Question

A single-effect evaporator with a heat transfer area of 70 m$^{2}$ concentrates a salt solution using steam. The salt solution feed rate and temperature are 10000 kg h$^{-1}$ and 40$^\circ$C, respectively. The saturated steam feed rate and temperature are 7500 kg h$^{-1}$ and 150$^\circ$C, respectively. The boiling temperature of the solution in the evaporator is 80$^\circ$C. The average specific heat is 0.8 kcal kg$^{-1}$ K$^{-1}$. The latent heat of vaporization is 500 kcal kg$^{-1}$. If the steam-economy is 0.8, the overall heat transfer coefficient is \(\underline{\hspace{2cm}}\) kcal h$^{-1}$ m$^{-2}$ K$^{-1}$ (rounded off to the nearest integer).

Hint:

Evaporator heat load always includes both sensible heating of the feed and the latent heat needed for evaporation.

Answer 1

Correct Answer: 675 - 680

Steam economy:
\[ \text{Economy} = \frac{\text{kg of water evaporated}}{\text{kg of steam used}} = 0.8 \] Steam used = 7500 kg/h, therefore evaporated water:
\[ W = 0.8 \times 7500 = 6000\ \text{kg/h} \] Total heat required in evaporator: \[ Q = \underbrace{F C_p (T_b - T_f)}_{\text{sensible}} + \underbrace{W \lambda}_{\text{latent}} \] Sensible heat: \[ Q_s = 10000 \times 0.8 \times (80 - 40) = 10000 \times 0.8 \times 40 = 320{,}000\ \text{kcal/h} \] Latent heat: \[ Q_l = 6000 \times 500 = 3{,}000{,}000\ \text{kcal/h} \] Total heat load: \[ Q = Q_s + Q_l = 3{,}320{,}000\ \text{kcal/h} \] Temperature driving force: \[ \Delta T = 150 - 80 = 70^\circ\text{C} \] Overall heat-transfer coefficient: \[ U = \frac{Q}{A\,\Delta T} = \frac{3{,}320{,}000}{70 \times 70} = \frac{3{,}320{,}000}{4900} = 677.55\ \text{kcal h}^{-1}\text{m}^{-2}\text{K}^{-1} \] Rounded to nearest integer: \[ U = 678 \]