Experiment 5 Heat

EXPERIMENT 5

SHELL AND TUBE HEAT EXCHANGER EXPERIMENT

OBJECTIVES

To evaluate and study the performance of the shell and tube heat exchanger at various

operating conditions

To determine the heat load, log mean temperature difference (LMTD), overall heat

transfer coefficient, U.

SUMMARY

This experiment is about the heat exchanger (HE). There are two types of heat exchanger;

double pipe and the shell and tube. Double pipe heat exchanger is the simplest one meanwhile

the shell and tube heat exchanger s the most commonly used. This experiment is held in Lab

6A. The objectives of this experiment is to evaluate and study the performance of the shell and

tube heat exchanger at various operating system and to determine the heat load, log mean

temperature difference (LMTD), overall heat transfer coefficient U. In this experiment, the

pressure, the flowrate of water should be taken care of because the pressure drop and water’s

flowrate contributes to the final result of this experiment. The type of the fluid in the shell and

tube heat exchanger is

NOTATION

INTRODUCTION & THEORY

To design a shell and tube heat exchanger, it is necessary to relate the total heat transfer rate to fluid temperature (inlet and outlet), overall heat transfer coefficient (U) and the total surface area for the heat transfer. Two such relations may be obtained by applying energy balance (sensible heat relationship)to the hot and cold fluid.

Sensible heat expression might be obtained by relating total heat transfer rate to the temperature difference between the hot and cold fluids and constant specific heats are assumed. These equations are independent of flow arrangement and heat exchanger type. Therefore, these equations are,

[1]

[2]

An expression may be obtained by relating the total heat transfer rate to the temperature difference between hot and cold fluid, where ∆T = Tn-Tc. This expression would be an extension of Newton’s law of cooling, with the overall heat transfer coefficient (U) used in placed of the singe convection (h) as Q = U A (∆T)m. The varies with position in the heat exchanger. It is necessaries to work with a rate equation of the following form,

Q = U A (∆T)m [3]

Where,

(∆T)m = Log mean temperature differenceQ = Total heat loadA = Total heat transfer area

This equation 3 with equations q and 2 would be performed a heat exchanger analysis. Before this can be done, however the specific form of (∆T)m must be established for the parallel flow and current-flow heat exchanger.

Parallel – Flow and Counter Flow Heat Exchanger

According to Figure 1, the hot and cold fluid temperature distribution associated with a parallel – flow heat exchanger would be summarized to determine expression for (∆T)m as,

1 = T hi – T ci and 2 = T ho – T co

Therefore,

Q hot = m (hot) × Cp (hot) × ∆T (hot)

Q cold = m (cold) × Cp (cold) × ∆T (cold)

HeatExchanger

Tci

Tho

Tci

Thi

(∆T)m = θ 1 – θ 2 ln( 1 / 2)

Equation 4 would give the log mean temperature difference (∆T)m.Equation 4 and equation 3 would be able to determine the rate of heat transfer for parallel flow.

Figure 1: Parallel Flow

The hot and cold fluid temperature distribution associated with the counter current – flow heat exchanger in Figure 2 in contrast to the parallel – flow exchanger, therefore, would be summarized to determine the expression for (∆T)m as,

Figure 2: Counter-current Flow

Therefore,

(∆T)m = 1 – 2ln( 1 / 2)

Equation 5 would give the log mean temperature difference (∆T)m. Also this equation and equation 3 would be able to determine the rate of heat transfer for counter flow.

