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GATE 2013 all thermodynamic questions solutions Hindi
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In this lesson we will discuss all GATE 2013 thermodynamic questions solutions

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

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amazing mcqs.....
1. GATE Previous years Thermodynamics questions GATE 2013 Thermodynamics solved questions set 1 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 cylinder contains 5 m 3 of an ideal gas at a pressure of 1 bar. This gas is compressed ina reversible isothermal process till its pressure increases to 5 bar. The work in kJ required for this process is (A) 804.7 (B) 953.2 (C) 981.7 (D) 1012.2

4. For tever sible isotheemal Process Pa

5. Specific enthalpy and velocity of steam at inlet and exit of a steam turbine, running under steady state, are as given below: Inlet steam condition Exit steam condition Specific enthalpy (kl/kg) 3250 2360 Velocity (m/s 180 The rate of heat loss from the turbine per kg of steam flow rate is 5 kW. Neglecting changes in potential energy of steam, the power developed in kW by the steam turbine per kg of steam flow rate, is (A) 901.2 (B) 911.2 (C) 17072.5 (D) 17082.5

6. sing SFEE ,we have 2%) 2000 2000 2000 w = 901.2 kw/kg

7. In a simple Brayton cycle, the pressure ratio is 8 and temperatures at the entrance of compressor and turbine are 300 K and 1400 K, respectively. Both compressor and gas turbine have isentropic efficiencies equal to 0.8. For the gas, assume a constant value of cp (specific heat at constant pressure) equal to 1 k]/kg K and ratio of specific heats as 1.4. Neglect changes in kinetic and potential energies. The power required by the compressor in kW/kg of gas flow rate is (A) 194.7 (B) 243.4 (C) 304.3 (D)270.1

8. 1-2 : Isentropic process Pr Ta = 81,4

9. Isentropic efficiencj 2. T1=604-12k T2-Ti Compressor p ae w = l (c04-12-300

10. .The thermal efficiency of the cycle in percentage (%) is (A) 24.8 (B) 38.6 (C) 44.8 (D) 53.1

11. Tu bine. Pow et given by 100-838-44 Heat supplied is q= cp (T3-T2)

12. Available ehersy is given by Sj-So l.oos-Insee -0.2e 'n.S- = ,.oos_ (soo-300-300 (0.05147)