Question

A concentric tube heat exchanger of length L = 2 m is used to thermally process a pharmaceutical product flowing at a mean velocity of u_m,c = 0.1 m/s with an inlet temperature of T_c,i = 20 degree C. The inner tube of diameter D_i = 10 mm is thin walled, and the exterior of the outer tube (D_0 = 20 mm) is well insulated. Water flows in the annular region between the tubes at a mean velocity of u_m,h = 0.2 m/s with an inlet temperature of T_h,i = 60 degree C. Properties of the pharmaceutical product are v = 10 * 10^-6 m^2/s, k = 0.25 W/m. K, p = 1100 kg/m^3, and c_p = 2460 J/kg middot K. Evaluate water properties at T_h bar = 50 degree C. Determine the value of the overall heat transfer coefficient U. Determine the mean outlet temperature of the pharmaceutical product when the exchanger operates in the counterflow mode. Determine the mean outlet temperature of the pharmaceutical product when the exchanger operates in the parallel-flow mode.

Answer 1

Concepts and reason

To solve the problem, relation between dimensionless numbers is to be understood. Also, the concept of heat exchanger analysis using effectiveness-number of transfer unit’s methods is also to be understood.

Convection:

Whenever the heat exchange is due to the movement of fluid particles it is called as convective heat transfer. The fluid particles transmit heat to a location at a lower temperature from a location at a higher temperature. Convection heat transfer requires a medium to propagate heat exchange.

Dimensionless numbers:

Dimensionless numbers are the numbers which do not have any physical dimension assigned to them.

Heat exchanger:

Heat exchangers are devices used for transfer of heat energy from one object to another. There are types of heat exchangers:

1.Parallel-flow heat exchanger.

2.Counter-flow heat exchanger.

Parallel-flow heat exchanger:

This is such a heat exchange in which the flow direction of the both the hot and cold fluids that flows inside the tube is unidirectional.

Counter-flow heat exchanger:

This is such a heat exchange in which the flow direction of the one of the fluid that flows inside the tube is opposite to the respective cold fluid.

To obtain solution for the given problem, initially calculate the Reynolds number for the cold fluid and determine the type of flow. Depending on the type of flow use the Nusselt number expression to calculate the convective heat transfer coefficient of the cold fluid. Similarly calculate the Reynolds number for the hot fluid and determine the type of flow.

Use the Nusselt number corresponding to the flow and determine the value of convective heat transfer coefficient of the hot fluid. Use the overall heat transfer coefficient expression, substitute the corresponding values and obtain the value of overall heat transfer coefficient.

Calculate the mass flow rate of cold and hot fluid. Further calculate the heat capacity of each fluid. Also determine the heat capacity ratio and values. Use the expression for effectiveness expression for counter flow heat exchanger substitute the respective values and obtain the value of effectiveness of the heat exchanger.

Later use the expression for effectiveness in terms of temperatures, substitute the values and calculate the mean outlet temperature of the cold fluid. Similarly repeat the effectiveness -procedure for parallel flow heat exchanger.

Fundamentals

The expression for liner interpolation to determine the value of an unknown data say, $y$ , at a given value of $x$ is given as

$y = (x - {x_1})\frac{{\left( {{y_2} - {y_1}} \right)}}{{\left( {{x_2} - {x_1}} \right)}} + {y_1}$

Here, $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ are the nearest data points taken from the given set of data.

Hydraulic diameter:

It is the effective diameter used for fluid flow in a pipe, conduit or duct of any shape. Mathematically it is given as,

Here, is the hydraulic diameter, is the cross-section area of the duct or pipe and is the wetted perimeter of the duct or pipe.

Hydraulic diameter of the annular space:

For a concentric tube with an outer diameter and inner diameter , the hydraulic diameter is calculated as:

Here, is the hydraulic diameter, is the outer diameter of the concentric tube and is the inner diameter of the concentric tube.

Reynolds Number:

For flow through pipe, Reynolds number is given by,

Here, is the Reynolds number for flow through pipe, is the mean fluid velocity, is the diameter of the pipe, is the density of the fluid and is the coefficient of dynamic viscosity of the fluid.

It is also expressed as,

Here, is the kinematic viscosity.

