Serial number: 6
Chemical Engineering Principles Experiment Report
Name: ×××
Department: School of Chemistry and Chemical Engineering
Class: ×××××× Student number: ×××××× Instructor: ××× Group members: ×××, ×××
Test project name: Convection feeding Determination of thermal coefficient
Experiment date: 2012.10.23 Laboratory: 7101
1. Experiment purpose
1. Observe the condensation of water vapor on the horizontal tube foreign currency Phenomenon; 2. Determine the heat transfer coefficient of forced convection of air in a round straight tube; 3. Master the method of thermal resistance temperature measurement.
2. Experimental Principle
In the sleeve heat exchanger, water vapor is passed through the annulus, and air or water is passed through the inner tube. The water vapor condenses and releases heat to heat the air. Or water, after reaching stability during the heat transfer process, there is the following formula:
V ρC p (t2-t 1) = α0A 0(T-Tw ) m = αi A i (tw -t) m
(T1-T w1)-(T2-T w2)
(T-Tw ) m = T I -T w1T 2-T w2(tw1-t 1)- (tw2-t 2)
(tw -t) m = t w1-t 1t w2-t 2
where:
V — is heated Fluid volume flow rate, m /s;
A 0, A i — heat transfer area of ??the inner tube and outer wall, m; ρ — density of the heated fluid, kg/m;
< p> (T-T w )m—logarithmic average temperature difference between water vapor and outer wall, C; C p—average heat of heated fluid, J/(kgC); (t w -t)m—relative relationship between inner wall and fluid Number average temperature difference;.
.
3
2
3
α0, αi - the flow rate of fluid to the inner wall of the inner tube and the flow rate of water vapor to the outer wall of the inner tube Flow heat transfer coefficient,
W/(m2.C);
.
t 2, t 1—the inlet and outlet temperatures of the heated fluid, C; . T 1, T 2—steam inlet and outlet temperatures, C;
. T w1, T w2, t w1, t w2—the inlet and outlet temperatures on the outer and inner walls, C.
When the thermal conductivity of the inner tube material is very good, that is, the λ value is large and the thickness of the cut tube wall is very thin, it can be considered that T w1= t w1, T w2= tw2, which is the measured value. point wall temperature.
V ρC p (t2-t 1) V ρC p (t2-t 1)
Then: α0 = , αi = A0(T-Tw )m A0(T- Tw )m
If the V, t2, t 1 of the heated fluid, the heat exchange area A 0 or A i of the inner tube, the water vapor temperature T, and the wall temperature T w1, T w2 can be measured , the measured average condensation heat transfer coefficient of water vapor and the measured average condensation heat transfer coefficient of the fluid in the tube can be calculated.
3. Experimental device and process
1. The experimental device is shown in Figure 1.
Figure 1 Water vapor-air heat exchange flow chart
2. Equipment and instrument specifications
(1) Copper tube specifications: diameter φ21×2.8mm , length L = 1000mm (2) Jacket glass tube specifications: diameter φ100×5mm, belly expansion L = 1000mm (3) Pressure gauge specifications: 0—0.1MPa
4. Experimental steps
< p> 1. Turn on the air switch of the main power supply, turn on the power switches of the instrument and inspection instrument, and power on the instrument. 2. Turn on the fan power switch on the instrument panel to let the fan work. At the same time, open valve 4 to allow a certain amount of air to flow into the casing heat exchanger.3. Open valve 1 and be careful to only open it to a certain degree. Opening it too much will cause the steam in the heat exchanger to escape. Opening it too small will cause the steam pressure in the glass tube to accumulate and produce glass. The pipe burst.
4. Before doing the experiment, the condensed water between the steam generator and the experimental device should be removed, otherwise the steam entrained with condensed water will damage the pressure gauge and pressure transmitter.
5. When steam is first introduced, the opening of valve 3 must be carefully adjusted so that the steam slowly flows into the heat exchanger, gradually heats up, and changes from "cold state" to "hot state". It must not Less than 10 minutes to prevent the glass tube from bursting due to sudden heat and pressure.
