Steam Turbines

Power Generation Through Steam Turbines

The basic process behind steam power generation is the “Rankine Cycle”. Water is heated until it is a saturated liquid. From there, it is compressed into steam. The steam is transferred to a turbine where the pressure of the steam is reduced (usually to sub atmospheric pressures) by expansion over the turbine blades. This process produces mechanical work which is latter converted into electricity through generator. The low pressure steam is condensed back to a liquid. The water, now referred to as return water, is mixed with new water, referred to as “feedwater”, and pumped back to the boiler. The figure below shows a common diagram used to describe the Rankine Cycle.

Steam Turbines have different classification criteria. These criterion are;

A-Steam Flow Direction in Steam Turbine

            i) Axial Steam Turbine

            ii) Radial Steam Turbine

B-Working Principle of Steam Turbine

            i) Impulse Steam Turbine

            ii) Reaction Steam Turbine

C-Based on Exit Steam and Application Based

            i) Condensing Steam Turbine

            ii) Non Condensing Back Pressure Steam Turbine

            iii) Extraction Steam Turbine

D-Based on Steam Pressure 

  

i)       Low pressure steam turbine, the turbine with pressure up to 2 ata. 

ii)      Middle pressure steam turbine, the turbine with pressure up to 40 ata. 

iii)     High pressure steam turbine, the turbine with pressure 40 – 170 ata. 

iv)     Very high steam turbine, the turbine with pressure exceeds 170 ata. 

 V)    Super critical pressure steam turbine, the turbine with pressure exceeds 225 ata.

A-Steam Flow Direction in Steam Turbine

i) Axial Steam Turbine 

ii) Radial Steam Turbine

In a radial-flow turbine, steam flows outward from the shaft to the casing.These steam turbines has direction of steam flow perpendicular to the axis of shaft. The unit is usually a reaction unit, having both fixed and moving blades. They are used for special jobs and are more common to European manufacturers, such as Sta-Laval (now ABB).

C-Three Classes of Steam Turbines on the Basis of Exit Steam and Application Based

1-Condensing Steam Turbines

2-Non Condensing (Back Pressure) Steam Turbines

3-Extraction Steam Turbines

1-Condensing Steam Turbine

These turbines Operate with an exhaust pressure less than atmospheric (vacuum pressure).

They experiences the maximum pressure drop through the turbine which results in greater energy extracted from each lbm or kg of steam input.ƒ Turbine efficiency ranges are  approx. 30-40%.ƒ The condenser can be either air or water cooled. Condenser cooling water can be utilized for processes or space heating loads. Condensing tubines areƒ Usually more expensive than Non-Condensing Back pressure turbines. These turbines are not used for Combined Heat and Power Applications. 

2-Non-Condensing Back Pressure Steam Turbine

Steam is expanded over turbine and exhaust low pressure steam is used to meat the thermal heating requirements. Steam is expanded to the extent providing required lower pressure of steam which facility can use. 

These turbines operate with an exhaust equal to or in excess of atmospheric pressure. Exhaust steam is used for lower pressure steam process loads. These turbines are available in smaller sizes and pass large amounts of steam per MW of output (low efficiencies). These turbines ƒ produce less useful work than a condensing turbine, but since the unused steam from the turbine is passed on to process loads, the lower turbine power generation efficiencies (15% to 35%) are not a concern. These turbines areƒ very cost effective when paralleled with pressure reduction valves (PRV), providing an efficient use of the pressure reduction requirements.ƒ Usually less costly than condensing turbines.

3-Extraction Steam Turbine

The other type of steam turbine used in CHP applications is called an extraction turbine. Basically these turbines are hybrid of Condensing and Non-Condensing turbines. In these turbines, steam in extracted from the turbine at some intermediate pressure. This steam can be used to meet the facilities steam need. The remaining steam is expanded further and condensed. Extraction turbines can also act as admission turbines. In admission turbines, additional steam is added to the turbine at some intermediate point. The figure below schematically shows the process of an extraction steam turbine.

