NEWS

ENERGY LOSSES IN STEAM INSTALLATIONS

Like everything in our lives, we are always faced with good, very good and bad or very bad events and things! Of course, we always try to do and give our best so that good or very good things prevail in our daily lives, unbalancing our mood and making us better and bigger!


If, on the contrary, we leave bad things and events at the mercy of everyday life, our lives will always be compromised by them. It may seem like an exaggeration, but the results we obtain always depend on each one of us so that things are done correctly and make our lives better.


These results depend on a dose of knowledge, a dose of predictability, a dose of common sense and a large dose of goodwill and optimism.


In a steam system, we have the involvement of mechanical, electrical or electronic, pneumatic, hydraulic products, various inputs such as water, water treatment, fuel and thermodynamic calculations and sizing. Finally, the workforce must coordinate and direct all of this and ensure that at the end of every working day, good things always prevail. This game must be a win-win for everyone involved in the industrial process, from the employee who takes care of cleaning to the chairman of the board or business owner. Each person is responsible within their responsibilities.


This article presents some small things considered bad or compromising, which are likely to happen on a daily basis in a steam system and which can be a source of large losses in a generic industrial process, directly interfering with everyone's success and the business results. Predicting and preventing them from happening is up to you!


We adopted the SYSTEM DATA indicated below, which served as the basis for the calculations, although they do not necessarily reflect the real situation and conditions found in the field.


It is a relatively small steam system with 8 common CASES of occurrences that result in energy losses and for which actions are relatively simple!


STEAM SYSTEM DATA ANALYZED:

*Steam Boiler: Saturated Dry

* Generation Capacity: 2,500 kg/h

* Efficiency: 90%

* Operating Regime: 16 h/d, 330 d/year = 5,280 h/year

* Generation Pressure: 8 kg/cm2

* Steam Enthalpy (Thermal E) P = 8.0 kg/cm2 : Ct = 662.0 kcal/kg, Cl = 485.6 kcal/kg and Cs = 176.4 kcal/kg

* Steam Enthalpy (Thermal E) P = 3.0 kg/cm2: Ct = 653.4 kcal/kg, Cl = 509.8 kcal/kg and Cs = 143.6 kcal/kg

* Feed Water Temperature: 60°C (60 kcal/kg)

* Fuel: Firewood 30% humidity, 450 kg/m3, (PCI)=3,000 kcal/kg, 1,350,000 kcal/m3

* Fuel: Firewood 12% humidity, 500 kg/m3, (PCI)=3,600 kcal/kg, 1,800,000 kcal/m3

* Fuel Cost: U$ 30.00/m3

* Required bottom discharge: 75 l/h (3% hypothetical)


COSTS

* Steam Production P = 8.0 kg/cm2 :1,350,000 kcal/m3 : (662-60) kcal/kg x 0.90 = 2,018 kg/m3

* Steam Generation Cost: P = 8.0 kg/cm2 => U$/kg steam = 0.0148

* Steam Production P= 3.0 kg/cm2 :1,350,000 kcal/m3 : (653.4-60) kcal/kg x 0.90 = 2,528 kg/m3 P = 3.0 kg/cm2 => U$ 0.0118/kg steam

* Saturated water generation cost: 1,350,000 kcal/m3: (176.4-60) kcal/kg x 0.90 = 10,438.1 kg/m3

* Water Generation Cost: P= 8.0 kg/cm² => U$/kg water = 0.0029


CASE 01 – See Figure 01

Example of a thermodynamic type trap that has an operational consumption of 3.5 kg/h of steam (real fact), without leakage (considered here a new trap).

3.5 kg/h x 5,280 h/year x U$ 0.0148/kg = LOSS = U$ 273,50 / year + cost of water and its treatment;

In just one trap!


CASE 02 – See Figure 02

Elimination of excess water through bottom or surface discharges in boilers Elimination of 10% instead of 4% of generation.

2,500 l/h x (10% - 4%) = (250 – 100) = 150 l/h x U$ 0.0029 x 5,280 h/year LOSS = U$ 2,280.96 / year and the cost of water treatment! LOSS = 150 l/h x 5,280 h/year = 792,000 liters of water per year!


