LESSON - 26 MEASUREMENT OF DRYNESS FRACTION: THROTTLING CALORIMETER AND COMBINED SEPARATING AND THROTTLING CALORIMETER AND THEIR NUMERICAL PROBLEMS

26.1. MEASUREMENT OF DRYNESS FRACTION

26.1.1. Throttling Calorimeter

Principle: The principle of the throttling calorimeter is to throttle the wet steam so that it becomes superheated.

Material: Refer Fig. 26.1. The Throttling Calorimeter consists of inner chamber and outer chamber. The pressure gauge attached to the outer casing and thermometer in inner chamber gives the pressure and temperature of steam after throttling, respectively. In order to prevent any heat transfer from and to the system, the outer chamber is insulated.

 Fig. 26.1. Throttling calorimeter along with its throttling process ‘1-2’ on h-s diagram

Procedure: A sample of wet steam at pressure ‘p1’ is taken from the main steam pipe of boiler into the inner chamber of the throttling calorimeter through a sampling tube and orifice throttling valve. This wet steam sample is throttled in orifice throttling valve before it enters into the calorimeter. Normally, pressure desired after throttled is few mm of Hg slightly above atmospheric pressure. The steam of the inner chamber is first allowed to flow down wards then flows upwards and finally turns downwards in the annular space between the inner and outer chamber. This is done so that the steam should flow long enough to ensure constant temperature of throttled steam.

During steam sampling: Note down the following observations:

  • Pressure of the wet steam before throttling = p2 bar

  • Pressure of the superheated steam after throttling = p2 bar

  • Temperature of the superheated steam after throttling = tsup,2 °C

For the accurate result from the throttling calorimeter, the necessary condition is that the steam should be in a superheated state after throttling. In general, 5°C of superheating is most desirable i.e. 5°C more than the saturation temperature corresponding to recorded pressure after throttling ‘p2’.

The process ‘1-2’ on h-s diagram in Fig. 26.2, represent the throttling process in orifice throttling valve of throttling calorimeter.

Calculations of dryness fraction of wet steam, x1:

Properties of wet steam before throttling at point ‘1’ by using Steam Tables:

Enthalpy of a saturated water corresponding to pressure p1 =   hf,1, kJ/kg.

Latent heat of evaporation corresponding to pressure p1 = hfg,1,kJ/kg

Properties of superheated steam after throttling at point ‘2’ by using Steam Tables:

       Enthalpy of a superheated steam corresponding to pressure p2 and temperature tsup,2 = hsup,2, kJ/kg.

Since during throttling process, the enthalpy remains constant, hence

Enthalpy of wet steam before throttling valve at point ‘1’ corresponding to pressure p1 = Enthalpy superheated steam after throttling valve at point ‘2’ corresponding to pressure p2

 or                       h1 =  hsup,2                                                                                                              ……….(26.1)

Where, Enthalpy of wet steam at point ‘1’ corresponding to pressure p1  is given by ,

h1= hf,1 + x1. hfg,1                                                                                                                   …….(26.2)

By using equation (26.2) in equation (26.1), we have

                   hf,1 + x1. hfg,1   =  hsup,2       

   or                                                                                                                         …….(26.3)                                     

An alternative method to find ‘x1 when superheated steam is treated as an ideal gas:

Then, enthalpy of a superheated throttled steam corresponding to pressure p2 and  temperature tsup,2   is calculated by  

                                          hsup,2 =   hg,2 + Cp(tsup,2 - tg,2)              , kJ/kg.                                       …….(26.4)      

                                                 where Cp is sp. heat of superheated steam  = 2.607 kJ/kg K

By using equations (26.2) and  (26.4) in equation (26.1), we have

           hf,1 + x1. hfg,1   =  hg,2 + Cp(tsup,2 - tg,2)

       or                                                                                         …….(26.5)                    

With the above equation (26.3) or (26.5) we can easily calculate the value of the dryness fraction.

