Skip to main content

Internal Boiler Water Treatment

In internal treatment, chemicals are added in the boiler water. Internal treatment to prevent scale deposition is usually carried out with some form of sodium phosphate (trisodium, disodium or monosodium). These chemicals react with dissolved salts to convert them into less harmful sludge which can stay in the boiler water without creating any harm to the boiler tube. Later on, this sludge can be removed from the boiler water by blowdown. The main aim of internal treatment is to precipitate the impurities present in the boiler water to get harmless salts or sludge.

Soda Ash (Sodium Carbonate or Na₂CO₃) Treatment

Soda ash (Na₂CO₃) treatment is done in smaller size boiler.

In this process, soda ash is added into the boiler water which reacts with chlorides and sulphates of magnesium and calcium to form insoluble sludge.

CaCl₂ + Na₂CO₃ → CaCO₃ + 2NaCl

MgSO₄ + Na₂CO₃ → MgCO₃ + Na₂SO₄

CaSO₄ + Na₂CO₃ → CaCO₃ + Na₂SO₄

Phosphate Treatment or High Pressure (HP) Dosing

In high pressure boiler, soda ash treatment cannot be used, as it forms caustic soda and increases with the increase in temperature. Hence, it is difficult to maintain pH of the boiler water. Also, at higher pH, solubility of calcium carbonate increases. So, in high pressure boiler, phosphate treatment is preferred as internal treatment.

Phosphate treatment is preferred in drum type boiler. Phosphate is added to the boiler water at the boiler drum.

If it is dosed at feed pipe, it may react with impurities and sludge may be deposited at the feed line. Phosphate dosing is done at boiler drum which is at higher pressure. So, this dosing is called as high pressure (HP) dosing.

Phosphate dosing is done in only drum type boiler. In once through boiler, phosphate treatment is not done.

Trisodium phosphate (Na3PO4), disodium phosphate (Na2HPO4) and monosodium phosphate (NaH2PO4) are used in boiler for dosing. These chemicals are also called as orthophosphates.

Trisodium phosphate (TSP) is highly alkaline. Disodium phosphate is less alkaline and monosodium phosphate is slightly acidic.

Blowdown

Concentrated solid precipitates at drum bottom (by internal treatment) are removed through blowdown, which may be either intermit tent or continuous, the amount of blowdown being governed by the allowable solids concentration and the purity of feedwater. It is usually expressed as a percentage. Thus,

% blowdown = (quantity of water blow down/quantity of feed water) x 100

For instance, 3% blowdown means that 3 per cent of the total amount of feedwater supplied is removed as liquid by blowdown of the drum. The continuous blowdown has the advantage of closer control of concentration and the recovery of part of the energy of the blowdown.

Since water purity is controlled by blowdown in drum type boilers, condensate polishing is not required in such boilers.

Example

Calculate the blow down rate for a boiler with an evaporation rate of 5 tons/hr, if the maximum permissible TDS in boiler water is 3000 ppm and with 18 % make up water addition. The feed water TDS is around 400 ppm.

Solution

Blow down in percentage

= (feed water TDS x % makeup)/(permissible TDS - feed water TDS)

= (400 x 18)/(3000 - 400) = 2.77%

If boiler evaporation rate is 5000 kg/hr then required blow down rate is

= 5000x2.77/100 = 138.5 kg /hr 

Subscribe to our newsletter for updates on Boiler Operation Engineering Exam 

Study material for BOE exams 


 


Comments

Popular posts from this blog

Mechanics of Fluids (Solved Numerical Problems)

Numerical The surface Tension of water in contact with air at 20°C is 0.0725 N/m. The pressure inside a droplet of water is to be 0.02 N/cm² greater than the outside pressure. Calculate the diameter of the droplet of water. (7 marks) (AMIE Summer 2023) Solution Surface tension, σ = 0.0725 N/m Pressure intensity, P = 0.02 N/m 2 P = 4σ/d Hence, the Diameter of the dropd = 4 x 0.0725/200 = 1.45 mm Numerical Find the surface tension in a soap bubble of 40 mm diameter when the inside pressure is 2.5 N/m² above atmospheric pressure. (7 marks) (AMIE Summer 2023) Answer: 0.0125 N/m Numerical The pressure outside the droplet of water of diameter 0.04 mm is 10.32 N/cm² (atmospheric pressure). Calculate the pressure within the droplet if surface tension is given as 0.0725 N/m of water. (AMIE Summer 2023, 7 marks) Answer: 0.725 N/cm 2   Numerical An open lank contains water up to a depth of 2 m and above it an oil of specific gravity 0.9 for a depth of 1 m. Find the pressure intensity (i) at t...

Geotechnical & Foundation Engineering (Solved Numerical Problems)

Numerical A 1000 cc core cutter weighing 946.80 g was used to find out the in-situ unit weight of an embankment. The weight of the core cutter filled with soil was noted to be 2770.60 g. Laboratory tests on the sample indicated a water content of 10.45 % and specific gravity of solids of 2.65. Determine the bulk unit weight, dry unit weight, void ratio, and degree of saturation of the sample. (AMIE Summer 2023, 8 marks) Solution Weight of soil in core cutter = 2770.60 - 946.80 = 1823.8 g Bulk unit weight, γ t = W/V = 1823.8/1000 = 1.82 g/ccDry unit weight, γ d = γ t /(1 + w) = 1.82/(1 + 0.1045) = 1.65 g/cc Void ratio, e = (G s γ w /γ d ) - 1 = (2.65 x 1.0/1.65) - 1 = 0.61 Degree of saturation, S = wG s /e = (0.1045 x 2.65)/0.61 = 0.4540 = 45.4% Numerical What is the theoretical height of the capillary rise and the capillary pressure in fine-grained soil with an effective size (D 10 ) of 0.002 mm? (AMIE Summer 2023, 4 marks) Solution D 10 = 0.002 mm;  Using the assumption that th...

Energy Systems (Solved Numerical Problems)

Wind at 1 standard atmospheric pressure and \({15^0}C\) has velocity of 15 m/s, calculate (i) the total power density in the wind stream (ii) the maximum obtainable power density (iii) a reasonably obtainable power density (iv) total power (v) torque and axial thrust Given: turbine diameter = 120 m, and turbine operating speed = 40 rpm at maximum efficiency. Propeller type wind turbine is considered. (AMIE Winter 2023) Solution For air, the value of gas constant is R = 0.287 kJ/kg.K 1 atm = 1.01325 x 105 Pa Air density \(\rho  = \frac{P}{{RT}} = \frac{{1.01325x{{10}^5}}}{{287}}(288) = 1.226\,kg/{m^3}\) Total Power \({P_{total}} = \rho A{V_1}^3/2\) Power density \(\begin{array}{l}\frac{{{P_{total}}}}{A} = \frac{1}{2}\rho {V_1}^3\\ = \frac{1}{2}(1.226){(15)^3}\\ = 2068.87{\mkern 1mu} W/{m^2}\end{array}\) Maximum power density \(\begin{array}{l}\frac{{{P_{\max }}}}{A} = \frac{8}{{27}}\rho A{V^3}_1\\ = \frac{8}{{27}}(1.226){(15)^3}\\ = 1226{\mkern 1mu} W/{m^2}\end{array}\) Assuming eff...