LESSON - 25 MEASUREMENT OF DRYNESS FRACTION: BUCKET OR BARREL CALORIMETER AND SEPARATING CALORIMETER AND THEIR NUMERICAL PROBLEMS

 25.1 MEASUREMENT OF DRYNESS FRACTION

There are four methods of determining the dryness fraction of steam experimentally. They are as follows:

(A)  Bucket or barrel calorimeter

(B)   Separating calorimeter

(C) Throttling calorimeter

(D) Combined separating and throttling calorimeter.

25.1.1.  Bucket or Barrel Calorimeter

Material: Refer Fig. 25.1. TheBucket/Barrel Calorimeter consists of a copper vessel which contains cold water. The copper vessel is insulated from the surrounding so as to prevent any heat transfer from or to the system. The top of the vessel is also covered by some detachable insulating medium. A thermometer is inserted in the water to record the temperature of it.      

 Fig. 25.1. Bucket Calorimeter

Procedure:

Before steam sampling: Note down the following observations

• Mass of empty copper vessel = m₁(say)

• Mass of cold water of vessel = m₂(say)

• Temperature of cold water and copper vessel = t₁(say)

• Specific heat of copper vessel = C_p,copper(say)

• Specific heat of water = C_p,water(say)

During steam sampling: Open the valve in sampling tube and allow steam sample from main steam pipe of boiler to discharge into cold water of vessel through fine exit holes. The steamcoming in contact with water gets condensed, giving out its entire latent heat and part of its sensible heat. This heat transfer results in an increase in the temperature of water. The mass of water in the calorimeter increases by the amount of steam condensed. As a precautionary measure, sufficient quantity of steam should be passed into the calorimeter so that there is a marked rise in the temperature so as to minimizing the error in the experiment. Now, the supply-of steam is closed.

• During sampling, pressure of steam in steam pipe recorder with pressure gauge = p (say)

• From Steam Table against pressure ‘p’ note down the saturation temperature of steam and latent heat of vaporization= t_sat and  h_fg(say) , respectively.

After steam sampling: Note down the following observations

• Mass of warm water in vessel = m₃ (say)

• Temperature of warm water and copper vessel after mixing of steam sample in cold water= t₂ (say)

Calculations to find dryness fraction, x:

 Mass of wet steam sample condensed in cold water = m₃ – m₂   = m (say)

Mass of dry steam in wet steam sample = x × m

 If there is no heat lost then according to conservation of energy,

Heat lost by mass of wet steam sample = Heat gain by copper vessel and cold water                                        …..….(25.1)

Where,

Heat lost by mass of wet steam sample =  Latent heat lost by mass of dry steam in wet steam sample + Sensible heat lost mass of wet steam sample

=                                                                                                              ……..(25.2)

Heat gain by copper vessel and cold water =  Sensible heat gain by mass of copper vessel + Sensible heat gain by mass of cold water  

                                                                                                        …..….. (25.3)

By using equations (25.2) and (25.3) in equation (25.1), we have

                

        

 

Drawback

• It is to be noted that this method for determining the dryness fraction is approximate.

• The increase in temperature of cold water after steam sampling should neither be too low nor too high. If the rise in temperature is large there is always a chance for heat losses due to radiation.

Problem 25.1: The following data refer to a test on bucket calorimeter. Pressure of steam from steam pipe = 6.5 bar, water equivalent of water at 20°C = 500 kJ/k, Water equivalent of bucket calorimeter = 30 kJ/k, Temperature of warm water after mixing of steam sample in cold water = 38°C, Mass of wet steam sample = 4 kg, Find the dryness fraction of steam.

Solution:

Given:  Pressure of steam from steam pipe, p = 6.5 bar;

       Water equivalent of water at 20°C, = 500 kJ/k;

      Water equivalent of bucket calorimeter, = 30 kJ/k;

      Temperature of cold water and copper vessel = t₁= 20°C

      Temperature of warm water after mixing of steam sample in cold water, t₂ = 38°C;

      Mass of wet steam sample, m = 4 kg

Determine the dryness fraction of steam:

        Formula:  The dryness fraction for bucket calorimeter is given  by

    

Finding unknown,    t_sat and h_fg:  

By using steam table (for dry saturated steam):

From steam tables for dry saturated steam at p = 6.5 bar, we have 

 t_sat = 161.9°C   and h_fg = 2077 kJ/kg  

Answer: Dryness fraction of steam,

                               

                                x =  = 0.8989  

25.1.2.  Separating Calorimeter

Principle: The principle of determining the dryness fraction from separating calorimeter is to separate out the moisture content from the wet steam.

Material: Refer Fig. 25.2. It consists of inner chamber having cross sectional area ‘A’ in which water separated from wet steam is collected. A gauge glass having a graduated scale is mounted on the inner chamber of the separating calorimeter to measure the amount of water separated after and before the sampling.It also consists of steam condensing chamber so that dry and saturated steam may be condensed and measured in it.

Procedure:

Before steam sampling: Note down the following observations

• Level of water in the inner chamber = L₁(say)

• Mass of water in the steam condensing chamber =  m₁(say)

During steam sampling: The wet steam enters the inner chamber of separating calorimeter through a sampling tube controlled by a valve. While testing for dryness the valve should be fully open avoiding any throttling or wiredrawing. In the inner chamber of separating calorimeter, separators or baffles plates are arranged in such a manner as to separate out the moisture from the steam. Hence, when steam enters the separating calorimeter, it undergoes a sudden reversal of direction of motion when it strikes the baffles plates. This causes the water particles which have greater inertia to separate from the wet steam. The water moves downwards(shown by blue drops) and collected at the bottom portion of the inner chamber of the separating calorimeter. The dry steam moves upwards (shown by firm arrows) and enters the annular space between the inner and outer surface of the separating calorimeter and goes downward.The dry and saturated steam is discharged from the separating calorimeter and finally condensed in the steam condensing chamber. The condensed steam in steam condensing chamber is then weighed.

Before steam sampling: Note down the following observations

• Level of water in the inner chamber = L₂ (say)

• Mass of water in the steam condensing chamber =  m₂   (say)

Calculations of dryness fraction, x:

Mass of water separated from the wet steam sample and collected at the bottom portion of the inner chamber     = m            ,kg    (say)

Mass of dry steam condensed in steam condensing chamber = (m₂ - m₁)  =  M   , kg         (say)

Dryness fraction of wet steam from main steam pipe, 

Drawback

This method is also approximate because of incomplete separation of moisture and inaccuracy of gauge.

Fig. 25.2. Separating Calorimeter

Problem 25.2: During a test on separating calorimeter the following observations were taken.

Mass of water separated from wet steam sample from main steam pipe = 0.3 kg/min, Mass of dry saturated steam passing through separating calorimeter and condensed in steam condensing chamber = 3 kg/min. Calculate the dryness fraction.

Solution:

Given: Mass of water separated from wet steam sample, m= 0.3 kg/min

           Mass of dry saturated steam condensing chamber, M = 3 kg/min

Determine the dryness fraction

        Formula:   Dryness fraction for separating calorimeter,

                              x = dryness fraction of wet steam = 

Answer:  dryness fraction of wet steam=  x =  = 0.909 = 90.9%