Saturday, 9 April 2011

Converting Textile Production in Kgs from Linear meter

Normally in Production in Textile Processing is recorded in linear meter. Sometimes the production in linear meters does not serve the purpose when it comes to specific energy consumption. For conversion of linear meter into kgs 3 parameters are required i) Linear Meters ii) Average Width iii) GSM

Production (kg) = 0.001 * Length (meter) * Width (meter) * GSM (gms / sq. meter)

Boiler Efficiency Calculations


The ASME method for Boiler Efficiency Calculation PTC 4 is elaborate and requires many input data points, measurements, and samples. With some practical assumptions, the calculations are made simple and are shown in this article.
The Losses method is used to calculate boiler efficiency. Each of the different Losses is calculated to determine the efficiency.

Data Required

For a coal fired boiler to calculate these losses we require the following data.
  • Higher Heating Value of coal on as Fired Basis HHV- kJ / kg
  • Proximate Analysis of Coal on as Fired basis which include
    • Fixed Carbon FC %
    • Volatile Matter VM %
    • Ash %
    • Moisture %
  • Hydrogen % in coal. This normally available only from an Ultimate analysis. This can be used from a past historic data for similar type of coal. The value normally is in the range of 2 - 4 %. 
  • Gas temperature at boiler exit - Tg °C
  • Ambient Temperature - Dry bulb - Ta °C
  • Oxygen in Flue gas on a dry basis - O2 %
    • The oxygen measurement should be from a location near to the Temperature point.
    • On line Oxygen, measurements are normally on wet basis.
    • Sampled Oxygen measurements are on dry basis. In coal fired units O2 % ( dry basis)= (O2 % ( wet basis)) / 0.9 
  • Unburnt Carbon in ash U %
    • In large coal fired plants, ash collection is in different locations. This is mainly at the Furnace bottom and the Precipitator Hoppers. The U % should be on a weighted average basis. A ratio of 85: 15 is practical between furnace bottom and precipitator ash collection.
    • U %=[U-fly ash x 85 + U-bot ash x 15] / 100
  • Carbon Monoxide in Flue gas - CO ppm
    • This is normally applicable in oil and gas fired units but can be applicable in Coal fired units if the combustion is very bad.
Most of these data is available readily in a power plant from online instruments and from daily analysis reports.
ASME or other codes require the ultimate analysis of Coal for finding the air and gas quantities to use in the efficiency calculations. This normally takes time. Here we make an assumption because the Stoichiometric air quantity lies within a small band for fossil fuels because of the interdependence of Carbon, Hydrogen and the Calorific value.
  • Stoichiometric Air qty
    • = 0.325 kg/ MJ for Coal Firing.

Controllable Losses

Losses itself can be categorised into three. First are the losses that the plant operators can control.
Following losses are in this category.
  • Loss 1. Exit gas loss or Dry gas Loss.
    • = 0.72 x [Tg - Ta ] / [21- O2]
  • Loss2. Unburnt Carbon loss in ash – normally for Coal fired units.
    • = U × Ash × 33810 / [100 - U ] / HHV 
  • Loss3. Unburnt Fuel as CO - normally for oil or gas fired boilers.
    • =0.0067 x CO / [21-O2 ]

Inherent Losses

Some Losses are due to the inherent characteristics of the fuel. The operator really has no control over these losses.
  • Loss 4 - due to the Hydrogen in the coal
  • Loss 5 - due to the Moisture in the coal
    • =[9 x H + M] x [1.88 × Tg + 2500 - 4.18×Ta] / HHV
Hydrogen on combustion forms water and together with the moisture in the coal evaporates and leaves with the flue gas. The vaporisation takes away some heat from the combustion and reduces boiler efficiency. This is part of the energy conversion process.

Minor Losses

Apart from the main losses mentioned above there are many minor losses. Since these are mainly uncontrollable linked to the main losses we assume the value of these losses.
  • Loss 6 -Radiation loss.
  • Loss 7 -Heat loss in ash.
  • Loss 8 -Heat loss in coal mill rejects.
  • Loss 9 -Loss due to moisture in air
    • 1 % for coal fired boilers with ash less than 20 %. 
    • 1.5 % for coal fired boilers with ash greater than 20 %.
Efficiency % = 100 - Sum of all Losses %

The Boiler engineer should really worry about category one. This is what the operator can adjust and reduce.


