Evaporation: Heat and Mass balance
Duhring graph is on the left side. RHS graph stands for the Enthalpy concentration chart for caustic soda.Solve - A given evaporator is to be fed with 10,000 lb/hr of a 20% solution of sodium hydroxide at 100 deg F. This is to be concentrated to 50% solution. The evaporator is supplied with saturated steam at 5 psig and operates with vapor space at a vacuum of 26 in. There is negligible entrainment. Condensate may be assumed to leave at steam temperature and losses by radiation are neglected. Calculate [1] Steam consumption [2] Heat transfer area, heat transfer coefficient U = 400 and [3], Steam economy
The post covers [1] General discussion of evaporators [2] Key controls [3] Heat and mass balance concept [4] Performance of evaporators [capacity] and steam economy [5] Boiling point elevation and Duhring graph and [6] Problem solution
General: Evaporation is a unit operation that separates a liquid from solids by means of heat transfer via vaporization or boiling. The purpose of evaporation is to concentrate a solution of a non-volatile solute (i.e., solids) and a solvent (i.e., liquid), which is typically water. Evaporating a portion of the solvent concentrates the solute into a more viscous liquid product
Evaporation process – short description
Single effect evaporator
All evaporators are comprised of two sections: a heating section, calandria (called a steam chest), and a vapor/liquid separation section. These sections can be located within a single vessel (body), or the heating section may be external to the vessel that houses the vapor/liquid separation section.
The solution containing the desired product is fed into the evaporator and passes across a heat source. The applied heat converts the water in the solution into vapor. The vapor is removed from the rest of the solution and is condensed while the now-concentrated solution is either fed into a second evaporator or is removed.
Double effect evaporator
Multiple Effect Evaporators
Evaporators are classified by the number of effects. In a single-effect evaporator [ above left side image], steam provides energy for vaporization, and the vapor product is condensed and removed from the system. In a double-effect evaporator [above center image], the vapor production of the first effect is used to provide energy for a second vaporization unit. The image on the right side is a triple effect evaporator. The vapor production of the second effect is used to provide energy for the vaporization of water in the third effect.
Multiple-effect evaporators can remove much larger amounts of solvent than is possible in a single effect.
Triple effect evaporator
In a multiple-effect arrangement, the latent heat of the vapor product of of an effect is used to heat the following effect. Effects are thus numbered beginning with the one heated by steam. It will have the highest pressure. Vapor from Effect I will be used to heat Effect II, which consequently will operate at lower pressure. This continues through the train: pressure drops through the sequence so that the hot vapor will travel from one effect to the next. Normally, all effects in an evaporator will be physically the same in terms of size, construction, and heat transfer area. Unless thermal losses are significant, they will all have the same capacity as well.
Evaporators are typically operated under a vacuum to reduce the temperature of boiling. Steam ejectors or mechanical vacuum pumps are often used to create a vacuum.
Key controls
Once through evaporator vs circulation evaporator
In once-through evaporation, the feed liquor passes through the calandria tubes only once and releases vapor, and leaves the unit as thick liquor. All evaporation is accomplished in a single pass. This is how multiple effect evaporators work where the total concentration is spread over several effects Once through evaporation is especially useful for heat-sensitive materials.
In circulation evaporation, a pool of liquid is held within the equipment. Incoming feed mixes with the liquid from the pool and the mixture passes through the tubes. Unevaporated liquid discharged from tubes returns to the pool so that only a part of total evaporation occurs in one pass. Circulation evaporators do not suit heat-sensitive materials. Circulation through tubes can operate by natural circulation because of density difference or by forced circulation by the pump.
Effect of feed temperature
The volume of liquid in an evaporator is always very large and in its final concentration and therefore at the boiling temperature of the final solution. The rate at which feed is added to such an evaporator should be such that the average temperature of the liquid boiling in the evaporator is not appreciably affected by such additions.
Choice of steam pressure
The general choice with some exceptions is about 1 atmosphere. The choice is low-pressure steam for large latent heat.
Pressure in vapor space
Evaporators don’t necessarily have to work in a vacuum. The reason for the vacuum is it is most economical to apply vacuum to feed steam at a modest pressure into an evaporator which gives an economical ?t.
Heat and Mass balance on a single effect evaporator
The capacity of an evaporator like any heat transfer process is given by
q = U A ?tm [ q is the rate of heat transfer, U is overall heat transfer coefficient, A is heat transfer area and ?tm is temperature difference].
Performance of an evaporator
The above diagram is a simple diagram for an evaporator. The heating surface is represented by a coil. F lb/hr of feed enters the evaporator with a solid content XF. Symbol X stands for weight fraction which is percentage divided by 100. Let the enthalpy of feed is Hf/lb. L is thick liquor taken out from evaporator with XL weight fraction of solute and hL enthalpy in Btu / lb. V lb is vapor with a solute concentration of y and enthalpy H Btu/lb. In most evaporators, the vapor is water, and therefore Y = 0.
Mass balance
F = L + V
For solute alone,
F*XF = L*XL + V*Y ------------ [1]
Let S lb of steam is supplied with enthalpy HS Btu/lb. Let S lb condensate leaves the heating surface with enthalpy hC Btu/lb.
One assumption is usually made in evaporators is that there is very little cooling of condensate. This is never more than a few degrees in practice and the sensible heat recovered from cooling the condensate is so small compared to latent heat of the steam supplied to the heating surface that this sensible heat is usually neglected. This amounts to assuming that the condensate will leave at the condensing temperature of the steam.