The critical Reynolds number corresponding to the onset of turbulence is given as:

For Reynolds number below the flow is considered to be laminar and above the critical Reynolds number the flow is considered as turbulent.

Nusselt number:

Nusselt number provides the measure of convective heat transfer occurring at the surface. Mathematical expression for Nusselt number for flow through pipe is given as,

Here, is the Nusselt number for the flow through pipe, is the convective heat transfer coefficient and is the thermal conductivity of the fluid.

Nusselt number varies with the type of the flow.

Nusselt number for fully developed laminar flow in a circular tube annulus with one surface insulated and the other at constant temperature and with the ratio of inner diameter to outer diameter of pipe as:

Here, is the Nusselt number of the inner surface of the pipe.

For turbulent flow through pipe the Nusselt number is a function of Reynolds number and Prandtl number and is given as,

Here, is the Nusselt number for flow through pipe and is the Prandtl number.

Overall heat transfer coefficient:

Overall heat transfer coefficient is related to total thermal resistance of heat transfer between two fluids. For tubular heat exchangers it is given by,

Here, is the overall hear transfer coefficient, is the heat transfer coefficient of the inner fluid and is the heat transfer coefficient of the outer fluid, is the length of the heat exchanger, is the fouling resistance of the inner surface of heat exchanger, is the fouling resistance of the outer surface of heat exchanger, is the surface area, is the area of inner surface of heat exchanger and is the area of outer surface of heat exchanger.

Neglecting the resistance due to fouling factors and conduction resistance, simplified form of overall heat transfer coefficient is given by,

Mass flow rate:

For flow through pipe, mass flow rate for a fluid is given as,

Here, is the mass flow rate of the fluid, is the density of the fluid, is the cross-sectional area of the pipe and is the mean velocity of the pipe.

Heat Capacity ratio:

It is defined as the ratio of minimum heat capacity of the fluid to the maximum heat capacity of the fluid and is given as:

Here, is the heat capacity ratio, is the minimum heat capacity of the fluid and is the maximum heat capacity of the fluid.

Number of transfer units:

Number of transfer units is a dimensionless parameter used for heat exchanger analysis and is given by,

Here, is the number of transfer units.

Effectiveness of the heat exchanger:

Effectiveness of a heat exchanger is defined as the ratio of actual heat transfer rate of a heat exchanger to the maximum possible heat transfer rate. It is given as,

Here, is the effectiveness of the heat exchanger, is the actual heat transfer rate of the heat transfer rate and is the maximum possible heat transfer rate.

It is also expressed in terms of temperatures as:

Or,

Here, is the outlet temperature of cold fluid, is the inlet temperature of cold fluid, is the outlet temperature of hot fluid, is the inlet temperature of hot fluid, is the heat capacity of cold fluid, is the heat capacity of hot fluid and is maximum heat capacity of either hot or cold fluid.

(a)

Write the expression of temperature conversion from Celsius to kelvin.

$T = T\left( {^\circ {\rm{C}}} \right) + 273$

Here, $T$ is the temperature in kelvin and $T\left( {^\circ {\rm{C}}} \right)$ is the temperature in degree Celsius.

Substitute $50^\circ {\rm{C}}$ for $T\left( {^\circ {\rm{C}}} \right)$ .

$\begin{array}{c}\\T = 50^\circ {\rm{C}} + 273\\\\ = 323{\rm{K}}\\\end{array}$

Understand that the value of properties of water given in the table are at $320{\rm{K}}$ and $325{\rm{K}}$ . To calculate the values at $323{\rm{K}}$ use the method of interpolation.

Consider ${T_x} = 325{\rm{K}}$ and ${T_y} = 320{\rm{K}}$ , and obtain the corresponding values of coefficient of dynamic viscosity and specific heat of water from the table of Thermophysical Properties of Saturated Water.