Reference: http://www.brighthub.com/engineering/mechanical/articles/54252.aspx#ixzz1J6E63jTG

Sunday, 3 April 2011

Heat Calculations Examples


EXAMPLE 1: A tubular Bushing Heater is required to heat a 10 gallon tank of water. Initial water temperature is at room temperature of approx. 70° F. Final water temperature required is 100° F. Water must be brought up to final temperature in 1 hour. No additional water is added to the system. Customer has an ideal system with zero heat loss. Calculate the wattage required to heat only the water (ignore the tank for this example.).
SOLUTION:
1. Basic equation for initial material heatup:
Wattage =weight (lbs) x spec. heat (btu/lb °F) x temp. change (°F)

3.412 (btu to watt/hr conversion) x heat up time in hours

2. From properties of liquids and engineering conversions charts:
a. 1 gallon of water = 8.3 pounds (lbs)
b. specific heat of water = 1.000 btu/lb
3. Plug numbers into equation and solve:
Wattage =(10 gal x 8.3 lbs/gal) x 1.000 btu/lb °F x (100 °F - 70 °F)

(3.412 btu/watt hr) x 1 hr

= 7.29.8 watts
Add 20% allowance = 730 watts x 1.20 = 875 watts
EXAMPLE 2: Customer requires Flanged Tubular Heater to heat a 200 gallon tank of water. Initial water temperature is at room temperature of approx. 70° F. Final water temperature required is 150° F. Water must be brought up to final temperature in 3 hours. No additional water is added to the system. Tank has approximately 25 square feet of 1Ú4Ó thick uninsulated steel walls and a total tank weight of 254 lbs. Water surface area is approximately 12 square feet and is open to the environment. Tank is installed in a room at 70° F.
SOLUTION:
1. Remember basic equations:
a. For material and tank heatup
Wattage =weight (lbs) x spec. heat (btu/lb °F) x temp. change (°F)

3.412 (btu to watt/hr conversion) x heat up time in hours

b. For heat loss from water surface:
average wattage loss =liquid surface area (sq. ft.) x watts/sq. ft. loss at final temp

2

c. For heat loss from walls of uninsulated steel tank
average wattage loss =tank wall surface area (sq. ft.) x watts/sq. ft. loss at final temp

2

2. From properties of liquids chart, engineering conversions chart, and heat loss graphs:
  1. 1 gallon of water = 8.3 pounds (lbs)
  2. Specific heat of water = 1.000 btu/lb
  3. Specific heat of steel = .12 btu/lb
  4. Heat loss from water surface at final temp = 300 w/sq. ft.
  5. Heat loss from uninsulated steel walls at final temp = approx. 25 w/sq. ft. x .83 (averaging factor) = 20.8 w/sq. ft.
3. Plug numbers into equations and solve:
wattage (to heat water) =(200 gal x 8.3 lbs/gal) x 1.000 btu/lb x (150° F - 70° F)

(3.412 btu/watt hr) x 3 hr

wattage (to heat water) =(254 lbs) x .12 btu/lb x (150° F - 70° F)

(3.412 btu/watt hr) x 3 hr

wattage (ave. loss from water surface) =12 sq. ft. x 300 w/sq. ft.

2

wattage (ave. loss from tank walls) =25 sq. ft. x 20.8 w/sq. ft.

