Showing posts with label pumps/compressors/fans. Show all posts
Showing posts with label pumps/compressors/fans. Show all posts

What is the dew point of compressed air???


In compressed air dryers, the dew point is a critical parameter to monitor and control. The dew point refers to the temperature at which moisture begins to condense out of the air as it is cooled. In the context of compressed air dryers, achieving a low dew point is essential to prevent moisture from causing damage to downstream equipment and processes.

 There are different types of compressed air dryers, such as refrigerated dryers, desiccant dryers, and membrane dryers, each with its own method of reducing the dew point:


Refrigerated Dryers:

 These cool the compressed air to reduce its temperature, causing the moisture to condense out and be drained away. The dew point achieved by refrigerated dryers typically ranges from 35°F to 50°F (1.7°C to 10°C).

 Desiccant Dryers:

 These use adsorbent materials such as silica gel or activated alumina to adsorb moisture from the compressed air. They can achieve much lower dew points, typically ranging from -40°F to -100°F (-40°C to -73°C), depending on the design and operating conditions.

 Membrane Dryers:

 These use a permeable membrane to selectively remove water vapor from the compressed air stream. They can achieve dew points ranging from -40°F to -100°F (-40°C to -73°C), similar to desiccant dryers.

 Monitoring and controlling the dew point in compressed air systems is crucial for maintaining the quality of the compressed air and preventing issues such as corrosion, contamination, and freezing in downstream equipment and processes. Instruments such as dew point sensors are used to measure the dew point accurately, allowing operators to adjust dryer settings as needed to achieve the desired dew point level.

 The recommended dew point temperature for compressed air depends on the specific application and industry standards. Different industries and applications have varying requirements for compressed air quality. Here are some general guidelines:

 ISO 8573 is an international standard that specifies compressed air quality classes based on particle concentration, oil content, and dew point temperature. The standard outlines different classes for various applications, ranging from Class 0 (the highest quality) to Class 6 (the lowest quality). Each class has specific limits for dew point temperature. For critical applications such as pharmaceuticals, food and beverage, electronics manufacturing, and certain types of machinery, lower dew point temperatures are typically required to prevent moisture-related issues.

For general industrial applications where moisture-sensitive equipment is not a concern, a dew point of around 35°F to 50°F (1.7°C to 10°C) may be sufficient.

For more demanding applications such as pneumatic control systems, painting processes, or instrument air in laboratories, dew points of around 35°F (1.7°C) or lower may be necessar.

In highly sensitive industries like pharmaceutical manufacturing or electronics assembly, dew points as low as -40°F (-40°C) or lower may be required to prevent contamination or damage to products and equipment.

Environmental factors such as ambient temperature and humidity levels can influence the dew point requirements. In hot and humid environments, lower dew points may be necessary to prevent condensation in the compressed air distribution system.

Calculating the dew point temperature of an air dryer involves understanding the operating principles of the dryer and the conditions of the compressed air being processed. There are several methods to calculate or estimate the dew point temperature, depending on the type of air dryer being used:

For refrigerated dryers, the dew point temperature can be estimated based on the design of the dryer and the temperature of the cooling medium (usually refrigerant).The dew point temperature achieved by a refrigerated dryer typically ranges from 35°F to 50°F (1.7°C to 10°C). It's often close to the outlet temperature of the refrigerated air.

Desiccant dryers adsorb moisture from the compressed air using a material like silica gel or activated alumina. The dew point temperature achieved by a desiccant dryer depends on factors such as the type and condition of the desiccant material, the design of the dryer, and the operating conditions.

The dew point can be calculated based on the inlet conditions of the compressed air (temperature and relative humidity), the type of desiccant used, and the design parameters of the dryer.However, precise calculation may require complex modeling or simulation.

The most accurate way to determine the dew point temperature of an air dryer is to use a dew point sensor.These sensors measure the moisture content of the air directly and provide real-time dew point readings. They are commonly used in industrial applications to monitor and control the performance of air dryers.


In practice, the dew point temperature of an air dryer is often monitored using a dew point sensor rather than calculated manually.

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How do you calculate the efficiency of pumps???


How do you calculate the efficiency of a pump?

Efficiency of the pump is the ratio of output power to the input power

Pump efficiency = Output power X 100 / Input power

Centrifugal pumps efficiency varies from 40% to 95% well.

Pump efficiency is equal to the power of the water/liquid produced by the pump divided by the pump’s shaft power input.

A pump’s output power is determined by how much water and how much pressure it delivers 

On what factors pumps efficiency depends on?