Drawback

We can not use this method for finding out the dryness fraction of very wet steam because the required 5°C superheated steam after throttling cannot be achieved with the very wet steam.

For example, the wet steam at 40 bar pressure requires minimum 93.5% dryness fraction before throttling to reach 5°C superheating at pressure slightly above atmospheric pressure after throttling (the minimum dryness fraction required for 40 bar pressure can be find by a process ‘1-2’ as shown in Fig. 26.2). This means if the wet steam at 40 bar pressure with dryness fraction less than 93.5% is throttled in throttling calorimeter to find the dryness fraction, the condition of steam after throttling will be either less than 5°C superheating or it will be wet as shown by process ‘3-4’ in Fig. 26.2 which is not desirable to find dryness fraction with throttling calorimeter. Thus, steam having dryness fraction less than 0.934 cannot be measured by a throttling calorimeter for steam pressure 40 bar.

Similarly, wet steam at 3 bar pressure requires minimum 98.5% dryness fraction before throttling to achieve desired condition after throttling. Likewise we can find minimum dryness fraction of other pressures, below which we cannot use throttling calorimeter method.

   Fig. 26.2. Throttling process on h-s diagram

Problem 26.1: The following data refer to a throttling calorimeter.

Pressure in the main steam pipe = 10 bar, Pressure after throttling = 1.2 bar, Temperature after throttling = 120°C. Assuming Cp = 2.303 kJ/kg K for steam after throttling. Calculate the

(a)    Dryness fraction.

(b)   Minimum degree of wetness which can be shown by this calorimeter for this steam. If the wetness is beyond the range of this calorimeter suggest another method.

Solution:

Given: Steam at state ‘1’:  Pressure in the main steam pipe before throttling, p1 = 10 bar:

By using steam table (for dry saturated steam):

       For state 1, From steam tables for dry saturated steam at p1 = 10 bar, we have 

       hf,1 = 763 kJ/kg,   hfg,1 = 2015 kJ/kg

Given:  Steam at state ‘2’: Pressure after throttling, p2 = 1.2 bar;   Temperature after throttling, t2 = 120°C;

 By using steam table (for dry saturated steam):

        For state 2, From steam tables for dry saturated steam at p2 = 1.2 bar, we have 

        ts,2 =104.8°C,

Since,  t2 = 120°C > ts,2 =104.8°C the condition of steam at state ‘2’ is superheated. Therefore, tsup,2 =  t2 = 120°C

By using steam table (for superheated steam)

       For superheated state 2, From steam tables for superheated steam at p2 = 1.2 bar and tsup,2 =  120°C, we have

       hsup,2 = 2714.81 kJ/kg

Given:  Assuming Cp = 2.303 kJ/kg K for steam after throttling. Calculate the

(a)   Determine the dryness fraction of steam in main steam pipe:

             Formula: During throttling process enthalpy remains constant.

                   i.e.       h1=  hsup,2

          Since the steam is wet at state ‘1’:    h1= hf,1+ x1. hfg,1          

         Therefore,       hf,1+ x1. hfg,1   =  hsup,2

         or                

    Answer:  Dryness fraction,  x1   =   =  0.9686

(b)   Determine the minimum degree of wetness

For minimum degree of wetness, the condition of steam after throttling should be at least 5°C superheated at 1.2 bar pressure.

Therefore superheat temperature at point ‘2’,

              tsup,2 = tsat °C + 5°C =104.8 + 5 = 109.8°C

    By using steam table (for superheated steam)

For superheated state 2, From steam tables for superheated steam at  p2 = 1.2 bar and tsup,2 =  109.8°C, we have

       hsup,2 = 2687 kJ/kg

       Formula: Since, during throttling process enthalpy remains constant.

                   i.e.       h1=  hsup,2

     or     hf,1+ x1. hfg,1  =  hsup,2   (as condition of steam at point ‘1’ is wet)

   or                

Answer:   The minimum degree of wetness

                                                      x1 =         = 0.9582

If the dryness fraction is lower than 95.28%, a combination of separating and throttling calorimeter will be used.

