Answer: The quantity of steam required is
0.0487kg/s = 175.38kg/h
Explanation: Quantity of steam required (Ms);
Ms = q ÷ He..............(1)
q = calculated heat transfer (kW)
He = evaporation heat of steam (kJ/kg)
STEP1:
CALCULATE THE HEAT TRANSFER
q = Cp × dT × m/s ............(2)
Where;
Cp is the specific heat of product = 4kJ/(kg8°C) = 0.5kJ/kg.°C
dT is the change in temperature (T2 - T1) = 1008°C - 308°C = 700°C
m/s is the mass flow rate = 1000kg/h = 0.278kg/s
Therefore using equation 2
0.5 × 700 × 0.278 = 97.3KW
q = 97.3KW
STEP2:
CALCULATE EVAPORATION HEAT OF STEAM
From the steam table we can't find steam at 1108°C. Therefore we will use interpolation to find the value.
From steam table:
At 1050°C, He=2006kJ/kg
At 1150°C, He=1991kJ/kg
Using interpolation formula;
Y = y1 + [(X-x1) ÷ (x2-x1)] × (y2-y1)
x1 = 1050
X = 1108
x2 = 1150
y1 = 2006
y2 = 1991
Therefore Y is
2006 + [(1108-1050) ÷ (1150-1050)] × (1991-2006)
2006 + 0.58 × (-15)
2006 - 8.7 = 1997.3
Y = 1997.3
Therefore the heat of evaporation He of the steam at 1108°C is 1997.3kJ/kg
STEP3:
CALCULATE THE QUANTITY OF STEAM REQUIRED
Using equation 1
Ms = q/He
Ms = 97.3 ÷ 1997.3 = 0.0487kg/s
Or convert to kg/h
0.0487kg/s × 3600s/h = 175.38kg/h
Therefore;
Ms = 0.0487kg/s = 175.38kg/h
