1) Consider an Ideal Gas Contained in Vessel

95HE-4 Sr. No. 6 EXAMINATION OF MARINE ENGINEER OFFICER Function: Marine Engineering at Operational Level HEAT ENGINES

M.E.O. Class IV (Time allowed - 3hours) India (2003) Afternoon Paper Total Marks 100

NB : (1) All Questions are Compulsory (2) All Questions carry equal marks (3) Neatness in handwriting and clarity in expression carries weightage (4) Illustration of an Answer with clear sketches / diagrams carries weightage.

1. Consider an ideal gas contained in vessel. If intermolecular interaction suddenly begins to act , which of the following happens. a) The pressure increases b) The pressure remains unchanged

c) The pressure decreases d) The gas collapses.

2. When a system is taken from a state A to state B along the path A-C-B, 180 KJ of heat flows into the system and it does 130 KJ work as shown in figure.

V How much heat will flow into the system along the path A-D-B, if the work done by it along the path is 40 KJ? a) 40 KJ b) 60 KJ c) 90 KJ d) 135 KJ

3. A gas expands from pressure P1 to pressure P2 (P2 = P1 / 10). If the process of expansion is isothermal the volume at the end of expansion is 0.55m3, If the process of expansion is adiabatic the volume at the end of compression will be closer to a) 0.45m3 b) 0.55 m3 c) 0.65 m3 d) 0.75 m3

4. A tank contains 25 kg of fresh water at 15. 15.5 Kg of ice at 0C are added then 2 Kg of steam at 4 bar 0.9 dry are blown into the mixture. Assuming no heat losses occur and taking the enthalpy of fusion of ice at 0C as 333.5 KJ/Kg,, What is the final temp of the mixture. a) 8.16C b) 10.2C c) 16C d) 6.12C 5. A steam pipe is 3.85m long when fitted at a temperature of 18C carrying steam at a temperature of 260C. Taking the coefficient of linear expansion of the pipe material as 1.25 x 10-5 / C , the increases in length will be. a) 10 mm b) 15 mm c) 11.65 mm d) 14.2 mm

6. A mass of 3.5 Kg of gas is cooled at constant volume from 150C to 25C and then expanded according to the law PV 1.4=C to -40C. If the initial pressure was 10 bar , determine a) The final pressure b) Final volume c) The heat extracted during cooling

Take Cv of the gas to be 730 J/Kgk and R=0.287 KJ/Kgk

7. The steam generation plant supplies 8500 Kg steam per hour at pressure 0.75 MN/m2, the steam is 0.95 dry. Feed water temp = 41.5C Coal consumption = 900 Kg/hr Calorific value of coal = 32450 KJ / Kg Determine a) The boiler efficiency b) The equivalent evaporation from and at 100C c) The saving in fuel consumption. If by installing an economizer it is estimated that the feed water temp could be raised to 100C assuming that other conditions remained unchanged and the efficiency of the boiler increases by 6%

8. A Refrigerating machine working on a reversed carnot cycle consumes 6KW for producing a refrigerating effect of 1000 KJ/min for maintaining a region at -40C, Determine the highest temp of the cycle and the cop of the Refrigerator machine, when this device is used as a heat pump.

9. Single acting single stage air compressor has an rpm of 240, has a bore of 200mm, the average piston speed is 180m/min. The air is sucked in at 1 bar and 15C and compressed to 5.67 bar. Assume R = 0.287 KJ/Kg K. Determine. a)The mass flow rate b)Rate of work done and exit temp for polytopic compression Where n=1.3

10. A steam leaves the nozzle and enters the blade wheel of a single impulse Turbine at a velocity of 900 m/sec and at an angle of 20 to the plane of rotation or plane of wheel. The blade velocity is 400m/sec and the exit angle of the blade is 25. Due to friction the steam losses 15% of its relative velocity across the blade. Calculate a) Inlet blade angle b) The magnitude and direction of absolute velocity of steam at exit.

------X------95HE-4 Sr. No. 6 EXAMINATION OF MARINE ENGINEER OFFICER Function: Marine Engineering at Operational Level HEAT ENGINES

M.E.O. Class IV (Time allowed - 3hours) India (2003) Afternoon Paper Total Marks 100

NB : (1) All Questions are Compulsary (2) All Questions carry equal marks (3) Neatness in handwriting and clarity in expression carries weightage (4) Illustration of an Answer with clear sketches / diagrams carries weightage.

