Power plant Maintenance: basics questions & answers.

1-Define the term maintenance.

Maintenance is keeping something in proper condition or good order.

Or
Action taken to prevent machine from its failure.

Or

It is a process of repair

2- What is the necessity of maintenance?

Necessity of maintenance:

• To avoid machine failure/damage

• To avoid harm to man

• To avoid production loss

• To reduce unnecessary inventory in store

• To increase equipment or component service life

• To increase equipment effectiveness

3-What are the different maintenance practices adopted in power plants?

• Breakdown maintenance

• Preventive maintenance

• Predictive maintenance/Condition based maintenance (CBM)

4-What is meant by breakdown and preventive maintenance?

In breakdown maintenance, maintenance of equipment is taken on its failure only. Machine runs continuously till its failure.

Preventive maintenance is time based activity, maintenance activity is taken to avoid machine from failure.

5-What is meant by predictive maintenance/CBM?

It is the maintenance practice based on condition of the equipments. This type of maintenance continuously monitors the equipments condition thoroughly and plans maintenance during stoppage or idle time of plant.

6-What are the activities involved in Condition based maintenance (CBM).

1-Equipments bearing temperature monitoring.

2-Equipments bearing vibration monitoring.

3-Thickness checking (pressure parts, storage tanks, cooling water line).

4-Testing (relay testing, lubricating oil).

5-Nondestructive tests...etc

6-Monitoring equipment performance (flow, pressure etc.).

Turbine oil and transformer oil standard testing parameters

7-Differentiate between Preventive maintenance and Breakdown maintenance.

Sl No.	Preventive Maintenance	Breakdown maintenance
1	Time or running based activity	Carried out as and when required
2	Will not lead to generation stoppage	Leads to generation stoppage
3	Work activities are already known to working persons	Due to breakdown, work activities are not known to working persons
4	Requires less man power	Requires more man power
5	Semi-skilled man power can handle preventive maintenance	Requires skilled man power
6	Consumes less or no spares	Consumes spares
7	More opportunities for improvement	Less or no opportunities for improvement
8	Maintenance activity consumes less time	Maintenance activity consumes more time
9	Less inventory can be maintained	More inventories to be maintained
10	No repetitive works	Involves repetitive works
11	Does not require experts	May require experts if machine is not restored within time.

8-List the advantages of implementation of TPM in power plant

• Increased equipment productivity

• Reduced equipment downtime

• Increased plant capacity

• Lower maintenance and production costs

• Approaching zero equipment-caused defects

• Enhanced job satisfaction

• Increased Return on Investment

9-What are the different forms of screw threads used in power plants and process plants?

It's all about HP heaters

Forms of Screw Threads:

1. British Standard Whit Worth (B.S.W.) Threads: It is a symmetrical V thread in which the angle between the flanks is 55°.

2. British Association (B.A.) Thread: It is the B.S.W. thread with fine pitches having thread angle 47.5°

3. American Standard Threads: Included angle is 60°

4. Unified Standard Thread: This thread has rounded roots and crests with included angle 60°

5. Square Thread: Used in power transmission, spindles of valves etc.

6. Acme Thread: It is a modified square thread. It is much stronger than square thread.

7. Butress Thread: Used for transmission of power in one direction only.

8. Metric Thread: It is an Indian standard thread similar to B.S.W, but has included angle 60°.

10-What is the function of shaft key?

Key is a piece of mild steel inserted between the shaft and hub or boss of the pulley/coupling to connect these together in order to prevent relative motion between them.

11-What are the various types of shaft keys used?

• Sunk keys: Rectangular, square, parallel, gib head and feather keys

• Saddle keys

• Round key

• Spines

• Woodruff key

• Gib head key

• Dowel

Power Transformers QnA

12-What is strut?

A machine part subjected to axial compressive force is called strut. Strut may be horizontal, vertical or inclined.

13-What is meant by pillar or column?

A vertical strut is called as pillar or column.

14-What are short and long columns?

The columns which have lengths less than 8 times of their diameter are called short columns. And columns which have lengths greater than 8 times of their diameter are called short columns.

15-What are the structural materials used in power plants?

Structural Materials:

Beam, Channel, angles, girder, purlin, bracings, gussets, plain and chequered plates, flats, corrugated plates, truss, trestles etc.

