MODEL QUESTION PAPER - 2006

Paper 4

Duration : 2 Hrs Max. Marks: 100

Section-1:Short QuestionsMarks: 10x1 = 10

(i)Answer all Ten questions ii) Each question carries One mark

S-1What are the losses not accounted for in the Indirect method of testing boiler efficiency?

  • Blowdown loss
  • Soot blower steam consumption.
  • Energy usage by auxiliary equipments like burners.

S-2The unit of the overall coefficient of heat transfer is:

a)k.cal/m2hr b)k.cal/ m2hr0C c)k.cal/kg0C

S-3What is renewable energy?

The renewable energy comes directly or indirectly from sun and wind and can never be exhausted. Hence they are called renewable.

S-4What is the difference between net present value method and internal rate of return method?

The net present value method is essentially a comparison tool which enables a number of projects to be compared whereas internal rate of return method is designed to assess whether or not a single project will achieve a target rate of return.

S-5What is anaerobic digestion?

Anaerobic digestion involves using bacteria to decompose organic matter in the absence of oxygen.

S-6What parameter in a psychometric chart is used for calculating refrigeration load in an air conditioning system?

Enthalpy (inlet and outlet air)

S-7What is the Capacity factor of wind turbine?

The capacity factor is the ratio of the wind turbines actual energy output for the year divided by the energy output if the machine operated at it’s rated power output for the entire year.

S-8The losses in a variable speed drive is:

a)12% b) 8% c) 5%

c) 5%

S-9The efficiency of a solar cell is a) 5% b) 15% c)30%.

b)15%.

S-10What is meant by “Solar window”?

The period of 4-5 hours in late morning and early afternoon (9am-3pm) is commonly called the solar window.

Section –IILong QuestionsMarks: 2x5 =10.

i)Answer all questions.

ii)Each question carries Five marks.

L-1The sensible heat loss in flue gas is very high in furnace applications. Explain?

For effective heat transfer from the gas to the furnace the temperature of the flue gas should be at least 500C higher than the operating temperature. Hence the flue gas temperatures are high leading to increased sensible heat loss.

L-2What are the advantages of the solar water heater?

  • There is saving in heating costs.
  • Saving in electricity/gas which are used for heating applications.
  • Solar energy is most readily available.
  • Solar energy is a renewable source of energy.

Section –III:Numerical QuestionsMarks: 4x20 = 80

i) Answer all Four questions.

ii) Each question carries Twenty marks.

N-1. A process plant has a back pressure turbine for power generation.The electrical power output from turbine is 1MW.The process data are as under:

  • The plant heat rate is 33,000 K.Cal/Kwh.
  • Enthalpy of steam at inlet of turbine is 700 K.Cal/Kg.
  • Enthalpy of feed water is 70K.Cal/Kg.
  • Enthalpy of steam at outlet of turbine is 650 K.Cal/Kwh.
  • Evaporation ratio of Boiler is 3.5 Kgs of steam/Kg of coal.

Calculate the following:

a)Flow rate of steam to turbine.

b)Power generation efficiency of turbo-alternator.

c)Overall plant fuel rate including boiler.

a)Overall plant heat rate

= Steam flow rate x (Enthalpy of steam –Enthalpy of feed water)/Power Output

Substituting values 33,000=Flow rate of steam x(700-70)/1000.

Flow rate of steam = 33000*1000/630 =52,381 Kgs/hr.

b)Power generation efficiency of turbo-alternator :

Heat energy input into turbine per Kg of inlet steam =700-650 =50 K.Cal/Kg.

Energy Input into turbine =52381*50 =26,19050 K.Cal/hr.

Power generation from turbine =1000KW.

Equivalent thermal energy output = 1000*860 =860000 K.Cal/hr.

Power Generation efficiency = 860000/26,19050 = 32.84%

c)Evaporation ratio of boiler =3.5 Kg of steam/Kg of coal.

Flow rate of steam from boiler = 52,381 kgs/hr.

Hence fuel flow rate to boiler = 52381/3.5 =14,966 Kgs of coal/hr.

Overall plant fuel rate including boiler = 14,966/1000 =14.97 Kgs of coal/Kw

N-2 The following parameters were observed during the testing of a cooling water pump:

  • Discharge pressure of pump : 5.0 Kg/cm2
  • Flow rate of water: 900 m3/hr.
  • Suction head : 2 metre below the pump centerline.
  • Height of cooling tower: 5 metre
  • Motor efficiency: 90%.
  • Density of water: 1000 Kg/m3
  • Current drawn by pump motor : 260 amps.
  • Voltage measured: 415volts.
  • Power factor measured: 0.9

Calculate the following:

a)Hydraulic power.

b)Pump Efficiency.

a)Hydraulic power = (900/3600)*(50-2)*1000*9.81/1000

= 117.72 KW.

b)Pump Efficiency :

Input Power to motor =1.732*0.415*260*0.9 =168.2KW

Pump shaft power = 168.2*0.9 = 151.38 KW.

Pump efficiency= (117.72/151.38)*100 = 77.76%

N-3Field tests were carried out on a chilling system to find out the energy performance ratios.The measurements data are as under:

* Temperature of the refrigerant entering the evaporator : (-2) 0C.

* Temperature of the refrigerant leaving the evaporator : (-5) 0C.

* Flow rate of the refrigerant: 15000Kgs/hr.

* Specific heat of the refrigerant: 2.3Kcal/Kg0C.

* Current input into compressor: 70A

* Voltage input measured: 415 V

* Power factor: 0.9

Find out the following :

a)Net refrigeration capacity.

b)Kw/ton rating.

c)Coeffient of performance (COP).

d)Energy Efficiency ratio.(EER).

a)Net refrigeration capacity = 15000*2.3*(-2 –(-5))/3024

= 34.23 TR.

b)Power input to compressor = 1.732*0.415*70*0.9 =45.28KW.

Kw/Ton rating = 45.28/34.23 = 1.32

c)COP =3.516/1.32 =2.66.

d)EER = 12/1.32 = 9.09

N-4 (i) What is the total weight of flue gas generated when 10Kg of butane(C4H10) is burned . The % O2 measured in the flue gas is 2.0.

ii) The flue gas is cooled from 600oC to 200 oC for generating steam. The total heat required per Kg of steam generation is 550 K.Cal/Kg. Find out the quantity of steam generated.

Additional Information:

Atomic Weights C=12, H=1, O=16;

Specific heat of flue gas =0.24 K.cal/Kg/ oC

Assume air is 77% Nitrogen (N2) & 23% Oxygen (O2)

2C4H10 +13O2 =8CO2 +10H2O

2(48+10) +13(32)=8(12+32)+10(2+16)

116Kg butane requires 416 Kg of Oxygen

10 Kg of butane requires 35.86 Kg of O2.

Air requirement is (100/23)*35.86

: 155.9Kg.

% O2 measured in the flue gas is 2.0%.

% Excess air: 2/(21-2) =10.53%.

Hence excess air quantity: (10.53/100)*155.9

: 16.42Kg.

Total flue gas quantity:155.9+16.42+10

: 182.32Kg.

Quantity of steam generated : 182.32*0.24*(600-200)/550

: 31.8 Kg of steam