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GATE 2014 previous year Thermodynamics solved questions of set 4
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In this lesson we will discuss GATE 2014 previous year Thermodynamics solved questions of set 4

Rutuja Bothara
Strengthen GATE /PSU/ESE/SSC JE/RRB JE /campus Placements preparation. A passionate mechanical engineer -VIT University

U
Unacademy user
mam can you upload these everyday strategy, so that we can do with you. simultaneously.
Sweta Kumari
4 months ago
No i cannot promise you...therefore i have not given any statements for courses to create on these topics...if you will have any probem with some topics then i aill create course on that
  1. GATE Previous years Thermodynamics questions GATE 2014 Thermodynamics solved questions set 4 By Rutuja Bothara


  2. RUTUJA BOTHARA . Perusing B.Tech in Mechanical Engineering from Vellore Institute of Technology, Vellore. (2015-2019) e Cleared JEE Mains 2015 Cleared VITEEE-2015 with AIR 8718 Cracked HITSEEE-2015 with AIR 91 . Worked as a Career Counsellor in Teachers Academy (2016- 2018) and BOLO app. https://unacademy.com/user/rurutujajb96- 9511


  3. A reversed Carnot cycle refrigerator maintains a temperature of-5 C. The ambient air temperature is 35 C. The heat gained by the nefrigerator at a continuous rate is 2.5 kl/s. The power (in watt) required to pump this heat out continuously is


  4. cop-12 268 Ti-T2308-268 CoP 67 CoP2 6,7 = 2.5x103 T2 26 8K


  5. A closed system contains 10 kg of saturated liquid ammonia at 10 C. Heat addition required to convert the entire liquid into saturated vapour at a constant pressure is 16.2 MJ. If the entropy of the saturated liquid is 0.88 k]/kg.K, the entropy (in kJ/kg.K) of saturated vapour is



  6. .Two identical metal blocks L and M (specific heat - 0.4 k]/kg.K), each having a mass of 5 kg, are initially at 313 K. A reversible refrigerator extracts heat from block L and rejects heat to block M until the temperature of block L reaches 293 K. The final temperature (in K) of block M is


  7. 2. univ -o- -) reversible P rocess Ti 2- 1 2 Tf 33.36K


  8. Steam with specific enthalpy (h) 3214 k)J/kg enters an adiabatic turbine operating at steady state with a flow rate 10 kg/s. As it expands, at a point where h is 2920 kj/kg, 1.5 kg/s is extracted for heating purposes. The remaining 8:5 kg/s further expands to the turbine exit, where h 2374 k]J/kg. Neglecting changes in kinetic and potential energies, the net power output (in kW) of the turbine is


  9. Net poee output 03214-29208 (29 20-2374


  10. 3 2 NV3 Proc ek8 1-2 , 0 V 2. Pi V2.


  11. Thank You For watching!! RATE REVIEW RECOMMEND