16-What is meant by truss?

A truss is a structure that consists of two-force members only, where the members are organized so that the assemblage as a whole behaves as a single object. A two-force member is a structural component where force is applied to only two points. A sample truss is shown in below figure.

17-What is the function of beam and channels in structural engineering?

A beam is a structural element that primarily resists loads applied laterally to the beam›s axis. Its mode of deflection is primarily by bending

18-What is the function of gusset plates in structural materials?

Gusset plates are triangular plates of steel that are used to connect beams and girders to columns.

19-List down the different types of materials used in power plant construction.

Following are the various materials used in power plant construction:

Low Carbon Steel: IS 2062, IS 1239: For structural steels like plates, angles, channels, beams, LP steam lines etc.

High Carbon Steel: SA 106 Gr. B/C, SA210 Gr, B/C, SA 516 Gr. 70: For Boiler pressure parts like water wall panels, headers, economiser coils, down comers, feed water lines, steam drum etc.

Alloy Steel: SA 213 Gr. P11/T11, P22/T22, P91/T91: Super heater coils and pipe lines.

Cast Iron: Travelling grate materials, pulley, coupling etc.

Copper: Air conditioning cooling lines.

Brass: Surface condenser and oil cooler tubes.

Stain Less Steel: SS 304, 316, 410 etc: Ejector tubes, surface condenser/oil cooler tubes.

Plastics: PVC pipe lines, valves, tanks etc.

20-Define the term spring in mechanical members.

Spring is an elastic body, whose function is to distort when loaded and recover its original shape when load is removed.

21-What are the applications of springs in power plant?

Springs are used in following equipments:

•  Safety valves
•  Relief valves
•   Control valves
•   Soot blowers cams
•    Limit switches
•    Steam lines hangers and supports
•    Turbine HP and LP valves
•    Turbine Emergency stop valve (ESV)
•     Solenoid valves
•     Pressure switches

22-What are the different types of springs used in power plants?

• Helical springs (Tension and compression helical springs)
• Conical and volute springs
• Torsion springs
• Leaf springs
• Disc springs

23-Write down the MOC of Turbine journal bearing liner (Babbitt metal), roller bearings, Turbine HP and LP casings, Turbine blades, Turbine rotor, labyrinth, ESV, Reduction gear box gears.

• Journal Bearing Liner: A tin base alloy containing 88% of tin, 8% of antimony and 4% of copper.

• Roller/Ball Bearing: Hardened chromium steel.

• Turbine HP Casing: Alloy steel.

• Turbine LP Casing: Carbon steel.

• Turbine Blades: Alloy steel Mo-13–15%, Chromium and stainless steel.

• Turbine Rotor: Forged alloy steel Cr, Mo and Vanadium.

• Labyrinth Packing: Nickel, lead, bronze, Carbon and SS.

• ESV: Alloy cast steel (1.25 Cr 0.5 Mo).

• Reduction Gear Box Gears (RBG): Forged alloy steel.

 24-What is the surface hardness of carbon steel pipes?

It is in the range of 25 to 30 HRC.

25-How do you convert hardness 200 BHN to hardness HRC?

As a thumb rule HRC 5 BHN/10.8

Therefore 200 BHN 5 18.5 HRC

26-What is the hardness of hard facing electrodes used for hard facing of sprocket teeth, fan impeller liners etc?

It is around 45 to 55 HRC.

27-What is shaft coupling?

A coupling is a mechanical device, which is used to connect driver and driven shaft permanently or

semi permanently.

28-What are the main functions of coupling?

Functions:

• Connects the shafts of two units, which are manufactured separately.
• Transfers motion, power and torque.
• To reduce transmission shock loads from one shaft to another.
• Protection against overloads.
• Introduces mechanical flexibility.

 29-What is the function of a shaft?

A shaft is rotating machine element which is used to transmit power from one place to another.

30-Define the term bearing

A bearing is a machine element which reduces friction between two members, having relative motion with respect to each other. It’s a device used to support and guide rotating, oscillating or sliding shaft.

Also read the 50-Questions & Answers on Bearings

31-What are the functions of bearings?

• It reduces heat developed, sound and wear out of the components, which are having relative motion. Thus saves energy.
• Supports a load.
• Guide moving/rotating parts.

