Ecole Centrale De Nantes

Master 2
Energy and Propulsion
2021-2022

Boiler and Heat Exchanger

By:
Rami HAJAR
Charbel SARRAF
Mohammed MASHAQI
Fabrizio SAVA

Presented to: Prof. Xavier TAUZIA

Contents
A. Introduction .......................................................................................................................................... 1
B. Boiler Study ........................................................................................................................................... 1
1. Combustion using Gas propane ....................................................................................................... 1
1.1 Energy Balance for Propane gas : ......................................................................................... 1
1.2 Domestic Boiler efficiency using gas propane: .................................................................... 2
2. Combustion using Liquid fuel........................................................................................................... 4
2.1 Boiler performance using Liquid fuel: .................................................................................. 6
2.2 Flame color ........................................................................................................................... 6
2.3 Exhaust Products .................................................................................................................. 7
3. Conclusion ........................................................................................................................................ 7
C. Heat Exchanger Study ........................................................................................................................... 8
1. Measurements ................................................................................................................................. 8
2. Co-current heat exchanger .............................................................................................................. 8
3. Heat transfer with counter-current flow ......................................................................................... 9
4. Discussion regarding heat transfer ................................................................................................ 10
A. Introduction
Boilers and heat exchangers are widely used in energy production (e.g. Thermal power station) and in the
building construction (fluid heating). The objective of this experiment is to study the boiler for two different fuels,
gas propane and liquid fuel, which is not known in composition, but through the CAPELEC gas analyzer we can find
the C/H ratio for this unknown fuel and compare it with propane gas as well. Different studies could be made to the
existing boiler like NOx production for all conditions (Lean, rich, stoichiometric). And since the efficiency of the boiler
is a function of many variables with their uncertainties, the error of the efficiency has been calculated numerically
by following the propagation of the uncertainty by using Mathematica Wolfram with (68% CL) (Confidence Limit).
The efficiency of the boiler using propane gas was of order of 80% and it is of order of 70% for liquid fuel. And the
difference between this efficiency and the usual domestic boilers efficiency is some losses could be discussed.
Moreover, a study for counter-current and co-current flows for heat exchanger could be made during this
experiment.

B. Boiler Study
1. Combustion using Gas propane
To find (A/F) Stoichiometric, we can write the stoichiometric reaction for the propane gas as follows:

𝐶𝐶3 𝐻𝐻8 + 5(𝑂𝑂2 + 3.76𝑁𝑁2 ) → 3𝐶𝐶𝑂𝑂2 + 4𝐻𝐻2 𝑂𝑂 + 18.8𝑁𝑁2

So, the (A/F)st in mass basis can be calculated as follows:

𝐴𝐴 𝑚𝑚𝑎𝑎𝑎𝑎𝑎𝑎 5(32 + 3.76 ∗ 28)
� � = = = 15.6
𝐹𝐹 𝑠𝑠𝑠𝑠 𝑚𝑚𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 (12 ∗ 3) + 8

Knowing the (A/F)st we can calculate the mass flow rate of air needed to burn 1.5g/s of propane as follows:
𝐴𝐴 𝑚𝑚̇𝑎𝑎𝑎𝑎𝑎𝑎
� � 𝑘𝑘𝑘𝑘
𝐹𝐹
λ=1= = 15.6 → 𝑚𝑚̇𝑎𝑎𝑎𝑎𝑎𝑎 = 84.24
𝐴𝐴 15.6 ℎ
� �
𝐹𝐹 𝑠𝑠𝑠𝑠
Dry volume concentration of CO2 in the exhaust can be calculated as follows:
3
[𝐶𝐶𝑂𝑂2 ]𝑑𝑑𝑑𝑑𝑑𝑑 = = 13.76%
18.8 + 3
1.1 Energy Balance for Propane gas :
To perform energy balance, we make a power balance of the boiler between the input power and the output power.
The input power is the amount of energy in the fuel, and the output power is the heat used to heat up the burnt
gases and the amount of heat absorbed by the water circuit due to combustion.