HeatExchanger

Tci

Tho

Tco

Thi

Results

Table Run 1

Set 1CW HW

Actual Flow : m3/ hr : USGPM

Temp, / , InletTemp, / , Outlet

FC : 1.4 m3/ hr : 6 USGPM

T13 : T1 : 25.8 ºCT14 : T2 : 48.9 ºC

FH : 2.3 m3/ hr : 10USGPM

T11 : T1 : 60.0 ºCT12 : T2 : 46.6 ºC

Pressure, psig, InletPressure Drop, mm H2O

PG-C : 7DP (Shell): 318

PG-H : 2.0DP(Tube): 922

CALCULATE FOLLOWING :

Temp Change, /

Average Temp, /

Q, Head Load, BTU / HR

T2-T1 : 23.1 ºCT2 + T1 : 37.35 ºC 2QC : 124, 395,6

T2-T1 : 13.4 ºCT2 + T1 : 53.5 ºC 2QH : 120,493.06

Computer ratio 1.032

Table Run 2

Set 2CW HW



FC : m3/ hr : 10 USGPM

T13 : T1 : 29.1 ºCT14 : T2 : 45.8 ºC

FH : m3/ hr : 10USGPM

T11 : T1 : 69.8 ºCT12 : T2 : 45.2 ºC


PG-C : 7.0DP (Shell): 926

PG-H : 2.0DP(Tube): 913


Temp Change, /

Average Temp, /


T2-T1 : 16.7 ºCT2 + T1 : 37.45 ºC 2QC : 150,316.5

T2-T1 : 24.6 ºCT2 + T1 : 57.5 ºC 2QH : 220,697.25


Table Run 3

Set 3CW HW




T13 : T1 : 32.2 ºCT14 : T2 : 50.6 ºC


T11 : T1 : 70.0 ºCT12 : T2 : 49.0 ºC


PG-C : 9.0DP (Shell): 891

PG-H : 6.0DP(Tube): 2022


Temp Change, /

Average Temp, /


T2-T1 : 18.3 ºCT2 + T1 : 41.4 ºC 2QC : 183,154

T2-T1 : 21.0 ºCT2 + T1 : 59.5 ºC 2QH : 283,227.84


Table Run 4

Set 4CW HW




T13 : T1 : 35.8 ºCT14 : T2 : 53.7 ºC


T11 : T1 : 61.2 ºCT12 : T2 : 51.4 ºC


PG-C : 9DP (Shell): 900

PG-H : 15DP(Tube): 3519


Temp Change, /

Average Temp, /


T2-T1 : 17.9 ºCT2 + T1 : 44.75 ºC 2QC : 161,018

T2-T1 : 9.8 ºCT2 + T1 : 56.3 ºC 2QH : 177,030


SAMPLE CALCULATION FROM DATA RUN 1.

COLD WATER (QC)

Convert 6 gal USGPM to m³ / Hr

6 gal / min x 1m³ / 264.17 gal x 60min/1 Hr

=1.36276 m³ / Hr

T° C Cp21.11 997.4 4.17926.67 995.8 4.179

∆T = T2 – T1 = 23.1 ° C Refer to the table of H2O

Interpolation to find at 23.1 ° C Interpolation to find Cp at 23.1 ° C

= 997.4 + ( 23.1-21.11) (992.1-994) = 4.179 + (23.11 – 21.11) (4.179 – 4.179) (26.67–21.11) (26.67-21.11) = 997.4 + 1.99 (-1.6) = 4.178 + 1.99 (0) 5.56 5.56 = 996.827 kg m3 = 4179 kJ/kg.° C

(m)cold = x flowrate

= 996.827 Kg / m³ x 1.36276 m³ / Hr = 1358.43 Kg / Hr

= (m)cold x Cp x ∆T = 1358.43 Kg / Hr x 4.179 KJ / Kg.° C x 23.1 ° C

= 131,136 KJ /Hr x 1000 J / KJ x 9.486x10-4 Btu / J

= 124,396 Btu/hr

HOT WATER (QH)

∆T = T2 – T1 = 60.46 – 46.6 = 13.4 ° C

Interpolation to find at 13.4 ° C Interpolation to find Cp at 13.4 ° C

=999.2 + (13.4 - 10.0) (998.6-999.2) = 4.195 + (13.4-10.0) (4.186-4.195) (15.56–13.4) (15.56-13.4) =998.256 kg/m3 = 4.1808 KJ/kg.° C