Cooling Tower: Performance calculation

First you should collect all the data as given below. Be sure that the data collected for these temperatures is most accurate because of lower absolute level of generally ~40°C average temperatures, an error of 0.5°C due to manual data collection & judgment will cause more than 1.2% error in the result at one calculation. Repeating such errors may result in cumulative errors of more than 10% in totality giving you totally absurd results.

So the basic point is that collect the data on regular basis, keep a watch to have a feel of real values & then proceed.

Actual Data
Cooling water flow rate - 4134 M3/hr
Cooling water inlet Temp - 44.0 °C
Cooling water exit Temp - 35.0 °C
Inlet air-wet bulb - 30.0 °C
Inlet air-dry bulb - 38.8 °C
Exit air-wet bulb - 40.7 °C
Exit air-dry bulb - 42.0 °C

Now follow step by step procedure for the calculation.

Step-1
Calculate waterside actual heat load, which is as below

Qw = 4134 x 1000 x (44 – 35) / 1000000
= 37.21 Gcal/Hr

Step-2
Calculate absolute humidity at wet bulb of inlet air, which is at 30°C in this case. This is a function of wet bulb temperature only.

The equation for the same is

1.4478310678E-10*(Tw^7)-2.6920*10e-8*(Tw^6)+1.99053*10e-6*(Tw^5)-6.65614*10e-5* (Tw^4)+0.00131879344*(Tw^3)+0.00125483272*(Tw^2)+0.291649083*Tw+3.802441

Where Tw is wet bulb temperature in °C. So,

H1 = 27.29 Kg/ ‘000Kg of dry air

Step-3
Calculate absolute humidity at dry bulb of inlet air, which is at 38.8°C in this case. It will give you saturation level of humidity, say H2.

Step-4
Find out &% Saturation. Of course it can be done from Psychometric charts but then you wont be able to use powerful Excel Tool for simulation of your cooling tower that’s why these equations are generated.

You can also use any good Excel Add-IN for Psycho properties if available.

Here, it will be %Sat = H1/H2

Step-5
Based on % Saturation find out the enthalpy content of moist air at inlet condition. Again I did it using self-developed equations ~10 years back.

I found it to be Hin = 26.196 Kcal/Kg of wet air.

Step-6
Similarly find out the moist air enthalpy at exit condition, which is

Hex = 41.630 Kcal/Kg of wet air

Step-7
Similarly, find out the absolute humidity at wet bulb for exit condition, which is 50.74 Kg/ ‘000 kg of dry air in this case.

Step-8
Calculate airflow based on heat load and enthalpy difference, which shall be as below

A = 4134000 x (44-35)/(41.630 – 26.196)
= 2410652 Kg/hr

Now based on Absolute Humidity difference, calculate amount of water evaporated as below

W = 2756000 X (50.74 – 27.29)/1000
= 64654 Kg/hr

Step-9
Now heat required for evaporation of this water can be calculated based on average latent heat of water evaporation at the inlet & exit temperature.

Average water temperature = 39.5 °C
Latent heat = 575.33 Kcal/Kg

Hev = 64654 x 575.33
= 37.20 Gcal/Hr

This is matching with the heat load of waterside hence, calculation is correct due to accurate temperature measurements.

So L/G comes out to be = 1.715 in this case.

Now we will see the NTU calculation & efficiency of tower, use of NTU method for predictions etc.

Step-1
First consider the cooling water exit temperature ‘twex’ in column A in excel sheet so i.e. 35°C in this case.

Put h’ in column B which is the enthalpy of saturated air at twex and can be calculated by the equation

h’=9.446443x10-13x(twex^8)-1.433603766x10-10x(twex^7)+5.39506924*10-9(twex ^6)+3.02962638*10-7(twex^5)-0.0000272854755*(B7^4)+0.00096596975*(B7^3)-0.005340108*(B7^2)+0.458708485*B7+2.219286635

Put tawet in column C starting with actual wet bulb temperature of entering air, which is 30°C in this case.