SOLUTION

1) The pressure increases

2) 90KJ

For the path A-C-B From 1st Law of TD Q = ΔU + W

Q = UB - UA + W 180 = UB - UA+ 130 UB - UA=180-130 UB-UA= 50 KJ

For the path A-D-B Q = ΔU + W Q = UB - UA + W

Since the initial and final states are same for both the process the change in internal energy will be same.

Q = UB - UA + W Q = 50 + 40 Q = 90 KJ of heat supplied

3) 0.45m3 soln :

When the gas is expanded isothermally PV=C i.e P1V1 = P2V2 P1 = V2 P2 V1 (10) = V2 V1 3 V1 = 0.55 = 0.055m 10 3 V1 = 0.055m

When the gas is expanded adiabatically PVr = C r r i.e P1V1 = P2V2 r ﴿P1 =﴾ V2 P2 ( V1) 1/r (P1) = V2 (P2 ) V1

1/1.4 (10) = V2 V1

Since the initial volume for both the process has to be same 1/1.4 3 (10) = V2 V2 = 0.284m 0.055

4) 8.16 o C

Amount of heat present in water = Q1 = m cpw ΔT

= 25 x 4.187 x ( 15 – 0)

= 1570.1 KJ

If we take the enthalpy or heat of water at 0 o C as zero Then the amount of heat present in ice at 0 o C is

Q2 = - ( 15.5 x 333.5 )

= - 5169.3 KJ Amount of heat present in steam

Q3 = ms ( nf + xnfg ) 4 bar

= 2 (605 + 0.9 x 2134 )

= 5051.2 KJ ice will gain heat , water and steam loses heat

:. Amount of heat present after mixing will be Q = Q 1 + Q2 + Q3 = 1570.1 – 5169.3 + 5051.2 = 1452 KJ

:. We know that Q = m mix cpw x (t-0)

1452 = (2 + 25 + 15.5) 4.187 ( t – 0)

t = 8.16 o C

5) 11.65 mm

Soln :

Force expansion = α x l (T2 – T1 ) = 1.25 x10 –5 x 3.85 ( 260 – 18) = 11.65 mm

6) Soln: 1

P = C P T 2 PV1.4 =C 3

V1 = mRT1 = 3.5 x 0.28 x ( 423) 2 P1 10 x 10

V1 = 0.425 m3

From 1-2

P = C T

P1 = P2 T1 T2 2 P2 = P1T2 = 10 x10 x 298 T1 423

2 2 P2 = 7.05 x 10 KN/m

From 2-3 PV1.4 = C

n-1 T2 = (V3 / V2)

1/ n-1 ) 1/1.4 – 1 V3 = V2 x ( T2 / T3 ) = 0.425 x ( 298 / 233

= 0.786 m3

n n P2 V2 = P3 V3

2 2 P3 = P2 = 7.05 x10 = 297 KN/m n 1.4 (V3 /V2 ) ( 0.786 / 0.425)

P3 = 2. 97 bar

From lst law of TD from 1 – 2 i.e cooling

Q =U2 – U1 + W

W = 0 constant volume

:. Heat capacity = U2 – U1 = m.C v (T1 – T2 )

= 3.5 x 0.730 (4.23 – 298) Heat extracted during cooling ( Q ) = 319 KJ

1) soln: ms = msteam mf

= 8500 = 9.44 Kg / Kg of coal 900

Sp Enthalpy of steam h = ( hγ + x hγg) 0.75

=[ 709.3+ 0.95 ( 2055.5 ) ]

= 2662.025 KJ/Kg

Sp enthalpy of water hw = cpw (T – 0) = 4.187 ( 41.5 – 0) = 173.9 KJ/Kg

Boiler efficiency = msteam ( h – hw) m x c1V

=ms ( h – hw ) = 9.44 (2662.025 – 173.9 ) c1 V 32450

= 72.38%

o Equivalent evaporation and at 100 C = ms (h – hw ) (hγ g )

= 9.44 ( 2662.025 – 173.9) 2256.9

= 10.40 Kg/Kg of coal

Energy required to generate steam under new condition

Sp. Enthalpy of BFW at 100oC hw = 4.87 (100 – 0) = 4187 KJ/Kg

Energy required to generate steam when economizer is incorporated is = ( h – hw )

