11. Solved Problem - 2

Solved Problem

Example 2: Evaporator An aqueous solution of a solute is concentrated from 5% to 20% (mass basis) in a single-effect short-tube evaporator. The feed enters the evaporator at a rate of 10 kg/s and at a temperature of 300 K. Steam is available at a saturation pressure of 1.3 bar. The pressure in the vapor space of the evaporator is 0.13 bar and the corresponding saturation temperature of steam is 320 K. If the overall heat transfer coefficient is 5000 W/(m\(^2\).K), calculate the

  • steam economy

  • heat transfer surface area.

Enthalpy Heat of vaporization
(kJ/kg) (kJ/kg)
Saturated steam (1.3 bar; 380 K) - 2000
Saturated steam (0.13 bar; 320 K) 2200 -
Feed (5% ; 300 K) 80 -
Concentrated liquor (20% ; 325 K) 400 -


Boiling point elevation is 5 K. (GATE-2000)

Solution: Mass balance:
Let us denote flow rates of feed as \(F\), vapor as \(V\), concentrated product as \(P\) ,steam as \(S\) and the mass fraction of solute as \(x\).
Overall mass balance: \[F = V + P \label{overmass}\] Balance on solute: \[Fx_F = Px_P \label{solutemass}\] From Eqs.(1) and (2) \[P = \frac{10\times 0.05}{0.2} = 2.5 \mbox{ kg/s}\] And \[V = F - P = 10 - 2.5 = 7.5 \mbox{ kg/s}\]

Energy balance: \[FH_F + S\lambda_S = VH_V + PH_P\] Given:

\(H_F\) 80  kJ/kg; \(\lambda_S\) 2000 kJ/kg
\(H_V\) 2200 kJ/kg; \(H_P\) 400 kJ/kg


Therefore, \[\begin{aligned} 10 \times 80 + 2000\;S &=& 7.5\times2200+2.5\times400 \\ S &=& 8.35 \text{ kg/s} \end{aligned}\]

\[\text{Steam Economy } = \frac{V}{S} = \frac{7.5}{8.35} = 0.898 = 89.8\%\] Estimation of heat transfer area: \[\text{Rate of heat transfer } Q = S\lambda_S = 8.35 \times 2000 = 16700 \text{ kJ/s}\] Also \[Q = UA\Delta T\] Therefore, \[A = \frac{16700\times1000}{5000\times(380-325)} = 60.73 \text{ m$^2$}\]