Pumps efficiency depends on

  • In put power
  • Output power
  • Type of liquid
  • Fluid temperature
  • Flow
  • Suction & discharge head
  • Losses & leakages
  • Fluid viscosity
  • Pump operating load
  • Pumps internal frictions

What do you mean by volumetric efficiency of a pump?

It is the ratio of actual flow delivered by the pump to the theoretical flow

How do you calculate the Mechanical or hydraulic efficiency of the pump?

Mechanical/Hyd.efficiency = Pump output power X 100 / Pump input power or pump shaft power

Mechanical efficiency is also calculated as

 = Theoretical torque required to drive the pump X 100 / Actual torque provided to drive the pump


A centrifugal pump delivers 0.2 m3/sec flow at total head 27 m, calculate its hydraulic power. Assume density of water 995 kg/m3

Pump hydraulic power = Pump flow in m3/sec X pump total head in meter X Fluid density in kg/m3 X 9.81 m/sec2 / 1000

Pump hydraulic power = 0.2 X 27 X 995 X 9.81 / 1000

Pump hydraulic power = 52.7 KW

A centrifugal pump of capacity 0.05 m3/sec flow has to lift the water from 3.5 m deep well & has to discharge the water at 45 meter height. Calculate the pump efficiency if its shaft power is 30 KW

Assume water density 998 kg/m3

Pump hydraulic power = Pump flow in m3/sec X pump total head in meter X Fluid density in kg/m3 X 9.81 m/sec2 / 1000

Total head = Discharge head + Suction lift

Total head H = 45 + 3.5 = 48.5 meter

Pump hydraulic power = 0.05 X 48.5 X 998 X 9.81 / 1000

Ph = 23.74 KW

Pump efficiency =Pump hydraulic power X 100 / Pump shaft power

Pump efficiency =23.74 X 100 / 30

Pump efficiency =79.13%

A centrifugal pump of delivering  0.65 m3/sec flow at 75 meter discharge head, pump has positive suction head around 2.7 meter from overhead tank.Pump efficiency is 59%, calculate the motor input power if efficiency power if efficiency of motor is 92.5%.

Assume water density 1000 kg/m3

Pump hydraulic power = Pump flow in m3/sec X pump total head in meter X Fluid density in kg/m3 X 9.81 m/sec2 / 1000

Total head = Discharge head -Suction head

Total head H = 75 -2.7 = 72.3 meter

Pump hydraulic power = 0.65 X 72.3 X 1000 X 9.81 / 1000

Ph = 461 KW

Pump efficiency =Pump hydraulic power X 100 / Pump shaft power = Ph X 100 / Ps

59 =461 X 100 / Ps

Ps = 781.35 KW

Motor efficiency = Pump shaft power X 100 / Motor input power =Ps X 100 / Pm

92.5 = 781.35 X 100 / Pm

Pm = 844.7 KW

A centrifugal pump of efficiency 63% develops hydraulic power 27.5 KW, calculate the motor efficiency if motor input power is 45 KW

Pump shaft power, Ps = Pump hydraulic power Ph / Efficiency of pump

Ps = 27.5 / 0.63 =43.65 KW

Motor efficiency = Pump shaft power X 100 / Motor input power

Motor efficiency= 43.65 X 100 / 45

Motor efficiency = 97%



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50-Interview questions & answers on centrifugal pumps


1. What are the centrifugal pumps?

Centrifugal pumps are the mechanical devices which pump or transport various fluids by converting their rotational kinetic energy into hydrodynamic energy.

2. Why the name centrifugal pump?

A centrifugal pump uses centrifugal force

3. Where the centrifugal pumps find applications in power plants? 

Read reference books for power plant O&M

  • Boiler feed water pump 
  • Auxiliary & main cooling water pumps 
  • Raw water transfer pumps  
  • Condensate extraction pump,  
  • Deaerator & feed water tank make up pumps 
  • Firefighting water pumps 
  • UF & RO feed water pumps 
  • MGF feed pump 
  • Degassed water transfer pumps 
  • Sometimes lube & control oil pumps 

4. How do you specify the centrifugal pumps? 

Centrifugal pumps are specified as bellow 

  • Flow in M3/Hr or M3/sec 
  • Head or discharge pressure in meter or bar or kg/cm2 
  • Shutoff head 

5.What are the various parts of centrifugal pumps? 

Centrifugal pumps have following parts 

  • Pump casing or diffuser 
  • Impeller 
  • Wear ring 
  • Shaft 
  • Lantern ring 
  • Stuffing box 
  • Inlet vertex 
  • Mechanical seal or gland packing 
  • Shaft sleeve 
  • Bearings 

6.What are the energy conversions take place in centrifugal pumps 

In centrifugal pumps hydraulic energy is being converted into kinetic energy  

7.What types of reducers are used at pump suction & discharge ends? 