Input power = Total output power

Input power = Heat absorbed by water circuit + Amount of Heat-to-heat flu gases
���
𝑚𝑚𝑓𝑓 × 𝐿𝐿𝐿𝐿𝐿𝐿 = �𝑚𝑚̇𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 × 𝐶𝐶𝑝𝑝𝑝𝑝 × ∆𝑇𝑇𝑤𝑤 � + �𝑚𝑚̇𝑎𝑎+𝑓𝑓 × 𝐶𝐶𝑝𝑝 × (𝑇𝑇𝑒𝑒𝑒𝑒ℎ − 𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎 )�…………………………….Eq(1)

The previous equation can be taken for simplifications.

Where

The ambient temperature (Tamb=16 °C)

Cpw: specific heat capacity for water at 20°C = 4.18 kJ/kg.K (and this value it will not change a lot at different
temperatures).

1
���
𝐶𝐶𝑝𝑝 =average specific heat capacity of products of the combustion, and we considered it as it was for air at T=620°C,
���
and so, 𝐶𝐶 𝑝𝑝 =1.008 kJ/kg.K. (assumption).

And after calculating doing the energy balance, it is observed from calculations in Table 1(next page), that sometimes
the balance is balanced and sometimes not balanced, and that’s because of the losses and the uncertainties of all
variables that measured. And also, the assumptions for Cpw and Cp of product gases.

1.2 Domestic Boiler efficiency using gas propane :

The domestic boiler efficiency can be calculated as the ratio between the useful energy needed to heat up the water
and the amount of energy available in the fuel, and that’s done by using the following formula:
𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 �𝑚𝑚̇𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 ×𝐶𝐶𝑝𝑝𝑝𝑝 ×∆𝑇𝑇𝑤𝑤 �
𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 = = …………………………….Eq(2)
𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑚𝑚𝑓𝑓 ×𝐿𝐿𝐿𝐿𝐿𝐿

The data we got for Gas propane, and by considering the uncertainty for each instrument used in measuring all the
data, we could calculate the amount of power in fuel (input power), amount of heat absorbed by the water circuit
due to combustion, and the efficiency of our domestic boiler. And Table1.1 below shows that.
Table 1: Propane Gas data measured and calculated power, and boiler efficiency considering all uncertainties.

Gas Propane
Stoich
Stoich Lean Lean approx rich

Twin (°C) 9.5 ± 0.5 9.5 ± 0.5 9 ± 0.5 9 ± 0.5 9 ± 0.5

Twout (°C) 48 ± 1 49 ± 1 47 ± 1 49 ± 1 44 ± 1
m dot water (g/s) 320 ± 3 320 ± 3 320 ± 3 320 ± 3 320 ± 3
Texh (°C) 680 ± 10 660 ± 10 640 ± 10 680 ± 10 620
m dot fuel (g/s) 1.5± 0.1 1.5± 0.1 1.5± 0.1 1.5± 0.1 1.5± 0.1
m dot air (kg/h) 85 ± 3 125 ± 2 155 ± 1 87 ± 3 72.5 ± 3
m dot air (kg/h) 85 125 155 87 72.5
CO(%vol) 1.04 0 0 0.249 4.683
CO2(%vol) 12.7 10.1 7.5 13.2 10.3
HC ppm vol 0 0 0 0 133
O2 (%vol) 0.06 5.1 8.9 0.16 0.06
λ 0.963 1.471 1.824 0.996 0.853
NOX ppm vol 48 44 30 51 40
m dot air (g/s) 23.61 34.72 43.06 24.17 20.14
(A/F) st,gas (A/F) st,
analysis calculated
15.7407 15.6
69525 ± 69525 ± 69525 ±
Input power[W] 4635 69525 ± 4635 69525 ± 4635 4635 4635
Amount of Heat to heat flu gases
[W] 16807 ± 304 23513 ± 384 28025 ± 456 17179± 308 13174 ± 266
51497 ± 53504 ± 46816 ±
Heat absorbed by water circuit [W] 1571 52835 ± 1575 50828± 1569 1577 1558
68304 ± 70683 ± 59990 ±
Total output power[W] 1600 76348 ± 1621 78853 ± 1634 1607 1581
Boiler Efficiency 0.74± 0.05 0.76 ± 0.05 0.73 ± 0.05 0.77 ± 0.05 0.67 ± 0.05

We could calculate the (A/F)st and it was 15.6 by using chemical reaction equation, and it is 15.74 by flowrates, and
the error is just 0.74, and that’s because the error in reading the flowrates.

