CONCENTRIC TUBE HEAT EXCHANGER

TABLE OF CONTENTS

No. Topics Page No.

1. Title 2

2. Objective 2

3. Theory 2

4. Equipment 3

5. Procedure 5

6. Data and Results 6

7. Sample Of Calculation 7

8. Analysis and Discussion 9

9. Conclusion 10

10. References 10

1

TITLE: CONCENTRIC TUBE HEAT EXCHANGER

OBJECTIVE: To demonstrate the effect of flow rate variation on the performance

characteristics of a counter-flow concentric tube heat exchanger.

THEORY:

The equations for calculating the performance characteristics: power emitted, power

absorbed, power lost, efficiency (), logarithmic mean temperature (), and overall heat

transfer coefficient (U).

The efficiency for the cold medium is:

ηc = (Tc,out – Tc,in) / (Th,in – Tc,in) × 100

The efficiency for the hot medium is:

ηh = (Th,in – Th,out) / (Th,in – Tc,in) × 100

The mean temperature efficiency is:

ηmean = ( ηc + ηh ) / 2

The power emitted is given below (where Vh is the volumetric flow rate of the hot

fluid):

Power Emitted = Vh ρh Cph ( Th,in – Th,out )

2

The power absorbed is given below (where Vc is the volumetric flow rate of the cold

fluid):

Power Absorbed = Vc ρc Cpc ( Tc,out – Tc,in )

The power lost is therefore:

Power Lost = Power Emitted – Power Absorbed

The overall efficiency (η) is:

η=(Power Absorbed / Power Emitted) × 100

The logarithmic mean temperature difference (ΔTm) is:

ΔTm = (ΔT1 – ΔT2) / ln (ΔT1/ΔT2)

= [ (Th,in – Tc,out) – (Th,out – Tc,in) ] / ln [(Th,in – Tc,out) / (Th,out Tc,in)]

The overall heat transfer coefficient (U) is:

U = Power Absorbed / As . ΔTm

Where the surface area (As) for this heat exchanger is 0.067 m²

EQUIPMENT: The experiment set-up consists of:a) Set of tube heat exchanger.b) Cold fluids supply and hot fluids supply.c) Digital stopwatch.

3

Figure 1: Set of tube heat exchanger. Figure 2: Volumetric Flow Rate.

Figure 3: Decade Switch. Figure 4: Flow Diagrams.

This experiment can be made using either parallel or counter flow operation. This

experiment was conducted as counter flow operation.

4

PROCEDURE:

1. Configure the experiment for counter flow heat exchanger operation such as turn

ON the heating elements to heat the fluids.

2. Set the required hot water inlet temperature to Th,in = 60º with the decade switch

and set the cold water volumetric flow rate (Vc) to run at a constant 2000

cm³/min.

3. Initially set the hot water volumetric flow rate Vh to 1000 cm³/min. Wait until 5

minutes before the six temperature readings are records.

4. Repeat this for volumetric flow rate,Vh of 2000, 3000 and 4000 cm³/min for hot

water. Record the temperature readings in the table.

5. After finish up the experiment, turn OFF the heating elements, close the valve for

hot and cold water.

5

DATA AND RESULTS:

Vh Th,in Th,mid Th,out Tc,in Tc,mid Tc,out

(cm³/

min)

(m³/s) (ºC) (ºC) (ºC) (ºC) (ºC) (ºC)

1000 1.667E-

5

60 53 47 30 31 35

2000 3.333E-

5

60 55 50.5 30 33 38

3000 5E-5 60 57 53 30 34 40

3900 6.5E-5 60 58 54 30 35 41

Vh Power

Emitted

Power

Absorbed

Power

Lost

Efficiency ΔT1 ΔT2 ΔTm U

cm³/

min

m³/s (W) (W) (W) (%) (ºC) (ºC) (ºC) W/(m². ºC)

1000 1.667E

-5

891.786 693.479 198.307 77.76 25 17 20.74 499.06

2000 3.333E

-5

1302.99 1109.57 193.42 85.16 22 20.5 21.24 779.69

3000 5E-5 1440.29 1386.96 53.33 96.3 20 23 21.46 964.63

4000 6.5E-5 1604.89 1525.65 79.24 95.02 19 24 21.4 1064.07

6

SAMPLE CALCULATION:

From table A-9 (Properties of saturated water):

At Tc,in = 30 ºC.

