hanoi university of mining and geology
Report of shell and tube heat exchanger experiment
ECH155B
Nguyễn Thành Trung – CTTTK7 – 1621090002
1.Abstract :
- Heat exchange is a process that allow the heat of a fluid transfer to
another without directly contact with each other. The process allow
hot flow transfer heat to cold flow, raise ones temperature while
decreasing other.
2.Introduction :
- In this experiment we will study the efficiency of shell and tube
heat exchanges. The objectives of this experiment are:
o To perform an energy balance of a shell and tube heat
exchanger.
o To investigatethe difference between cocurrent and
countercurrent flows.
o To determine the overall heat transfer coefficient for a shell
and tube heat exchanger.
o To investigate the effect flow rates on heat exchanger
efficiency and overall heattransfer coefficient.
3.Equations :
3.1. Mass flow rate :
Where :
ρ = Density of fluid (kg/m3)
= Volumetric flow rate (m3/s)
3.2. Heat transferred:
Q = m.Cp. ΔT
Where
m = Mass flow rate (kg/m3)
Cp = Heat capacity of fluid (kJ/kg.K)
ΔT = Different of temperature of in/out fluid.
1
3.3. Log Mean Temperature Difference:
ΔT 1−ΔT 2
ΔTltmd = ln (
ΔT 1
)
ΔT 2
Where :
ΔT1 = (T2-T3)
ΔT2 = (T1-T4)
3.4. Total heat transfer area:
dm = 0.5(dod + did)
Where :
dod,did = external/internal diameter of the tube respectively (m)
L = n.l
Where :
L = total heat transfer length (m)
n = number of tube
l = length of an individual tube (m)
A = π.dm.L
Where :
A = total heat transfer area (m2)
3.5. Global heat transfer coefficient:
Q
U = A .∆
lmtd
Where:
A = total heat transfer area (m2)
ΔTltmd = Log Mean Temperature Difference
2
3.6. Effectiveness coefficient:
Qreal
ε= Q
maximum
Where:
Qreal = Actual heat exchanged (W)
Qmaximum = Maximum heat exchanged (W)
Or
T h , i−T h ,o
ε = T −T if mh.Cp,h < mcCp,c
h ,i c ,i
T c , o−T c ,i
ε = T −T if mh.Cp,h > mcCp,c
h, i c, i
3.7. Number of transferred unit:
U.A
NTU = (m .C ¿¿ p) ¿
min
Where
U = Global heat transfer coefficient (W/m2.K)
A = total heat transfer area (m2)
m = Mass flow rate (kg/m3)
Cp = Heat capacity of fluid (kJ/kg.K)
3.8. Capacity coefficient:
(m .C p )min
Cr =
¿¿¿
3.9. Effectiveness coefficient calculated with NTU:
ε = 1−e¿ ¿ ¿ for parallel flow
ε = 1−e¿ ¿ ¿ for counter-current flow
3.10.Outlet temperature with NTU:
- If mh.Cp,h < mcCp,c
3
Th,o = Th,i – ε(Th,i – Tc,i)
Tc,o = Tc,i – Cr(Th,i – Th,o)
- If mh.Cp,h > mcCp,c
Th,o = Th,i – Cr(Tc,o – Tc,i)
Tc,o = Tc,i + ε (Th,i – Tc,i)
Where:
Th,I, Th,o, Tc,I, Tc,o = Temperature of hot/cold flow inlet/outlet.
4.Experiment method:
1) Check that the valves are opened and that we have co/counter-
current flow pattern.
2) Check that the heating tank is filled with water above the level switch.
3) Switch on pump and the resistor (equipment supply).
4) Set the tank temperature in 60°C (ST16).
5) Fix the hot water flow (SC1) to your group’s assigned flow rate.
Mr.Tuan will adjust the cold water flow rate (SC2).
6) Once steady state operating conditions have been obtained write
down the temperature and flow measurements on the experiment sheet
for the countercurrent flow pattern on the next page.
7) Set the valves appropriately in order to invert the cold water flow
direction to produce a cocurrent flow pattern.
8) Make sure that the cold water flow rate is not altered and that the hot
water flow rate and temperature is maintained at around 65°C when you
change to the cocurrent flow pattern.
9) Once steady state operating conditions have been obtained again
write down the temperature and flow measurements.