Put w as absolute humidity at tawet in column D that is calculated from the same formula as shown in Part-I of this post.

Put hcal as humidity at tawet using the formula given above for h’ in column E.

Put ha as humidity at actual wet bulb temperature of entering air, which is 30°C in this case. Yes, that means initially in the first row of calculation sheet hcal & ha will be same. This is in column F.

Now put calculation of difference of h’ – ha in column G.

Step-2
In first row G will be automatically zero.
Now in second row consider the twex 2 = (Twin – Twex)/19 + twex 1
i.e. twex 2 = (44 – 35 ) / 19 + 35
= 0.474 + 35 = 35.474°C

Copy this formula in column A for next 19 rows. This gives you incremental evaluation of tower step by step along the total tower height from 35° at exit at bottom to 44° at inlet at the top.

Copy h’ formula in column B for the same no of rows.

Step-3
Now put any assumed figure for tawet in column C, w in column D, hcal in column E.

Now calculation for ha will change which will come from actual L/G ratio of tower calculated in Part-I.

Use the following formula for ha in second row onwards.

ha 2 = ha1 + L/G * (twex 2 - twex 1) + (w 2 – w 1) / 1000 * twex 1
= ha1 + 1.715 * (35.474 – 35.00) + (w 2 – w 1) / 1000 * 35.0

Based on other figures it will vary.

Now since you have assumed tawet, hcal will be different from ha. Put this difference in next column G.

Now either change tawet manually to make the difference Zero in column G or use goal seek from excel. This will give you tawet, which is supposed to be the actual wet bulb temperature of air exiting from the tower at the top finally.

This will complete first part of NTU calculation after completing all the rows.

Step-4
Now in next column i.e. H; put (h’ – ha) value which is Column B – Column F and copy it down till the last row.

Put reciprocal of column H in column I. This will give you 1/ (h’ – ha) value and copy it down till the last row.

Now in next column J, leave first row blank & start from second row where you should put average of first & second row in column I. This will give you average of 1 / (h’ – ha) for first & second value. Copy this formula also down till the end of rows.

Step-5
Now in column K, put NTUL as calculated below (From second row as column J starts from second row).

NTUL = Column J x (twex 2 - twex 1)
= Column J x (35.474 – 35.0)

Copy this formula in all rows.

In column L, put progressive summation of NTUL calculated in column K i.e. in each row of column L, use previous row of column L + same row of column K.

This value at the end of last row will give your towers total NTU for liquid side.

Step-6
Repeat all calculations in next two columns for NTUG similar to Step-5 above and find out final value of gas transfer units. The only difference is to use the following formula to calculate NTUG in column M.

NTUG = Column J x (ha 2 - ha 1) ha is in column F.

Use progressive sum again in column N.

Now I will give you guidelines on using these calculations for prediction of performance, prediction of new conditions, calculation of existing system and how to improve it in the next part of this post.

Ways to Save Energy in Pumps

Here is my experience based on energy audits of pumping systems in various chemical, metal, textile & petrochemical units.

• Design systems with lower capacity and total head

1. Do not assume these requirements are fixed.
2. Calculate flow requirement based on actual mathematical nos without margins in each stage & then add 10-20% straightforward as Normal capacity of the pump. For example if process side heat load in an exchanger is based on normal flow of say 100 M3/hr then do not consider cooling water requirement for peak condition of 120 or say 140 M3/hr. Just calculate it based on normal flow of 100 M3/hr at this stage.
3. Total head requirements can be reduced by: lowering process static gage, pressure, minimizing elevation rise from suction tank to discharge tank, reducing static elevation change by use of siphons, lowering spray nozzle velocities, lowering friction losses through use of larger pipes and low-loss fittings, and eliminating throttle valves.
4. After calculating total requirement of Flow & Head this way, simply add 10-20% in both parameters as design margin based on your judgement about process variation. This should be your normal capacity. You still have higher margins becasue rated condition are further higher than these values.
5. Also keep in mind that worst conditions dont come all simultaneously. You can still meet few peak demands.
6. Dont worry about undersizing of your pump. You can add later & this approach is beneficial in overall longer run as you can switch your additional capacity ON & OFF.
7. This can give you a saving of ~10-20% in your pumping system. A thorough review is must.

