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Problem / Discussion

Eva oration - Refri eration 

Robi Andoyo, STP., M.Sc., Ph.D

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Tomato juice at 5.5 kg/s feed flow rate and 60   0 C inlet temperature is concentrated in a double-effect forward feed evaporator using steam in the first effect at a pressure of 198.5 kPa (absolut). The heat transfer area, the overall heat transfer coefficient, and the pressure in each effect are shown in the following table. The solids content and the heat 

capacity of the feed are 11% and 3900 J/kg   0 C respectively.Calculate the steam flow rate, the solids content at the exit of each effect, the steam economy, and the flow rate of cooling water in the condenser.

Assume : The condensate in each effect exits at the condensation temperature.

The operation is steady state.

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Step 1

Draw the process diagram

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Step 2Combine the heat transfer rate equation with an enthalpy

balance in the first effect to calculate the steamconsumption

= 198.5 kPa

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• The saturation temperature in the 1st effect, Tsat1,

is found from steam tables forPsat = 90 kPa to be equal to 97

0C = Tb1•

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Step 3

Calculate the solids content

Hv1 equal to the enthalpy of saturated steam,Hv1 saturated. Thus:

Hv1 at 970C = 2671 kJ/kg (from Steam tables)

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• Find mv1

: A mass balance on the product gives

mf = mo1 + mv1 or mv1 = mf - mo1

• Find the enthalpy of liquid streams

Using 0 0C as a reference temperature, the

enthalpy of the liquid streams is:

 H = cpT

Thus :

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Solve mo1, substitute values, and calculate the

mass flow rate at the exit of the first effect

mo1 = 3,6 kg/s

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Calculate the solids content at the exit of the

first effect from a solids mass balance

3,60,16

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At the exit of the second effect• Write an enthalpy balance around the second

effect

3,6

3,6 1,9

1,9

970

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• The saturation temperature in the second effect,

Tsat2, is found from steam tables for Psat = 17.9

kPa to be equal to 57.6 0C.

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3.6 1.9

1,83.6

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• Substitute values and calculate the evaporationrate in the second effect

• Calculate the solids content at the exit of the

second effect from a solids mass balance

mv2

= mi2

- mp

= 3,6 – 1,8 = 1,8 kg/s

 

0,163,6

1,8

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Step 4

Calculate the steam economy

1,9

,

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It wished to freeze 15 ton of fish /dayfrom an initial temp. of 10 0 C to a finaltemp. of -8 0C using a stream of cold air.

Estimate the max capacity of ref. Plantrequired. If the heat transfer-coefficientfrom the air to evap. Coil (over all) is 22

.-

.-

.-

. u uevaporator coil required if the logarithmicmean temp drop across the coil is 12 0C.

Spesific heat of fish is 3.18 kJ/kg.0

C abovefreezing and 1.67 kJ/kg.0C bellow freezing& latent heat is 276 kJ/kg

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Enthalpy change in fish:

(10 x 3.2) + (8 x 1.67) + 276 = 31.8 + 13.4 + 276CpL.(TL-Tref) + Cpf.(Tf-Tref) + λ = 321.2 kJ kg

-1

Heat removed in freezin : 

15 x 1000 x 321.2 = 4.82 x 106 kJ day-1

Average rate of heat removal:(4.82 x 106)/(24 x 60 x 60) = 55.8 kJ s-1

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Step 6

Calculate the cooling water flow rate in thecondenser

• Write an enthalpy balance in the condenser

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Rate of heat transfer q = U A ∆T LM 

A = q / U ∆T LM= (55.8) / (22 x 12)

= 0.21 x 103 m2

 = 210 m2

.