Suction side: Eccentric type & Discharge side: Concentric 

8.What are the two main types of centrifugal pumps?

Axial flow & Radial flow

9.What is the function of impeller in centrifugal pumps?

It converts kinetic energy of pump into hydrodynamic energy by rotary motion

10.What is the function of pump casing?

Casing converts velocity head from impeller into pressure head & also guides the flow to the discharge end.

11. What are the types of pump casing?

Volute & diffusers are two different types of pump casing

12. What do you mean by volute?

A volute is a spiral-like geometry with an increasing through-flow area, reducing the velocity of the fluid and increasing the static pressure

13. What are the different types of volutes?

Single volute & Double volute

15. Write down the working principle of centrifugal pumps

In centrifugal pumps, fluid enters the impeller through inlet eye & exists along the circumference between the vanes of impeller. This impeller is connected to shaft & in turn to motor, this rotary motion of the impeller converts kinetic energy of the fluid into hydrodynamic energy.

16.What are the types of impellers?

Open impeller: As its name suggests, an open impeller has vanes that are open on both sides without any protective shroud. These are structurally weak.

These are used for low flow & low head applications. Generally used for pump solids or sludge. These require much NPSH.

Semi open impeller: Semi-open impellers have a back-wall shroud that adds mechanical strength to the vanes.

Closed impeller: Are very robust & require low NPSH

Impellers are also classified as single suction & Double suction

17.What are the rotary & stationary parts of the pumps?

Rotary parts:

  • Shaft
  • Impeller
  • Shaft sleeve
  • Bearings

Stationary Parts

  • Pump casing
  • Gland packing or mechanical seal
  • Lantern ring

18.Why eccentric reducers are used at pump suction side? 

To avoid air locking & cavitation eccentric reducers are used at suction side 

19. What do you mean by the NPSH in pumps? 

It is the net positive head required at pump suction to avoid cavitation 

20. What do you understand by the term cavitation? 

Cavitation is the formation & collapsing of vapor bubbles at pump’s suction 

21. How the cavitation does affect the pump’s life? 

  • Cavitation causes 
  • Vibrations in pump 
  • Damage of impellers 
  • Heavy noise 

22. What are the factors considered for centrifugal pumps design? 

  • Flow required 
  • NPSH available & NPSH required 
  • Total head 
  • Pump efficiency 
  • Fluid used 

23. What are the materials used for pump casing? 

Generally cast steel or cast iron are used for single stage centrifugal pumps 

24. What are the materials used for Impellers? 

Impellers are made up of cast iron, gun metal & stain less steel 

25. What is the function of wear ring? 

As the name indicates it protects the wear & tear of impeller 

26. What do you mean by static suction head in pump?

Therefore, the static suction head is the vertical distance from the center line of the pump to the free level of the liquid to be pumped.

27. What do you mean by static suction head in pump?

Static discharge head is the vertical distance between the pump centerline and the point of free discharge or the surface of the liquid in the discharge tank.

28. What do you mean by total static head?

Total static head is the vertical distance between the free level of the source of supply and the point of free discharge or the free surface of the discharge liquid.

29. What do you mean by total head?

It is total dynamic discharge head plus total dynamic suction head

Note: If source water level is below the pump center line, then

Total head = Discharge head Suction lift

If source Water level is above the pump suction line, then

Total head = Discharge head-Suction head

30. What are the problems associated with centrifugal pumps? 

Following are the common problems associated with pumps 

  • Low discharge pressure 
  • Low delivery 
  • Cavitation 
  • High vibrations 
  • Pump seize 
  • Over load 
  • More suction lift 
  • Air locking & No priming 

 31. What are the reasons for no delivery or no discharge in centrifugal pumps?

  •  Probable reasons are
  • Air lock in pump suction
  • Suction valve closed
  • Low tank level
  • 32. What are the reasons for low delivery?
  • Suction valve partially opened
  • Reverse rotation of pump
  • Low speed of pump
  • Suction strainer is chocked

 33. What are the reasons for over load of pump?

  •  More flow
  • High speed
  • Reverse rotation of pump
  • Pump discharge kept open to atmosphere
  • Internal friction in impeller & wear ring or impeller & casing
  • More tightened gland packing
  • No lubricant in bearing or bearing seized

 34. What are the potential reasons for pump vibrations?

  •  Overloading of pump
  • Reverse rotation of pump
  • Impeller rubbing inside the casing
  • Misalignment
  • Damaged bearing
  • Shaft run out
  • Shaft imbalance