32-What are the major types of bearings?

Journal bearings and rolling contact bearings are the two main types of bearings.

33-Classifry the rolling contact bearings.

 Ball Bearings:

• Deep groove ball bearing
• Angular contact ball bearings
• Self-aligning ball bearing
• Thrust ball bearing

B. Roller Bearings:

• Spherical roller bearing
• Cylindrical roller bearing
• Taper roller bearing
• Thrust Roller and Needle Bearings.

34-What are the different parts of rolling contact bearings?

• Outer race
• Inner race
• Cage
• Rolling elements

 35-What are the reasons for bearing failure?

Reasons for Bearings Failure:

• Lack of lubrication
• Excessive vibrations
• Dirt in lubrication
• Misalignment
• Overload
• Fatigue, repeated jerk loads
• Excessive clearance or interference fit

36-List down the Do’s and Don’ts for bearings handling and storing.

Following are the Do’s and Don’ts for bearing:

DO’s

• Keep the bearings in protective packing till they are mounted.
• Cover the assembled bearings appropriately to avoid dust ingress.
• Store the bearings in dry area.
• Use tubes, Puller, Hydraulic nuts, and other tools for removal/fixing of the bearings.
• Use correct amount of specified lubrication at right time with correct procedure.

Don’ts

• Over lubrication is a silent killer for bearings. Do not over lubricate.
• Never allow welding at bearings that will damage the contact surfaces.
• Don’t let the bearing stand upright, store them flat on their sides.
• Do not use water to clean the bearings.

37-What is meant by bearing clearance?

Bearing clearance is the clearance or gap between outer race or inner race and rolling elements. And in journal bearings it is the clearance between shaft and bearing liners.

38-How do you calculate the bearing clearance of bearing having bore diameter D?

As a thumb rule Bearing clearance minimum = 0.00185 X D

Bearing clearance maximum =0.00254 X D

39-What is the hardness of rolling contact bearings?

Bearings are usually made up of hardened chromium steel having hardness 55–60 HRC.

40-What are the reasons for bearing seize?

• Lack of lubrication
• Less clearance
• Over speed
• Contaminant in lubricant

41-What are the reasons for bearing scoring and smearing?

• Incorrect lubrication
• Contamination
• Misalignment
• Sharp speed ramp up

42-What is the major difference between a deep groove ball bearing 6205 ZZ and 6305 ZZ?

A deep groove ball bearing 6305 ZZ is bigger in size having more outer diameter and width, whereas 6205 ZZ is smaller in size with less OD and width. Both the bearings have same bore diameter.

 43-What is the shaft size on which a deep groove ball bearing of size 6215 C3 fits?

For deep groove ball bearing

Shaft size = 15 X 5 = 75 mm…(Last two digits of bearing number X 5)

Note: In above bearing 6 indicates type of ball bearing (deep groove, angular contact etc), 2 indicates the size of a bearing (OD and width) and last two digits indicate the bore of bearing. In any bearing last two digits indicate the bore diameter.

In tapered bore spherical roller bearings 22222 K/C3,

Shaft size 5 (Last two digits) 22 X 5 + Sleeve thickness

44-If a machine has higher vibration in only vertical/horizontal/axial direction then what does it indicate?

• If there is more vertical vibration and other direction vibrations are minimum, then indicates looseness in machine components.
• If there is more horizontal vibration and other direction vibrations are minimum, then indicates unbalance of rotating part.
• If there is more axial vibration and other direction vibrations are minimum, then indicates misalignment.

45-Define the term lubrication

Lubrication is any procedure that has the effect of reducing friction and wear

46-What is the purpose of lubrication?

Purpose of Lubrication:

• To reduce friction
• To minimize wear & tear
• To cool the parts
• To disperse contaminants
• Acts as sealant
• To transfer the power
• Provides electrical isolation rotor to ground

47- What are the main contents of lubricants?

•    90% Base oil Synthetic or mineral oil
•     7–8% Additives
•    2–3% Thickeners 

48-What are the different types of lubrication?

Following are the some lubrication types:

• Boundary lubrication
• Full film lubrication
• Hydrodynamic lubrication

49-What are the different properties of lubricants especially of turbine lube oil which is very must to check before use?