2

Total wattage = 12,974 w + 238 w + 1800 w + 260 w = 15,272 w (15.3 kw)
Total wattage with 20% contingency = 18,326 watts (18.3 kw)
EXAMPLE 3: Same conditions as example 1 except tank and water are already up to final operating temperature. No additional water or other material is added to system. How much wattage is required to maintain the tank and water at the 150° F temperature in the 70° F room?
SOLUTION:
1. The tank and water are already up to the final operating temperature, therefore, no heat is required to bring the system up to temperature.
Heatup wattage = 0
2. The only factors affecting heat requirements are heat losses from the water surface and tank walls at operating temperature.
3. Heat equations to counteract heat losses at operating temperature
a. For heat loss from water surface:
wattage loss = Liquid surface area (sq. ft.) x watts/sp. ft. loss at final temp
b. For heat loss from walls of uninsulated steel tank:
wattage loss = tank wall surface area (sq. ft.) x watts/sq. ft. loss at final temp
4. Use same heat loss - watts/sq. ft. values from graphs as in example 2:
a. Heat loss from water surface at final temp = 300 w/sq. ft.
b. Heat loss from uninsulated steel walls at final temp = approx. 25 w/sq. ft. x 8.3 (averaging factor) = 20.8 w/sq. ft.
5. Plug numbers into equations and solve:
wattage (ave. loss from water surface) = 12 sq. ft. x 300 w/sq. ft.
wattage (ave. loss from tank walls) = 25 sq. ft. x 20.8 w/sq. ft.
Total wattage (to maintain temp) = 3600 w + 520 w = 4120 w (4.1 kw)
Total wattage with 20% contingency = 4944 watts (4.9 kw)

Thursday, 31 March 2011

Cavitation in Pumps


Cavitation means that cavities or bubbles are forming in the liquid that we're pumping. These cavities form at the low pressure or suction side of the pump, causing several things to happen all at once:
The cavities or bubbles will collapse when they pass into the higher regions of pressure, causing noise, vibration, and damage to many of the components.
We experience a loss in capacity.
The pump can no longer build the same head (pressure)
The pump's efficiency drops.

The cavities form for five basic reasons and it's common practice to lump all of them into the general classification of cavitation. This is an error because we'll learn that to correct each of these conditions, we must understand why they occur and how to fix them. Here they are in no particular order :
Vaporization
Air ingestion (Not really cavitation, but has similar symptoms)
Internal recirculation
Flow turbulence
The Vane Passing Syndrome

Vaporization .

A fluid vaporizes when its pressure becomes too low, or its temperature too high. All centrifugal pumps have a required head (pressure) at the suction side of the pump to prevent this vaporization. This head requirement is supplied to us by the pump manufacturer and is calculated with the assumption that fresh water at 68 degrees Fahrenheit (Twenty degrees Centigrade) is the fluid being pumped.

Since there are losses in the piping leading from the source to the suction of the pump, we must determine the head after these losses are calculated. Another way to say this is that a Net Positive Suction Head is Required (N.P.S.H.R.) to prevent the fluid from vaporizing.

We take the Net Positive Suction Head Available (N.P.S.H.A.) subtract the Vapor Pressure of the product we are pumping, and this number must be equal to or greater than the Net Positive Suction Head Required.

To cure vaporization problems you must either increase the suction head, lower the fluid temperature, or decrease the N.P.S.H. Required. We shall look at each possibility:

Increase the suction head
Raise the liquid level in the tank
Raise the tank
Pressurize the tank
Place the pump in a pit
Reduce the piping losses. These losses occur for a variety of reasons that include :
The system was designed incorrectly. There are too many fittings and/or the piping is too small in diameter.
A pipe liner has collapsed.
Solids have built up on the inside of the pipe.
The suction pipe collapsed when it was run over by a heavy vehicle.
A suction strainer is clogged.
Be sure the tank vent is open and not obstructed. Vents can freeze in cold weather
Something is stuck in the pipe, It either formed there, or was left during the last time the system was opened . Maybe a check valve is broken and the seat is stuck in the pipe.
The inside of the pipe, or a fitting has corroded.
A bigger pump has been installed and the existing system has too much loss for the increased capacity.
A globe valve was used to replace a gate valve.
A heating jacket has frozen and collapsed the pipe.
A gasket is protruding into the piping.
The pump speed has increased.
Install a booster pump

Lower the pumping fluid temperature
Injecting a small amount of cooler fluid at the suction is often practical.
Insulate the piping from the sun's rays.
Be careful of discharge recirculation lines. They can heat the suction fluid.

Reduce the N.P.S.H. Required
Use a double suction pump. This can reduce the N.P.S.H.R. by as much as 25%, or in some cases it will allow you to raise the pump speed by 40%
Use a slower speed pump.
Use a pump with a larger, impeller eye opening.
If possible, install an Inducer. These inducers can cut N.P.S.H.R. by almost 50%.
Use several smaller pumps. Three half capacity pumps can be cheaper than one large pump plus a spare. This will also conserve energy at lighter loads.