(m)hot = x flowrate=998.256 Kg/m³ x 2.2713m³/Hr=2267.3 kg/hr

= (m)hot x Cp x ∆T= 22367.3 Kg/Hr x 4.1808 KJ/Kg.° C x 13.4 ° C= 127,022 KJ/Hr x 1000 J / KJ x 9.486 x 10-4 Btu / J

= 120,493.06 Btu/hr

THEN FIND RATIO OF

= 124,395.6 Btu/hr = 1.03 120,493.06 Btu/hr

THEN FIND Q AVERAGE

= [(124,395.6 + 120,493.06)] / 2= 122,444

CALCULATION PARALLEL FROM DATA RUN 1.

T° C Cp10 999.2 4.195

15.56 983.6 4.186

ΔLMTD for parallel:

=60.0 °C – 25.8°C =34.2 °C

= 48.9 °C – 46.6 °C = 2.3 °C ΔT=(34,2 °C – 2.3 °C)/ln (34.2 / 2.3) =31.9 °C / 2.699 = 11.8178 + 273.15K = 284.96 K

Then, find U,

=(122,444) / [31.67 ft² x 1m² / (3.2808 ft²) x 284.96 K] = 146.037 m -2 .K -1

CALCULATION COUNTER CURRENT FLOW FROM DATA RUN I

ΔLMTD for counter flow:

=60.0 °C- 48.9 °C =11.1°C

=46.6 °C – 25.8 °C =20.8 °C

= (11.1 – 20.8) / ln (11.1 / 20.8) =15.44 +273.15K = 288.59 K Then, find U,

= (122,444) / [31.67 ft² x 1m² / (3.2808 ft²) x 284.96 K] =144.2 m -2 K -1

DISCUSSION

A heat exchanger is a device designed to efficiently transfer heat from one fluid to

another. In a heat exchanger hot and cold fluids enter separate chambers or tubes of the heat

exchanger unit. The hot fluid transfers its heat to a conductive surface (solid partition) between

it and the cold chamber, subsequently the partition transfers the heat to the cold fluid. The hot

and cold fluids are never combined

According to Newton's Law of Cooling heat transfer rate is related to the instantaneous

temperature difference between hot and cold media

in a heat transfer process the temperature difference vary with position and time

From our result, the ratio get nearest to one is RUN I. The factors that influence the

result to get nearest to one because of the flow rate. Differential pressure in shell more lower

than differential pressure in tube because cold water are came into shell and the hot water are

come into the tube. In our calculation, the value U of parallel flow and counter current flow are

differences. The values of U in parallel flow are higher than counter current flow because it

exchange of heat transfer in shell and tube heat exchanger. The values of U in parallel flow is

146.037 m-2.K-! and the values of U in counter current flow is 144.2 m-2K-1

The equation to find heat hot and cold is:

QH = (m)hot (Cp)hot (ΔT)hot and Qc = (m)cold (Cp)cold (ΔT)cold

The value for flow rate has to convert in m3/hr. Base on the calculation, to find density

and Cp we have to used interpolation. Other we use formula which Q = total heat

load, A = Total heat transfer area, (ΔT)m = log mean temperature differrence . we use this

formula to find overall heat transfer coefficient, U.

In this experiment also we have to determine if the heat exchanger is a parallel or

counter current flow. Different flow with different formula we use. For parallel flow we use

(ΔT)m = where θ1 = Thi - Tci. And θ2 = Tho – Tco . For counter current flow we use same

above but θ1 = Thi - Tco and θ2 = Tho – Tci . Then we compare the log mean temperature, (ΔT)m to

determine which flow rate is better. From the result, we can see that (ΔT)m C.C..P(counter current

flow) always give higher value than parallel. This is because the heat exchange in the same

time so the temperature will increase.

