On the vapor pressure data for different NaCl dilutions at different temperatures

 The data presented in this post were extracted from the International Critical Tables.

As is usual in evaporation operations the boiling temperature elevation (BPE) is a key data for engineering calculations. Then, the vapor pressure for different combinations of solute concentration and temperatures are to be combined with the Duhring approximation (Duhring's lines).

The data was just rewritten from the source previously mentioned.

	Vapor pressure, mm Hg
Wt %	0.0	2.5	5.0	7.5	10.0	12.5	15.0	17.5	20.0	22.5	25.0	27.5
t °C
0	4.579	4.5	4.4	4.4	4.3	4.2	4.1	4.0	3.8	3.7	3.5
10	9.21	9.1	8.9	8.8	8.6	8.4	8.2	8.0	7.7	7.4	7.1
20	17.54	17.3	17.0	16.7	16.4	16.1	15.7	15.3	14.8	14.2	13.6
30	31.83	3.4	30.9	30.4	29.8	29.2	28.5	27.7	26.8	25.8	24.7
40	55..34	54.5	53.6	52.7	51.7	50.7	49.5	48.1	46.6	44.9	43.0
50	92.54	91.2	89.7	88.1	86.4	84.7	82.8	80.5	78.1	75.3	72.2
60	149.46	147.2	144.8	142.3	139.7	136.8	133.7	130.0	126.0	121.7	116.8
70	233.79	230.2	226.4	222.4	218.3	213.9	208.9	203.5	197.5	190.7	183.1
80	355.47	350	344	338	332	325	318	309.5	300.5	290.2	278.9	266
90	526	517	509	500	491	481	470	458	445.1	430	414	395
100	760	748	736	723	710	695	680	665	643	622	599	572
110	1075.4	1057	1040	1022	1003	983	961	936	911	881	849	810
B. P., °C	100	100.44	100.9	101.4	101.93	102.51	103.16	103.89	104.72	105.68	106.78	108.12

On the heat capacity for NaCl solutions

 This post is based on the results published by James C. S. Chou and Allen M. Rowe Jr. in their paper Desalination, 6(1969) 105-115.

For enthalply calculations the heat capacity is key. However, this last parameter depends on temperature and pressure, and in the case of dilutions on the concentration of solute.

From the thermodynamical point of view the formal relationship between enthalpy and heat capacity is expressed as,

$h=h_0+\int_{T_0}^Tc_P\,dT+\int_{P_0}^P\left[ v-T\left( \dfrac{\partial v}{\partial T} \right)_P \right]dT$    Eq. (01)

where the subscript $0$ indicates a reference data or condition that must be known in order to estimate the data at another set of conditions (without subscripts). $c_P$ and $v$ are the heat capacity at constant pressure and the specific volume, respectively. Chou and Rowe provide math expressions for these two parameters (fortunately):

$c_P=1.3041791 - 8.1519942x + 16.203997x^2 -\left( 0.19159475\times 10^{-2}\right.$

$\left. -0.029952864x+0.0037589577x^2\right)T+\left( 0.29944976\times 10^{-5} \right.$

$\left. -0.498581\times 10^{-4}x-0.89329066\times 10^{-6}x^2 \right)T^2$        Eq. (02)

where $x$ is the mole fraction and the temperature is given in $K$. The units of $c_P$ are $cal/g\,C$. The specific volume is:

$v=A(T)-P\, B(T)-P^2\, C(T)+w\, D(T)+w^2\, E(T)-wP\,F(T)$

$-w^2P\, G(T)-\dfrac{1}{2}wP^2\, H(T)$        Eq. (03)

where $w$ is the salt weight fraction in the solution and the remperature $T$ must be given in $K$. The $A$ through $H$ temperature functions are defined as,

$A(T)=5.916365-0.010357941T+0.92700482\times 10^{-5}T^2$

$-\dfrac{1127.5221}{T}+\dfrac{100674.1}{T^2}$

$B(T)=0.52049144\times 10^{-2}-0.10482101\times 10^{-4}T+0.83285321\times 10^{-8}T^2$

$-\dfrac{1.1702939}{T}+\dfrac{102.27831}{T^2}$

$C(T)=0.11854697\times 10^{-7}-0.65991434\times 10^{-10}T$

$D(T)=2.5166005+0.011176552T-0.17055209\times 10^{-4}T^2$

$E(T)=2.8485101-0.015430471T+0.22398153\times 10^{-4}T^2$

$F(T)=-0.0013949422+0.77922822\times 10^{-5}T-0.17736045\times 10^{-7}T^2$

$G(T)=0.0024223209-0.13698670\times 10^{-4}T+0.20303356\times 10^{-7}T^2$

$H(T)=0.55541298\times 10^{-6}-0.36241535\times 10^{-8}T+0.60444040\times 10^{-11}T^2$

Finally, for the purpose of a reference situation we may take the data of enthalpy at $25\,C$ and pressure of $1\,atm$. Again, Chou and Rowe provide an expression for it in the range of salt weight fraction $w$: $28.8524\%$ to $0.0006\%$. This is,

$h_0=24.953(1-w)+30.805561w^{1.5}-161.50632w^2$

$+79.059598w^{2.5}+114.83149w^3$        Eq. (04)

where the enthalpy $h_0$ is given in $cal/g\,solution$.

