Skip to main content

Direct Method of Boiler Efficiency

Boiler Efficiency

Thermal efficiency of a boiler is defined as the percentage of heat input that is effectively utilized to generate steam. There are two methods of assessing boiler efficiency.

  • Direct method: Where the energy gain of the working fluid (water and steam) is compared with the energy content of the boiler fuel. 
  • Indirect or Loss method: Where the efficiency is the difference between the losses and the energy input.

Join our WhatsApp Channel - The Indian Boiler Engineer for regular updates 

Direct Method

This is also known as 'input-output method' due to the fact that it needs only the useful output (steam) and the heat input (i.e. fuel) for evaluating the efficiency. This efficiency can be evaluated using the formula

Boiler efficiency = (heat output/heat input) x 100  

Parameters to be monitored for the calculation of boiler efficiency by direct method are :

Quantity of steam generated per hour (Q) in kg/hr.

Quantity of fuel used per hour (q) in kg/hr.

The working pressure [in kg/cm2(g)] and superheat temperature (QC), if any

The temperature of feed water (0C)

Type of fuel and gross calorific value of the fuel (GCV) in kcal/kg of fuel

η = [Q(hg - hf)/(q x GCV)] x 100                       

Where,           

hg - Enthalpy of saturated steam in kcal/kg of steam

hf - Enthalpy of feed water in kcal/kg of water

The fuel calorific value may be gross or net and accordingly, the efficiency reported is referred to as efficiency on GCV or NCV basis.

Example

Find out the efficiency and evaporation ratio of the boiler by direct method with the data given below:

Type of boiler  : Coal fired

Quantity of steam (dry) generated : 8TPH

Steam pressure / temp : 10kg/cm2(g)/180 °C

Quantity of coal consumed : 1.8 TPH

Feed water temperature : 85 °C

GCV of coal : 3200 kcal/kg

Enthalpy of saturated steam at 10 kg/cm2 pressure : 665 kcal/kg

Enthalpy of feed water : 85 kcal/kg

Solution

η = [8(665 - 85)/(1.8 x 3200)] x 100 = 80% (On GCV basis)
Evaporation = (Qty of steam/Qty of fuel consumed) x 100
= 8000/1800 = 4.44

It should be noted that boiler may not generate 100% saturated dry steam, and there may be some amount of wetness in the steam.

Subscribe to our newsletter for updates on Boiler Operation Engineering Exam 

Study material for BOE exams 

Labels:

Comments

Post a Comment

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...
Post a Comment
Read more

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...
Post a Comment
Read more

Solved Numerical Problems from Boiler Operation Engineer Exams

Properties of Steam
Post a Comment
Read more