2
And since the efficiency is a function of many variables with their uncertainties, the error of the efficiency has been
calculated numerically by following the propagation of the uncertainty by using Wolfram Mathematica with (68%
CL). Figure 1 shows the boiler efficiency as function of lambda
λ = 1, the max efficiency is

Figure 1: Boiler efficiency as function of 𝜆𝜆 (excess of air) with 68% CL using Gas propane.

We can note from Figure 1 that the maximum efficiency of the boiler happens at the stoichiometric at has a value of
0.77%. and if we look at table1, we found Texh at its maximum value also at the stoichiometric condition (Texh=680°C).
And that’s why the efficiency of the boiler has the maximum value. And because we had an uncertainty in reading
measurements, the efficiency at the stoichiometric could be between 72% and 83%.

When the mixture is lean where ( λ > 1), the efficiency of boiler is decreasing, as the excess of air reduces the
max. temperature after combustion, and when the mixture is rich at ( λ < 1), the efficiency also decreasing.
So the maximum efficiency happens at the stoichiometric condition when we just the enough amount of air
to burn the all fuel. More air can decrease the final temperature, and more fuel means that we don’t have
enough oxidizer(air) to burn the fuel.
And the following shape illustrate that the final temperature happens just at the stoichiometric.

And After performing the energy balance using Eq(1), we found after calculation(in Table1) that the input power is
in the range of total output power, but this balance has some errors and differences due to the uncertainty in reading
measurements, and as we see in Table1, for example, at the stoichiometric we have total output power of 68304 ±

3
1600 W, and input power is 69525 ± 4635, and we think that the differences also because the losses going to outside
the walls of the boiler as the boiler is not insulated.

Moreover, in the lab the sensor we to measure the flame temperature is not directly at the flame position, but it was
a little bit farther of 1 m, So this is also a loss!

2. Combustion using Liquid fuel

The same analysis we did to liquid fuel, but this time we don’t know the formula of the fuel. But we can compute
the C/H ratio, as in the chemical reaction this the most important parameter concerns the excess of air ratio(λ) or
the equivalence ratio (∅) . And we don’t care about how many atoms of carbon or hydrogen. Just we look at the
ratio.

Depending on Eq1 and Eq2, the efficiency of the boiler could calculated, and (A/F)st found to be 14.56 (calculated
with mass flows when we adjust the λ to be 1 in gas analysis during the experiment. And Table2 shows all measured
values and calculated ones considering the uncertainties.

Table 2: Liquid flued data measured and calculated power, and boiler efficiency considering all uncertainties.

Liquid Fuel
Lean rich Lean Lean Lean Stoich
Twin(°C) 11.5 ± 0.5 10.5 ± 0.5 10 ± 0.5 10 ± 0.5 9 ± 0.5 8 ± 0.5
Twout(°C) 43 ± 1 40 ± 1 40.5 ± 1 35 ± 1 32 ± 1 48 ± 1
m dot water (g/s) 325 ± 3 330 ± 3 325 ± 3 325 ± 3 325 ± 3 300 ± 3
Texh(°C) 663 ± 10 570± 10 625± 10 590± 10 560± 10 680± 10
mf [g/s] 1.48± 0.1 1.46± 0.1 1.5± 0.1 1.5± 0.1 1.5± 0.1 1.5± 0.1
m dot air [kg/h]
m dot air [kg/h] 81± 3 67± 2 125± 1 160± 3 179± 3 80 ± 3
CO2 [%] 13.1 11.1 8.9 6.3 5 14.8
CO [%] 0 5.253 0 0.061 0.505 0.302
O2[%] 2.22 0.02 8.1 11.8 13 0.42
HC [ppm] 0 169 0 0 28 0
NOx [ppm] 85 50 44 15 9 87
λ 1.12 0.870 1.589 2.034 2.276 1.017
ma[g/s] 22.50 18.611 34.722 44.444 49.722 22.222
(A/F) st
14.567
68945 ± 68076 ± 69525 ± 69525 ± 69525 ± 69525 ±
Input power[W] 4635 4635 4635 4635 4635 4635
Heat release by combustion to flu gas 15644 ± 11213 ± 22235 ± 26583 ± 28087 ± 15877 ±
[W] 292 226 373 485 534 292
42792 ± 40692 ± 41434 ± 33962 ± 31245 ± 50160 ±
Heat absorbed by water circuit [W] 1569 1585 1566 1550 1545 1489
58436 ± 51905 ± 63670 ± 60545 ± 59333 ± 66037 ±
Total output power[W] 1596 1602 1610 1625 1635 1517
0.59 ± 0.45 ±
Efficiency 0.62 ± 0.04 0.59 ± 0.04 0.045 0.48 ± 0.04 0.037 0.72 ± 0.05