• Viscosity
• Flash point
• Pour point
• Moisture content
• Acid No
• Neutralization value
• Sediments content
• Mineral test
• Demulsibility
• Dissolved gas analysis

 50-A 6205 2Z bearing has OD 52 mm and width 15 mm, what amount of grease is required for first lubrication?

We have,

Bearing lubrication quantity in grams = Bearing OD (D) X Width (B) X 0.05

                                                                   = 52 X 15 X 0.05 = 39 grams.

51-What are the effects of misalignment?

• Effects of misalignment
• Excessive vibration and noise
• Bearings failure
• Seals failure
• Excessive heat generation
• Higher power consumption
• No equipment reliability

52-What are the possible reasons for misalignment?

• Loose/broken foundation bolts
• Loose shim packs/dowel pins
• Loose coupling bolts
• Jerk loads due to frequent start/stop

Online leakage arresters & sealants

Calculation of raw water requirement and Reservoir size for a power plant

In order to calculate the raw water requirement for a power plant, first of all we should know the losses in power plant.
As per new law raw water consumption should be 2.5 liters/kwh, presently it is around 4.5 liters/kwh.

Different losses in power plants :

A-Boiler

• Boiler blow down : 2% of steam generation

• Deaerator vent loss : 0.1% of Deaeration

• Soot blower steam : 0.8 to 1% of Boiler steam generation per day

B-Turbine & Process steam 

• Process steam lines losses due to drain & traps : 1% of steam generation

• Cooling water evaporation loss : 2% of cooling water circulation

• Losses in process return condensate : 10% of steam given to process

C-Water treatment plant (WTP)

• Clarifier : 2% of total raw water input or 5% of DM water generation

• MGF back wash water loss : 40% of MGF inlet water (Back wash frequency : once in every 8 hours of operation)

• Reverse Osmosis (RO) plant reject water : 25% of total input water to RO plant or 35% of RO capacity

• Water losses in DM plant regeneration process : 30% of DM plant capacity for every regeneration process

D-Other miscellaneous losses & cunsumption

• Losses due to leakages : 10 MT/day

• Water used for drinking & service : 10 MT/day

• Fire fighting system : 40 MT/day...Considered Jockey pump is running for 4 hours in a day at 10 M3/hr flow

Let us discuss the above lossed by considering an example of 20 MW Cogeneration based power plant.This plant has Boiler of capacity 130 TPH,Steam given to process application is 65 TPH. Calculate the RO plant , DM plant capacity & Raw water consumption per day.

Calculation of total water requirement by considering various losses

Step-I

Calculate DM water requirement per hour

• Blow down loss = 130 X 2% = 2.6 TPH

• Deaerator vent loss = 130 X 0.1% = 0.13 TPH

• Soot blower steam loss = 130 X 1% / 24 = 0.054 TPH

• Process steam loss = 130 X 1% = 1.3 TPH

• Losses in process steam condensate = 65 X 10% = 6.5 TPH

•  SO total DM plant capacity = 2.6 + 0.13 + 0.054 + 1.3 + 6.5 =10.58 TPH

• DM plant is being stopped for 4 hours in a day for regeneration purpose,

Therefore, DM plant capacity = 10.58 X 24/20 = 12.696 TPH

Step-II

Calculate the cooling water make up.

• From the above example, quantity of steam going to the condenser is 130-65 = 65 TPH

• Cooling water circulation flow is = 65 X 60 = 3900 TPH

• Note: Considered 60 TPH of CW is required to condensate the 65 TPH of steam

• So total cooling tower loss = 3900 X 2% = 78 TPH 

Step-III

Calculate WTP losses

• Clarifier blow down losses = 12.96 X 5% = 0.65 TPH

• RO plant reject water = 12.96 X 35% = 4.53 TPH

• DM water regeneration losses (SAC+SBA+MB) = (3 X (12.96 X 30% ))/24 = 0.486 TPH

Step-IV

Other miscellaneous

• Leakage losses = 10 TPD =10/24 =0.24 TPH

• Drinking & Service water = 10 TPD =10/24 = 0.24 TPH

• Fire fighting water = 10 TPD =10/24 =0.24 TPH

So total raw water requirement =12.696 + 78 + 0.65 + 4.53 + 0.486 +0.24 + 0.24 +0.24 = 97.08 =97 TPH