 35. Too much noise coming from pump inside, what does this mean?

  •  Air lock in pump
  • Overloading of pump
  • Pump discharge line is less than actual required
  • Cavitation
  • No lubricant in bearings

 36. What are the common mistakes done during pump installation?

  • Choosing poor foundation
  • Note: Pump foundation weight should be 3 to 4 times the pump weight
  • Lesser size suction pipe line
  • Lesser size discharge pipe line
  • Interchanging concentric & eccentric reducers

37. What are the safety protections & interlocks given for a centrifugal pumps?

  • Over load
  • Low load
  • High bearing vibrations
  • High bearing temperature
  • High suction DP
  • Source water level low

38. How do you increase the head & flow of pump by modifying impeller size?

By increasing the impeller diameter head & flow can be increased

By increasing the impeller width flow can be increased

 39. What are the reasons for reduction of pump efficiency?

  • Operating the pump at lower capacity
  • Operating the pump at higher load
  • Throttling the discharge valve
  • Increase in impeller & wearing clearance
  • Lower suction head
  • High suction lift

Calculation part

 40. How do you calculate NPSHA?

 NPSHA is Net positive suction available

NPSHA = Atmospheric pressure + static head - vapor pressure - pressure loss in the suction piping - pressure loss due to the suction strainer.

 41. A centrifugal pump of rated capacity 75 M3/Hr & total head 35 meter is supplying water to fill a tank in 2 hours, calculate the total power consumption. Consider pump & motor efficiency 50% & 85% respectively

 Power consumption = Pump flow in m3/sec X Pump total head in meter X fluid density X g / (1000 X Pump eff. X Motor eff)

Power consumption = (75/3600) X 35 m X 1000 kg/m3 X 9.81m/s2 / (1000 X 0.5 X 0.85)

Power consumption = 16.83 KWH

Power consumption in 2 hours = 16.83 X 2 = 33.66 KW

 42. A centrifugal pump having hydraulic power 22 KWH, discharge & suction head 55m & 12m respectively

Calculate the pump flow in m3/hr, assume density of water 990 kg/m3

Pump flow = Pump hydraulic power X 1000 / (Pump total head X density of fluid kg/m3 X 9.81 m/s2)

Pump flow = 22 X 1000 /( (55-12) X 990 X 9.81)

Pump flow = 0.052 m3/sec

Pump flow in M3/hr = 0.052 X 3600 = 189.6 M3/hr

 43. A centrifugal pump having hydraulic power 15KWH & pump efficiency 65% calculate the pump shaft power

 Pump shaft power = Pump hydraulic power / Pump efficiency = 15 / 0.65 = 23 KW

 44. A centrifugal pump produces flow 20M3/hr (Q1) flow at rated speed 1500 RPM (N1) , then calculate the flow of pump at 1000 RPM(N2)

 We have pump affinity law

 Q1/Q2 = N1/N2

20 / Q2 = 1500 / 1000

Q2 = 13.33 M3/hr

45. A centrifugal pump consumes power of 25KW (P1) at speed of 1500 RPM (N1), after reducing certain RPM its power consumption reduces by 5 KW (P2), calculate that speed

 We have pump affinity law

 P1/P2 = (N1/N2)3

25 / 5 = (1500 / N2)3

N2 = 877.2 RPM

 46. A centrifugal pump produces 150 m (H1) head at 3000 RPM (N1), calculate the head produced if its speed reduced to 50%

We have pump affinity law

H1 / H2 = (N1/N2)2

N2 = N1 X 50% = 3000 X 0.5 = 1500 RPM

150 / H2 = (3000 / 1500)2

H2 = 37.5 meter

47. A centrifugal pump having impeller diameter 250 mm produces flow 250 M3/hr, calculate the diameter of impeller to produce flow 300 M3/hr

We have

Q1 / Q2 = D1 / D2

250 / 300 = 0.250 / D2

D2 = 0.35 m = 350 mm

48. A centrifugal pump having impeller diameter 300 mm produces 250 m head & what could be the diameter if we want to reduce the head by 30m

Reduced head = 250 – 30 = 220 m

We have

H1 / H2 = (D1/D2)2

250 / 220 = (300 / D2)2

D2 = 281.4 mm

49. A centrifugal pump having impeller diameter 150 mm (D1) consumes 15 kw (P1), what is the size of impeller if we want reduce power by 4 KW

P2 = P1-4 = 15-4 = 11 KW

We have

P1 / P2 = (D1 / D2)3

15 / 11 = (150 / D2)3

D2 = 135.2mm


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