Question

A concentric tube heat exchanger for cooling lubricating oil is comprised of a thin-walled inner tube...

A concentric tube heat exchanger for cooling lubricating oil is

comprised of a thin-walled inner tube of 25 mm diameter carrying water and an outer tube

of 45 mm diameter carrying the oil. The mass flow rates of both fluids are 0.1 kg/s. The

exchanger operates in counter-flow with an overall heat transfer coefficient of 60 W/m2

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Answer #1
Concepts and reason

Heat Exchanger:

This device is used to implement the process of heat exchange between two fluids that are at different temperatures and separated by a solid wall.

When the two fluids are separated by concentric tubes and two fluids flow in opposite direction the device is termed as counter flow concentric tube heat exchanger.

Apply overall energy balance on the hot fluid to find the heat transfer. Then apply the energy balance on the cold fluid to find the outlet temperature of cold fluid. Use the LMTD method and calculate the length of the exchanger.

Fundamentals

Write the expression for energy balance of hot fluid.

q=mc (T.-Two)

Here, the mass flow rate of hot fluid is ,the specific heat constant of hot fluid is , the inlet temperature of hot fluid is , and the outlet temperature of the hot fluid is .

Write the expression for energy balance of cold fluid.

q=mc (T. -Teo)

Here, the mass flow rate of cold fluid is ,the specific heat constant of cold fluid is , the inlet temperature of cold fluid is , and the outlet temperature of the cold fluid is .

Write the expression for LMTD for counter flow heat exchanger.

AT - AT, - AT
(Tho -T.:)-(Th1-T..)
In
The -
The-T

Find the length of the Counter flow heat exchanger using the following relation:

q=UAATIM.CF
=U (FDL) AT..CF

Here, the overall heat transfer coefficient is , the diameter of the tube is , and the length of the tube is .

Draw the schematic diagram of the counter flow heat exchanger.

( Oil,h
m Water,c
>
AT
X

Properties for Hot Fluid:

Density of the fluid,p=800kg/m

Specific heat of hot fluid,Cpu = 1.9 kJ/kg-K

Viscosity of hot fluid isVr = 1x10 m/s

Thermal conductivity of hot fluid isk
=134x10-W/m.K

Prandtl number of hot fluid isProin = 140

Properties for Cold Fluid:

Density of the fluid, p=1000kg/mº;

Specific heat of cold fluid isCpe = 4.2 kJ/kg-K

Viscosity of cold fluid is ve = 7x10-7m/s

Thermal conductivity of cold fluid is kçe = 640x10 W/m-K

Prandtl number of cold fluid is Prze = 4.7

a.1

Calculate the heat transfer using the overall energy balance on the hot fluid.

q=mc (T.-Two)

Substitute 0.1kg/s
for ,1900J/kg.K
for , 100°C
for , and for .

q=0.1x1900(100-60)
= 7600 W

a.2

Apply energy balance to the cold fluid.

q=mc (T.-T.)

Substitute 7600 W
for ,0.1kg/s
for ,4200 J/kg.
for , for .

7600=0.1x 4200(T., – 30)
Teco = 48.1°C

b)

Write the expression for LMTD for counter flow heat exchanger.

AT
(Tho -T.)-(T-1).
In
(T -T.
T-TA

Substitute for , for ,48.1°C
for , and for .

(60-30)-(100 - 48.1)
60-30
In
(100-48.1)
= 40°C

Find the length of the Counter flow heat exchanger using the following relation:

q=UAATIM.CF
=U (FDL) AT..CF

Substitute for ,60 W/m²K
for ,0.025 m
for ,40°Cork
for AT...CF
.

7600 = 60((0.025)L)40
L = 40.3m

Ans: Part a.1

Therefore, the total heat transfer is 7600 W
.

Part a.2

Therefore the outlet temperature of the cold fluid is 48.1°C
.

Part b

Therefore the length of the heat exchanger is 40.3 m
.

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