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 25^oC and convert into 100^oC 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 = 25^OC

Water Boiling Temp = 100^OC

Steam Tem = 100^OC

Steam Pressure = Atmospheric Pressure

Heat require to rise the temp from 25^OC to 100^OC 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 100^OC to 100^OC 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 25^OC to 100^OC 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)

Q_50kg = 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.

Comments

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