It's a general rule of thumb that hot water and gas free hydrocarbons can use up to 50% of normal cold water N.P.S.H. requirements, or 10 feet (3 meters), whichever is smaller. I would suggest you use this as a safety margin, rather than design for it.

Air ingestion (Not really cavitation, but acts like it)

A centrifugal pump can handle 0.5% air by volume. At 6% air the results can be disastrous. Air gets into as system in several ways that include :
Through the packing stuffing box. This occurs in any packed pump that lifts liquid, pumps from a condenser, evaporator, or any piece of equipment that runs in vacuum.
Valves located above the water line.
Through leaking flanges.
Pulling air through a vortexing fluid.
If a bypass line has been installed too close to the suction, it will increase the temperature of the incoming fluid.
Any time the suction inlet pipe looses fluid. This can occur when the level gets too low, or there is a false reading on the gauge because the float is stuck on a corroded rod.

Both vaporization and air ingestion have an adverse affect on the pump. The bubbles collapse as they pass from the eye of the pump to the higher pressure side of the impeller. Air ingestion seldom causes damage to the impeller or casing. The main effect of air ingestion is loss of capacity.

Although air ingestion and vaporization can both occur, they have separate solutions. Air ingestion is not as severe as vaporization and seldom causes damage, but it does lower the capacity of the pump.

Internal Recirculation

This condition is visible on the leading edge of the impeller, close to the outside diameter, working its way back to the middle of the vane. It can also be found at the suction eye of the pump.

As the name implies, the fluid recirculates increasing its velocity until it vaporizes and then collapses in the surrounding higher pressure. This has always been a problem with low NPSH pumps and the term Suction Specific Speed to guide you in determining how close you have to operate to the B.E.P. of a pump to prevent the problem.

The higher the number the smaller the window in which you can operate. The numbers range between 3,000 and 20,000. Water pumps should stay between 3,000 and 12,000. Here is the formula to determine the suction specific speed number of your pump:


rpm = Pump speed

Capacity = Gallons per minute, or liters per second of the largest impeller at its BEP

Head= Net positive suction head required (feet or meters) at that rpm
For a double suction pump the flow is divided by 2 since there are 2 impeller eyes
Try to buy pumps with a suction specific speed number lower than 8500.(5200 metric ) forget those over 12000 (8000 metric) except for extreme circumstances.
Mixed hydrocarbons and hot water at 9000 to 12000 (5500 to 7300 metric) or higher, can probably operate satisfactorily.
High specific speed indicates the impeller eye is larger than normal, and efficiency may be compromised to obtain a low NPSH required.
Higher values of specific speed may require special designs, and operate with some cavitation.
Normally a pump operating 50% below its best efficiency point (B.E.P.) is less reliable.

With an open impeller pump you can usually correct the internal recirculation problem by adjusting the impeller clearance to the manufacturers specifications. Closed impeller pumps present a bigger problem and the most practical solution seems to be to contact the manufacturer for an evaluation of the impeller design and a possible change in the design of the impeller or the wear ring clearances.

Turbulence

We always prefer to have liquid flowing through the piping at a constant velocity. Corrosion or obstructions can change the velocity of this liquid, and any time you change the velocity of a liquid, you change its pressure. Good piping layouts would include :
Ten diameters of pipe between the pump suction and the first elbow.
In multiple pump arrangements locate the suction bells in separate bays so that one pump suction will not interfere with another. If this is not practical, a number of units can be installed in a single large sump provided that :
The pumps should be positioned in a line perpendicular to the approaching flow.
There must be a minimum spacing of at least two suction diameters between pump center lines.
All pumps are running.
The upstream conditions should have a minimum straight run of ten pipe diameters to provide uniform flow to the suction bells.
Each pump capacity must be less than 15,000 gpm..
Back wall clearance distance to the centerline of the pump must be at least 0.75 of the suction diameter.
Bottom clearance should be approximately 0.30 (30%) of the suction diameter
The minimum submergence should be as follows:

