Heating area calculation in a two-stage evaporator
Here a double effect evaporator is analized from the point of view of the heating area. Heat and mass balances are used again but with a different twist.
This example was adapted from the book of Principles of Unit Operations by Foust A. S. et al.
2.3 The approach based on equal heat transfer
1 The situation
A two effects evaportator is to be used for the concentration of an aqueous solution from $2.00\%$ to $25.00\%$. In its nature, the suspension is made of water and some organic solids, and the overall heat transfer coefficient $U$ is $500.00\,Btu/hr\,ft^2\,F$ in the first effect while in the second effect it is $700.00\,Btu/hr\,ft^2\,F$. Also, the evaporator will run in feedback mode with a heating surface of $2,000.00\,ft^2$ in each effect.
For simplicity of the analysis, it shall be considered that the boiling point elevation is zero. The suspension enters the evaporators at $100.00\,F$, the steam is fed at a pressure of $100.00\,psia$. Also, the condenser operates at $2.00\, in\,abs\, Hg$.
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| Fig. 01 A sketch showing the two stage evaporator with relevant variables and notation. |
The following questions arise:
- What is the liquor mass flow rate $S_2$?
- What is the feeding mass flow rate $F$?
- What is the steam mass flow rate $W$?
- What are the solvent vapor mass flow rate leaving each effect: $L_1$ and $L_2$?
- What are the latent heats of the steam and solvent vapors: $\lambda_W$, $\lambda_{L1}$ and $\lambda_{L2}$?
- What are the equations representing the mass balance?
- What are the equations representing the heat balance?
- What is the composition $x_{S2}$ of the liquor leaving the effect #2?
- What are the heat flows $q_1$ and $q_2$?
- What is the heating area in effect #1, $A_1$?
- What is the heating area in effect #2, $A_2$?
Hints
- As base for calculations consider that $1,000.00\, lb_m/hr$ of liquor are to be produced
- The solvent vapor streams mass flow rate from both effects are different
- The pressure and temperature of the solvent vapor stream should be estimated rather than supposed
- No thermodynamical data for the solution is available. You can use the data for liquid water instead
- As approach to the solution take the heat transported by both solvent vapor streams are equal. This is: $q_1=q_2$.
2 Answer to questions
Following the same process for previous examples, we should first try to determine data from tables and from the simple application of available formulas. Since it is known that the steam has pressure $p_W=100.00\,psia$, its temperature $T_W$ and latent heat $\lambda_W$ can be easily found from steam tables to be:
Also, since the vapor entering the condenser is at $2.00\,in\,abs\,Hg$ it is straight to think that the solvent vapor stream leaving the effect #2, also known as $L_2$, would be $p_{L2}=2.00\,in,abs\,Hg$ too. Therefore, the corresponding thermodynamical data for $L_2$ would be:
2.1 From the mass balances
Some further information can be obtained from the mass balances. Since we already know that $S_1=1,000.00\,lb_m/hr$, $x_{S1}=0.25$ and $x_{F}=0.02$ we may first consider a solute global balance as follows,
$S_1\,x_{S1}=F\,x_F$ Eq. (01)
Then, the feeding mass flow rate is found to be,
Also, another mass balance can be made so that the solvent vapor streams are considered. This is as follows,
$L_1+L_2=F-S_1$ Eq. (02)
Therefore,
$L_1+L_2=11,500.00\,lb_m/hr$ Eq. (03)
Other equations useful to determine unknowns are the solute mass balances on each effect, these are:
$S_2\,x_{S2}=S_1\,x_{S1}$ Eq. (04)
$F \,x_F = S_2\,x_{S2}$ Eq. (05)
which at this moment can not be used since there are several unknowns.
2.2 From the heat balances
Let us look at the streams shown in Fig. 01 above. Based on this, the following heat balance equations can be written,
$W\, \lambda_W+S_2\, h_{S2} = L_1\, H_{L1}+S_1\,h_{S1}$ Eq. (06)
$L_1\, \lambda_{L1} + F\,h_F = L_2\, H_{L2} + S_2\, h_{S2}$ Eq. (07)
for the effects #1 and #2, respectively. Equations (06-07) tell us that so many variables are unkwon. Flow rates and thermodynamical properties are unknown so that we need to keep digging for more formulas to use.
One direction we may look to is that of the net heat supplied to each effect. The net heat supplied to effect #1 is expressed as,
$q_1=U_1A_1\left( T_W - T_{S1} \right)$ Eq. (08)
while for effect #2, this would be,
$q_2=U_2A_2\left( T_{L1}-T_{S2} \right)$ Eq. (09)
Again, a dead end is found since some data is known but some temperatures remain unknown. Here, a different analysis must be followed.
