experiment 5 heat
TRANSCRIPT
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
Actual Flow : m3/ hr : USGPM
Temp, / , InletTemp, / , Outlet
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
Pressure, psig, InletPressure Drop, mm H2O
PG-C : 7.0DP (Shell): 926
PG-H : 2.0DP(Tube): 913
CALCULATE FOLLOWING :
Temp Change, /
Average Temp, /
Q, Head Load, BTU / HR
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
Computer ratio 0.681098
Table Run 3
Set 3CW HW
Actual Flow : m3/ hr : USGPM
Temp, / , InletTemp, / , Outlet
FC : m3/ hr : 10 USGPM
T13 : T1 : 32.2 ºCT14 : T2 : 50.6 ºC
FH : m3/ hr : 15USGPM
T11 : T1 : 70.0 ºCT12 : T2 : 49.0 ºC
Pressure, psig, InletPressure Drop, mm H2O
PG-C : 9.0DP (Shell): 891
PG-H : 6.0DP(Tube): 2022
CALCULATE FOLLOWING :
Temp Change, /
Average Temp, /
Q, Head Load, BTU / HR
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
Computer ratio 0.6467
Table Run 4
Set 4CW HW
Actual Flow : m3/ hr : USGPM
Temp, / , InletTemp, / , Outlet
FC : m3/ hr : 10 USGPM
T13 : T1 : 35.8 ºCT14 : T2 : 53.7 ºC
FH : m3/ hr : 20USGPM
T11 : T1 : 61.2 ºCT12 : T2 : 51.4 ºC
Pressure, psig, InletPressure Drop, mm H2O
PG-C : 9DP (Shell): 900
PG-H : 15DP(Tube): 3519
CALCULATE FOLLOWING :
Temp Change, /
Average Temp, /
Q, Head Load, BTU / HR
T2-T1 : 17.9 ºCT2 + T1 : 44.75 ºC 2QC : 161,018
T2-T1 : 9.8 ºCT2 + T1 : 56.3 ºC 2QH : 177,030
Computer ratio 0.90955
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)
Q cold = m (cold) × Cp (cold) × ∆T (cold)
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)
Q cold = m (cold) × Cp (cold) × ∆T (cold)
=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
Q hot = m (hot) × Cp (hot) × ∆T (hot)
= 2267.3 kg/hr
Temperature Change, ∆T = 49.75 ˚C (see Result)
Q hot = m (hot) × Cp (hot) × ∆T (hot)
= 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)
Counter Flow: θ1 = T hi –T co
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)
Counter Flow: θ1 = T hi –T co
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)
Counter Flow: θ1 = T hi –T co
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
Given area, A = 31.67 ft2 converts to m2
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
Given area, A = 31.67 ft2 converts to m2
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
Given area, A = 31.67 ft2 converts to m2
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