26.1.2. Combined Separating and Throttling Calorimeter

A combined separating and throttling calorimeter is used to remove the drawback of Separating Calorimeter (i.e. water particles from wet steam are not fully separated) and the drawback of Throttling Calorimeter (i.e. not suitable for very wet steam).

Principle: In this calorimeter

  • The moisture from wet steam sample is first removed in separating calorimeter, so that dryness fraction of wet steam sample is increased above 0.95 before steam sample is entering into throttling calorimeter. During this process pressure and temperature remains constant.

  • The sample coming out from separating calorimeter is then passed through throttling calorimeter where it expands to superheated steam.

Material: Refer Fig. 26.2. It consists of a combined separating calorimeter and throttling calorimeter as discussed in previous sections.

Procedure: A sample of wet steam from main steam pipe (represented by point ‘A’ in Fig. 26.2) is first enters the separating calorimeter through a sampling tube. While passing through separating calorimeter, most of the moisture from wet steam is separated out and collected in the inner chamber of the separating calorimeter where it is measure with scale. Now, comparatively dry steam having dryness fraction above 0.95 from separating calorimeter (represented by point ‘B’ in Fig. 26.2) is passed into throttling calorimeter through throttling valve where it expands to superheated steam (represented by point ‘C’ in Fig. 26.2). The superheated steam after passing through throttling calorimeter is then exhausted from throttling calorimeter into a condenser, provided at the bottom of the throttling calorimeter, where it is finally condensed and measured.

The representation of separating-throttling processes of combined separating and throttling calorimeter on T-s and h-s diagrams is shown in Fig. 26.3.

During steam sampling: Note down the following observations:

  • Pressure before throttling  = p1  (say)  bar

  • Pressure after throttling  = p2     (say)  bar

  • Temperature of a throttled superheated steam =  tsup,C  (say),      oC

After steam sampling: Note down the following observations:

  • Mass of condensate collected in the container provided at the bottom of throttling calorimeter (i.e. mass of steam coming from separating calorimeter into throttling calorimeter through throttling valve) = M (say),kg

  • Mass of water collected in the inner chamber of the separating calorimeter (i.e. mass of water separated in the separating calorimeter =  m (say),kg

 

Fig. 26.2. Combined Separating and Throttling Calorimeter

Fig. 26.3.: Representation of two processes of separating-throttling calorimeter on T-s and h-s diagrams.

Calculations of dryness fraction of wet steam, x:

  • Partial dryness fraction of wet steam coming from main steam pipe, measured by separating calorimeter alone,  

                                                                                                ……………..(26.6)

  • Dryness fraction of steam entering into throttling calorimeter from separating calorimeter through throttling valve, measured by throttling calorimeter alone (by using equation 26.3)

                     

Where,

hf,B is enthalpy of saturated water at p1 (from steam tables)   , kJ/kg.

hfg,B is Latent heat of vapouization at p1 (from steam tables),kJ/kg

           hsup,C is enthalpy of a throttled superheated steam corresponding to pressure p2 and  temperature tsup,C (from superheated steam table),kJ/kg .

  • Mass of dry steam in the wet steam sample collected from main steam pipe = the mass of dry steam entering the throttling calorimeter = x2 . M

Finally, The total dryness fraction of wet steam collected from main steam pipe measure by Combined Separating and Throttling Calorimeter

     

By using equation (26.6) in above equation, we have

            x = x1.x2

This shows that the dryness fraction of steam measured from combined separating and throttling calorimeter is the product of dryness fraction x1 and x2 measured by separating and throttling calorimeter respectively.

Among all methods, this method of determining the dryness fraction of steam is the most satisfactory.

