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Posted Date: 13 Dec 2010 Posted By:: VEERAIYAN BOSE Member Level: Gold Points: 5 (Rs. 1)

2010 Anna University Chennai B.E Mechanical Engineering Me 2301 — thermal engineering  november/december 2010 Question paper
B.E./B.Tech. DEGREE EXAMINATION, NOVEMBER/DECEMBER 2010 Fifth Semester
Mechanical Engineering ME 2301 — THERMAL ENGINEERING (Regulation 2008)
Time : Three hours Maximum : 100 Marks Answer ALL questions
PART A — (10 × 2 = 20 Marks)
1. Why is Carnot cycle not used in real applications? 2. Draw the PV diagram for a dual cycle. 3. Draw the valve timing diagram for a CI engine. 4. What is the indicated power of four cylinder engine if BP with 4cylinder working is 18.75 kW and BP with 3cylinder working is 13.06 kW. 5. What are the factors reducing the final velocity of steam in nozzle flow? 6. What is the difference between impulse and reaction turbine? 7. How is the inter cooler used to reduce the power consumption of compressor? 8. List the advantages of multistage compressor over single stage compressor. 9. What is dew point temperature? 10. Define the COP of refrigerators.
PART B — (5 × 16 = 80 Marks)
11. (a) Derive an expression for the air standard efficiency of Diesel cycle and then deduce it for mean effective pressure. [Marks 16] Or (b) A six cylinder four stroke petrol engine has a swept volume of 300 cubic cm per cylinder, a compression ratio of 10 and operates at a speed of 35000 rpm. If the engine is required to develop an output of 73.5 kW at this speed, calculate the cycle efficiency, the necessary rate of heat addition, the mean effective pressure, maximum temperature of the cycle and efficiency ratio. The pressure and temperature before is entropic compression are 1.0 bar and 15°C respectively, take Cv = 0.72 and ? = 1.4. [Marks 16]
12. (a) (i) Explain the working principle of 4stroke engine. (Marks 8) (ii) With a neat diagram explain the working of battery ignition system. (Marks 8) Or (b) (i) Describe the working of Diesel fuel pump. (Marks 8) (ii) Explain the pressure feed lubrication system with a neat diagram. (Marks 8)
13. (a) In a steam nozzle, the steam expands from 4 bar to 1 bar. The initial velocity is 60 m/s and initial temperature is 200°C. Determine the exit velocity if the nozzle efficiency is 92% and the dryness fraction at exit. [Marks 16] Or (b) A single row impulse turbine develops 132.4 kW at a blade speed of 175 m/s using 2 kg of steam per sec. Steam leaves the nozzle at 400 m/s. Velocity coefficient of the blade is 0.9. Steam leaves the turbine blades axially. Assuming no shock determine the nozzle angle, blade angles at entry and exit. [Marks 16]
14. (a) A single acting 14 cm × 10 cm reciprocating compressor is operating at P1 = 1 bar, T1 = 20°C, P2 = 6 bar and T2 = 180°C. The speed of compressor is 1200 rpm and shaft power is 6.25 kW. If the mass of air delivered is 1.7 kg/min, calculate the actual volumetric efficiency, the indicated power, the isothermal efficiency, the mechanical efficiency and the overall efficiency. [Marks 16] Or (b) A single stage reciprocating air compressor has clearance volume 5% of stroke volume of 0.05 m3/sec. The intake conditions are 95 kN/m2, 300 K. The delivery pressure is 720 kN/m2. Determine the volumetric efficiency referred to (i) intake conditions (ii) atmospheric conditions of 100 kN/m2 and 290 K (iii) FAD and (iv) power required to drive the compressor, if the ratio of actual to indicated power is 1.5. Take index of compression and expansion as 1.3. [Marks 16]
15. (a) One kg of air at 35°C DBT and 60% RH is mixed with 2 kg of air at 20°C DBT and 13°C dew point temperature. Calculate the vapour pressure and dew point temperature of stream one, enthalpy of both the streams and specific humidity of the mixture. [Marks 16]
(b) The temperature range in a Freon12 plant is –6°C to 27°C. The compression is is entropic and there is no cooling of the liquid. Find the COP assuming that the refrigerant (i) after compression is dry and saturated (ii) leaving the evaporator is dry and saturated. The properties of F12 are given in the table :
Sl.No. t°C hf hg sf sg Cp 1 –6 413 571 4.17 4.76 0.641 2 27 445 585 4.28 4.75 0.714 [Marks 16]
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