$\begin{array}{l}\\{\mu _x} = {\rm{577}} \times {10^{ - 6}}{\rm{ N}} \cdot {\rm{s/}}{{\rm{m}}^2}\\\\{\mu _y} = {\rm{528}} \times {10^{ - 6}}{\rm{ N}} \cdot {\rm{s/}}{{\rm{m}}^2}\\\\{c_{{p_x}}} = {\rm{4}}{\rm{.180 kJ/kg}} \cdot {\rm{K}}\\\\{{\rm{c}}_{{p_y}}} = {\rm{4}}{\rm{.182 kJ/kg}} \cdot {\rm{K}}\\\end{array}$

Here, ${\mu _x}$ is the coefficient of dynamic viscosity at $320{\rm{K}}$ , ${\mu _y}$ is the coefficient of dynamic viscosity at $325{\rm{K}}$ , ${c_{{p_x}}}$ is the specific heat of water at $320{\rm{K}}$ and ${c_{{p_y}}}$ is the specific heat of water at $325{\rm{K}}$ .

Further obtain the thermal conductivity and Prandtl Number corresponding to ${T_x}$ and ${T_y}$ from the table of Thermophysical Properties of Saturated Water.

$\begin{array}{l}\\{k_x} = {\rm{640}} \times {10^{ - 3}}{\rm{ W/m}} \cdot {\rm{K}}\\\\{k_y} = {\rm{645}} \times {10^{ - 3}}{\rm{ W/m}} \cdot {\rm{K}}\\\\{\rm{P}}{{\rm{r}}_x} = {\rm{3}}{\rm{.77}}\\\\{\rm{P}}{{\rm{r}}_y} = {\rm{3}}{\rm{.42}}\\\end{array}$

Here, ${k_x}$ is the thermal conductivity at $320{\rm{K}}$ , ${k_y}$ is the thermal conductivity at $325{\rm{K}}$ , ${\rm{P}}{{\rm{r}}_x}$ is the Prandtl Number at $320{\rm{K}}$ and ${\rm{P}}{{\rm{r}}_y}$ is the Prandtl Number at $325{\rm{K}}$ .

Write the expression of interpolation for coefficient of dynamic viscosity of water at $323{\rm{K}}$ .

${\mu _h} = (T - {T_x})\frac{{\left( {{\mu _y} - {\mu _x}} \right)}}{{\left( {{T_y} - {T_x}} \right)}} + {\mu _x}$

Here, is the dynamic viscosity of the hot fluid.

Substitute $323{\rm{K}}$ for $T$ , ${\rm{577}} \times {10^{ - 6}}{\rm{ N}} \cdot {\rm{s/}}{{\rm{m}}^2}$ for ${\mu _x}$ , ${\rm{528}} \times {10^{ - 6}}{\rm{ N}} \cdot {\rm{s/}}{{\rm{m}}^2}$ for ${\mu _y}$ , $325{\rm{K}}$ for ${T_y}$ and $320{\rm{K}}$ for ${T_x}$ .

$\begin{array}{c}\\\;{\mu _h} = (323{\rm{K}} - 320{\rm{K}})\frac{{\left( {{\rm{528}} \times {{10}^{ - 6}}{\rm{ N}} \cdot {\rm{s/}}{{\rm{m}}^2} - {\rm{577}} \times {{10}^{ - 6}}{\rm{ N}} \cdot {\rm{s/}}{{\rm{m}}^2}} \right)}}{{\left( {325{\rm{K}} - 320{\rm{K}}} \right)}} + {\rm{577}} \times {10^{ - 6}}{\rm{ N}} \cdot {\rm{s/}}{{\rm{m}}^2}\\\\ = \left( 3 \right)\left( {\frac{{ - 49 \times {{10}^{ - 6}}}}{5}} \right) + {\rm{577}} \times {10^{ - 6}}\\\\ = - 29.4 \times {10^{ - 6}} + {\rm{577}} \times {10^{ - 6}}\\\\ = 547.6 \times {10^{ - 6}}{\rm{ N}} \cdot {\rm{s/}}{{\rm{m}}^2}\\\end{array}$

Write the expression of interpolation for specific heat of water of water at $323{\rm{K}}$ .

${c_{p,h}} = (T - {T_x})\frac{{\left( {{c_{{p_y}}} - {c_{{p_x}}}} \right)}}{{\left( {{T_y} - {T_x}} \right)}} + {c_{{p_x}}}$

Here, is the specific heat of the hot fluid.