Total raw water required in a day = 97 X 24 =2328 TPD

Take 10% Extra Margin = 2328 X 110% = 2561 TPD

Size of raw water reservoir by considering water requirement for 7 days

Total storage capacity = 7 X 2561 =17927 Tonnes or M3.....As density of water is 1000 kg/m3

Take reservoir  depth of 6 meter

Then tank size 

17927 M3 = Length X Breadth X 6

L X B =2987.3 M2

Take length of the reservoir = 60 meters

then breadth of the reservoir will be = 2987.3/60 = 49.8 =50 M3

So size of the reservoir will be = 60 X 50 X 6 meters

'

10-Tips to reduce LOI in Boilers

Calculation of coal handling plant and bunker capacity

Coal handling plant (CHP) consists of ;

• Coal unloader from vehicles

• Coal storage yards

• Coal vibrofeeders

• Coal vibroscreen

• Magnetic seprators

• Coal crusher

• Coal weigher

• Series of Conveyors

• Water sprinkling system

• Coal shuttle conveyor

• Coal storage bunkers

Inorder to calculate the coal handling and coal storage bunkercapacity, first we need to know the coal requireent per day for a plant, coal handling maintenance stratergy

We shall calculate the coal handling & bunker capacity by taking an example of 25 MW thermal power plant consuming coal of GCV 4900 kcal/kg & having heat rate 3200 kcal/kwh running at 100% PLF.

Plant running load = 50 MW X 100/100 = 25 MW

We know that coal consumed = Heat rate X Power generation / (Coal GCV)

Coal consumed =3200 X 25 / 4900 = 16.32 MT/hour

Coal consumed in a day = 16.32 X 24 =391.68 MT take round figure 392 MT/day

A-Coal handling capacity calculation

Consderations:

• Coal handling plat is operated 14 hours in a day

          Note: Some designer consider operating hour of the plant is 10-12hours

• 10 hours used for cleaning and preventive maintenance

CHP capacity =392/14 =28 Tonnes / hour

Take 50% extra margin on capacity

CHP capacity = 28 X 150% = 42 Tonnes/hour

Power plant Maintenance Calculations

B-Coal yard design

For 25 MW power plants coal can be supplied through truckes & lories .It is sufficient to keeping the stock of 7-8 days coal in coalshed.

Coal shed capacity = Lenght 50 meter X Width 30 meter X Avg Height of coalheap 4 meters X coal density 800 kg/m3 = 4800 MT

Which han hold the coal for 12 days of full load operation days

If the plant is of bigger size, then the coal is supplied through railway & Wagons & stored at yard..

Yard capacity =Length X Width X avg.Height of 6 meter X Coal densitykg/3

Coal from Shed & yard is then shifted & dozed to vibrofeeder hopper through chain & tractor dozers

C-Selection of bunker capacity

Buker capacity is generally designed for 12-14 hours of full load operation

Let us calculate with 12 hours of operation hours

Coal consumption in 12 hours = 392 X 12 /24 =196 MT

Volume of bunker = 196 X 1000/800 = 245 M3

Take 10% of extra margin on bunker as some of coal stucked on bunker wall surface cannot be used for Boiler. Which is unaccountable

So total water volume of Bunker = 245 X 110% = 269.5 take round figure =270 M3

Basics terms used in Thermodynamics & related calculations

                                Thermodynamics:

It is an axiomatic science which deals with the relations among heat, work and properties of equilibrium system.It is a science, which deals with interaction between energy and material system.

Terms used in Thermodynamic science:

System:It is a finite quantity of matter or a prescribed region of space.

Boundary:It is the actual or hypothetical envelope enclosing the system.

Closed system: If the boundary of the system is impervious to the flow of matter is called  closed system.

Open system: there is  flow of matter into and out of the system

Isolated system:An isolated system doesnot exchange energy  & matter with any of the other system

Homogeneous system: A system which contains single phase is called homogeneous system

Hetergeneous system : Here system consists of more than two phases

Pure substance: A substance which is homogeneous in compossition &  chemical aggregation

State: It is the condition of the system at an instant of time as described or measured by its properties

Cycle:Any process or series of processes whose end states are identical is termed as a cycle

Process: It occurs when the system undergoes a change in state or n energy transfer at a steady state.