From the result we can show that the cold water and hot water is increase from run I to run

IV. So we can see that if the flow rate increase the temperature also will increase. Hence if the

temperature cold and hot water increase, the average temperature also will increase

CONCLUSION

From this experiment, the Run that get nearest ratio to 1 is Run I. The Factor that might be

influence this result is the flow rate of cold and hot water. The counter current flow always give

highest reading than the parallel flow. This is because counter current flow is a efficient

circulating in heat exchanger. To determine which flow is better we have to calculate log mean

temperature, (ΔT)m overall heat transfer coefficient, U for parallel and counter current flow.

TUTORIAL

1. Compare and calculate the valves of Q hot and Q cold and select the set of temperature and

flow rates data where calculated values of Q hot and Q cold are close to each other (ratio Q cold

÷ Q hot ≈ 1).

Run I-Cold water

m (mass flowrate) = ρ (density) × FR (flowrate)---------- (1)

To find ρ (density) and Cp (specific heat):

Interpolation: (from Table A-9 Properties of saturated water)


ρ (density) = 996.827 kg/m 3

Cp (specific heat) = 4.179 KJ/kg. ˚C

FR (flowrate) = 1.3627 m 3 /hr (see Result)

m (mass flowrate) = ρ (density) × FR (flowrate)

= 996.827 kg/m3 × 1.3627 m3/hr

= 1358.43 kg/hr

Temperature Change, ∆T = 23.1 ˚C (see Result)


=1358.43 Kg / Hr x 4.179 KJ / Kg.° C x 23.1 ° C = 131,136 KJ /Hr x 1000 J / KJ x 9.486x10-4 Btu / J

= 124,396 Btu/hr

Run IV- Hot Water

m (mass flowrate) = ρ (density) × FR (flowrate)---------- (1)

To find ρ (density) and Cp (specific heat)

Interpolation: (from Table A-9 Properties of saturated water)

ρ (density) = 988.256 kg/m 3

Cp (specific heat) = 4.1808 KJ/kg. ˚C

FR (flowrate) = 2.2713 m 3 /hr (see Result)

m (mass flowrate) = ρ (density) × FR (flowrate)

= 988.256 kg/m3 × 2.2713 m3/hr


= 2267.3 kg/hr

Temperature Change, ∆T = 49.75 ˚C (see Result)


= 22367.3 Kg/Hr x 4.1808 KJ/Kg.° C x 13.4 ° C= 127,022 KJ/Hr x 1000 J / KJ x 9.486 x 10-4 Btu / J

= 120,493.06 Btu/hr

Ratio Q = 124,395.6 / 120493.8

= 1.032

Compare value Q hot = 120,493.06 Btu/hr and Q cold = 124,395.6 Btu/hr, the differential of heat

transfer between Q hot and Q cold is 3902.6 Btu/hr. We can see that cold water in the tube

transferred more heat than water in the shell and this have make the heat transfer between hot

and hot water in run 2 are most efficient than others.

2. Determine the type of fluid flow in this heat exchanger (Parallel/Counter Flow). Calculate

the log mean temperature difference (LMTD) for this shell and tube heat exchanger.

This experiment we using Counter Current flow.

Calculations of the log mean temperature difference (LMTD) for this shell and tube

heat exchanger:

RUN 1

∆ Tm C.C.F = (θ1- θ2) ÷ ln (θ1 / θ2)

Counter Flow: θ1 = T hi –T ci

θ2 = T ho – T co

θ1 = 60.0 ˚C – 48.9 ˚C = 11.1˚C

∆ Tm C.C.F = (θ1- θ2) ÷ ln (θ1 / θ2)

Counter Flow: θ1 = T hi –T co

θ2 = T ho – T ci

θ2 = 46.6 ˚C – 25.8 ˚C = 20.8˚C

∆ Tm C.C.F = (11.1˚C – 20.8˚C) ÷ ln (11.1˚C / 20.8˚C)

= 15.44

RUN 2

∆ Tm C.C.F = (θ1- θ2) ÷ ln (θ1 / θ2)


θ2 = T ho – T ci

θ1 = 69.8 ˚C – 45.8 ˚C = 24.0 ˚C

θ2 = 45.2 ˚C – 29.1 ˚C = 16.1 ˚C

∆ Tm C.C.F = (24.0˚C – 16.1 ˚C) ÷ ln (24.0 ˚C / 16.1 ˚C)