Substitute $323{\rm{K}}$ for $T$ , ${\rm{4}}{\rm{.180 kJ/kg}} \cdot {\rm{K}}$ for ${c_{{p_x}}}$ , ${\rm{4}}{\rm{.182 kJ/kg}} \cdot {\rm{K}}$ for ${{\rm{c}}_{{p_y}}}$ , $325{\rm{K}}$ for ${T_y}$ and $320{\rm{K}}$ for ${T_x}$ .

$\begin{array}{c}\\\;{c_{p,h}} = (323{\rm{K}} - 320{\rm{K}})\frac{{\left( {{\rm{4}}{\rm{.182 kJ/kg}} \cdot {\rm{K}} - {\rm{4}}{\rm{.180 kJ/kg}} \cdot {\rm{K}}} \right)}}{{\left( {325{\rm{K}} - 320{\rm{K}}} \right)}} + {\rm{4}}{\rm{.180 kJ/kg}} \cdot {\rm{K}}\\\\ = \left( 3 \right)\left( {\frac{{0.002}}{5}} \right) + {\rm{4}}{\rm{.180}}\\\\ = 0.0012 + {\rm{4}}{\rm{.180}}\\\\ = {\rm{4}}{\rm{.181 kJ/kg}} \cdot {\rm{K}}\\\end{array}$

Write the expression of interpolation for thermal conductivity of water at $323{\rm{K}}$ .

${\left( {{k_h}} \right)_o} = (T - {T_x})\frac{{\left( {{k_y} - {k_x}} \right)}}{{\left( {{T_y} - {T_x}} \right)}} + {k_x}$

Here, is the thermal conductivity of hot fluid flowing in annular region of the tube.

Substitute $323{\rm{K}}$ for $T$ , ${\rm{640}} \times {10^{ - 3}}{\rm{ W/m}} \cdot {\rm{K}}$ for ${k_x}$ , ${\rm{645}} \times {10^{ - 3}}{\rm{ W/m}} \cdot {\rm{K}}$ for ${k_y}$ , $325{\rm{K}}$ for ${T_y}$ and $320{\rm{K}}$ for ${T_x}$ .

$\begin{array}{c}\\\;{\left( {{k_h}} \right)_o} = (323{\rm{K}} - 320{\rm{K}})\frac{{\left( {{\rm{645}} \times {{10}^{ - 3}}{\rm{ W/m}} \cdot {\rm{K}} - {\rm{640}} \times {{10}^{ - 3}}{\rm{ W/m}} \cdot {\rm{K}}} \right)}}{{\left( {325{\rm{K}} - 320{\rm{K}}} \right)}} + {\rm{640}} \times {10^{ - 3}}{\rm{ W/m}} \cdot {\rm{K}}\\\\ = \left( 3 \right)\left( {\frac{{5 \times {{10}^{ - 3}}}}{5}} \right) + {\rm{640}} \times {10^{ - 3}}\\\\ = \left( {3{\kern 1pt} \times {{10}^{ - 3}}} \right) + \left( {{\rm{640}} \times {{10}^{ - 3}}} \right)\\\\ = {\rm{643}} \times {10^{ - 3}}{\rm{ W/m}} \cdot {\rm{K}}\\\end{array}$

Write the expression of interpolation for Prandtl Number at $323{\rm{K}}$ .

${\rm{Pr}} = (T - {T_x})\frac{{\left( {{\rm{P}}{{\rm{r}}_y} - {\rm{P}}{{\rm{r}}_x}} \right)}}{{\left( {{T_y} - {T_x}} \right)}} + {\rm{P}}{{\rm{r}}_x}$

Substitute $323{\rm{K}}$ for $T$ , ${\rm{3}}{\rm{.77}}$ for ${\rm{P}}{{\rm{r}}_x}$ , ${\rm{3}}{\rm{.42}}$ for ${\rm{P}}{{\rm{r}}_y}$ , $325{\rm{K}}$ for ${T_y}$ and $320{\rm{K}}$ for ${T_x}$ .

