Chapter 16, Problem 16.1P

(a)

Interpretation Introduction

Interpretation:

The steam consumption, evaporator economy, and surface area are to be determined when the temperature of the feed is ${51.7}^o\text{ C}$ .

Concept Introduction :

First and foremost, the mass flow rate of vapor and product stream coming out of the evaporator are to be determined. Then based on temperature-pressure conditions, enthalpy of vaporization of liquid and steam is needed. Then, steam consumption can be determined as the product of vapor flow rate and enthalpy ratio of vaporization enthalpy over steam enthalpy (based on steam temperature). On the other hand, evaporator economy, and surface area requirements depend on the feed temperature and overall material and enthalpy balance calculations.

Answer to Problem 16.1P

The steam consumption rate is $17782.430\text{kg/h}$ , the economy is $0.925$ , and the heating area is $56.41{\text{m}}^{\text{2}}$ .

Explanation of Solution

Given,

Feed flow rate $(F)$ is $20000\text{kg/h}$ .

Solid fraction in feed $(x_F)$ is $0.02$ .

Solid fraction in the product $(x_P)$ is $0.45$ .

Solid fraction in vapor $(x_V)$ is $0$ .

Feed temperature $(T_F)$ is $51.7{}^{o}C$ .

Steam temperature $(T_s)$ is $120.5{}^{o}C$ .

Specific heat capacity of feed $(c_{p,F})$ is $3.77\text{J/g}\cdot {}^{o}C$ .

Overall heat transfer coefficient $(U)$ is $2800{\text{W/m}}^{\text{2}}{\cdot }^{\text{o}}\text{C}$ .

An overall material balance on the evaporator is given by the following expression.

  $P+V=F$   ...... (1)

Here, $P$ and $V$ are the mass flow rates of product and vapor stream respectively.

Overall solid balance on the evaporator is given by the following expression.

  $x_{P}P+x_{V}V=x_{F}F$

Putting values of known variables gives the following.

  $\begin{array}{l}\text{(0}\text{.45)}P=(0.02)(20000\text{}\text{kg/h)}\\ =\frac{0.02}{0.45}(20000\text{}\text{kg/h)}\\ =3555.55\text{kg/h}\end{array}$

Using the value of product flow rate in equation (1) gives the following.

  $\begin{array}{l}V=F-P\\ \text{=(20000-3555}\text{.55})\text{kg/h}\\ \text{=16444}\text{.45}\text{kg/h}\end{array}$

Given the pressure inside the evaporator is $102\text{mmHg}\text{abs}$ , the corresponding pressure $\left(p_s\right)$ in kPa units will be calculated in the following manner.

  $\begin{array}{l}{p}_s=\frac{102\text{mmHg}}{760\text{mmHg}}\times 101.325\text{}\text{kPa}\\ \text{=13}\text{.6}\text{kPa}\end{array}$

Using a steam table shows that at this pressure the corresponding saturation temperature using the steam table is approximately $52{}^{o}C$ .

Boiling point rise (BPR) is neglected. Assume that the vaporized stream exhibits the same properties as water.

Using the temperature of $52{}^{o}C$ , the vaporization enthalpy of vapor or gaseous stream ${\text{(λ}}_v)$ is $2379\text{kJ/kg}$ .

Given steam temperature is $120.5{}^{\text{o}}\text{C}$ , the corresponding vaporization enthalpy of steam ${\text{(λ}}_s)$ using the steam table is $2200\text{kJ/kg}$ .

Overall enthalpy balance on the evaporator is given by the following expression.

  $F{h}_F+S{\lambda }_s=P{h}_P+V{\lambda }_V$

Here, $S$ is the mass flow rate of steam or the so-called steam consumption rate, $h_F$ and $h_P$ are the enthalpies of feed and product stream respectively.

In order to compute the steam consumption, no enthalpy contribution by feed and product stream is taken into consideration.

Then the above equation changes in the following manner.

  $S{\lambda }_s=V{\lambda }_V$  ......(2)

Using equation (2) as the basis for further calculations, the following is obtained.

  $\begin{array}{l}S=V\left(\frac{{\lambda }_P}{{\lambda }_s}\right)\\ =16444.45\text{kg/h}\left(\frac{2379\text{kJ/kg}}{2200\text{kJ/kg}}\right)\\ =17782.430\text{kg/h}\end{array}$

The steam economy $(E)$ is obtained in the following manner.

  $\begin{array}{l}E=\frac{V}{S}\\ =\frac{16444.45\text{kg/h}}{17782.4\text{kg/h}}\\ =0.925\end{array}$

The heat transfer area $(A)$ is obtained in the following manner.

  $\begin{array}{l}A=\frac{V{\lambda }_V}{U(T_s-T_F)}\\ =\frac{16444.45\text{kg/h}\times \frac{1\text{h}}{3600\text{s}}\times 2379\text{kJ/kg}\times \frac{1000\text{J}}{1\text{kJ}}}{2800{\text{W/m}}^{\text{2}}{\cdot }^{\text{o}}\text{C}\times {\left(120.5-51.7\right)}^{\text{o}}\text{C}-}\\ =56.41{\text{m}}^{\text{2}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The steam consumption, evaporator economy, and surface area are to be determined when the temperature of the feed is ${21.1}^o\text{ C}$ .

Concept Introduction :

First and foremost, the mass flow rate of vapor and product stream coming out of the evaporator are to be determined. Then based on temperature-pressure conditions, enthalpy of vaporization of liquid and steam is needed. Then, steam consumption can be determined as the product of vapor flow rate and enthalpy ratio of vaporization enthalpy over steam enthalpy (based on steam temperature). On the other hand, evaporator economy, and surface area requirements depend on the feed temperature and overall material and enthalpy balance calculations.