Mechanical energy added to a fluid - A pump

 An important component of the general mechanical energy equation

$\dfrac{p_1}{\gamma}+z_1+h_A-h_R-h_L+\dfrac{v_1^2}{2g}=\dfrac{p_2}{\gamma}+z_2+\dfrac{v_2^2}{2g}$        Eq. (01)

is $h_A$ since it represents the mechanical energy added to the fluid so that it can continue its trip through the pipe. At this point, two cases are visualized,

• the energy $h_A$ is one the system requires in order to the pipe system to do its work, and
• the energy $h_A$ is brought from the technical features of a real life pump, for example, so that the conditions of the pipe system are changed in turn.

Of course, in most technical problems both cases presented above are part of the solution in an iterative procedure.

What is the relationsip between $h_A$ and a pump

Since the pump takes electrical energy and transforms it into mechanical energy to be later transferred into the fluid, these  two parameters should be related. This is done through the effeciency of the motor,

$\mathbf{e_M}=\dfrac{Power\, output\, from \, the\, motor}{Power\, delivered\, by\, fluid}=\dfrac{P_O}{P_R}$    Eq. (02)

where $P_O$ is the mechanical power the motor so that the impeller may turn at certain speed with certain force and $P_R$ is the power the fluid actually receives. Since part of the power $P_O$ is wasted as heat or lost due to wearing of mechanical parts, $P_O>P_R$ and as a consequence $\mathbf{e_M}<1$. If $\mathbf{e_M}=1$ you would have an impossible thermodynamical machine. $P_O$ is a parameter measured and supplied, in the name plate, by pump manufacturers so that this data is easy to get.

Notice that Eq. (02) is the same if instead of a pump the case were that of a turbine (actioned by a mechanical energy of the fluid). We would be talking about $h_R$ instead of $h_A$ too.

For a centrifugal pump the power transferred into the fluid and the energy added $h_A$ are related as follows,

$P_R=h_A\gamma Q$        Eq. (03)

where $Q$ is the volumetric flow rate and $\gamma$ a property of the fluid. Also, from Eq. (03) it is obvious that $P_R$ would be very hard to measured. Therefore, it is more common to speak of the efficiency of a pump which we know is smaller than 1 but with present technological advances could be in the range of 0.8 - 0.9 for new equipments. Then, the usual case would be that $P_R$ is unknown, so that,

$P_R=\mathbf{e_M}P_O$        Eq. (04)

and consequently, Eq. (01) becomes,

$\mathbf{e_M}P_O=h_A\gamma Q$        Eq (05)

From Eq. (05), $h_A$ is isolated to be,

$h_A=\dfrac{\mathbf{e_M}P_O}{\gamma Q}$        Eq. (06)

In this way $h_A$ can only be estimated if the efficiency of the pump $\mathbf{e_M}$ and power of the motor of the pump $P_O$ are known.

Fitting data to a power law equation

 This is a common task for experimentalists looking for relationship among data sets and groups of these.

The common feature of data behaving according to a power law equation,

$y=\alpha\,x^\beta$        Eq. (01) 

where $\alpha$ and $\beta$ are free parameters or parameters for adjustment is that once $\ln y$ vs $\ln x$ is plotted, a straight line is obtained.

Schematic representation of data behaving as power law Eq. (01) (left) and data producing a straight line in log log plot (right).

For short, rather than having to use logarithms to get data from a log-log plot, $\alpha$ and $\beta$ can be determined so that all data can be easily represented by Eq. (01).

How to determine the free parameters

Vey easy. Since in the log-log plot the curve becomes a straight line, it should have an equation like,

$\ln y=m\, \ln x + b$        Eq. (02)

where $m$ is the slope and $b$ is the ordinate at which the line cuts the axis. On the other hand, if operate with $\ln$ on Eq. (01) we obtain,

$\ln y=\ln\left| \alpha \, x^\beta \right|$

$\ln y = \ln \alpha + \beta\ln x$        Eq. (03)

Next, comparison of Eqs. (02-03) leads to,

$m=\beta$ and $b=\alpha$        Eq. (04)

Well, this is not all because the real question is not being answered, how can we estimate $\alpha$ and $\beta$? Equation (4) is just a partial answered. A reliable way of finding $m$ and $b$ is by a linear regression using the minimum squares method so that the reader is then referred to,

Minimum squares method

I hope you can find this useful.

Economy in an evaporator

 Steam is the most expensive factor in evaporators operation. From this, follows that the economy of such an equipment has to be estimated or measured somehow.

For short, the economy of an evaporator, of single stage or multiple effect, is just defined as,

        Eq. (01)

In single stage evaporators the economy will usually be smaller than 1 while in multiple effect evaporators, this parameter could be greater than 1. This is because for every effect the heat in the solvent vapor is used as much as possible. On the other hand, single stage evaporators are easy to use but with low economy while multiple effect evaporators are difficult to used but of higher economy.

Notice, that the economy is given in terms of the mass but if mass flow rates of solvent vapor and steam are the only available data, these can be used without any issue.

Evaporator capacity

This is defined as the mass flow rate of solvent vapor extracted from the solution, fiven in mass/hour units. 

        Eq. (02)

Of course, in a multiple effect evaporator all solvent vapor streams would contribute.

Steam consumption

The total steam consumption can be easily calculated from division of the evaporator capacity over its economy,

        Eq. (03)

In this case, the steam consumption is given as mass flow rate.

Any question? Write in the comments and I shall try to help.

Other stuff of interest

• Understanding the law of Boyle
• Measuring atmospheric pressure
• Some examples of temperature instruments
• Minor losses  - Formulas
• What is a process variable?
• What are the most important process variables?
• Time dependence of process variables
• A list of process variables
• Composition variables for mixtures

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Ildebrando.