Question

A forward feed triple effect evaporator, where each effect has 150 m2
of heating surface is used to
concentrate a solution containing 5% solids to a final concentration of 35% solids. Steam is available at
97 kPa (gauge), and the boiling point at the last effect is 40 °C, The overall heat transfer coefficients, U in
W/m2
°C are 2900 in effect 1, 2600 in effect 2 and 1300 in effect 3. The feed enters the evaporator at
90 °C. Calculate the flow rate of feed and the steam consumption. Assume boiling point elevation is
negligible.

Answer 1

Ans:

Solid feed rate = 97*150*0.05

                             = 727.5 kg/hr = 0.202 kg/s

Bottom outlet from third effect = 727.5/0.35 = 254.62 kg/hr = 0.0707 kg/s

Total evaporation = 727.5 – 254.62 = 472.87 kg/hr = 0.1313 kg/s

Assuming equal evaporation in all three effects.

W1 = Evaporation rate in first effect

W2 = Evaporation rate in second effect

W3 = Evaporation rate in third effect

W1 = W2 = W3 = 157.62 kg/hr = 0.0437 kg/s

Outlet from first effect = Wf – W1

                                      = 0.202 – 0.0437 = 0.1583 kg/s

Outlet from second effect = Wf – W2 – W1

                                      = 0.202 – 0.0437 – 0.0437 = 0.1146

Outlet from third effect = Wf – W3 – W2 – W1

                                      = 0.202 – 3(0.0437) = 0.0709