= 19.78

RUN 3

∆ Tm C.C.F = (θ1- θ2) ÷ ln (θ1 / θ2)


θ2 = T ho – T ci

θ1 = 70.0 ˚C – 50.6 ˚C = 19.4 ˚C

θ2 = 49.0 ˚C – 32.2 ˚C = 16.8 ˚C

∆ Tm C.C.F = (19.4 ˚C – 16.8 ˚C) ÷ ln (19.4 ˚C / 16.8 ˚C)

= 18.06

RUN 4

∆ Tm C.C.F = (θ1- θ2) ÷ ln (θ1 / θ2)


θ2 = T ho – T ci

θ1 = 61.2 ˚C – 53.7 ˚C = 7.5 ˚C

θ2 = 51.4 ˚C – 35.8 ˚C = 15.6 ˚C

∆ Tm C.C.F = (7.5 ˚C - 15.6 ˚C) ÷ ln (7.5 ˚C / 15.6 ˚C)

= 11.06

3. Compute the overall heat exchanger coefficient, U for this heat exchanger if given

A= 31.67 ft2 (Total heat transfer area for heat exchanger).

RUN 1

U C.C.F = Q average ÷ Area × ∆ Tm C.C.F

Q average = (124,395 Btu/hr + 120,493 Btu / hr) ÷2

= 122.444 Btu/hr

Given area, A = 31.67 ft2 converts to m2

31.67 ft2 × 1 m2/ 10.764 ft2 = 2.942 m2

∆ Tm C.C.F = 15.44˚C

= 15.44 ˚C + 273.15 K

= 288.59 K

U C.C.F = 122,444 Btu/hr ÷(2.942 m2 × 288.59 K)

= 144.2 Btu/m 2. hr.K

RUN 2

U C.C.F = Q average ÷ (Area × ∆ Tm C.C.F )

Q average = (150,316 Btu/hr + 220,697 Btu/hr ) ÷2

= 185,506 Btu/hr


31.67 ft2 × 1 m2/ 10.764 ft2 = 2.942 m2

∆ Tm C.C.F = 19.78 ˚C + 273.15 K = 292.93K

Q average = U C.C.F × ∆ Tm C.C.F × Area

U C.C.F = 185,506 Btu/hr ÷(2.942 m2 × 292.93 K)

= 215.25 Btu/m 2 .hr.K

RUN 3

U C.C.F = Q average ÷ (Area × ∆ Tm C.C.F )

Q average = (183,154 Btu/hr + 283227 Btu/hr ) ÷2

= 233,190.5 Btu/hr


31.67 ft2 × 1 m2/ 10.764 ft2 = 2.942 m2

∆ Tm C.C.F = 18.06˚C + 291.21 K

U C.C.F = 233,190.5 Btu/hr ÷(2.942 m2 × 291.21K)

= 272.18 Btu/m 2 .hr.K

RUN 4

U C.C.F = Q average ÷ Area × ∆ Tm C.C.F

Q average = (161,018 Btu/hr + 177,029.9 Btu/hr ) ÷2

= 169,023.95 Btu/hr


31.67 ft2 × 1 m2/ 10.764 ft2 = 2.942 m2

∆ Tm C.C.F = 11.06˚C +273.15 = 284.21 K

U C.C.F = 169,023.95 Btu/hr ÷(2.942 m2 × 284.21 K)

= 202.146 Btu/m 2 .hr.K

4. How overall heat transfer coefficient, U varies with the flow rate of the water?

Note: The Q (total heat load) is approximately equal to the average value between

calculated value of Qc and Qh.

When there are changes in flow rate of water, the heat transfer for hot and cold water

will be different from before changes take place. So, with the heat transfer of both hot and

cold water will change, heat transfer will affect the value of overall heat transfer

coefficient, U

Experiment 5 Heat

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