• Emphasize on Efficiency first

1. Despite the tendency to emphasize initial cost, you will save cost in the long run by selecting the most efficient pump type and size at the onset.
2. The choice of a pump depends on the service needed from the pump. Considerations are flow and head requirements, inlet pressure or net positive suction head available, and the type of liquid to be pumped.
3. Maximum attainable efficiency of a centrifugal pump is influenced by the designer's selection of pump rotating speed as it relates to "specific speed." Purchasers need to be aware of this, as well as the decision criteria for determining the type of pump to use.
4. Consider LCC (Life Cycle Costing) option instead of initial cost only. Click here to learn more about LCC Analysis.
5. Remember ENERGY is the most expensive "commodity" today.
6. People generally loose ~80% more money due to non LCC approachover a period of its service life.

• Divided Use

1. Design or select no of pumps based on different possible scenarios & always follow the operation philosophy of bulk & makeup supply for any system.
2. This approach saves at least 10-15% over conventional selection of equal size pumps.
3. This helps in putting the smaller pump on auto mode with header pressure switch so that excess pump capacity can be turned on/off.
4. Two pumps can be operated in parallel during peak demand periods, with one pump operating by itself during lower demand periods. Energy savings result from running each pump at a more efficient operating point and avoiding the need to throttle a large pump during low demand.
5. Analternative is to use one variable-speed pump and one constant-speed pump. Use or selection depends on the process behaviour e.g. how fast the demand is changing? How many time it is changing? Is my process critical? etc.

• Avoid end of curve operation

1. Generally in case of cooling water pumps head & flow both are selected with plenty of margins, comfortable to the cushion needed by selection manager. This results in near end curve operation without throttling. This is worst operation of pumps in almost every situation. Avoid it.

• Use pumps as drives

1. Use them as drivers / turbines to recover pressure energy that would otherwise be wasted.
2. Practically all centrifugal pumps will perform as turbines when operated in reverse.
3. A hydraulic power recovery turbine can recover pressure energy when used to drive a generator, or assist the driver of a pump or a compressor.

Useful Tips on Use of ASD/VFD for pumps

ASDs are ideally suited for variable-torque loads from centrifugal pumps, fans, and blowers when the system load requirements (head, flow, or both) vary with time. Conditions that tend to make ASDs cost-effective include the following:

1. High horsepower (greater than 15 to 30 hp)—the higher the pump horsepower, the more cost-effective the ASD application.
2. Load type—Centrifugal loads with variable-torque requirements (such as centrifugal pumps or fans) have the greatest potential for energy savings. ASDs can be cost-effective on positive displacement pumps, but the savings will generally not be as great as with centrifugal loads.
3. Operating hours—In general, ASDs are cost-effective only on pumps that operate for at least 2,000 hours per year at average utility rates.
4. High utility rates—higher utility energy charges provide a more rapid payback on an investment in an ASD.
5. Availability of efficiency incentives—where they are available, electric utility incentives for reducing energy use or installing energy-saving technologies will reduce payback periods.
6. Low static head—ASDs are ideal for circulating pumping systems in which the system curve is defined by dynamic or friction head losses. They can also be effective in static-dominated systems—but only when the pump is carefully selected. A thorough understanding of pump and system interactions is critical for such applications.
   
   
   
   
   Courtesy of  Profmaster