$\begin{array}{c}\\\;{\rm{Pr}} = (323{\rm{K}} - 320{\rm{K}})\frac{{\left( {{\rm{3}}{\rm{.42}} - {\rm{3}}{\rm{.77}}} \right)}}{{\left( {325{\rm{K}} - 320{\rm{K}}} \right)}} + {\rm{3}}{\rm{.77}}\\\\ = \left( 3 \right)\left( {\frac{{ - 0.35}}{5}} \right) + {\rm{3}}{\rm{.77}}\\\\ = \left( { - 0.21} \right) + {\rm{3}}{\rm{.77}}\\\\ = 3.56\\\end{array}$

Obtain the density of water at $323{\rm{K}}$ from its property table.

$\rho = 988.1{\rm{ kg/}}{{\rm{m}}^3}$

Write the expression for Reynolds number for flow through pipe.

Here, is the mean velocity of the cold fluid, is the inner diameter of the tube, is the kinematic viscosity of the cold fluid and is the Reynolds number of the cold fluid.

Substitute for , for and for .

Understand that, the value is less than . Hence, the nature of flow through the inner tube is laminar.

Write the expression of Nusselt number for inner pipe.

$N{u_i} = \frac{{{h_i}{D_i}}}{{{{\left( {{k_c}} \right)}_i}}}$

Here, is the heat transfer coefficient for fluid in the inner pipe of concentric tube, is the diameter of the inner pipe, is the thermal conductivity of cold fluid flowing in inner pipe and is the Nusselt number of the inner surface of the pipe.

Write the expression of ratio of inner diameter to outer diameter of pipe.

$n = \frac{{{D_i}}}{{{D_o}}}$

Here, $n$ is the ratio of inner diameter to outer diameter of pipe.

Substitute $10{\rm{ mm}}$ for ${D_i}$ and $20{\rm{ mm}}$ for ${D_o}$ .

$\begin{array}{c}\\n = \frac{{10{\rm{ mm}}}}{{20{\rm{ mm}}}}\\\\ = 0.5\\\end{array}$

Write the value for Nusselt number for fully developed laminar flow in a circular tube annulus with one surface insulated and the other at constant temperature and with the ratio of inner diameter to outer diameter as .

Substitute for .

Substitute for and for .

$\frac{{{h_i}\left( {{\rm{10 mm}}\left( {\frac{{0.001{\rm{ m}}}}{{1{\rm{ mm}}}}} \right)} \right)}}{{\left( {0.25{\rm{ W/mK}}} \right)}} = 5.74$

Solve for ${h_i}$ .

$\begin{array}{l}\\\left( {{h_i}} \right)\left( {0.01} \right) = 1.435\\\\{h_i} = 143.5{\rm{ W/}}{{\rm{m}}^2}K\\\end{array}$

Write the expression for hydraulic diameter for the annular space.

Substitute for and for .

Write the expression for Reynolds number for flow through annular region of the pipe.

Here, is the mean velocity of the hot fluid, is the hydraulic diameter of the tube, is the density of the hot fluid and is the Reynolds number of the hot fluid.

Substitute for , for , for and for .

As the Reynolds number is greater than 2300, the flow is considered to be turbulent.

Write the expression of Nusselt number for annular region of concentric tube.

$N{u_o} = \frac{{{h_o}{D_o}}}{{{{\left( {{k_h}} \right)}_o}}}$

Here, $N{u_o}$ is the Nusselt number for annular region of concentric tube, is the heat transfer coefficient for fluid in the annular region of concentric tube and is the hydraulic diameter of the annular region.

Write the expression for Nusselt number for turbulent flow.

Substitute for .

Substitute for , for , for and for .

Write the expression for overall heat transfer coefficient.

Substitute for and for .

$\begin{array}{c}\\U = \frac{1}{{\frac{1}{{\left( {143.5{\rm{ W/}}{{\rm{m}}^2}{\rm{K}}} \right)}} + \frac{1}{{\left( {1723.2{\rm{ W/}}{{\rm{m}}^2}{\rm{K}}} \right)}}}}\\\\ = \frac{1}{{6.96864115 \times {{10}^{ - 3}} + 5.803156917 \times {{10}^{ - 4}}}}\\\\ = \frac{1}{{7.54895 \times {{10}^{ - 3}}}}\\\\ = 132.5{\rm{ W/}}{{\rm{m}}^2}{\rm{K}}\\\end{array}$

(b)

Write the expression to calculate the area of the inner tube.

