Question

A single-effect evaporator is used, which concentrates 15,000 kg/hr of 15% to 45% solids sodium hydroxide solution at a temperature of 30 degree C. The steam flow has a manometric pressure of 160 kpa and the vacuum pressure is 80 mmhg, the overall coefficient of 1,000 W/m^2. Calculate the amount of water vapor required and the required heating surface. Enthalpy of solution at 15% and 30 degree CH = 50 Btu/lb, Boiling point of the solution at 45% T = 175 degree F Enthalpy of solution at 45% and 175 degree F H = 180 Btu/lb.

Answer 1

(a) The amount of water vapor required which will be given as :

using a formula, we have

q = _s (h_s - h_c)

(a) The amount of water vapor required which will be given as: using formula, we have q = m_s (h_s - h_c) where, m_s = flow rate of steam = 15000 kg/hr = 4.16 kg/s h_s = specific enthalpy of steam = 180 Btu/lb = 418.6 kJ/kg h_c = specific enthalpy of condensate = 50 Btu/lb = 116.3 kJ/kg then, we get q = (4.16 kg/s) [(418.6 kJ/kg) - (116.3 kJ/kg)] q = (4.16 kg/s) (302.3 kJ/kg) q = 1257.5 W (b) The required heating surface will be given by - we know that, q = U A Delta T where, U = overall heat transfer coefficient = 1000 W/m^20 C Then, we get (1257.5 W) = (1000 W/m^20 C) A [(79.4^0 C) - (30^0 C)] A = (1257.5 W)/[(1000 W/m^20 C) (49.4^0 C)] A = (1257.5 W)/(49400 W/m^2) A = 0.0254 m^2

where, _s = flow rate of steam = 15000 kg/hr = 4.16 kg/s

(a) The amount of water vapor required which will be given as: using formula, we have q = m_s (h_s - h_c) where, m_s = flow rate of steam = 15000 kg/hr = 4.16 kg/s h_s = specific enthalpy of steam = 180 Btu/lb = 418.6 kJ/kg h_c = specific enthalpy of condensate = 50 Btu/lb = 116.3 kJ/kg then, we get q = (4.16 kg/s) [(418.6 kJ/kg) - (116.3 kJ/kg)] q = (4.16 kg/s) (302.3 kJ/kg) q = 1257.5 W (b) The required heating surface will be given by - we know that, q = U A Delta T where, U = overall heat transfer coefficient = 1000 W/m^20 C Then, we get (1257.5 W) = (1000 W/m^20 C) A [(79.4^0 C) - (30^0 C)] A = (1257.5 W)/[(1000 W/m^20 C) (49.4^0 C)] A = (1257.5 W)/(49400 W/m^2) A = 0.0254 m^2

h_s = specific enthalpy of steam = 180 Btu/lb = 418.6 kJ/kg

h_c = specific enthalpy of condensate = 50 Btu/lb = 116.3 kJ/kg

then, we get

q = (4.16 kg/s) [(418.6 kJ/kg) - (116.3 kJ/kg)]

q = (4.16 kg/s) (302.3 kJ/kg)

q = 1257.5 W

(b) The required heating surface will be given by -

we know that, q = U A $\Delta$ T

where, U = overall heat transfer coefficient = 1000 W/m²⁰C

Then, we get

(1257.5 W) = (1000 W/m²⁰C) A [(79.4 ⁰C) - (30 ⁰C)]

A = (1257.5 W) / [(1000 W/m²⁰C) (49.4 ⁰C)]

A = (1257.5 W) / (49400 W/m²)

A = 0.0254 m²