Reversible process: A process is one which can be stopped at any stage and reversed so that the system & surroundings are exactly restored to their initial states.

Example:Exapnsion & compression of springs, Electrolysis

Irrreversible process: In thisprocess heat is transferred through a finite temperature.

Example:Combustion, free expansion, heat transfer

Temperature: It is state of a body  which distinguishes a hot body from a cold body.It is measured in Fahrenheit, Degree centi gare, Kelvin etc

Pressure: It is a force per unit area.Pressure is exerted by gases, vapours & liquids.It is measured in Kg/cm2,Bar, Pascal, N/m2, N/mm2, mm of water column.

Thermal Equilibrium: Temperature of the system does not change with time and has same value at all the points of the system.

Mechanical Equilibrium: It is the condition where there are no unbalanced forces within the system, pressure in the system is same at all the points.

Chemical Equilibrium: Composition of chemicals does not change with time and no chemical reaction takes place in the system.

Enthalpy:It is the total heat content of the steam. Expressed in h…kcal/kg.

Entropy: is a function of a quantity of heat, which shows the possibility of conversion that heat into work. That is, if Entropy is more then there is minimum availability for conversion of heat into work and for minimum entropy there is maximum availability for conversion of heat into work. It is measured in kcal/kg.

Boyle’s law:At constant temperature, pressure of a perfect gas is inversely proportional to its volume.

P = 1/V or PV = Constant

Charles law:At constant pressure, volume of a given mass of a gas is directly varies as its temperature.

V = T

(V1/T1) = (V2/T2) = …… Constant

Gay-Lussac law:The pressure of a given mass of a perfect gas varies directly as its temperature, when the volume remains constant.

P = T

(P1/T1) = (P2/T2) = Constant

Relation among pressure, temperature, enthalpy and specific volume of steam

• At constant temperature enthalpy decreases with increase in pressure
• At constant pressure enthalpy increases with increase in temperature.
• Specific volume increases with decrease in pressure and increase in temperature
• Enthalpy of evaporation decreases with increase in pressure and temperatures.

Specific heat: Amount of heat required to raise the temperature of unit mass of substance by 1 degree centigrade.It is measured in constant pressure Cp & Constant volume Cv.Mesured in kcal/kg

Specific volume: It is the volume occupied by the unit mass of the system.It is measured in m3/kg.

Intensive property:It is the property of steam, whose value for the entire system is not equal to the sum of their values for the individual parts of the system.

Example: Temperature, pressure and density.

Extensive property:It is the property of steam, whose value for the entire system is equal to the sum of their values for the individual parts of the system.

Example: Volume and mass.

Calculations for BOE exam

Zeroth law of thermodynamics:The law states that, when two bodies are in thermal equilibrium with a third body, they are also in thermal equilibrium with each other.

First law of thermodynamics: Law states that when a system undergoes a thermodynamic cycle then the net heat supplied to the system from surroundings is equal to net work done by the system on its surroundings.

Second law of thermodynamics:It states that, it is impossible to construct an engine working on a cyclic process, to transfer heat from a body at a lower temperature to a body at a higher temperature without the aid of an external energy.

Different types of heat given to water to convert it into steam

Sensible heat : Amount of heat given to water to bring water from its normal temperature to Boiling temperature or saturation temperature. It is denoted by hf & measured in kcal/kg.

Latent heat: Amount of heat iven to saturated water to convert into saturated steam.It is denoted by hf g& measured in kcal/kg.

Total heat : It is the quantity of heat required to convert 1 kg of water into wet steam at constant pressure.It is the sum of total heat of water and the latent heat.It is also called as enthalpy.Denoted as h for wet steam and hg for dry steam

h = hf + x hfg

x is the dryness fraction of steam.

Dryness fraction of steam (x) : It is the ratio of the mass of actual dry steam to the mass of the steam containing it.

x = Ms/(Ms + Mw)


Superheated steam: when steam is heated after it has become dry and saturated.It is called superheated steam.

hsup = hf + hfg + Cps (Tsup-Ts)

Whwre, Cps = Specific heat of super heated steam = 0.5 kcl/kg
Tsup = Temperature of superheated steam
Ts = Saturated temperature at saturated pressure
(Tsup-Ts) is called as Degree of superheat.