Vc = 2000 cm³/min = 2000 cm³/min × 1 min/60 s × 1 m³/100³ cm³

=3.333E-5 m³/ s

ρc = 996 kg / m³

Cpc = 4178 J/kg.K

At Th,in = 60 ºC.

ρh = 983.3 kg / m³

Cph = 4185 J/kg.K

a) Power Emitted = Vh ρh Cph ( Th,in – Th,out )

= (1.667E-5 m³)(983.3 kg / m³)(4185 J/kg.K)(333K – 320K)

= 891.786 W

b) Power Absorbed = Vc ρc Cpc ( Tc,out – Tc,in )

= (3.333E-5 m³/ s)( 996 kg / m³)(4178 J/kg.K)(308K – 303K)

= 693.479 W

c) Power Lost = Power Emitted – Power Absorbed

= 891.786 W – 693.479 W

= 198.307 W

d) Overall Efficiency, η =(Power Absorbed / Power Emitted) × 100

= (693.479 W / 891.786 W) × 100

= 77.76 %

7

e) Logrithmic Mean Temperature Difference, ΔTm= (ΔT1 – ΔT2) / ln (ΔT1/ΔT2)

ΔT1 = Th,in – Tc,out

= 60 ºC - 35 ºC

= 25 ºC

ΔT2 = Th,out – Tc,in

= 47 ºC – 30 ºC

= 17 ºC

ΔTm = (ΔT1 – ΔT2) / ln (ΔT1/ΔT2)

= (25 ºC – 17 ºC) / ln [(25 ºC/17 ºC)

= 20.74 ºC

f) Overall Heat Transfer Coefficient, U = Power Absorbed / As . ΔTm

U = Power Absorbed / As . ΔTm

= 693.479 W / (0.067 m²×20.74 ºC)

= 499.06 W/ m² .ºC

8

ANALYSIS AND DISCUSSION

Heat exchanger are commonly used in practice, and an engineer often finds

himself or herself in a position to select a heat exchanger that will achieve a specified

temperature change in a fluid stream of known flow rate, or to predict the outlet

temperatures of the hot and cold fluids stream in a specified heat exchanger.

The variation of temperature of hot and cold fluids in a counter-flow heat

exchanger is given in figure 4. Note that the hot and cold fluids enter the heat exchanger

from opposite ends, and the outlet temperature of the cold fluid in this case may exceed

the outlet temperature of the hot fluid. In the limiting case, the cold fluid will be heated to

the inlet temperature of the hot fluid. However, the outlet temperature of the cold fluid can

never exceed the inlet temperature of the hot fluid, since this would be a violation of the

second law of thermodynamics.

9

CONCLUSION

It can be concluded that the experiment is done successfully. The power emitted

and power absorbed are increased when we compared the effect of changing the

volumetric flow rate of the hot fluid. Besides, the power lost that we get shows decreasing

value unless the last reading give some increased value. This is maybe because of the

error while doing the experiment that may cause by conduction and convection between

hot and cold fluid while doing counter flow operation. The overall efficiency are

reasonable and doesn’t exceed the 100%. From our experiment, the overall heat transfer

coefficient will increase when the volumetric flow rate of the hot fluid are increase. So

that, the conclusion that can be done is the overall heat transfer coefficient, the power

emitted and power absorbed are influenced by the changing of volumetric flow rate of the

hot fluid.

REFERENCE

1. Heat and Mass Transfer (A Practical Approach) – 3rd EditionYunus A. CengelMcGraw Hill (2006).

2. Class Note KJM531-Heat Transfer.

10