Anna University Marine Heat Engines and Applied Thermodynamics Exam Paper
Time : Three hours Maximum : 100 marks
Answer ALL questions.
(Use of Steam Tables and Charts allowed)
PART A — (10 ´ 2 = 20 marks)
1.Efficiency of Carnot power plant cycle depends on temperature range of the cycle and not on the particular fluid used. Explain.
2.State the reasons, as to why a Carnot cycle, though most efficient, is not used in practice for steam power plants.
3.Draw a line diagram of the components of steam power plant working on Rankine cycle.
4.List the methods of governing of steam engines.
5.What is ‘‘Missing Quantity’’ in a steam engine? How it can be reduced?
6.State the reasons for exit velocity drop/losses in a steam nozzle.
7.What are the effects of frictional losses in a steam nozzle?
8.What are clearance volume and clearance ratio of a reciprocating compressor?
9.Define the following for reciprocating compressor.
(a) Pressure Ratio (b) Free Air Delivery.
10.What is an Air Motor? Where is it used in marine application?
PART B — (5 ´ 16 = 80 marks)
11.(i) Derive the expression for relationship between Area, Velocity and Pressure in a nozzle flow. (10)
(ii)What is Mach Number? Explain accelerated and decelerated flow in a nozzle.
(a)(i) Derive the expression for Rankine efficiency. (8)
(ii) Steam at 10 bar is supplied to a prime mover and exhaust takes place at 0.2 bar. Find the Rankine efficiency in the following cases :
(aa) Steam is dry saturated at inlet (bb) steam is 0.9 dry at inlet.
(cc) Steam is 50 C superheat at inlet. Neglect pump work. (8)
(b) (i) What are reheating and regeneration in a Rankine cycle? How efficiency of the Rankine cycle can be improved by them? (8)
(ii)Steam at 100 bar and 500°C enters the ideal engine that has one stage of reheat. The steam is exhausted from the engine at 0.07 bar and 85% dry. The net work developed by the engine is 1600 kJ/kg of steam. Calculate the thermal efficiency of the engine. (8)
- (a) (i)Define mechanical efficiency, thermal efficiency and overall efficiency of a reciprocating steam engine. (6)
(ii) A double acting single cylinder steam engine with cylinder diameter 15 cm and stroke 20 cm, is to develop 20 KW of indicated power at 100 rpm, with a cut off at 20% of the stroke. The back pressure is 0.28 bar. Determine the admission pressure, if the diagram factor is 0.72. Also, calculate the indicated thermal efficiency of the engine, if it receives 222 kg of dry steam per hour. (10)
(b) (i) How are compound steam engines classified? (2)
(ii)Explain the working ‘Woolfe compound steam engine’ with a sketch and details of torque crank rotation diagram. (7)
(iii) Explain the functioning of D–slide valve of a reciprocating steam engine. (7)
(a)(i)What is compounding of steam turbines and what are the different types? (6)
(ii)Explain any one type of compounding of steam turbine with a diagram. (10)
(b)(i) Draw the velocity diagram of an impulse turbine blade, indicating all notations. (6)
(ii) In a single stage impulse turbine the blade angles are equal and the nozzle angle is 20 degree. The velocity coefficient for the blade is 0.83. Find the maximum blade efficiency possible. If the actual blade efficiency is 90% of maximum blade efficiency, find the possible ratio of blade speed to steam speed. (10)
(a)(i)Indicate the methods to improve the thermal efficiency of a simple open cycle Gas Turbine plant, with a line diagram. (6)
(ii) In an open cycle constant pressure gas turbine, air enters the compressor at 1.02 bar and 27 degree C. The pressure of air after the compressor is 4.08 bar. The isentropic efficiencies of compressor and turbine are 80% and 85% respectively. The air to fuel ratio is 80%. Find the thermal efficiency. Assume Cp = 1 kJ/kg and r = 1.4 for air and gases. C.V. for fuel is 141720 kJ/kg. (10)
(b) (i)What is an indicator diagram for a reciprocating compressor? Explain with a PV diagram. (6)
(ii) Determine the size of the cylinder of a double acting air compressor of 32 KW of I.P. in which air is drawn in at 1 bar and compressed
to 16 bar according to law pv1.25 = const. RPM = 300, Piston
speed = 180 m/min; volumetric efficiency is 0.8. (10)