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

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)

Reference:http://www.nphheaters.com