4. To calculate the inside (tube side) film coefficient, hi (W/m².˚C), the following empirical correlation can be used:
Nu = (hi*di)/kf = 0.027* Re 0.8 * Pr 0.33 *(μ/μw)0.14
Once again, assume μ = μw
Thus the equation can be simplified to (hi *di)/kf = (0.027* Re0.8 * Pr1/3 )
Making hi the subject of the formula:
hi = (kf* 0.027* Re0.8 * Pr1/3 )/ di
Where:
-
di = inside tube diameter (m)
- Pr = Prandtl’s number (dimensionless)
- Re = Reynold’s number (dimensionless)
-
kf = (W/m.°C)
5. To calculate the experimental shell side pressure drop across the heat exchanger, simply use the equation:
ΔPshell = ρ*g*Δh
Where:
- ρ = density of fluid flowing through the shell. (kg/m³)
- g = acceleration due to gravity (m/s²)
- Δh = height difference of mercury between the left hand side and the right hand side of the manometer corresponding to the shell side.
Similarly, the experimental tube side pressure drop across the heat exchanger can be found by:
ΔPtube = ρ*g*Δh
Where:
- ρ = density of fluid flowing through the tubes. (kg/m³)
- g = acceleration due to gravity (m/s²)
- Δh = height difference of mercury between the left hand side and the right hand side of the manometer corresponding to the tube side.
6. To calculate the theoretical shell side pressure drop across the heat exchanger, the following empirical correlation can be used:
ΔPshell = 8*jf* (Ds/de)* (L/lb)* (ρ*us² /2)* (μ/μw) -0.14
Where:
-
Ds = shell outside diameter (m)
-
de = equivalent diameter (m)
- L = length of 1 tube (m)
- lb = baffle spacing (m)
- ρ = density of fluid flowing through the shell. (kg/m³)
-
us = velocity of flowing fluid through the shell (m/s)
- Assume μ = μw
-
jf = shell side friction factor obtained from figure D4 (see appendix) for the given baffle cut percentage.
7. To calculate the theoretical tube side pressure drop across the heat exchanger, the following empirical correlation can be used:
ΔPtube = Np [8*jf* (L/di)* (μ/μw) -m + 2.5]* (ρ*ut² /2)
Where:
- Np = number of tube passes
- L = length of 1 tube (m)
-
di = tube inside diameter (m)
-
Assume μ = μw
- ρ = density of fluid flowing through the tubes (kg/m³)
-
ut = velocity of fluid flowing through the tubes (m/s)
Sample Calculations (for Run 1)
Mean Temperature: shell side
Tmean= Tinlet+T(outlet)2
= 21.1+32.82
=29.65 OC
Specific Heat Capacity of Water:
4.173-4.17932.22-26.67=Y-4.17929.65-26.67
Y= 4.176 KJ/Kg oC
Density of water:
994.9-995.832.22-26.67=x-995.829.65-26.67
X= 995.32 kg/m3
Heat transfer (shell):
Q= V x ρ x Cp x ∆ T
Q=4.176 x 0.995 x 0.26(32.8-21.2)
=12.53 kW
Experimental overall heat transfer co-efficient:
Area : A=π x 24 x 0.015x1.95
A= 2.205 m2
Log mean temperature: △Tlm=T1- t2-T2- t1ln(T1- t2)T2- t1
△Tlm=41.8-32.8-(36.4-21.1)ln(41.8-32.8)/(36.4-21.1)
= 11.87 oC
Q=U x A x ΔTm
Thus, U = Q/ A x ΔTm
=12.53/ (2.205 x 11.87)
=478.73 W/m2 oC
Shell side Heat Transfer Coefficient:
Re = G x De/µ
G= total massarea= 995.32*0.260.003695*1000 = 70.05 kg.m-2.s-1
µ was interpolated from table c-2 : properties of water (Holman, 2002)
Re= 70.05*0.0081/8.09*10-4
Re= 701.37
Pr= Cp x µ /Kf
Pr= 4.176*103*8.09*10-4/0.619
Pr= 5.46
Nu= h x De /Kf = jh x Re x Pr1/3 x(µ/µw)0.14
=0.02 * 701.37*(5.46)0.33
Nu = 24.70
Thus, heat transfer coefficient (h) = Nu x Kf/De
24.70*0.6190.00810 = hs= 1887.62 w/m2 OC
Tube side
Tmean= 41.8+36.42
Tmean = 39.10 0C
Specific Heat Capacity of Water:
x-4.17439.1-37.78=4.174-4.17443.33-37.78
x = 4.174 KJ/Kg0C
Density of water:
x-99339.1-37.78=990.6-99343.33-37.78
x=992.43Kg/m3
Heat transfer (Tube):
Q= V x ρ x Cp x ∆ T
0.65 x992.43x4.174(41.8-36.4)1000
=14.54 KW
Area : A=π x 24 x 0.014x1.95
= 2.058m2
Log mean temperature: △Tlm=T1- t2-T2- t1ln(T1- t2)T2- t1
△Tlm=41.8-32.8-(36.4-21.1)ln(41.8-32.8)/(36.4-21.1)
= 11.87 oC
Q=U x A x ΔTm
Thus, U = Q/ A x ΔTm
=14.54(1000)/ (2.058 x 11.87)
=595.21 W/m2 oC
tube side Heat Transfer Coefficient:
Re = G x De/µ
G= total massarea= 992.43*0.650.003*1000 = 215.03 kg.m-2.s-1
µ was interpolated from table c-2 : properties of water (Holman, 2002)
Re= 215.03 *0.0081/6.66*10-4
Re= 2615.23
