Indian BOE Coaching Services
Gujarat BOE-2011 Paper-1 (Boiler Engg-1) Solution
Q.1(a) : Describe any one method of calibrating horizontally installed cylindrical
fuel tank.
Answer : Volumetric calibration : It is performed by adding liquid, usually cold water to
a fuel tank in small volumes . It depends of the volume of the tank and may vary from 50
to 400 liters. The liquid can either be metered in a calibrated flow meter or poured from a
calibrated volumetric prover. Basically, in fuel tank calibrated flow metering systems are
used. After the new volume is added the level is measured and recorded by electronic
level gauge. All temperature corrections, provoked by the water temperature change
must be indicated in the graduation table. If the fuel tank service conditions for
hydrostatic pressure are different from the calibrating conditions, the volume error will
occur. Thus corrections for hydrostatic pressure must be made. For those fuel tanks
which were installed on different angles from which they were calibrated (installation
angles may vary because of geological reasons or wrong installation) must be made one
more correction in order to obtain a precise data. If the tank inclination is different from
the calibrated condition, there will be errors in the reported volume. The volumetric
calibration has the limitations which with some corrections gives satisfactorily results.
Therefore it is important to mention, that volumetric calibration process requires a certain
amount of water subject to the fuel tanks capacity and when the process is finished the
fuel tanks must be cleaned again to avoid further incorrect tank usage.
Q.1(b) : Find the surface area and volume of a hexagonal tank with the given
Apothem length 2m, side 3m and height 4m.
Answer :
For a hexagonal tank :
Surface area of base A =1.5x√3xl2 =1.5x1.73x2x2=10.38 sqm
Surface area of lateral M =hxlx6=4x2x6=48 sqm
Total Surface area S =2xA+M=2x10.38+48=68.96sqm
Volume of tank=Axh=10.38x4=41.52 cubm
Q.2(a) : Describe wet back three pass package boiler in detail with sketch. Explain
why and where it is suitable to use.
Answer :
Silent features of wet back three pass package boiler :
1. Low cost heating, trouble-free operations & long-lasting.
2. Available with fixed grate and rocking grate furnace.
3. Available in shell & smock tube types.
4. Works on natural proper water circulation.
5. Provided with ID fan for maintaining to drought.
6. Safety valve, steam stop valve, water level gauge glass & control.
7. Switches mounted on the boiler.
8. Multi-fuel option – coal/lignite/wood/briquettes.
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Gujarat BOE-2011 Paper-1 (Boiler Engg-1) Solution
Q.2(b) : Calculate boiler pipe thickness for boiler having given the working
pressure using following formula from IBR-350 : WP = 2 f e ( t – c ) / ( D – t + c )
Where WP is working pressure 100kg/sq cm(g), t=minimum required thickness
mm, D=Pipe outside diameter 508mm, f=pipe metal allowable stress
1200.0kg/sqcm, e=joint efficiency 1.00, c=0.25.
Solution : Working pressure WP =2fe(t–c)/(D–t+c)
Or 100 =2x1200x1.00x(t-0.25)/(50.8-t+0.25)
Or (t-0.25)/(50.8-t+0.25) =100/2400
Or 2400(t-0.25) =100(50.8-t+0.25)
Or 2400t-600=5080-100t+25
Or 2400t+100t=5080+25+600
Or 2500t=5705
Boiler pipe thickness t=5705/2500=2.282cm=22.82mm
Q.3(a) : A circular beam of 105mm diameter is subjected to shear force of 500kg.
Calculate the value of maximum shear stress and sketch the variation of shear
stress along the depth of the beam.
Solution :
Diameter of beam=105mm =0.105m
Area A=π/4xd2=3.14/4x0.105x0.105=0.00865m2
Shear force S=500kg=500N=0.5kN
Mean shear stress
ζmean=S/A=0.5/0.00865=57.8kN/m2
Using the relation,
Maximum shear stress ζmax =4/3xζmean
=4/3x57.8=77.07kN/m2
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Gujarat BOE-2011 Paper-1 (Boiler Engg-1) Solution
ζ
m
a
x
105mm
Beam cross-section Shear stress distribution
Q.3(b) : Find the torque which a shaft of 250mm diameter can safely transmit, if
shear stress(due to torsion) is not to exceed 460kg/cm2.