To find the C/H ratio, we assume the fuel CxHy, and at stoichemetric condition (approximately) the products
composition from Table2 is as follows in, assuming the NOx is NO2

4
 Table 3: dry concentration at stocihemtric condition for Liquid fuel.

CO (%) CO2(%) HC(ppm) O2 (%) NOx(ppm)

0.302 14.8 0 0.42 8.7×10-5

By substituting the values in the chemical reaction, we got

𝐶𝐶𝑥𝑥 𝐻𝐻𝑦𝑦 + 𝑑𝑑 (𝑂𝑂2 + 3.76𝑁𝑁2 ) → 0,148𝐶𝐶𝐶𝐶2 + 0.00302𝐶𝐶𝐶𝐶 + 0,0042𝑂𝑂2 + (8,7 × 10−5 )𝑁𝑁𝑁𝑁2 + 𝑎𝑎𝐻𝐻2 𝑂𝑂 + 𝑏𝑏𝑁𝑁2

As the analysis of the gas done without water(dry concentrations in Table 3) as the water extracted from products
before entering the gas analyzer , so the concentration of N2 from dry products can be calculated as follows:

To find b:

𝑁𝑁2 (%𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣) = 𝑏𝑏 = 1 − (%𝐶𝐶𝐶𝐶 + %𝐶𝐶𝐶𝐶 + %𝑂𝑂2 + %𝑁𝑁𝑂𝑂𝑥𝑥 )

= 1 − (0.00302 + 0.148 + 0.0042 + 8.7 × 10−5 )

𝑏𝑏 = 0.848893

To find (x):

Carbon balance:

x=0.148 +0.00302 x= 0.15102

To find (a):

H balance :

y= 2a  a=y/2

To find (d) :

Oxygen balance :

2d = (0.148*2) + 0.00302 + (0.0042*2) + (2 * 8,7 × 10−5 ) + y/2  d = 0.153797 + y/4

The chemical reaction becomes

𝑦𝑦
𝐶𝐶0.15102 𝐻𝐻𝑦𝑦 + (0.153797 + )(𝑂𝑂2 + 3.76𝑁𝑁2 )
4
𝑦𝑦
→ 0,148𝐶𝐶𝐶𝐶2 + 0.00302𝐶𝐶𝐶𝐶 + 0,0042𝑂𝑂2 + (8,7 × 10−5 )𝑁𝑁𝑁𝑁2 + 𝐻𝐻2 𝑂𝑂 + 0.848893𝑁𝑁2
2
To find (y) :

Nitrogen balance:

(0.153797+y/4)*2*3.76 = (8,7 × 10−5 )+ (2*0.848893) y=0.2879

So C/H = x/y = 0.15102/0.2879 = 0.524

5
2.1 Boiler performance using Liquid fuel:
As before, by considering the uncertainties with 68% CL, the efficiency curve as function of lambda coefficient is
shown in Figure 2.

Figure 2: Boiler efficiency as function of λ (excess of air) with 68% CL, using liquid fuel.

We can note from Figure 2, that as before the maximum of domestic boiler efficiency is at its maximum when λ =
1, at stoichiometric condition, and it decreases when the mixture is lean and rich.
Moreover, we can note by comparing with gas propane, the order of efficiency of the boiler using gas
propane is more than when using liquid fuel. The maximum efficiency using liquid fuel is around 75%.
And that’s because of losses, and we know the efficiency for domestic boiler is of order 95% in life. But here is less
because of the losses in the pipes and wall of boiler by convection and conduction heat transfer, and instruments
errors.