Pr= Cp x µ /Kf
Pr=4.174 *103*6.66*10-4/0.632
Pr= 4.40
Nu= h x De /Kf = jh x Re x Pr0.33 x(µ/µw)0.14
=0.0095 *2615.23*(4.40)0.33
Nu = 40.51
Thus, heat transfer coefficient hi = Nu x Kf/De
hi = (40.51*0.632)/0.00810
= hi= 3160.78 w/m2 OC
Theoretical Overall Heat-Transfer Coefficient:
h0= 1887.62 w/m2OC
hi= 3160.78 w/m2 OC
hid= 5680 w/m2 OC
d0= 0.015 m
di= 0.014 m
kw= 377 w/m OC
UO = 955.86 w/m2 OC
Experimental Pressure Drop (shell and tube side):
ΔPshell = ρ x g x Δh
ΔPshell = 0.995 x 1000 x 9.81 x (6 / 100)
=585.66 Pa
ΔPtubes= 0.992*9.81*0.21*1000
= 2043.62 pa
Shell side Theoretical Pressure Drop:
ΔP = 8 x jf x (Ds/ de) x (L / lb) x (ρ x Us2 /2) x (µ/µw)0.14
=8 x 0.089 x (0.16/0.0081) x (1.95/0.1818) x (995.32 x 0.07042/2)
= 372.08 Pa
Tube side Theoretical Pressure Drop:
ΔP =Np x (8 x jf x (L /di) x (µ/µw)-m + 2.5) x (ρ x Ut2 /2)
=(8 x 0.068 x 1.95/0.015 +2.5) x (992.43 x 0.2172/2)
=1710.88 Pa
Results
Mean Temperature:
Specific Heat Capacity of Water:
Density of water:
Heat transfer (shell):
Experimental overall heat transfer co-efficient:
Shell side Heat Transfer Coefficient:
Tube side Heat Transfer Coefficient:
Theoretical Overall Heat-Transfer Coefficient:
Experimental Pressure Drop (shell and tube side):
Shell side Theoretical Pressure Drop:
Tube side Theoretical Pressure Drop:
Conclusion
-
In accordance with the 2nd Law of Thermodynamics, heat was indeed transferred from the hot water in the tubes to the cold water is the shell.
- With counter current flow, it was noted in the raw data that the temperature decreased from T8 (tube inlet) to T7 (tube outlet) and the temperature increased from T6 (shell inlet) to T1 (shell outlet).
- With the heat exchange noted above, the data and calculations also reflected that heat lost in the tubes were not equivalent to the heat gained by the water in the shell. Some of this heat was lost to the surrounding environment due to poor insulation of the heat exchanger.
- The amount of heat lost/gained also decreased with increasing flow rate. This is because an increase in flow rate, essentially, decreases the amount of contact time between fluid and the heat transfer area – resulting in less heat being transferred.
- The use of the log mean temperature difference is employed rather as it provides a linear relation for the temperature profile across the heat exchanger.
- The percentage errors between the heat transfer coefficients obtained experimentally and those obtained with correlations are very high. This is owing to the considerable differences in the 2 methods.
-
The use of correlations is time consuming and tedious, however, they do account for the pressure/temperature dependent properties such as density, viscosity, thermal conductivity.
- It is crucial to note that the correlations are not specific to the heat exchanger used and are general correlations, thus experimental analysis (which is of greater pertinence to the actual, specific system) is often more reliable.
- The inclusion of baffles leads to increased heat transfer, due to greater contact time between heat transfer area and turbulent fluid in the shell.
- Irregular data (as thus, irregular results) can be contributed to the fluctuating hot water flow rates throughout the experiment.
Recommendations
- Manometers should be inspected and air must be removed prior to the commencement of the practical.
- Insulation for the heat exchanger should be considered to decrease the heat losses to the surrounding environment.
Errors
- A log mean temperature must be used in the correlations, since the change in temperature across the heat exchanger is not constant. This increases accuracy.
- Leaks, lack of insulation and the presence of air bubbles are possible sources of error.
- Heat was lost to the surrounding environment due to poor insulation of the heat exchanger and could have affected the accuracy of the results.
- Air bubbles that aren’t bled out of the system decreases accuracy.
- Accuracy and reliability depend on manometer quality.
by
skzn (student)