Problem 26.2: In a test on a combined separating and throttling calorimeter, the following data were obtained:-

Pressure in the steam main = 14 bar,

Pressure of steam after throttling = 13.2 cm of Hg gauge,

Mass of steam collected in the separator = 0.8 kg/min ,

Discharge from throttling calorimeter = 10 kg/min,

Temperature of steam after throttling = 115°C,

Barometer = 76 cm of Hg.

Calculate (a) the dryness fraction of steam in the main, (b) the velocity of flow in the steam main, if the diameter is 12 cm and the flow of steam is 130 kg/min, (c) the mass of water required to condense the steam after the throttling calorimeter if the temperature rise of the cooling water is l2°C.

Solution:

Given: Steam before throttling (state ‘1’):  Pressure in steam main before throttling, p1 = 14 bar:

By using steam table (for dry saturated steam):

For state 1, From steam tables for dry saturated steam at p1 = 14 bar, we have 

hf,1 = 830 kJ/kg,  hfg,1 = 1960 kJ/kg, vg,1 = 0.1408 m3/kg

Given:  Steam after throttling (state ‘2’):

             Pressure of steam after throttling, p2 = 13.2 cm of Hg gauge

                                                                  p2 = 76+13.2 = 89.2 cm of Hg.

                                                                       =   bar

          Temperature after throttling, t2 = 115°C;

   By using steam table (for dry saturated steam):

 For state 2, From steam tables for dry saturated steam at p2 = 1.1889 bar, we have 

 ts,2= 104.8°C,   hg,2 = 2683 kJ/kg,

          Since,  t2 = 115°C > ts,2 =104.8°C the condition of steam at state ‘2’ is superheated. Therefore, tsup,2 =  t2 = 115°C

  By using steam table (for superheated steam)

       For superheated state 2, From steam tables for superheated steam at p2 = 1.1889 bar and tsup,2 =  115°C, we have

       hsup,2 = 2704.3 kJ/kg

Given:   Mass of steam collected in the separator, m = 0.8 kg/min;

              Discharge from throttling calorimeter, M = 10 kg/min,

(a)  Determine the dryness fraction of steam in the main:

Formula: The dryness fraction of steam from separating and throttling calorimeter is calculated by the equation:

                                x = x1. x2

   where, x1 = Dryness fraction of steam measured by separating calorimeter,

               x2 = Dryness fraction of steam measured by throttling calorimeter

Finding unknown, x1 and x2:

                         Dryness fraction of steam measured by separating calorimeter,

                          x1 = 0.9259 

                          Dryness fraction of steam measured by throttling calorimeter,  

                          x2 =   = 0.956

     Answer:    x = x1.x2  = 0.9256 x 0.956 = 0.885

(b)  Determine the velocity of flow in the steam main, if the diameter is 12 cm and the flow of steam is 130 kg/min

       Given:

               Diameter of pipe, d = 12 cm = 0.12 m

Formula: From continuity equation, 

           or   velocity  = C1 =  

            Finding unknown, v1

                   Sp. volume of steam in the main pipe for wet steam,   v1 = x vg,1

                                                                                                                   = 0.885 x 0.1408

                                                                                                                   = 0.1246 m3/kg

Answer:     Velocity = C1 =  =1432.2   m/min

(c) Determine the mass of water  required to condense the steam after the throttling calorimeter if the temperature rise of the cooling water is l2°C.

            Given:  Temperature rise of the cooling water, Δt =  l2°C

 Formula:  If  mass of water be required in the condenser to remove the superheat and latent heat of M mass of steam from throttling calorimeter, then

                   M (hsup,2 – ts,2)  =  x Cp x Δt

Answer:               M (hsup,2 – ts,2)  =  x Cp x Δt

                        10 (2704.3 – 104.8) =  x 12 x 4.18

                or    = 518.24  kg/min

Last modified: Wednesday, 11 September 2013, 9:39 AM