GATE Previous years Thermodynamics questions GATE 2011 Thermodynamics solved questions set 1 By Rutuja Bothara

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

In an experimental set-up, airflows between two station P and Q adiabatically. The direction of flows depends on the pressure and temperature conditions maintained at P and Q. Thee condition at the P is 150 kPa and 350 K. The temperature at station Q is 300 K. The following are the properties and relations pertaining to air: Specific heat at constant pressure, Cp 1.005 kJ/kg K; Specific heat at constant volume, Cv 0.718 kJ/kgK; Characteristic gas constant, R = 0.287 kJ/kg K Enthalpy, h CpT Internal energy, u CVT If the air has to flow from station P to station Q, the maximum possible value of pressure in kPa at station Q is close to (A) 50 (B) 87 (C) 128 (D) 150

Station P Staton 2 -The flow betae e n +co stations s adiabatic Pi 0:4 soo 3 So From oluing this equation P2 87.45 kPa.

If the pressure at station Q is 50 kPa, the change in entropy (SQ SP) in k]/kg K is (A)-0.155 (B) 0 (C) 0.160 (D) 0.355

change in entropy is iven as Pi 3S0 Kg.k

The temperature and pressure of air in a large reservoir are 400k and 3 bar respectively. A converging-diverging nozzle of exit area 0.005m2 is fitted to the wall of reservoir shown in figure. The static pressure of air at the exit section for isentropic flow through the nozzle is 50kPa. The characteristic gas constant and the ratio of specific heats of air are 0.287 kJ/kg K and 1.4 respectively Flow from the reservoir reservoir Nozzle exit

The density of air in kg/m3 at the nozzle exit is (A) 0.560 (B) 0.600 (C) 0.727 (D) 0.800

Tr an isentropic Process,we hav Pa Pt RT 28 400

The mass flow rate of air through the nozzle in kg/s is (A) 1.30 (B) 1.77 (C) 1.85 (D) 2.06

C P 400 using povo Po 0.28 233-73 OGK3s

For constaht ehtropy Ta STi 0-1 72 # gl.kp 3 00

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RUTUJA Bothara

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Mohd sarfaraz alam

6 months ago

very nice video mam really

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Mona Choudhary

6 months ago

thank you 😊