Answer to Problem 16.1P

The steam consumption rate is $18830.6\text{kg/h}$ , the economy is $0.873$ , and the heating area is $59.73{\text{m}}^{\text{2}}$ .

Explanation of Solution

Before proceeding, consider the following.

Reference feed temperature $(T_{F,1})$ is $51.7{}^{o}C$

New feed temperature $(T_{F,2})$ is $21.1{}^{o}C$

Net heat $(q_F)$ that must be added to the feed is obtained in the following manner.

  $\begin{array}{l}{q}_F=F{c}_{p,F}(T_{F,1}-T_{F,2})\\ =\frac{20000\text{kg/h}\times \text{1000}\text{g/kg}}{3600\text{s/h}}\times 3.77\text{J/g}{\cdot }^{\text{o}}\text{C}\times {\text{(51}}^{\text{.7-21}}\text{C}\\ =640900\text{W}\end{array}$

Net heat in the vapor stream $\left(q_V\right)$ is obtained in the following manner.

  $\begin{array}{l}{q}_V=V\left({\lambda }_v\right)\\ =\frac{16444.45\text{kg/h}}{3600\text{s/h}}\left(2379\text{kJ/kg}\right)\times \text{1000}\text{W/kW}\\ =10866700\text{W}\end{array}$

The total heat $\left(q\right)$ obtained is as below.

  $\begin{array}{l}q=q_F+q_V\\ =\left(640900+\text{10}866700\right)\text{W}\\ =11507600\text{W}\end{array}$

Therefore, steam consumption is obtained as below.

  $\begin{array}{l}S=\left(\frac{q}{{\lambda }_s}\right)\\ =\frac{11507600\text{W}\times \frac{1\text{J/s}}{1\text{W}}}{2200\text{kJ/kg}\times 1000\text{J/kJ}}\\ =5.23\text{kg/s}\times 3600\text{s/h}\\ \text{=18830}\text{.6 kg/h}\end{array}$

The steam economy $(E)$ is obtained in the following manner.

  $\begin{array}{l}E=\frac{V}{S}\\ =\frac{16444.45\text{kg/h}}{18830.6\text{kg/h}}\\ =0.873\end{array}$

The heat transfer area $(A)$ is obtained in the following manner.

  $\begin{array}{l}A=\frac{q}{U(T_s-T_F)}\\ =\frac{11507600\text{W}}{2800{\text{W/m}}^{\text{2}}{\cdot }^{\text{o}}\text{C}\times {\left(120.5-51.7\right)}^{\text{o}}\text{C}}\\ =59.73{\text{m}}^{\text{2}}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The steam consumption, evaporator economy, and surface area are to be determined when the temperature of the feed is ${93.3}^o\text{ C}$ .

Concept Introduction :

First and foremost, the mass flow rate of vapor and product stream coming out of the evaporator are to be determined. Then based on temperature-pressure conditions, enthalpy of vaporization of liquid and steam is needed. Then, steam consumption can be determined as the product of vapor flow rate and enthalpy ratio of vaporization enthalpy over steam enthalpy (based on steam temperature). On the other hand, evaporator economy, and surface area requirements depend on the feed temperature and overall material and enthalpy balance calculations.

Answer to Problem 16.1P

The steam consumption rate is $16356.13\text{kg/h}$ , the economy is $1.005$ , and the heating area is $51.88{\text{m}}^{\text{2}}$ .

Explanation of Solution

Before proceeding, consider the following.

Reference feed temperature $(T_{F,1})$ is $51.7{}^{o}C$

New feed temperature $(T_{F,2})$ is $93.3{}^{o}C$

Net heat $(q_F)$ that must be added to the feed is obtained in the following manner.

  $\begin{array}{l}{q}_F=F{c}_{p,F}(T_{F,1}-T_{F,2})\\ =\frac{20000\text{kg/h}\times \text{1000}\text{g/kg}}{3600\text{s/h}}\times 3.77\text{J/g}{\cdot }^{\text{o}}\text{C}\times {\text{(51}}^{\text{.7-93}}\text{C}\\ =-871288.89\text{W}\end{array}$

Net heat in the vapor stream $\left(q_V\right)$ is obtained in the following manner.

  $\begin{array}{l}{q}_V=V\left({\lambda }_v\right)\\ =\frac{16444.45\text{kg/h}}{3600\text{s/h}}\left(2379\text{kJ/kg}\right)\times \text{1000}\text{W/kW}\\ =10866700\text{W}\end{array}$

The total heat $\left(q\right)$ obtained is as below.

  $\begin{array}{l}q=q_F+q_V\\ =\left(-871288.89+\text{10}866700\right)\text{W}\\ =9995411.11\text{W}\end{array}$

Therefore, steam consumption is obtained as below.

  $\begin{array}{l}S=\left(\frac{q}{{\lambda }_s}\right)\\ =\frac{9995411.11\text{W}\times \frac{1\text{J/s}}{1\text{W}}}{2200\text{kJ/kg}\times 1000\text{J/kJ}}\\ =4.543\text{kg/s}\times 3600\text{s/h}\\ \text{=16356}\text{.13 kg/h}\end{array}$

The steam economy $(E)$ is obtained in the following manner.

  $\begin{array}{l}E=\frac{V}{S}\\ =\frac{16444.45\text{kg/h}}{16356.13\text{kg/h}}\\ =1.005\end{array}$

The heat transfer area $(A)$ is obtained in the following manner.

  $\begin{array}{l}A=\frac{q}{U(T_s-T_F)}\\ =\frac{9995411.11\text{W}}{2800{\text{W/m}}^{\text{2}}{\cdot }^{\text{o}}\text{C}\times {\left(120.5-51.7\right)}^{\text{o}}\text{C}}\\ =51.88{\text{m}}^{\text{2}}\end{array}$