CASE 03 – See Figure 03

Lack of thermal insulation in Ø 2” steam piping.

Example: For a 5 meter long section => Loss 306 kcal/m.h

306 kcal/m.h x 5m x 5,280 h/year = 8,078,400 kcal/year: (662-60) kcal/kg = 13,419.27 kg/year of steam x U$ 0.0148/kg LOSS = U$ 198,60 / year

Plus the inevitable erosion problems in pipes and accessories!


CASE 04 – See Figure 04

Reducing vapor pressure in equipment:

Ex.: (8.0 kg/cm2 x 3.0 kg/cm2 ), heating 1,000 l/l water, T1 = 25°C and T2 = 75°C:

Consumption P = 8.0 kg/cm2 = 1,500 l/h x 1 kcal/kg x (75 – 25)°C / 485.6 kcal/kg = 155.4 kg/h 155.4 kg/h x U$ 0.0148/ kg = U$ 2,30/h

Consumption P = 3.0 kg/cm2 = 1,500 l/h x 1 kcal/kg x (75 – 25)°C / 509.8 kcal/kg = 147.1 kg/h 147.1 kg/h x U$ 0.0118 / kg= U$ 1,74/h

LOSS = (11.50 – 8.68) R$/h x 5,280 h/year = U$ 14,889.60/year

CASE 05 – See Figure 05

Dry firewood 12% moisture x Wet firewood 30% moisture.

* Dry Firewood: 1,800,000 kcal/m3: (662-60) kcal/kg x 0.90 = 2,691 kg/m3

* Wet Firewood: 1,350,000 kcal/m3: (662-60) kcal/kg x 0.90 = 2,018 kg/m3

* Generation Economy: 2,500 kg/h: 2,691 kg/m3 = 0.929 m3 /h

                                      2,500 kg/h: 2,018 kg/m3 = 1,239 m3 /h

* (1.239 - 0.929) m3 / h x 5,280 h/year x U$ 30.00/m3

LOSS = U$ 49,104.00 / year


CASE 06 – See Figure 06

Leakage through 2mm hole with pressure of 8.0 kg/cm2.

* Hole Area = 0.00000314 m2

* Flow Coefficient Kv = 0.07993

* Steam Loss = 8.33 kg/h

* 8.33 kg/h x 5,280 h/year x U$ 0.0148/kg

LOSS = U$ 650,94 / year


CASE 07 – See Figure 07

4 mm scale on boiler tubes (thermal insulation) => 10% increase in fuel consumption:

* Firewood 30% humidity: 1,350,000 kcal/m3: (662-60) kcal/kg x 0.90 = 2,018 kg/m3

* Annual generation: 2,500 kg/h: 2,018 kg/m3 = 1.24m3 /h x 5,280 h/year = 6,541 m3 /year

* 6,541 m3 / year x 10% = 654.1 m3 / year x R$ 30.00/m3

LOSS = U$ 19,623.00 / year!

 

CASE 08

Condensate not returned.

For every 6°C increase in temperature in the boiler feed water, we save 1% in fuel! If instead of feeding the boiler with water at 40°C, we feed it at 60°C:

* Steam Production P=8.0 kg/cm2 : 1,350,000 kcal/m3 : (662-40)kcal/kg x 0.90 = 1,953.3 kg/m3

* Steam Production P=8.0 kg/cm2 : 1,350,000 kcal/m3 : (662-60)kcal/kg x 0.90 = 2,018.3 kg/m3

* (1,953.3 : 2,018.3 – 1) x 100 = 3.22%

* Annual generation: 2,500 kg/h : 1,953.3 kg/m3 = 1.28 m3 /h x 5,280 h/year x 3.22% = 217.6 m3 /year

LOSS = 217.6 m3/year x U$ 30.00/m3 = U$ 6.528,00 / year!


Article kindly provided by Engineer Diego Bevilacqua from Termyka Rationalization, Recovery and Use of Energy.


Contacts:

diego@termyka.com.br

+55 (11) 99797-7297