Solution :
Given, D=250mm=25cm, Fs=460kg/cm2
To find moment of resistance of shaft
MR =0.1962xfsxd3
=0.1962x460x25x25x25
=1410187.5 kgcm
=14101.875kgm
Q.4(a) : What is the function of condenser in steam power plant. List out the merits
and demerits of surface condensers over jet condensers.
Answer :
Function of condenser in steam power plant : Low exhaust pressure is necessary to
obtain low exhaust temperature. But the steam cannot be exhausted to the atmosphere
if it is expanded in the engine or turbine to a pressure lower than the atmospheric
pressure. Under this condition, the steam is exhausted into a vessel known as
condenser where the pressure is maintained below the atmosphere by continuously
condensing the steam by means of circulating cold water at atmospheric temperature.
ADVANTAGES OF A SURFACE CONDENSER
The various advantages of a surface condenser are as follows:
1. The condensate can be used as boiler feed water.
2. Cooling water of even poor quality can be used because the cooling water does not
come in direct contact with steam.
3. High vacuum (about 73.5 cm of Hg) can be obtained in the surface condenser. This
increases the thermal efficiency of the plant.
DISADVANTAGES OF A SURFACE CONDENSER
The various disadvantages of' the surface condenser are as follows:
1. The capital cost is more.
2. The maintenance cost and running cost of this condenser is high.
3. It is bulky and requires more space.
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Gujarat BOE-2011 Paper-1 (Boiler Engg-1) Solution
Q.4(b) : In a small capacity of thermal power plant, the condenser is provided with
a separate air cooling section. The temperature of steam entering the condenser is
56*C and temperature at the air pump suction is 46*C. The barometer reads 76cm
of Hg. Find
[1] The vacuum in condenser.
[2] If the discharge of dry air pump is 90m3/min. Find the air leakage in condenser
kg/hr.
[3] Loss of condensate due to the water vapour carried with air through air-pump.
Solution :
[1] Considering no ingress of air at the inlet, the pressure in the condenser will be the
same as the pressure of the steam entering at that temperature.
From the steam table at 56*C, the corresponding pressure
= 0.1653 bar absolute = 0.1653 x 750 = 123.98 mm of Hg,
Vacuum in the condenser = 760 – 123.98 = 636.02 mm of Hg
[2] Now at the outlet the partial pressure of the steam = 0.10101 bar abs. (from steam
table at 46*C)
Therefore the partial pressure of air = 0.1653 – 0.10101 = 0.0643 bar
The discharge of air pump = 90 m3/ min = 5400m3/ hr.
Here T = 46*C + 273 = 319 *K,P = 0.0643 bar = 0.0643 x 102 kPa, V = 5400 m3, R of air
=0.294 kJ/ kg* K
PV = mRT
Therefore mass of air leakage in the condenser per hour
m= (0.0643 x 102 x 5400)/ (0.294 x 319)=35416.44/93.786 = 377.63 kg/hr
[3] Loss of condensate due to water vapour carried with air through air pump
Suppose dryness fraction of exhaust steam is 0.85.
It means 15% moisture content in existing air.
Mass of water vapour = Total mass of air leakage x % moisture content.
=377.63x0.15 =56.6445kg/hr
Q.5(a)-[1] : Explain the classification of combustion system used for coal burning
giving short notes on each.
Answer : Three T’s of combustion :Homogeneous mixture of carbon and oxygen
content is important. To ensure good combustion is controlling the "three T's" of
combustion which are
(1) Temperature high enough to ignite and maintain ignition of the fuel,
(2) Turbulence or intimate mixing of the fuel and oxygen, and
(3) Time, sufficient for complete combustion.
