The system was allowed to reach study state (when outlet temperatures no longer vary), then the water temperature in and out of the lower exchanger was measured, the steam temperature, the condensate temperature, the water flowrate and the condensate flowrate. The condensate flowrate was determined by measuring the time of collection of a know volume of condensate.
The procedure was carried out foe six different cold water flowrates over a wide range of values; - for each run was ensured that the study state has been reached before taking readings.
THEORY:
A major resistance to heat transfer from the hot steam to the cold water is due a stagnant layer of steam and water on the surface on each side of metal wall of the inner tube. If the flowrate of the cold water is increased, the cold water becomes more turbulent and the stagnant layer on the cold water side of the inner becomes thinner, thus reducing its resistance to heat transfer.
An energy balance shows that, at study state.-
Loss of heat from the hot steam, Qh= gain of heat by the cold water, Qc
For the water side:
QC=mc.Cpc. (Tc out – Tc in) ……………………………… (1)
For the steam side:
Qh =mh.λv+mh.Cph. (Ts-Th out) …………………………. (2)
Where mh can be calculated from mh = Fh* ρc
And mc = Fc * ρc
Symbols:-
Cpc= specific heat capacity of cold water, J/kg.K
Cph= specific heat capacity of hot condensate, J/chg.
Fc = volumetric flowrate of cold water, m3/s
Fh = volumetric flowrate of hot condensate, m3/s
mc = mass flowrate of cold water, kg/s
mh = mass flowrate of hot condensate, kg/s
Qc = energy gained by cold water, J/s
Qh = energy lost by steam, J/s
Tc in = temperature of cold water entering
Ts = temperature of steam entering exchanger, K
Th out = temperature of hot condensate leaving exchanger, K
λv = latent heat of vaporisation of steam, J/kg
ρc = density of cold water, kg/m3
ρh = density of hot condensate, kg/m3
RESULTS:
The results which took in the laboratory:
CALCULATION:
Note:
Conversion:
(Litters/min) = (0.001/60) (m3/s)
(Millilitres/min) = (0.000001/60) (m3/s)
C= 273 K
Data given:
Cpc = 4180 J/kg.K
Cph = 4220 J/kg.k
λv =2.22 * 106 J/kg.k
ρc = 996 Kg/m3
ρh = 958 Kg/m3
- To worked out the value of Qc we used equation ( 1), but first we must got the value of mc from this equation :
mc = Fc * ρc
To get values of Fc we convert the values of cold water flowrate from (Litters/min) to (m3/s), after we did all that steps we can work out the value of Qc easily. (As we know the value of ρc and Cpc given)
Example: (I used the first run in this example):
Fc = 17 * (0.001/60) = 0.000283 (m3/s)
mc=0.000283 * 996= 0.2822 kg/s
Qc =0.2822 * 4180 (314 – 280) = 40106.26 J/s
- To worked out the value of Qh we used equation (2), but we had first to got the value of mh from this equation:
mh = Fh* ρc
To get value of Fh we took the average of the condensate floweate values and then we converted it from (millilitres/min) to (m3/s).
Example: (I used the first run in this example):
Fh = (1800+1840)/2 * (0.000001/60) = 3.03333E-05(m3/s)
mh=3.03333E-05 * 958 = 0.029059333 kg/s
Qh = 0.029059333*2.22 * 106 + 0.029059333 * 996 (386 – 385) =64634.35 J/s
DISSCUSSING THE RESULT& CONCOLUTION:
- When the flowrate is increased the % difference between the energy lost by steam and the energy gained by cold water will decrease.
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The relation between Qc and Fc is direct proportion, therefore to transfer heat faster, we must increase the flowrate.
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The value of the % difference in the first run is 125.7, which I think it is wrong and that happened because the flowarete value was too small, and that make the value of Qc small as well, therefore the difference is high.
BIBLIOGRAPHY:
Laboratory worksheet
