Question

A fruit juice is being heated in an indirect heat exchanger using steam as a heating medium. The product flows through the heat exchanger at a rate of 1000 kg/h and the inlet temperature is 308C. Determine the quantity of steam required to heat the product to 1008C when only latent heat of vaporization (2230.2 kJ/kg) from steam at 1108C is used for heating. The specific heat of the product is 4 kJ/(kg 8C).

Answer 1

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