Too much, or too little fuel with the available combustion air may potentially result in
unburned fuel and carbon monoxide generation. A very specific amount of O2 is needed
for perfect combustion and some additional (excess) air is required for ensuring
complete combustion. However, too much excess air will result in heat and efficiency
losses. Usually all of the hydrogen in the fuel is burned and most boiler fuels, allowable
with today's air pollution standards, contain little or no sulfur. So the main challenge in
combustion efficiency is directed toward unburned carbon (in the ash or incompletely
burned gas), which forms CO instead of CO2.
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Gujarat BOE-2011 Paper-1 (Boiler Engg-1) Solution
Perfect, good and incomplete combustion :
Below mention three comparative figures are showing the difference combustion process
with changing of air & fuel ratio .
Combustion Controls : Combustion controls assist the burner in regulation of fuel
supply, air supply, (fuel to air ratio), and removal of gases of combustion to achieve
optimum boiler efficiency. The amount of fuel supplied to the burner must be in
proportion to the steam pressure and the quantity of steam required. The combustion
controls are also necessary as safety device to ensure that the boiler operates safely.
Various types of combustion controls in use are:
1. On/Off control: The simplest control, ON/OFF control means that either the burner is
firing at full rate or it is OFF. This type of control is limited to small boilers.
2. High/low/off control: Slightly more complex is HIGH/LOW/OFF system where the
burner has two firing rates. The burner operates at slower firing rate and then switches to
full firing as needed. Burners can also revert to the low firing position at reduced load.
This control is fitted to medium sized boilers.
3. Modulating control: The modulating control operates on the principle of matching the
steam pressure demand by altering the firing rate over the entire operating range of the
boiler. Modulating motors use conventional mechanical linkage or electric valves to
regulate the primary air, secondary air, and fuel supplied to the burner. Full modulation
means that boiler keeps firing, and fuel and air are carefully matched over the whole
firing range to maximize thermal efficiency.
Q.5(a)-[2] : Determine fundamentally the minimum quantity of air required for
complete combustion of 1kg of (i) Carbon C (ii) Hydrogen H (iii) Methane gas CH4.
Solution :
(i) C +O2 =CO2
12 2x16 44
It means 12 kg of carbon required 32kg of oxygen which is equivalent to (32/0.23)
139.13kg of air.
Hence 1 kg of carbon required (139.13/12) 11.59 Kg of air.
(ii) 4H +O2 =2H2O
4 2x16 36
It means 4 kg of hydrogen required 32 kg of oxygen which is equivalent to (32/0.23)
139.13kg of air.
Hence 1 kg of hydrogen required (139.13/4) 34.78 kg of air.
(iii) CH4 +2O2 =CO2 + 2H2O
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12+4 2x16x2 =44 36
It means 16kg of methane required 64 kg of oxygen which is equivalent to (64/02.3)
278.26 kg of air.
Hence 1 kg of methane required (278.26/16) 17.39 kg of air.
Q.5(b) : In a boiler test the following quantities were obtained.
Mean temperature of feed water 59*C
Mean boiler pressure 12bar
Mean steam dryness fraction 0.95
Mass of coal burnt per hour 250kg
Calorific value of coal 32400Kj/Kg
Mass of water supplied to the boiler in 7hour and 14minutes 16500kg
Mass at the end of the test was less than that at the commencement by 1000kg
Calculate the thermal efficiency of the boiler and actual equivalent evaporation
from and at 100*C per kg of coal.