5.Calculation and result:
x (m) Position Parallel Flow Counter Flow
4
o
K 4 6 8 10 4 6 8 10
Tank temperature ST 16 337.25 337.25 337.25 337.25 337.35 337.35 337.05 336.85
331.7
Hot flow in 0 ST 1 331.35 331.55 332.05 330.85 330.95 331.45 331.45
5
320.0
Hot flow out 0.5 ST 2 317.15 318.75 321.25 315.95 317.45 319.15 320.35
5
303.5
Cold flow in/out 0 ST 3 303.25 303.45 303.65 312.85 313.55 314.55 315.05
5
0.125 ST 4 307.65 307.95 308.25 308.55 308.85 309.55 310.45 310.85
0.25 ST 5 310.15 310.75 311.25 311.55 308.45 309.15 309.95 310.35
0.375 ST 6 310.85 311.65 312.25 312.75 306.25 306.85 307.45 307.75
312.8
Cold flow out/in 0.5 ST 7 311.25 312.15 313.55 303.45 303.55 303.65 303.65
5
Hot flow rate SC 1 (l/min) 1.8 2.2 2.6 3 1.8 2.2 2.6 3
Cold flow rate SC 2 (l/min) 3 3 3.1 3.1 3 3.1 3 3.1
Hot mass flow 0.03666 0.04333 0.03666 0.04333
SC 1 (kg/m3) 0.03 0.05 0.03 0.05
rate 7 3 7 3
Cold mass flow 0.05166 0.05166 0.05166 0.05166
SC 2 (kg/m3) 0.05 0.05 0.05 0.05
rate 7 7 7 7
Hot Heat
4.18 4.18
capacity 4.184 4.184 4.184 4.184 4.184 4.184
4 4
(kJ/kg.K)
Cp (kJ/kg.K)
Cold Heat
4.17 4.17
capacity 4.179 4.179 4.179 4.179 4.179 4.179
9 9
(kJ/kg.K)
Number of tubes 21
Length of tube (m) 0.5
Ther. Conductivity Staintless Steel (W/m.K) 20
Internal diameter (Dint) (m) 0.008
External diameter (Dext) (m) 0.01
Internal tube Thickness (m) 0.001
2
Heat transfer Inter Area (m ) 0.0126
Heat transfer Exter Area (m2) 0.0157
Internal diameter (m) 0.148
Shell tube External diameter (m) 0.16
Thickness (m) 0.006
Density (kg/m3) 1000
5
Total area for heat
transfer (m2) 0.29673
- The calculation and result are shown as follow:
Heat
1782.38 1963.69 2121.28 1870.24 2230.07
transferred Qh (W) 2259.36 2071.08 2322.12
4 1 8 8 2
by hot water
Heat
transferred 1817.86 2137.55 2277.55 2461.43
Qc (W) 1671.6 2008.01 1964.13 2159.15
by cold 5 9 5 1
water
145.825 113.278 121.801
Heat Lost Ql (W) 110.784 -93.882 -88.07 -47.483 -139.311
7 5 5
Log mean
14.2233 14.8408 15.8600 15.0832 15.5845 16.1899 16.5495
temperature ΔTltmd 15.3819
2 9 2 4 5 1 5
different
Global heat U
422.317 445.914 464.759 480.087 417.871 447.858 464.208 472.864
transfer (W/m2.K
3 5 4 4 8 8 3 9
coefficient )
Hot capacity 153.413 181.306 153.413 181.306
125.52 209.2 125.52 209.2
rate 3 7 3 7
m.Cp
Cold
208.95 208.95 215.915 215.915 208.95 215.915 208.95 215.915
capacity rate
Efficiency 0.50533 0.45551 0.41489 0.38028 0.54379 0.49270 0.44244 0.39928
ε
coeff. 8 6 4 2 6 1 6 1
Global heat
transfer
125.314 132.316 137.908 142.456 123.995 132.893 137.744 140.313
coefficient * U.A
2 2 1 3 1 1 5 2
Total
transfer area
Number of
0.99836 0.86248 0.76063 0.68095 0.98785 0.86624 0.75973 0.67071
Transfer NTU
1 2 4 8 1 2 2 3
Units
Capacity 0.60071 0.73421 0.83971 0.60071 0.71052 0.86770
Cr 0.9689 0.9689
coeff. 8 1 3 8 7 4
Efficiency
0.44741 0.40943 0.37500 0.54772 0.49610 0.44420 0.40396
coeff. by εNTU 0.49835
6 5 5 1 1 5 6
NTU
Temperatur 317.346 318.977 320.203 321.399 315.842 317.356 319.101 320.219
Th,o
e at outlet 4 6 9 9 4 8 1 7
calculated 311.780 312.847 313.374 314.114 312.400 313.142 314.322 314.404
by ε Tc,o
2 9 6 1 7 1 8 8
6.Graphs:
6
Graph of Temperature & Position of Parallel
flow.
45
40 Hot flow level 4
35 Hot flow level 6
Hot flow level 8
Temperature (oC)
30
Hot flow level 10
25
Cold flow level 4
20 Cold flow level 6
15 Cold flow level 8
10 Cold flow level 10
5
0
0 0.1 0.2 0.3 0.4 0.5 0.6
Position (m)
Graph of Temperature & Position of Counter-
flow.
45
Hot flow level 4
40
Hot flow level 6
35
Temperature (oC)
Hot flow level 8
30
Hot flow level 10
25
Cold flow level 4
20 Cold flow level 6
15 Cold flow level 8
10 Cold flow level 10
5
0
0 0.1 0.2 0.3 0.4 0.5 0.6
Position (m)
7.Summary:
7
- The heat transferred rate of both hot and cold flow are much
higher in counter-current flow and so do the efficiency.
- The efficiency of counter-current flow is higher than the parallel
flow due to the stable difference in temperature between 2 fluids
over the entire length of the fluid path.
0.45551 0.41489 0.38028 0.54379 0.49270 0.44244
ε
0.505338078 6 4 2 6 1 6 0.399281
0.44741 0.40943 0.37500 0.54772 0.49610 0.44420
εNTU
0.498350292 6 5 5 1 1 5 0.403966
Error 1.77816 1.31573 1.38756 0.72183 0.69020 0.39755
1.382794305 1.173537
(%) 4 9 5 9 3 6
- Considering the error in efficiency of both real and NTU method,
the gap is from 0.39~1.77%, which is very low, proof the high
accuracy.
Parallel flow Counter flow
Level 4 6 8 10 4 6 8 10
Hot out
(Real) 44 45.6 46.9 48.1 42.8 44.3 46 47.2
(oC)
Hot out
(Simulate 45.34 46.56 47.88 48.71 42.06 44.16 45.74 46.92
) (oC)
Gap (oC) 1.34 0.96 0.98 0.61 0.74 0.14 0.26 0.28
3.0454545 2.10526315 2.08955 1.26819 1.72897 0.31602 0.56521 0.5932
Error (%)
5 8 2 1 2 7 7 2
- The error is small enough to accept, the error caused maybe
because of the disturbance that can’t be controlled during the real
experiment. For more detailed information, please take a look at
the aspen HYSYS file attached.