2.2 Flame color

We could also note the color of the flame in rich and lean mixtures, and we took photos as below

Flame in Rich Flame in Lean

We note from those images that in Lean the color is more brighter, and in rich the flame is more dark, because in
rich condition there is more soot, and we can note that also from the smell of excess HC in rich.

6
2.3 Exhaust Products
From Table 1 and 2, it is obvious that when we are in stoichiometric condition, the amount of CO2 is maximum,
and CO is minimum, and also there is no oxygen in products, where in lean the oxygen becomes appear.

Moreover, the NOx are maximum at the stoichiometric because the NOx depends on the temperature of
exhaust gases, and the temperature is maximum at the stoichiometric, and in rich mixture and lean, the
temperature decreases, so the NOx decreases. And that’s shown Figure 3 below.

NOx ppm Vs λ
Liquid fuel
100

80
NOx [ppm]

60

40

20

0
0.00 0.50 1.00 1.50 2.00 2.50
λ

Figure 3: NOx in ppm as function of excess of air for liquid fuel.

3. Conclusion
After studying two types of fuels for domestic boiler. We conclude that boiler uses propane gas is more efficient
than boiler uses liquid fuel. So it is better to use propane gas. And that’s because the C/H ratio. C/H for propane is
3/8 = 0.375 , so we have less percentage of hydrogen that the liquid fuel(C/H=0.524). so increasing the hydrogen
content in the fuel will increase the efficiency of combustion and the amount of heat release. So less C/H for the fuel
more efficient heat release.

Also, the efficiency for both fuels was underestimated, because of the losses went to wall of the boiler and escaped
to outside and the amount of heat went to heat the flue gases that we didn’t use. And also because the uncertainties
in the instruments used for the measurements. And also unexpected losses could be exist.

7
C. Heat Exchanger Study
From the mesuremenets with different cases, we get these values for co-current and counter Current (Tables
below)

1. Measurements
Co-Current:

case Q_c(L/min Q_H(L/min T_H Q_C Q_H T_H,ou T_H,in T_C,in T_C,out
) ) t
1 1.25 1.25 30 1.24 1.25 27.6 30 14.4 17.7
2 1.25 2.5 30 1.24 2.52 28.2 29.8 14.3 18.5
3 2.5 1.25 30 2.53 1.25 27 29.9 14 16.1
4 2.5 1.25 50 2.49 1.26 41.4 53 13.2 18.3
5 1.25 2.5 50 1.26 2.51 45.4 50.8 12.4 23
6 1.25 1.25 50 1.26 1.24 43.4 51.1 12.7 21.2
Counter-Current:

Case QC(L/min) QH(L/min) TH (°C) QC QH TH-in TH-out TC-in TC-out

1 1.25 1.25 30 1.25 1.25 29.9 26.1 11.7 15.3
2 1.25 2.5 30 1.25 2.5 29.8 27 11.7 16.5
3 2.5 1.25 30 2.5 1.25 30 25.5 11.5 13.9
4 2.5 1.25 50 2.5 1.25 50 39.9 11.3 16.3
5 1.25 2.5 50 1.25 2.5 50 44.6 11.4 22.1
6 1.25 1.25 50 1.25 1.25 50 41.6 11.4 19.7

2. Co-current heat exchanger

We have water properties C = 4180 J/kg-1. K-1 and 𝜌𝜌= 1000 kg.m-3 and Pa = (PH + PC)/2.

Heat flow of the hot side can be expressed by:

Q
PH =m wH ⋅ C pmean ⋅ (THin − THout ) = H ⋅ C pmean ⋅ (THin − THout )
60
On the other side, the heat flow on the cold side can be expressed by:
Q
PC =m wC ⋅ C pmean ⋅ (TCout − TCin ) = C ⋅ C pmean ⋅ (TCout − TCin )
60
So the mean Heat flow will be equal to:

Pm =
(φH + φC )
2
Then making use of LMTD method we computed the experimental global heat exchange coefficient.
(T − T ) − (THout − TCout )
∆TLn =Hin Cin
T −T
Ln Hin Cin
THout − TCout
S = π de L
Pm
U=
∆TLn ⋅ S