2.3 The approach based on equal heat transfer
Following the idea in this section title and using Eqs. (08-09), another equation with reduced unknowns appear,
$U_1A_1\left( T_W-T_{S1} \right)=U_2A_2\left( T_{L1}-T_{S2} \right)$ Eq. (10)
Equation (10) can now be reorganized with the known variables in mind,
$\dfrac{\left( T_W-T_{S1} \right)}{\left( T_{L1}-T_{S2} \right)}=\dfrac{U_2A_2}{U_1A_1}=1.4$ Eq. (11)
Also, since the BPE is zero, Eq. (11) can also be rewritten as,
On the other hand, there is another fact we should notice. If consider, for now, that the temperature of the heating media (steam) going into each effect is an indicator of the energy being used we may just find it to be,
from the steam inlet in effect #1 to the solvent vapor stream outlet in effect #2. From Eq. (12), we may observe that part of the $226.85\,F$ range is used in effect #1 and the rest in effect #2. In other words,
Next, since our main concern is with the unknown: $T_{L1}$. Therefore, $T_W$ can be isolated from Eq. (12),
$T_W=1.4\left( T_{L1}-T_{S2} \right)+T_{L1}$ Eq. (13)
and substitute it into Eq. (13) to give,
$1.4\left( T_{L1}-T_{S2} \right)+\left( T_{L1}-T_{S2} \right)=226.85$ Eq. (14)
from which, it follows that the temperature of the solvent vapor stream $L_1$ and that of the liquor leaving the effect #2, $S_1$, are,
From this single temperature $T_{L1}=T_{S1}$, the missing thermodynamical data in Eqs. (06-07) can be now estimated and as a consequence. This is,
$H_{L1}=1,143.99\,Btu/lb_m$
$\lambda_{L1}=980.42\,Btu/lb_m$
About the thermodynamical data of the suspension. It is supposed to be very similar to that of water, therefore, from the steam table we find that,
2.4 The mass flow rates of steam
Using the data estimated in the previous section, the mass flow rates: $L_1$, $L_2$ and $W$ can now be found from Eqs. (06-07). You may think that the data for $h_{S2}$ have not been determined but since the term $S_2\,h_{S2}$ repeats in bouth equation we may just proceed with a simple substitution. The result is,
$W\, \lambda_W + L_1\, \left( \lambda_{L1}-H_{L1}\right) + F\, h_F - L_2\, H_{L2}=S_1\,h_{S1}$ Eq. (15)
Also, since from the beginning we take $q_1=q_2$ it follows that,
too. Thus, Eq. (15) becomes
$L_1\, \left( 2\lambda_{L1}-H_{L1}\right) + F\, h_F - L_2\, H_{L2}=S_1\,h_{S1}$ Eq. (17)
In Eq. (17), $L_2$ is unknown but it can be written in terms of $L_1$ using the mass balance presented in Eq. (03). In this way, Eq. (17) can still be reduced to,
$L_1\, \left( 2\lambda_{L1}-H_{L1}\right) + F\, h_F - \left(11,500.00-L_1\right)\, H_{L2}=S_1\,h_{S1}$ Eq. (18)
From Eq. (18), $L_1$ is now easily isolated. The result is,
$L_1=\dfrac{S_1\,h_{S1}+11,500.00\, H_{L2}-F\, h_F}{2\lambda_{L1}-H_{L1}+H_{L2}}$ Eq. (19)
and whose numerical evaluation gives,
and $L_2$ can be determined from Eq. (03). This is,
From the estimations for $L_1$ and $L_2$, $S_2$ can also being determined from a mass balance on effect #2. This is,
$F=L_2+S_2$ Eq. (20)
and as a consequence, the composition of stream $S_2$ as,
We may also go back to Eq. (16) and determine $W$. It was found,
2.5 Heat transferred and heating areas
Finally, the flow of transported heat by the heating media and the heating areas can be estimated. This is as follows,
and the heating areas are,
$A_1 = \dfrac{q_1}{U_1\left( T_W - T_{S1} \right)} = \dfrac{6,132,558.39\,lb_m/hr}{500\,Btu/hr\,ft^2\, F\left( 327.82\,F-195.49\,F \right)} = 92.69\,ft^2$
$A_2=\dfrac{q_2}{U_2\left( T_{L1} - T_{S2} \right)} = \dfrac{6,132,558.39\,lb_m/hr}{700\,Btu/hr\,ft^2\, F\left( 195.49\,F-100.97\,F \right)} = 92.69\,ft^2$
As you can see, $A_1=A_2$, practically. Then, this is the end of the problem. If, the areas were not the same we should try to make them as close as possible by choosing some relationship between $q_1$ and $q_2$. This is, $q_1$ can be made a multiple of $q_2$, and follow the process presented above.
I hope you will find this post interesting and useful.
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Ildebrando.
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