Heat balance
Heat in = Heat out
[Heat in feed] + [ Heat in steam] = [Heat in thick liquor] + [Heat in vapor] + [Heat in condensate] + [ Heat loss by radiation]
Neglecting heat loss by radiation,
F*hF + S*HS = L*hL + V*H + S*hC ------------ [2]
Performance of evaporators [capacity] and steam economy
The performance of a steam-heated evaporator is measured in terms of its capacity and economy. Capacity is defined as the number of kilograms of water vaporized per hour. Economy (or steam economy) is the number of a kilogram of water vaporized from all the effects per kilogram of steam used. For a single effect evaporator, the steam economy is about 0.8 (<1). The capacity is about n-times that of a single effect evaporator and the economy is about 0.8n for an n-effect evaporator. However, pumps, interconnecting pipes, and valves are required for the transfer of liquid from one effect to another effect which increases both equipment and operating costs.
Boiling point elevation (BPE) Most evaporators produce concentrated liquor having a boiling point considerably higher than that of pure solvent (or water). This phenomenon is called boiling point elevation (BPE). BPE occurs as the vapor pressure of a solution (usually aqueous solution) is less than that of the pure solvent at the same temperature. The boiling point of a solution is a colligative property. It depends on the concentration of solute in the solution for a pair of solute and solvent. BPE of the concentrated liquor reduces the effective temperature driving force compared to the boiling of the pure solvent. Equilibrium vapor generated from a solution exhibiting boiling point elevation is superheated with respect to vapor generated during boiling of the pure solvent. The vapor is generated at the solution boiling point, which is higher than the pure component boiling point. The vapor, however, is solute-free, so it won’t condense until the extra heat corresponding to the elevation is removed, thus it is superheated. Therefore, the BPE of the concentrated solution must be known for evaporator design. Determination of BPE: For strong solutions, the BPE data is estimated from an empirical rule known as the Dühring rule. This states that the boiling point of a given solution is a linear function of the boiling point of pure water at the same pressure. Thus if the boiling point of the solution is plotted against the corresponding boiling point of pure water at the same pressure, a straight line is generated. Different lines are obtained if such plots made for solutions of different concentrations. The main advantage is that Dühring lines can be drawn if boiling points of a solution and water at two different pressures are known. This line can be used to predict the boiling point of a solution at any pressure. A Dühring plot for the NaOH-water system is given below
Duhring graph
Duhring graph is on the left side. RHS graph stands for the Enthalpy concentration chart for caustic soda.
Solution of problem
Problem statement: A given evaporator is to be fed with 10,000 lb/hr of a 20% solution of sodium hydroxide at 100 deg F. This is to be concentrated to 50% solution. The evaporator is supplied with saturated steam at 5 psig and operates with vapor space at a vacuum of 26 in. There is negligible entrainment. Condensate may be assumed to leave at steam temperature and losses by radiation are neglected. What is the steam consumption? If overall heat transfer coefficient U = 400, what is the heating surface required?
Given [ For symbols refer to heat and mass balance diagram above]
F = 10,000 lb/hr, XF = 0.20, tF = 100 degF, XL= 0.50, y = 0.
From steam table, HS = 1156 Btu/lb, hC = 196 Btu/lb, tS = 228 deg F
Vapor space at a vacuum of 26 in.
The boiling point of water at 4-in abs pressure [26 -in vacuum] = 125.4 deg F
Enthalpy of saturated steam at 125 deg F = 1116 Btu/lb [ from steam table]
The boiling point of concentrated solution = 198 deg F at the boiling point of water at 125.4 deg F from Duhring graph below on the left side. RHS graph stands for the Enthalpy concentration chart for caustic soda.
From Enthalpy concentration chart for caustic soda, on RHS,
hF = 56.5 Btu/lb, hL = 222 Btu/lb
Material balance
F*XF = L*XL + V*Y
[10,000]*[0.2] = 0.5*L
L= 4000 lb / hr, therefore V = 6000 lb/hr
Enthalpy balance
Steam consumption
From equation [2]
F*hF + S*HS = V*H + L*hL + S*hC
Or, F*hF + S*[ HS – hC] = V*H + L*hL --------------- [3]
In calculating H, the enthalpy of vapor leaving the solution it must be considered that the vapor is in equilibrium with the boiling solution at a pressure of 4 in absolute and therefore is superheated in comparison with vapor in equilibrium with water at the same pressure. The specific heat of superheated steam at this region has been taken as 0.46, therefore
H = 1116 [ enthalpy of saturated vapor at 125.4 degF] + 0.46 [ specific heat of superheated steam] * [198-125.4][degree of superheat] = 1149 Btu/lb
Substituting value of H in equation [3]
F*hF + S*[ HS – hC] = V*H + L*hL
[10,000]*[56.5] + S*[1156 (enthalpy of steam)- 196( enthalpy of condensate)]
= [6000 (vapor)*1149(enthalpy of vapor) + [4000 (liqor)*222] [image 4]
S = 7520 lb / hr
Steam economy:
Water evaporated:
Initial water = 10000 x 0.80 = 8000 lb/hr [ 2000 lbs solid and 8000 lb water, 20% solution]
Final water = 4000 x 0.5 = 2000 lb solid and 2000 lb/ hr water, 50% solution
Water evaporated = 8000 – 2000 = 6000 lb/hr
Steam economy = 6000/7520 = 0.8
Heating surface
Q = U*A*?t
[7520]*- [1156-196] = [400] (U = 400) * A *[228(steam temperature) – 198] [image 3, Duhring lines for sodium hydroxide]
A = 600 sq ft.