Hr
Finally 1 TR = 3020.07 Kcal/Hr
Q 2.0: 25 Pascal = How much mm Water Gauge ( mm WG) Pressure !
Ans2.0
We Know 1 atm = 10.33 m of water gauge = 10.33x1000 mmWG
Also 1 atm = 760 mm of Hg = 1.013x10⁵ N/m² = 1.013x10⁵ Pascal
Now 1 Pascal = 1/1.013x10⁵ atm
1 Pascal = (10.33x10^3)/(1.013x10⁵) = 0.10197 mmWG
So the 25 Pascal = 25 x 010197 mm WG = 2.5493 mmWG
Q3.0: What is the weight of 40 litre of Water in Kg ? MCQ Type
Ans3.0
Since the density of water at 4 degree centigrade is 1000 Kg/m3, So the here itself we are calculating the weight of water in Kg by taking the specific weight of water.
Description of Questions Remarks
Q4.0: What will be the 1 Air Change Per Hour (ACPH) for a cubical room of
25 Meter of side dimension ?
Ans4.0
Volume of the Room = 25 x 25 x 25 = 15625 m3
Volume of the Room = 25 x 25 x 25 = 15625 x 35.00 Cubic ft = 546875 Cubic feet
So the 1 ACPH = (Volume in Cubic feet/60) = (546875/60) = 9114 CFM
Q5.0: An Air stream at 30° C mixes with another stream of air (equal mass)
at 20° C. What is the mixing temperature ?
Ans5.0:
We know Air mixing Equation is as follow-
T mix air = ( CFM1 X Tm1 + CFM2 x Tm2)/(CFM1 + CFM2)
Since the air quantity of equal mass So CFM1 = CFM2
Temperature of Mixed Stream = (Tm1+Tm2)/2 = (20+30)/2 = 25 ° C
Q6.0: Find the mixture condition of two equal mass of air. ( Use P-Chart)
Condition -1: DBT = 80 Degree F And Staturated, Condition 2: DBT =
80 Degree F and Dew point temp = 60 Degree F .
Ans6.0:
Since the mass of the air stream are equal = m₁=m₂
Temperature of the Mixture (Tm ) = (m1T1+m2T2)/(m1+m2)
Tm = ( 80 + 80 ) /2 = 80° F ------------------------(a)
Humidity of the Mixture (Hm) = (m1H1+m2H2)/(m1+m2)
Hm = ( 100+30) /2 = 65% ------------------------(b)
Similarly, a question No- 7 can also be as
Description of Questions Remarks
Q7.0: Find the Mixture Conditions of two air stream at given two
conditions. Mass of first air stream is 10 times the mass of second air
stream ?
Ans7.0: Consider m2 = m so m1= 10m and solve from earlier tricks.
Q8.0: Find the Cooling Load on Cooling Coil at given following conditions (In
TR). Use P=Chart
Air Entering - 10000 CFM, DBT= 85°F, WBT=70°F
Air Leaving - 10000 CFM, DBT= 53°F, RH = 88%
Ans8.0
Using above given two conditions, we can find enthalpy from
Psychrometric Chart
We will find from P-chart,
h1 = 34 Btu/lba of dry air
h2 = 21 Btu/lba of dry air
so the ΔH= h1-h2 = 13 btu/lba of dry air
also from P-Chart, Taken Specific Volume (v) = 14 cubic ft /lba
also we know 1 TR = 12000 Btu/Hr
So the Load on the Cooling Coil in TR = (CFM x 60 x ΔH) / (Specific
Volume v) TR
TR Load = (10000 x 60 x 13 )/(14 x 12000) TR
TR Load = 46.42 TR ------------Load on Cooling Coil
Q9.0: Sun rays fall on the following surfaces of equal mass. Which surface
temperature will be highest to las temperature ? Materials are
Glass,Wood,Silver,Black Iron, Concrete, water.
Ans9.0:
Since the mass of all the components are equal so the temperature
raise will only depend on the specific heat capacity of the material.
Q = mCpΔT --------------Heat Generated in the body
Where Cp is the specific heat capacity of the material, is the amount
of heat required to raise the temperature of body by 1°C
From Above equation we can say that the Specific heat capacity Cp is
inversally propotional to raise in tempearature of body.
Silver (0.23) > Black Iron (.451) > Glass (.84) > Concrete (.88)>Wood
(1.76)>Water (4.184), Where value in the ( ) represents the Specific
Heat Capacity in J/g ° C for above material body.
Descriptive Type but in Interview, Interviewer can ask for the trick to read P-Chart
Description of Questions Remarks
Q10: Can High Pressure Compressed air be divided into two different
streams - one is hot and other is cold ?
What is the name of device which can achieve this ?
Ans10.0
Division of two different air streams - one is hot and other is cold
from a single high pressure compressed air stream is called vorticity.
Vorticity is a pseudo vector field that describes the local spinning
motion of continuum near some point (the tendency of something to
rotate ), as would be seen by observer located at that point and
travelling along with the flow.
The Vortex tube also known as the "Ranque Hilsch Vortex tube "is a
mechanical device that separate a compressed gas into hot and cold
streams. The gas emerging from hot end can reach upto 200°C and
the gas emerging from the cold end can reach upto -50°C.
Regards
Deepak Sharma
Funda of Mechanical Engineerig Team
Monday, 19 October 2020
Top HVAC Technologies around the world in Modern Age
Hi friends today I want to introduce the top latest technology in HVAC around the world. This is absolutely interesting and mesmerizing for your learning in HVAC for both who has industrial experience and also who is persuing refrigeration and airconditioning engineering or mechanical engineering either BTech / Diploma.
So without delay lets start the watching of the videos.
Regards
Deepak Sharma
Funda of Mechanical Engineering
https://www.youtube.com/watch?v=sfRadKxZhx8&t=165s
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Hi friends today I want to introduce the top latest technology in HVAC around the world. This is absolutely interesting and mesmerizing for ...