6. When everything is ready, open the steam inlet valve 3, adjust the steam pressure to 0.01MPa, and keep the steam pressure unchanged.
7. When manually adjusting the air flow, you can change the flow of cold fluid to a certain value by adjusting the air inlet valve 4, and record the experimental value after stabilization; change different flow rates and record the experiments at different flow rates. numerical value.
8. Record 3 to 5 sets of experimental data, complete the experiment, close the steam inlet valve 3 and air inlet valve 4, and turn off the instrument power supply and fan power supply.
9. Turn off the steam generator.
5. Data recording and processing
1 Fill in the experimental data in the following table:
Table 1 Temperature of each temperature measurement point under different flow rates
Flow rate m /h
3
Cold air enters
.
Cold air comes out
.
The inner wall of the tube enters
.
Out of the inner wall of the tube
.
The outer wall of the tube enters
.
The outer wall of the tube enters
.
Port temperature/C port temperature/C port temperature/C port temperature/C port temperature/C port temperature/C port temperature/C
12.0 15.0 18.0 21.0 24.0
40.1 40.9 41.5 41.7 42.2
78.9 78.6 78.0 77.5 77.2
102.2 102.0 101.7 101.9 101.9
98.6 98.4 98.2 98.4 98.3
102.1 101.9 101.6 101.7 101.7
102.3 102.2 101.9 102.0 102.1
2. Calculate the heat transfer coefficient of cold fluid, compare the experimental value with the theoretical value list, calculate the error at each point, and analyze and discuss.
Taking the first set of data as an example
It is known that at 40℃, ρ=1.128 kg/m3 CP =1.005×103 J/(kg ℃)
< p>102.3-102.1.(T-T w )m = C
102.3In 102.178.9-40.1.
(t w -t )m = = 57.33C
78.9In 40.1
α0 =
V ρC p (t2-t 1) 12.0×1.128×1.005×57.33
2. C )
A0(T-Tw )m 1×2×3.14×0.0154×102.20
Theoretical value α= 81.14 W/(m2. C ) Relative error η=
p>81.14-78.91
×100% = 2.75%
81.14
Table 2 Comparison of cold fluid heat transfer coefficients under different flow rates
p>Flow rate m /h 12.0 15.0 18.0 21.0 24.0
3
All results are listed in the table below:
α0/ W/(m2. C )
78.91
99.43 119.94 139.60 159.99
α
Theoretical value
W/(m2. C) Relative error η/%
2.75 4.91 15.44 23.17 28.49
81.14 94.78 103.90 113.34 124.52
As can be seen from the above table, as the cold air flow rate increases, , α0 increases, and as the condensed gas flow rate increases, the measurement error also increases.
3. Determine the correlation expression
N u m
The values ??of constants A and m in 0.4. Pr
Check "Principles of Chemical Engineering, Volume 1" 》It can be obtained: At 40℃, λ=2.756×10-2 W/(m ·℃ )
N u =
α ×d
= 1.629×104×17×10-3/2.746×10-2 = 1.008×104 λ
Check "Principles of Chemical Engineering Volume 1" to get: At 40℃, μ=1.91×10-5Pa·s Re =
ρ ×d ×u
1.132×17×10-3×18.38/1.90×10-5 = 1.862×104
μ
Pr = P r
0.4
C P ×μ
=1.005×103×1.90×10-5/2.746×10-2 = 0.695 λ= 0.695
0.4
= 0.865
N u 440.4 =1.008×10/0.865 = 1.165×10 Pr
Each The set of data are calculated and listed in the following table
Table 3 Nu and Re under different flow rates α0/
23
m /h W/(m.C )
Nu×10-4 Re P r
N u
Pr 0.4, 11650 16030 17640 18840 1.948< /p>
lnRe
In
9.83 10.04 10.14 10.23 10.29
N u
Pr 0.4
< p> 12.0 15.0 18.0 21.0 24.078.91
99.43 119.94 139.60 159.99
1.008 1.387 1.512 1.630 1.687
18620 23 030 25220 27590 2.936
0.695 0.695 0.697 0.696 0.697
9.36 9.68 9.77 9.84 9.88
After calculation, it can be seen that for the heated air in the tube, the Plant's accuracy number Pr does not change much. , can be considered as a constant, P r =
0.695×3+0.696×2+0.697×2+0.698
= 0.6961
8
P r 0.4= 0.69610.4=0.865
According to
N u m
0.4 =ARe, taking the logarithm of both sides at the same time can be obtained: Pr< /p>
N u
ln = mlnRe+lnA
Pr0.4
with
N u
is the ordinate, lnR e is the abscissa, and we get a straight line with slope m and intercept lnA. Draw Pr0.4
9.9
as follows:
N u P r 0. 4
9.8
9.7
9.5
9.4
9.3
lnRe
Figure 2
N u
0.4—lnR e curve Pr
9.6
It can be concluded from the figure: m=0.06276 lnA = 1.36392 A = 3.9115 4. Experimental discussion< /p>
(1). It can be seen from the experimental images that it basically conforms to a linear relationship.