Solution :
Ms =Mass of steam generation per hour
=Net mass of water supplied per hour
=(16500-1000)/7.25=2138 kg/hr
From steam table at 12 bar hf=798.4kj/kg & hfg=1984.3kj/kg
Hs =Specific enthalpy of steam at 12 bar
= hf+x.hfg=798.4+0.95x1984.3=2683.485kj/kg=641.68kcal/kg
Thermal efficiency of boiler = Heat output/Heat Input
=Msx(Hs-hf)/QxGCV
=[2138x(641.68-59)]/[250x(32400/4.182)]
=1245769.84/1936872.31=0.6432=64.32%
Mact =Mass of steam generation/Mass of fuel consumption=2138/250=8.552
Factor of equivalent evaporation “F” =(Hs-hf)/539=(641.68-59)/539=1.081
Actual equivalent evaporation from and at 100*C per kg of coal =
=MactxF =8.552x1.081=9.245
Q.6(a) : Briefly explain working of the following :
1. Feed injector.
2. Feed regulators.
3. Combined high pressure steam and low-water safety valve.
4. Water softeners.
Answer :
1. Feed water injector : The injector is a simple appliance for feeding a boiler with
water by the direct use of steam from the same boiler or even of steam of a lower
pressure as in the exhaust steam injector which utilizes the steam discharged by
a non-condensing engine. The injector was invented by Giffard, a French
engineer, in 1958. A well-known modern form of injector is illustrated by figure
which is a longitudinal section of Holden and Brooke’s automatic restarting
injector. Steam enters at A and flows into the steam cone B. In rushing across
the opening at the lower end of B a partial vacuum is formed around that opening
and water entering at C mixes with the steam in the combining cone DE, the
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Gujarat BOE-2011 Paper-1 (Boiler Engg-1) Solution
steam being condensed. This condensation of the steam augments the vacuum
and increases the flow of water.
2. Boiler feed water regulators :
The boiler feed water regulator valve used in many power plants is required to
transition from feed pump recirculation to operation of the unit. Not only is the
valve used to initially fill the steam drum, it is also used to control flow during
normal operation when the steam drum is under pressure. This valve, therefore,
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Gujarat BOE-2011 Paper-1 (Boiler Engg-1) Solution
must address cavitation during initial operation and provide adequate range
ability to address the entire feed water requirements. During initial operation the
service conditions for the feed water regulator valve nearly match that of the feed
pump recirculation valve with the exception of the flow rates. The regulator valve
will begin to transition the flow from the recirculation valve and will open as the
recirculation valve closes. The valve must have adequate cavitation protection
during initial filling of the drum and then transition to flow control mode.
3. Combined high pressure steam and low-water safety valve :
This is a twin purpose valve, which service as a safety device against drum level low and
the drum pressure very high The valve consists of two mechanisms, out of which, one
operates, to alert the boiler operator, whenever the drum water level in the boiler goes
down below the safe working level and the second valve operates, when the steam
pressure in boiler drum increases more than the working pressure. This kind of design of
valve was first manufactured and introduced by M/s Hopkinson Co. of Hedersfield of
London and hence known by their name. The mechanism of this valve consists of a lever
fitted inside the boiler shell, hanging from top of the drum. The lever is suspended on a
pivoted point to one end and the other end of the valve lever caries ceramic slabs
attached to it, which normally floats over the water level inside the drum. The other end
of the lever carries balanced weight and there is a knife edge in between the pivot and
the balanced weight which rests on the washer attached to the vertical spindle to the
bottom end of which a dead weight is suspended and the top end of which keeps the
inner valve in required position. The outer big valve as shown is having one vertical
spindle which is extended through the top blind flange cover of the valve and a horizontal
lever is fitted running through the fork fitted on to the vertical spindle. The horizontal
lever is pivoted near the vertical valve spindle and at one end of the lever one balanced
weight and to the other further end a dead weight is attached to balance the steam
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pressure exceeds than the working pressure the lever is fitted and valve opens steam is
released out. The valves are very useful for tank boilers like Lancashire, Cornish and
marine etc. The setting of the float is very important so that the low water valve operates
at exact position to give caution to the operator. . The construction of such valves is as
illustrated in figure.
4. Water softeners : Water softening is the removal of calcium, magnesium, and
certain other metal cations in hard water. The resulting soft water is more
compatible with soap and extends the life time of plumbing. Water softening is
usually achieved using lime softening or ion-exchange resins. The presence of
certain metal ions in water causes a variety of problems. These ions interfere
with the action of soaps. They also lead to buildup of lime scale, which can foul
plumbing, and promote galvanic corrosion. In industrial scale water softening
plants, the effluent flow from the re-generation process can precipitate scale that
can interfere with sewage systems.