Here, is the cross-sectional area and is the inner diameter of the tube.

Substitute for .

Write the expression to calculate the area of the annulus of tube.

Here, is the cross-sectional area is the outer diameter of the tube and is the inner diameter of the tube.

Substitute for and for .

Write the expression for mass flow rate of cold fluid, that is pharmaceutical product.

Here, is the density of the cold fluid, is cross sectional area of the inner tube, is the mean velocity of the cold fluid and is the density of the cold fluid.

Substitute for ,for and for .

Write the expression for mass flow rate of hot fluid, that is water.

Here, is the density of the hot fluid, is cross sectional area of the annulus of tube, is the mean velocity of the hot fluid and is the density of the hot fluid.

Substitute for ,for and for .

Write the expression for heat capacity of the cold fluid.

Here, is the heat capacity of the cold fluid, is the mass flow rate of the cold fluid and is the specific heat of the cold fluid.

Substitute for and for .

Write the expression for heat capacity of the hot fluid.

Here, is the heat capacity of the hot fluid and is the mass flow rate of the hot fluid.

Substitute for and $4.181{\rm{ kJ/kgK}}$ for .

$\begin{array}{c}\\{C_h} = \left( {0.0466{\rm{ kg/s}}} \right)\left( {4.181{\rm{ kJ/kgK}}\left( {\frac{{1000{\rm{ J}}}}{{1{\rm{ kJ}}}}} \right)} \right)\\\\ = 194.83{\rm{ W/K}}\\\end{array}$

Write the expression for heat capacity ratio.

Substitute for and for .

$R = \frac{{{C_c}}}{{{C_h}}}$

Substitute for and for .

Write the expression of surface area of the tube.

$A = \pi {D_i}L$

Here, is the length of the tube.

Write the expression of .

$NTU = \frac{{UA}}{{{C_{\min }}}}$

Substitute for and $\pi {D_i}L$ for $A$ .

$NTU = \frac{{U\left( {\pi {D_i}L} \right)}}{{{C_c}}}$

Substitute for ,for ,for and for .

Write the expression for effectiveness in terms of for counter flow heat exchanger.

Substitute for and for .

Write the expression for effectiveness in terms of temperature.

Substitute for .

Substitute for , for and for .

$\begin{array}{l}\\0.32 = \frac{{\left( {{T_{co}}} \right) - \left( {293{\rm{ K}}} \right)}}{{\left( {333{\rm{ K}}} \right) - \left( {293{\rm{ K}}} \right)}}\\\\{T_{co}} - \left( {293{\rm{ K}}} \right) = 12.8{\rm{ K}}\\\\{T_{co}} = 305.8{\rm{ K}}\\\end{array}$

Convert the unit of temperature from Kelvin to Celsius.

$\begin{array}{c}\\{T_{co}} = 305.8{\rm{ K}}\\\\{\rm{ = }}\left( {305.8 - 273} \right)^\circ {\rm{C}}\\\\ = 32.{\rm{8}}^\circ {\rm{C}}\\\end{array}$

(c)

Write the expression for effectiveness in terms of for parallel flow heat exchanger.

Substitute for and for .

Write the expression for effectiveness in terms of temperature.

Substitute for .

Substitute for , for and for .

$\begin{array}{l}\\0.318 = \frac{{\left( {{T_{co}}} \right) - \left( {293{\rm{ K}}} \right)}}{{\left( {333{\rm{ K}}} \right) - \left( {293{\rm{ K}}} \right)}}\\\\{T_{co}} - \left( {293{\rm{ K}}} \right) = 12.72{\rm{ K}}\\\\{T_{co}} = 305.72{\rm{ K}}\\\end{array}$

Convert this temperature from Kelvin to degree Celsius.

$\begin{array}{c}\\{T_{co}} = 305.72{\rm{ K}}\\\\{\rm{ = }}\left( {305.72 - 273} \right)^\circ {\rm{C}}\\\\ = 32.72^\circ {\rm{C}}\\\end{array}$

Ans: Part a

The value of overall heat transfer coefficient is .

Part b

The mean outlet temperature of pharmaceutical product when operating in counter flow mode is .

Part c

The mean outlet temperature of pharmaceutical product when operating in parallel flow mode is .