heat transfer for gate

 

Hello Aspirants!

This article is designed specifically for aspirants of GATE ME 2017. Based on our experts’ analysis of the previous GATE papers, we have devised this article which will highlight the important sub-topics of Heat Transfer which the aspirants must focus on to excel in their preparation.Also questions from GATE has been discussed.

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Important Topics in Heat Transfer for GATE 2017

  • Generally, 5-10 marks are allotted to this section.
  • You should thoroughly study the topics like Conduction, Convection and Radiation.
  • In Radiation, you must focus on topics like shape factor and heat exchange between non- black bodies
  • Problems on Conduction and Critical thickness are frequently covered in the exam. You must thoroughly prepare them
  • You must thoroughly prepare topics like Unsteady Conduction (Lumped Heat Analysis) as this is very important from the examination perspective.

must read : How to Prepare Refrigeration and Air- Conditioning For GATE|| important topics

  • You must prepare Problems in Heat Exchangers like LMTD and NTU.
  • Numbers in convection like Pr No, Gr No, Re No, etc carry maximum weight age from GATE perspective
  • You must possess brief knowledge of other topics like Efficiency Calculation, Biot Number, Prandtl Number, Heat transfer through slabs, shells, cylinders and condensers, etc.

 

Solved Examples of Heat Transfer from GATE

 Question :  An amount of 100 kW of heat is transferred through a wall in steady state. One side of the

wall is maintained at 127OC and the other side at 27OC. The entropy generated (in W/K) due

to the heat transfer through the wall is _______

Answer :    80 to 85

Solution :       ∆S1 = Q/T1                                                                                                                                            

∆S2 = Q/T2                                                                                          

(∆S)generated  = ∆S1 + ∆S2                                                                                   

= Q/400 -  Q/300                                                                                  

= (100 x 103/100) (1/4 – 1/3)

= 103 X – 0.0833                                                                                                                                          

= - 83.33 W/K

Entropy Generated = 83.33 W/K                                                                                    

 

     

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 Question : A double-pipe counter-flow heat exchanger transfers heat between two water streams. Tube side water at 19 liter/s is heated from 10OC to 38OC. Shell side water at 25 liter/s is entering at 46OC. Assume constant properties of water, density is 1000 kg/m and specific heat is 4186 J/kg K. The LMTD (in OC) is ________

Answer     10.8 to 11.2

Explanation :                    

                Given mh = 25 L/S, Th,i = 46º C, Th,o = ?                      

            mc = 25 L/S, Tc,i = 10º C , Tc,o = 38º C                       

            Density = 1000 Kg/m3, C= 4186 J/Kg.K

           

Energy Balance

mc C (Tc,o – Tc,i) = mn C (Th,i – Th,o)                                                                                                           

19 (38-10) = 25 (46 – Th,o)

Th,o = 24.72º C

LMTD = ϴ1 - ϴ2 /ln ϴ12                                                                                           

Tc,o

ϴ1 = 46 – 38 = 8º C

ϴ2 = 24.72 – 10 = 14.72º C

LMTD = 8 – 14.72 /ln(8/14.72)

= 11.0206º C

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