Q.6(b) : Explain the working principle of natural draught. Theoretically derive
formula to calculate height of chimney for given parameters.
Answer : Working principle of natural draught : It is the draft system in which suction
of combustion air is created with the natural effect of height of the chimney without any
external aid. Drawl of hot gases is effected by creation of negative pressure inside the
furnace by the density difference between the column of hot gases and the column of
atmospheric air. The hot gases are lighter because of higher temperature, whereas the
atmospheric air is heavier, and hence the movement of hot gases takes place upwards
continuously. Atmospheric air enters the combustion zone, while the hot gases are
sucked out and discharged through the chimney. Hotter the flue gas, lighter it becomes,
but the atmospheric air almost remain constant, without any significant variation. Hence
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higher the temperature of the flue gas at entry to chimney better is the draught effect in
the furnace. The height of the chimney also matters in creating the draft effect. If the
chimney height is more, better draught will be induced due to high difference in weight of
column of hot flue gas inside the chimney and the cold column of atmospheric air outside
the chimney. When the created draft effect is high inside furnace, more combustion air
will be induced into the furnace, assisting in improved combustion of fuel.
Chimney Height calculation derivation :
Lets,
w = Weight of air required in kg for burning 1 kg of fuel
T = Average absolute temperature of gas passing through chimney in deg. K
Ta = Absolute temperature of air outside the chimney in deg. K
h =Furnace Draught in mm of water.
H = Height of chimney in meters
w + 1 = Weight of the flue gas produced in kg per kg of fuel burnt.
The volume of chimney gases produced may be taken as equal to the volume of air
supplied. Since the volume of the solid or liquid fuel burnt is so small as compared with
the volume of air supplied that it may be neglected.
Taking the volume of 1 kg of air at 0 oC and 760 mm of Hg as 0.7734 m3,
According to Avogadro’s principle, at NTP, the volume of gas will be the same as the
volume of air at NTP.
Volume of gases at NTP = 0.7734 ×w m3 per kg of fuel burnt.
Since volume of gas is proportional to its absolute temperature (by Charle’s law),
Volume of chimney gases at T K = 0.7734 w × T/ 273 m3/kg of fuel burnt,= wT/ 353 m3/
kg of fuel
And density of chimney gases at temperature T K = (w+1)/ wT/ 353 = 353 (w+1)/wT
kg/m3.
Similarly, the density of atmospheric air at Ta k =w/ wT/ 353 = 353/ Ta kg/m3
Let H be height of the chimney required in metres measured from the level of furnace
grate.
Pressure exerted by a column of hot chimney gas of H meters height,
= 9.81 × (ρg AH)/ A = 9.81 ρgH N/m2
where ρg is in kg/ m3 = 9.81 × 353 (w+1)/wT × H N/m2
Similarly, pressure due to column of outside (cold) air of same area and H metres height
= 9.81 × 353/ Ta × H N/m2
Let P be the pressure causing the draught in N/m2. Since the pressure causing the
draught is due to the difference of pressure due to column of hot gases within the
chimney and the pressure due to an equal column of outside (cold) air,
P = {9.81 × 353/ Ta × H} – {9.81 × 353 (w+1)/wT × H}
= 9.81 × H × 353 { 1/ Ta – (w+1)/wT } N/m2.......(Equation-1)
Since the density of water is 1,000 kg/m3, a water column of 1 m or 1000 mm height
will exert a pressure of 1000 × 9.81 N/m2.
Let h be draught pressure in mm of water, then the draught pressure, P in N/m2 is given
as
P = ρgh = 1000 × 9.81 × h / 1000 = 9.81h N/m2
h = P/9.81 mm of water
Substituting the value of P we have in (Equation-1)
h = 353 H [1/Ta – (w + 1) / wT] mm of water
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