Heating surface of a process plant evaporator
- Case of distilled water -
Here, a process plant evaporator used for distillation of water is being considered. Notice that chemical and process plant evaporators are different. In process plant evaporators blowdown is used.
This example was adapted from the old-timer Process Heat transfer by Donald Q. Kern. The situation focuses on a key design parameter of evaporators: surface heating, which would later become relevant for the commissioning of such an equipment for a specific purpose.
CONTENTS
1.1 Interesting questions to ask
2.1 Available data and comments
1 The situation
Technical insight for an evaporator aimed to output distilled water from untreated water is required. An estimated mass flow rate of $10,000\, lb_m/hr$ of distilled water is the goal. The plant where this evaporator is to installed has steam available at $300, F$. A low cost scenario in which the condenser, attached to the evaporator, vents to the atmosphere is to be considered.
From experiences with condensers operating in this mode it is estimated a presure loss of about $5\, psi$, including the effect of pipes. Another condition to take into account is that the developed pressure inside evaporator chamber could be about $19.7\, psia$, which is equivalent to $226\, F$.
Based on the presented situation, estimate the corresponding heating surface $A$ if a reasonable overall heat transfer coefficient of $605\,Btu/hr\cdot ft^2\cdot F$ could be enough for operation.
1.1 Interesting questions to ask
Notice that in the posed situation manometric and absolute pressures are presented so that careful handling of these parameter are required. Also, since steam is both in the product and as the heating medium temperatures and pressures are used as synonym but do not hesitate to check on your steam tables.
Also, it is known that the evaporator is for plant process operations purposes but no blowdown is mentioned, so what is the purpose of such a label in this case? What about the type or configuration of the evaporator?
2 Process of solution
As you read above, the main question here is: what is the magnitud of $A$? Then, instead of going around looking for useless data and formulating equations of any nature, it could be helpful to look at the formula involving $A$. This is,
$A=\dfrac{q}{U\cdot \Delta T}$ Eq. (01)
where $q$ is the heat transferred from the steam to the untreated water so that vapor can be form from it and later passed to the condenser, $U$ is the overall heat transfer coefficient and that, partly, carries information of the evaporator configuration or mechanical design and $\Delta T$ is the temperature difference so that heat transferred from the steam to the untreated water (in the evaporator). In other words,
$\Delta T=T_W-T_S$ Eq. (02)
From Eqs. (01-02) it is straight that $T_W$ and $T_S$ are already known from the situation presentations while $q$ neeeds to be calculated. However, data for enthalpies are available, for steam, so that all we need is to carefully check on the energy related to the streams.
2.1 Available data and comments
All provided data are presented in the sketch of Fig. 01 along with the unknowns. Of course, not all variables are of interest to answer the question: what is the heating surface $A$?
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| Fig. 01 Sketch of the evaporator and condenser along with known data and unkowns variables. |
Some of the unknowns can be found directly from a water steam table. Based on the formula for the heating surface $A$, the heat transferred $q$ is just the available energy in the steam $W$.
The temperature gradient $\Delta T$ is the force driving the heat transfer. In this case, this would be from the steam to the fluid in the evaporator.
2.2 The energy $q$
This energy $q$, in the form of the enthalpy of vaporizaton $H_{W-vap}$, is instantaneously transferred to the process fluid (untreated water). Therefore, after looking in a steam table you can easily find that if the steam has $T_W=300\,F$ then,
$H_{W-vap}=910.24\,Btu/lb_m$ Eq. (03)
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| Fig. 02 Detail of calculation of $H_{W-vap}$ from data in a steam table at $T_W$. |
One may argue that you can not be sure that the whole energy in the steam may not be completely transferred to the process fluid but there is one thing you need to know. The steam will always transfer its enthalpy of vaporization (no matter what).
Also, the energy $q$ can be expressed as,
$q=W\,H_{W-vap}$ Eq. (04)
since it should change according to amount of steam entering the evaporator. If you reduce the $W$, the energy supplied would be reduced too. However, $W$ remains unknown.
At this point, we should be creative in the way of looking for $W$. This can be calculated from a heat balance: the energy in the stream of solvent vapor $L$ should be the same transferred from $W$. This is,
$L\,H_{L-vap}=W\,H_{W-vap}$ Eq. (05)
Being honest about Eq. (05), this is just a rough approximation but in this case it is all we have. The enthalpy of vaporization $H_{W-vap}$ takes the fluid in $F$ from $T_F$ to its boiling temperature $T_S$ so that pure water evaporates from the untreated water. Of course, the boiling temperature of pure water is smaller than that of untreated water but if perhaps the untreated water and the pure water were ver similar in their physical properties the boiling temperatures would be the same or very close (of pure and untreated water). Such can be this situation that, in steady conditions (after certain time of operation), that $S$ and $W$ would be at the same temperature and then the vapor $L$ spliting from $S$ would carry the same energy so that Eq. (05) holds.
Here, $W$ is not necessary because, based on our previous arguments,
$q=L\,H_{L-vap}$ Eq. (06)
too. Anyway, its calculation is presented for the sake of completeness. Isolating $W$ from Eq. (05) gives,
$W=\dfrac{L\,H_{L-vap}}{H_{W-vap}}$ Eq. (07)
In Eq. (07), $H_{L-vap}$ remains unknown but since it was provided that the vapor stream $L$ is at $226\,F$, this unknown can be found in a steam table as $H_{L-vap}=961\,Btu/lb_m$, approximately. Thus, from Eq. (07) it follows,
$W=10,557.66\,lb_m/hr$ Eq. (08)
And from here, it follows that,
$q=9,610,000.00\,Btu/hr$ Eq. (09)
2.3 The gradient of temperatures $\Delta T$
It is straight to find out that $T_W=300\,F$ but the other temperature may be a little hard to understand. Why choosing $T_S$ instead of $T_F$?
The temperature gradient can not start at $T_F$ since the untreated water in evaporator is not always at this temperature but in the feeding pipe. Once the untreated starts boiling the temperature will rise to $T_S$ so that the true driving temperature gradient is $T_W-T_S$. Therefore,
$\Delta T=T_W-T_S$ Eq. (10)
2.4 The heating surface $A$
In this case, finding $A$ simply reduces to substitue the known data into Eq. (01). This is as follows,
$A=214.65\,ft^2$ Eq. (11)
This is the end. I hope, you find this post very useful.
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Ildebrando.
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