=2662.025– 418.7

= 2243.3 KJ/Kg new boiler efficiency = 72.38 + 6 = 78.38% :. Boiler efficiency = msteam ( h – hw ) mγ C1 V

0.7838 = 8500 x 2242.925 mγ x 32450

m γ = 749.57 Kg/ hr

saving in full = 900 – 749.57 = 150.43 Kg of coal/hr

2) Soln:

Work done= W = 6KW Q = 1000 KJ/min o TL = - 40 + 273 = 233 K

PAPER – 2

1) The length of each steel rail is 10m in winter. If the coefficient of linear expansion of steel is 12x10-6/OC and temp increases by 15C in summer the gap to be left between the rail is

a) 0.0018m b) 0.0012m c) 0.0022m d) 0.05m

2) In the figure shown, E is a heat engine with efficiency of 0.4 and R is a refrigerator, Given that Q2+Q4=3Q1. The cop of the refrigerator is

a) 2.5 b) 3.0 c) 4.0 d) 5.0

3) 80gm of water at 30C are poured on a large block of ice at 0C. The mass of the ice that melt is a) 1600mgm b) 30gm c) 150gm d) 8gm

4) Three moles of an ideal gas are compressed to half the initial volume at a constant temp of 300K. The work done in the process is a) 5186J b) 2500J c) –2500J d) –5186J 5) A heat pump operating between high temperature T1 and lower temperature T2 has its cop expressed as a) T1 T1-T2

b) T1-T2 T1+T2 c) T2 T1-T2

d) T1+T2 T1-T2

6) Six cylinder, two stroke cycle heavy oil operated main engine has cylinder diameter of 580mm and a piston stroke of 1700mm. When the engine speed is 116rev/min it uses 1204 kg of heavy fuel oil of calorific valve 42700 Kj/Kg in one hour. The cooling water amounts to 66.62 tonnes/hour entering at 15C leaving at 63C. The torque transmitted at the engine coupling is 583 KN-M and the indicated mean effective pressure is 15 bar, Determine, 1) The indicated power 2) The brake power 3) Amount of water required per Kg of level 4) Brake Thermal efficiency 5) Brake mean effective pressure 6) Fuel used per Kw hour on Brake power bases.

7) In a Frean-12 refrigerator, the freon leaves the condenser as a saturated, liquid at 20C and the evaporator temperature is -10C and the freen leaves the evaporator as a vapour of 0.97dry calculate

a) The dryness fraction at the evaporator inlet b) The cooling effect per Kg of refrigerator c) The volume flow of refrigerant entering the compressor if the mass flow is 0.1 Kg/sec.

8) In a single stage impulse turbine steam leaves the nozzle at a velocity of 600m/sec at 18  to the plane of rotation of the blades, the linear velocity of the blades is 230 m/sec neglecting friction across the blades and assuming the steam leaves the blade wheel in an axial direction. Calculate 1) The inlet angle of the blades so that the steam enters without shock 2) The outlet black angle 3) The Black or Diagram efficiency 9) A cold storage compartment is 45 m long by 4m wide by 2.5m high. The four walls ceiling and flour are covered to a thickness of 150mm with insulating material which has a thermal conductivity of 5.8x10-2 W/mK. Calculate the quantity of heat leaking through the insulation per hour when the outside and inside the temperature of the material is 15C and -5C respectively.

10) 3 Kg of wet steam at 14 bar and dryness fraction of 0.95 are blown into 100Kg of water at 22C. Find the resultant temperature of water.

SOLUTION

1) Ans- A

Increase in length = 2xlx(T2-T1) = 12x10-6x10x(15) = 0.0018m 2) Ans- D W= Q2-Q1 Also W = Q3-Q4 Q2-Q1= Q3-Q4 Q2+Q4= Q1+Q3 3 Q1= Q3+Q1 2 Q1= Q3

efficiency of heat engine E = W = Q1

0.4 = W

Q1 W = 0.4 Q1

Cop (Refrigeration) = Q3 = Q3 = 2Q1 = 5 W Q4 Q1 O1 4Q1 Cop = 5

3) Ans – B Heat gained by ice = Heat lost by water from 35C to C Mice x latent heat = mwater