(2). During the experiment, the interval between each group should be maintained at 5 to 6 minutes to ensure that the data obtained is in a relatively stable state.
6. Experimental Error Analysis
1. During the measurement of the heat transfer coefficient of cold fluid, recording the data before changing the flow rate before it is stable will cause a certain error.
2. The instability of cold fluid flow causes errors in readings, which inevitably causes errors when processing data.
3. Changes in temperature cause the density of the cold fluid to change, causing errors in calculations.
7. Questions for thought
1. What impact does the flow direction of cold fluid and steam have on the heat transfer effect in the experiment?
Answer: When the cold fluid and steam flow in parallel, the heat transfer temperature difference is smaller than the heat transfer temperature difference in countercurrent flow. At the same inlet and outlet temperature, the countercurrent heat transfer effect is greater than the parallel heat transfer effect.
2. If steam with different pressures is used for experiments, what impact will it have on the α correlation?
Answer: Basically no impact. Because α∝(ρ2g λ3r/μd0△t)1/4, when the steam pressure increases, both r and △t increase, and other parameters remain unchanged, so (ρ2g λ3r/μd0△t)1/4 does not change much, so It is believed that vapor pressure has no effect on the α correlation.
3. During the experiment, what will be the impact if the condensed water is not drained away in time? How to drain the condensed water in time? Answer: The condensed water is not drained away in time and adheres to the outer wall of the tube, increasing the thermal resistance and reducing the heat transfer rate. Set a drain outlet at the lowest part of the outer pipe to drain away the condensed water in time.
4. In the experiment, is the wall temperature measured close to the vapor side or the cold fluid side? Why? Answer: Close to the steam temperature; because the steam condensation heat transfer film coefficient is much larger than the air film coefficient. 5. During the steam condensation process, if there is non-condensable gas, what impact will it have on heat transfer? What measures should be taken? Answer: Non-condensable gas will reduce the circulation volume of refrigerant and reduce the cooling capacity. And the non-condensable gas will stay in the upper pipe of the condenser, causing the actual condensation area to decrease, the condensation load to increase, the condensation pressure to increase, and the cooling capacity to decrease. Moreover, due to the increase in condensation pressure, the exhaust pressure will increase, which will also reduce the service life of the compressor. The entry of air and the quality of the air should be controlled.
8. Experimental experience
This experiment is to measure the convection heat transfer coefficient. This is the second chemical engineering principle experiment, and it is also a relatively successful experiment. During the experiment, the students cooperated with each other, completed the experimental requirements seriously and patiently, and achieved the expected results.
The shortcomings:
1. In the process of measuring the heat transfer coefficient of cold fluid, everyone patiently waited patiently for each temperature to stabilize before recording the data. When it came time to test the next few groups, I saw that other students had finished the experiments, so I couldn't wait to finish them early, which resulted in larger errors in the subsequent groups of experiments.
2. Perhaps because of the cold weather, students often wandered around the steam generator during the experiment, which was very dangerous.
In short, this experiment is generally satisfactory. Regarding the shortcomings of this experiment, I will correct them in the next experiment and work hard if they are not.