8
 Then the efficiency of the heat exchanger can be expressed by:
Pa
E=
Pmax
Where
Pmax Cmin (THin − TCin )
=
Results:
P_H P_C P_a delta t ln S U C_c C hot C min P max Efficiency
209 285.076 247.038 12.53474 0.019697786 1000.532 86.38667 87.08333 86.38667 1347.632 0.183313
280.896 362.824 321.86 12.37428 0.019697786 1320.473 86.38667 175.56 86.38667 1338.993 0.240375
252.5417 370.139 311.3403 13.24306 0.019697786 1193.52 176.2567 87.08333 87.08333 1384.625 0.224855
1018.248 884.697 951.4725 30.69659 0.019697786 1573.579 173.47 87.78 87.78 3493.644 0.272344
944.262 930.468 937.365 29.68479 0.019697786 1603.088 87.78 174.8633 87.78 3370.752 0.278088
665.1773 746.13 705.6537 29.56392 0.019697786 1211.747 87.78 86.38667 86.38667 3317.248 0.212723

3. Heat transfer with counter-current flow

Same as the co-current heat exchanger, we used the same formulas to compute the heat flow
Q
PH =m wH ⋅ C pmean ⋅ (THin − THout ) = H ⋅ C pmean ⋅ (THin − THout )
60
Q
PC =m wC ⋅ C pmean ⋅ (TCout − TCin ) = C ⋅ C pmean ⋅ (TCout − TCin )
60
(φ + φ )
Pm = H C
2
In Counter current Heat exchanger, ΔTLN is expressed by:
(T − T ) − (THout − TCin )
∆TLn =Hin Cout
T −T
Ln Hin Cout
THout − TCin
S = π de L
Pm
U=
∆TLn ⋅ S
Then the efficiency of the heat exchanger can be derived by using the formula 14.
Pa
E=
Pmax
Results:
P_h P_c P_a delta ln S U C_c C_hot Cmin Pmax Efficiency
330.9167 313.5 322.2083 14.49977 0.019697786 1128.128 87.08333 87.08333 87.08333 1584.917 0.203297
452.8333 418 435.4167 14.37195 0.019697786 1538.056 87.08333 174.1667 87.08333 1576.208 0.276243
391.875 418 404.9375 15.02555 0.019697786 1368.17 174.1667 87.08333 87.08333 1611.042 0.251351
879.5417 870.8333 875.1875 31.08029 0.019697786 1429.548 174.1667 87.08333 87.08333 3370.125 0.25969
940.5 931.7917 936.1458 30.47322 0.019697786 1559.58 87.08333 174.1667 87.08333 3361.417 0.278497
731.5 722.7917 727.1458 30.24997 0.019697786 1220.335 87.08333 87.08333 87.08333 3361.417 0.216321

9
4. Discussion regarding heat transfer
• To compare the influence of hot and cold flow rate on some parameters we should compare cases with fixed
TCin and THin. so we do compare cases (1,2,3) or cases (4,5,6).

Thermal power exchanged is higher when we have a difference between cold flow and hot flow rate.

And it has the highest value when the higher flow rate multiplies the higher ∆T. And this is due to the difference
between the inlet and outlet temperature. ∆T (hot) =THin-THout ,

∆T (cold)=TCout –TCin.

The global heat exchange coefficient and the heat exchanger efficiency varied the same way as the power
exchanged.it have same variation law according to flow variation.

• To compare the influence of hot water temperature on some parameters we should compare cases with fixed
hot and cold flow rates. So we do compare cases (1, 6) or cases (2, 5) or cases (3, 4).

It’s obvious that in the three cases that the power exchanged, the global heat exchange coefficient and the heat
exchanger efficiency increase when the hot water inlet temperature increase. Because we increased the temperature
difference between cold and hot water.

• Counter flow heat exchangers are inherently more efficient than co current heat exchangers because they
create a more uniform temperature difference between the fluids, over the entire length of the fluid path. The
power exchanged and the global heat exchange coefficient are higher in counter-current flow than the co-
current flow.

• The difference between Ph and Pc is due to the thermal power dissipated to the ambient.

Not all the heat lost by the hot flow is gained by the cold flow but some of it is dissipated to the ambient.

The thermal power dissipated = Ph -Pc.

• Theoretical Global heat exchange coefficient is higher than the real one, because of fouling.

10