FLOW MINIMUM SUBMERGENCE

20,000 GPM
4 FEET

100,000 GPM
8 FEET

180,000 GPM
10 FEET

200,000 GPM
11 FEET

250,000 GPM
12 FEET

The metric numbers are :
FLOW MINIMUM SUBMERGENCE

4,500 M3/HR
1.2 METERS

22,500 M3/HR
2.5 METERS

40,000 M3/HR
3.0 METERS

45,000 M3/HR
3.4 METERS

55,000 M3/HR
3.7 METERS


The Vane Passing Syndrome

This type of cavitation damage is caused when the OD of the impeller passes too close to the pump cutwater. The velocity of the liquid increases as it flows through this small passage, lowering the fluid pressure and causing local vaporization. The bubbles then collapse at the higher pressure just beyond the cutwater. This is where you should look for volute damage. You'll need a flashlight and mirror to see the damage, unless it has penetrated to the outside of the volute.

The damage is limited to the center of the impeller vane. If it's a closed impeller, the damage will not extend into the shrouds. You can prevent this problem, if you keep a minimum impeller tip to cutwater clearance of 4% of the impeller diameter in the smaller impeller sizes (less than 14' or 355 mm.) and a 6% clearance in the larger impeller sizes (greater than 14" or 355 mm.).

To prevent excessive shaft movement, some manufacturers install bulkhead rings in the suction eye. At the discharge side, rings can be manufactured to extend from the walls to the impeller shrouds.
(Ref: mcnallyinstitute)

Pump Efficiency

Multiply the pump head by the pump's capacity, and then use a simple conversion number.

Flow = 300 gallons per minute of fresh water as measured coming from the pump discharge.
Head = 160 feet. We measured it at the discharge side of the pump and corrected it for the fact that the gage was two feet above the pump center line.
Note:If there were any positive head on the suction side of the pump that head would have to be subtracted. A negative suction head would be added to the discharge head.

 The centrifugal pump pumps the difference between the suction and the discharge heads. There are three kinds of discharge head:

Static head. The height we are pumping to, or the height to the discharge piping outlet that is filling the tank from the top. Note: that if you are filling the tank from the bottom, the static head will be constantly changing.

Pressure head. If we are pumping to a pressurized vessel (like a boiler) we must convert the pressure units (psi. or Kg.) to head units (feet or meters).

System or dynamic head. Caused by friction in the pipes, fittings, and system components. We get this number by making the calculations from published charts

Suction head is measured the same way.

If the liquid level is above the pump center line, that level is a positive suction head. If the pump is lifting a liquid level from below its center line, it is a negative suction head.
If the pump is pumping liquid from a pressurized vessel, you must convert this pressure to a positive suction head. A vacuum in the tank would be converted to a negative suction head.
Friction in the pipes, fittings, and associated hardware is a negative suction head.
Negative suction heads are added to the pump discharge head, positive suctions heads are subtracted from the pump discharge head.

Here is the formula for measuring the horsepower out of the pump:
 




Remember that we are using the actual horsepower or kilowatts going into the pump and not the horsepower or kilowatts required by the electric motor. Most motors run some where near 85% efficient.

An 85% efficient motor turning a 76% efficient pump, gives you a real efficiency of 0 .85 x 0.76 = 0.65 or 65% efficient.

A survey of popular pump brands demonstrates that pump efficiencies range from 15% to over 90%. The question then arises, "Is this very wide range due to poor selection, poor design, or some other variable which would interfere with good performance?" The best available evidence suggests that pump efficiency is directly related to " the specific speed number " with efficiencies dropping dramatically below a number of 1000 . Testing also shows that smaller capacity pumps exhibit lower efficiencies than higher capacity designs.

Ref: mcnallyinstitute 

 
 
 
 



 





 

Pumping Power Requirement in Horse Power





GPM = Gallons per Minute
Head = Height of Water (ft)
Efficiency of Pump = %/100
 PSI = Pounds per square Inch

Specific Gravity of Water = 1.0
 

1 CuFt per Sec. = 448 GPM
1 PSI = A Head of 2.309 ft (water weight)
62.36 lbs per CuFt at 62°F





Equation for Measuring flow through rectangular wier

The Kindsvater-Carter rectangular weir equation

Rectangular Weir


Rectangular Weir Equation.GIF (3293 bytes)