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50 litter of water is heated with Electric Heater at Atmospheric Pressure from a temperature of 25C and convert into 100C Steam.


50 litter of water is heated with Electric Heater at Atmospheric Pressure from a temperature of 25oC and convert into 100oC Steam. How much heat and what capacity of Electric heater is required to do so. The evaporation will be done within 10 hours (10 x 3600 = 36000 seconds). 



SOLUTION
 
Water to be evaporate = 50 kgs or 50 liters
Feed Water Temp = 25OC
Water Boiling Temp = 100OC
Steam Tem = 100OC
Steam Pressure = Atmospheric Pressure

Heat require to rise the temp from 25OC to 100OC of 1000 gm water
Q1 = m C (T2 – T1)
Q1 = 1000 x 1 x (100 – 25)
Q1 = 1000 x 1 x 75
Q1 = 75 x 103 Cal

Now heat require to convert 1000 gm of water at 100OC to 100OC Steam at Atmospheric pressure
Q2 = mL
Q2 = 1000 x 540
Q2 = 5.4 x 105 Cal

Total Heat required to convert 1000 gm (1Kg) of water at 25OC to 100OC Steam.
Q = Q1 + Q2
Q = 75 x 103 + 5.4 x 105
Q = 615 x 103 Cal
Q = 615 x 103 x 4.184 = 2.57 x 106 Joule                             (1cal = 4.184 J)
Q50kg = 2.57 x 106 Joule x 50 = 128.5 x 106 Joule

Now calculation for Electric Heater required capacity
We know that, P = V x I
Also we know, V = I x R
And,  H = I2 x R x T
So, Q = P x T

Where,
P = Power in watt
V = Voltage in volt
I = Current in amp
R = Resistance in omh
T = Time in seconds
Q = Heat in joules

Now put the value in (Q = P x T) this equation
128.5 x 106 = P x 10 x 3600
P = 128.5 x 106/36000
P = 3569.44 W

If Electric Heater efficiency is about 80% then
Required Electric Heater Power = 3569.44/80%
Required Electric Heater Power =4461.8 W


Therefore we use 4500 W rating Electric Heater to evaporate the water within 10 hours.




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