Basic calculations:

1-Convert the pressure 80 cm of Hg to Kpa

Pressure at 760 mm of Hg = Density X g X h

                                           = 13.596 X 1000 X 9.81 X 760/1000

                                           = 101325 Pa =101.325 Kpa

2-Convert 2mm of water column pressure to Pascal

Pressure due to 2 mm of water column =1000 X 9.81 X 2 =19620 Pa

3-On a piston of 15 cm diameter of a force of 2000 N is applied uniformely caluclate the pressure on piston

Pressure = Force applied/Area of the piston

P = 2000 / ((3.142 X 0.152/4).....Area of piston =Pi X D2/4

P = 113636.363 N/M2

3-Atube contains an oil of specific gravity 0.88 to a depth of 1500 mm, find the gauge pressure at this depth

SG of oil =0.88

Depth of the oilin tube =1500 mm

We know that,

P = Density X g X h

P = 0.88 X1000 X  9.81 X 1.5...As SG =Density of oil / Density of water, Density of water is 1000 kg/m3

P = 12,949.2 N=/m2

1 N/M2 =1 X 10-5 kg/cm2

P =0.13 kg/cm2

4-Determine the dryness fraction of steam which has 2 kg of water in suspension with 60 kg of steam

Mass of dry steam Ms = 60 kg

Mass ofwater in suspension = 2 kg

Dryness fraction of steam x = 60/(60+2) =0.967

5-Determine the amount of heat given to 3 kg of water at 25 deg c to convert it into steam at 7 kg/cm2G and 0.9 dry.

Mass of water =Mw = 3 kg

Temperature of water = tw = 25 deg c

Pressure of steam = 7 kg/cm2

Fom the steam table at 7 kg/cm2 pressure (take absolute pressure i.e at 8 kg/cm2),

hf =172.73 kcal/kg

hfg = 488.22 kcal/kg

We have the formula for enthalpy of 1 kg of steam at 0 deg c is

h = hf + xhfg =172.73 + 0.9 X 488.22 =612.12 kcal/kg

Sensible heat of 1 kg of water at 25 deg c is = Mw X Cpw X (tw-t0)

=3 X 1 X (25-0) =75 kcal/kg

Net quantity of heat supplied per kg of water =612.12-75 =537.12 kcal/kg

Therefore total amount of heat to be supplied =3 X 537.12 =1611.38 kcal

6-Determine the mass of 0.2 m3 of wet steam at a pressure of 5 kg/cm2G and dryness fraction of 0.85. Also calculate heat of 1 m3 of steam.

Steam pressure P = 5 kg/cm2

Dryness fraction x =0.85

Fro steam table at 5 kg/cm2G

hf = 160.64 kcal/kg

hfg =497.47 kcal/kg

 Vg = 0.312 m3/kg

Density = 1/(x X Vg) =1/(0.85 X 0.312) =3.77 kg/m3

Mass of 0.2 m3 of steam = 0.2 X 3.77 =0.754 kg

Total heat of 1m3 of steam which has 3.77 kg of mass

=3.77h

=3.77 X (hf + xhfg) = 3.77 X (160.64 + 0.85 X 497.47) = 2199.7 kcal

7-What amount of heat would be required to produce 100 MT of steam at pressure of 67 barA and temperature 490 deg c from 150 deg c? Consider Cp of super heated steam =0.5 kcal/kg

Mass of steam = 100 MT =100 X 1000 =100000 kg

Steam pressure P=67 kg/cm2 absolute

Temperature of steam tsup =490 deg C

Temperature of water t =150 deg C

Refer steam table at 67 kg/cm2A & 490 deg c 

hf =300 kcal/kg

hfg =362.91 kcal/kg

ts = 283.75 kcal/kg

hsup = hf + hfg + Cps X (tsup-ts) =300 + 362.91 + 0.5 X (490-283.75) =766.035 kcal

Amount of heat already associated with 1 kg of water = 1 X 1 X (150-25) =125 kcal..Specifc heat of water is 1 kcal/kg

So net heat required = 766.035-125 =641.035 kcal

Total amount of heat required =641.035 X 100000 =64,103,500‬ kcal