Saturday, 26 July 2025

Why do high pressure Boilers have higher efficiency??

 

High pressure Boilers will always have higher efficiency and lower fuel consumption as compared to low pressure Boilers.

In high pressure Boilers, heat transfer is more & losses are minimum, which has been proved theoretically in subsequent steps. Working of steam at higher pressure and higher temperature does more work than low pressure and low temperature steam.

Following are the major justified reasons for why high-pressure Boilers have higher efficiency??

Read more>>>>>Boiler and calculations













1-At higher pressure, steam can be generated at higher temperature.

Higher operating pressure of the Boiler facilitates higher steam temperature, which does more work than low pressure Boilers

2-More heat recovery options in High pressure Boilers.

High-pressure boilers often incorporate regenerative feedwater heating like HP and LP heaters more effectively, improving overall efficiency.

3-High pressure boilers have lower fuel consumption:

Because of high team generation is at higher pressure, feed water at economiser also enters at higher temperature leading into lower fuel consumption.

That is sensible heat addition is lesser in high pressure Boilers

4-High pressure and temperature boilers will have more enthalpy

5-Stam generated at higher pressure will have lower specific volume, hence Boiler will be compact in size.

6-Turbine consuming high pressure steam will generate more power at lesser quantity of steam input, that is lesser specific steam consumption & fuel consumption. High enthalpy steam means more energy per kg, so less steam is needed to produce the same power output.

7-Heat losses in High pressure Boilers are generally less as compared to low pressure Boilers

8-Operate at higher temperatures and pressures, improving the Rankine cycle efficiency, that is More work output from the same amount of fuel (better fuel economy).

9-High pressure leads to faster heat transfer and quicker steam production, improving responsiveness.

10-Advanced high-pressure systems allow better instrumentation and automation, improving safety, monitoring, and control.

 

Let us have discussions through calculation.

Let us discuss the heat content in 1 kg of steam at pressure 65 Kg/cm2 and 125 Kg/cm2.

For 65 kg/cm2 pressure Boiler, steam temperature will be 490 deg C (Tsup)

And feed water temperature at economiser inlet will be around 130 deg C

Therefore,

Steam enthalpy = 812 kcal/kg

Enthalpy of feed water Hfw= 132 kcal/kg

Enthalpy at operating pressure Hf =296 kcal/kg

Saturation temperature of water at steam drum, Ts= 284 deg C

Enthalpy of evaporation Hfg =367 kcal/kg

Therefore heat content in 1 kg of steam Hg = Hf + Hfg + 0.5 X (Tsup-Ts) –(1 X Hfw)

Hg = (296 + 367 + 0.5 X (490-284))-(1 X 132) = 634 kcal/kg

For 125 kg/cm2 pressure Boiler, steam temperature will be 550 deg C

And feed water temperature at economiser inlet will be around 210 deg C (Tsup)

Therefore,

Steam enthalpy = 832 kcal/kg

Enthalpy of feed water Hfw= 212 kcal/kg d

Enthalpy at operating pressure Hf =364 kcal/kg

Saturation temperature of water at steam drum, Ts= 327 deg C

Enthalpy of evaporation Hfg =271 kcal/kg

Therefore heat content in 1 kg of steam Hg = Hf + Hfg + 0.5 X (Tsup-Ts) –(1 X Hfw)

Hg =( 364+ 271 + 0.5 X (550-327))-(1 X 212) = 534 kcal/kg

 

From the above two calculation, it shows that steam generated at higher pressure needs lesser (534 kcal/kg) heat addition than steam generated at lower pressure (634 kcal/kg)


Following chart gives the heat required to generate 1 kg of steam at various pressure and feed water temperature.

SL NO.

BOILER DESCRIPTION

UOM

1

2

3

4

5

6

1

Operating pressure (P)

Kg/cm2

21

45

67

87

110

121

2

Operating temperature (T)

Deg C

350

450

490

515

540

550

3

Eco. Inlet feed water temperature

Deg C

105

125

125

125

125

125

4

Enthalpy at operating pressure (Hf)

Kcal/kg

223

271

302

325

349

360

5

Enthalpy of evaporation at operating pressure (Hfg)

Kcal/kg

445.3

397

361

331

296.9

280

6

Saturation temperature (Ts)

Deg C

214.07

259.00

286.10

305.41

323.85

331.66

7

Heat supplied to generate 1 kg of steam at rated parameters

Kcal/kg

631.27

638.50

639.95

635.80

628.97

624.17

 

 Read more>>>>>powerplant and calculations


Friday, 18 July 2025

How to calculate the belt conveyor speed?

 

Belt conveyors have been using in industries like steel, pharmaceutical, food, textile, sugar, distillery etc. Knowing or deciding the speed of the conveyors is very vital for designing the capacity of the conveyor.

How belt speed is related to the capacity of the conveyor?

Higher the belt speed, more is the conveyor capacity and Vice Versa.

Generally, the speed of the conveyors is around 0.8 to 2 m/sec.

Actually, there are two methods followed to calculate the belt speed

 1-Practical method

In this case, time taken to travel the belt from head pulley to tail pulley or from Head to Head.

Belt Speed (m/s)=Time Taken (s) / Distance Travelled (m)

For example: A belt conveyor of length 125 meter takes 4 minutes for complete rotation, the conveyor speed is calculated as













Belt speed = Distance traveled / Time elapsed in one complete rotation

Belt speed =

Belt speed = (250) / (4 X 60 sec)

Belt speed = 250 / 240 = 1.04 m/sec.

2-Theoretical method.

In this method Belt conveyor speed is calculated by using theoretical calculation method.

Input required:

1-Head pulley (D) in meter

2-Gear box input and out put speed in RPM or Gear box reduction ratio

3-Motor speed in RPM

Conveyor speed / Belt speed in m/sec = π X D X N / 60

Where, D is Diameter of Head pulley in meter

N is the speed of the Pulley in RPM

Speed of the pulley = Motor speed / Gear box reduction ratio.










Examples:

1-A belt conveyor of length 350 meter, takes around 740 seconds for complete rotation. Calculate the speed of the belt.

Belt speed = Distance traveled / Time elapsed in one complete rotation

Belt speed = (350 X 2) / 740

Belt speed = 700 / 740

Belt speed = 0.95 m/sec.

2-Calculate the speed of the belt conveyor having following details.

Motor output speed: 1450 RPM

Gear box reduction ratio:45: 1

Head Pulley Diameter D = 500 mm

Now, calculate the Gear box out put speed = Motor speed / Reduction ratio

Gear box output speed = 1450/45 = 32.22 RPM

Since gear box out shaft is directly connected to conveyor pulley. Hence speed of the Gear box out put shaft and Head pulley are same.

Gear box out put speed = Head Pulley speed

Therefore, Conveyor speed V in m/sec = π X D X N / 60

Therefore, Conveyor speed V in m/sec = π X 0.5 X 32.22 / 60

Conveyor speed V in m/sec = 0.84 m/sec

What are the factors affecting the belt conveyor speed??

Following are the major factors, which affect the conveyor speed.

  • Head pulley diameter
  • Motor speed
  • Gear box reduction ratio
  • Load on the conveyor
  • Friction between rotating idlers and belt
  • Conveyor inclination
  • Environmental factors like rain, dust etc


Read more>>>>on >>>power plant and calculations


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