Experiment Is Performed To Determine The Friction Loss Biology Essay
In this practical, experiment is performed to find the clash loss in consecutive pipes with unvarying cross sectional country. On the other manus, the mirror loss due to pipe adjustments, like valves, articulations and decompression sicknesss is besides analyzed.
Besides, the usage of venture metre to mensurate the flow rate is learnt.
In order to analyze the heat transportation of heat money changers, two type flow agreement, analogue and counter flow are constructed and different speed of the flow rate is operated to detect the affect of the public presentation of a heat money changer. The temperature informations are collected during the experiment and computation and secret plans based on the informations are obtained. The consequences enable the experimenter to analyze the different efficiency when runing the heat money changer at a peculiar agreement and flow rate.
When a fluid flows through a pipe, the fluid atoms near the pipe wall have comparative low speed and the one near the lope move with comparative high speed. Because of this comparative gesture and the viscousness of the fluid, shear emphasis are produced. This syrupy action causes dissipation, which is usually referred as pipe clash loss.In order to analyze friction loss of flow in pipe, the Bernoulli ‘s equation Sing the two surfaces normal to the way of the fluid flow, an overall energy balance equation that governs the fluid flow can be derived:Equation 1: Overall Energy Balance for incompressible fluidwhere each term has the unit of Joule/kg ( m2s-2 ) .The major caput loss was contributed by the frictional effects that arose due to fluid flow and intermolecular forces of attractive force, whereas the minor caput loss occurred as a consequence of other physical obstructors to the flow. Examples are the presence of opening home bases, Venturi metres, decompression sicknesss, flux control valves, pipe adjustments, and so on.
The major and minor caput loss are represented by the undermentioned equations:Equation 2: Major caput loss Equation 3: Minor caput lossThe clash factor, degree Fahrenheit, in Equation 2 is mostly dependent on the flow government, and therefore, dependant on the Reynolds figure. In add-on, for turbulent flow, degree Fahrenheit besides becomes dependent on the comparative raggedness Iµ/D for unsmooth pipes.Equation 4: Clash factor for laminar flowEquation 5: Colebrook ‘s equation for clash factor in disruptive flow Equation 6: Blasius expression for clash factor for smooth pipesFor this experiment, appropriate equations from Equation 2 to 6 were chosen to unite with the overall energy balance equation, which were reduced consequently, for rating of the specified measures.
4. Equipment and Materials
A H2O to H2O turbulent flow concentric tubing heat money changer manufactured by P. A. Hilton Ltd. is used for the probe.
The heat money changer is a dual pipe type with hot H2O fluxing through the cardinal tubing while chilling H2O flows through the annulate infinite.A warmer ( electrical opposition type ) is used to heat up hot H2O and the H2O cools as it flows through the heat money changer, and on go forthing base on ballss through a pump which provides the go arounding caput. Then, it passes the warmer once more to acquire reheat. On the other manus, cold H2O base on ballss through a flow control valve and flow metre to the unit. Two way control valves can trade the chilling H2O inlet/outlet of the money changer. Therefore, the flow can be either a analogue or antagonistic way.
The below shows the exposure of the operating unit:
Photo – Water to H2O turbulent flow concentric tubing heat money changer
The heat money changer was connected in parallel flow constellation. The H2O degree in the warmer armored combat vehicle was checked at the right degree. The whole unit was turned on and the hot H2O flow valve was to the full unfastened. These stairss were done by the lab staff. The undermentioned stairss are the stairss that experimenters performed during lab:The hot H2O flow rate was reduced to 50g/s and the cold H2O flow rate was set around 15g/s.After waiting the system to be stabilized, the temperature from T1 to T10 and flow rates were recorded.The chilling H2O way control valve was switched to “ counter flow ” place instantly.
Then, step 2 was repeated.Fully opened the hot H2O flow valve.Adjusted the cold H2O to convey the average hot H2O temperature, [ T3 +T6 ] /2 to around 65°C. Then, step 2 was repeated.
The hot H2O flow rate was reduced to 80 % of the initial value without altering the cold H2O flow rate.Adjusted the warmer control to convey the average hot H2O temperature, [ T3 +T6 ] /2 to around 65°C. Then, step 2 was repeated.Repeated measure 6 and 7 with hot H2O flow rate about 60 % , 40 % and 20 % of the initial value.
The consequences of the experiment are tabulated below in Table 1 and Table 2:TrialParallel FlowCounter FlowMetal wall at recess, T1 ( °C )57.767.7Metal wall at issue, T2 ( °C )57.345.8Hot watercourse at recess, T3 ( °C )70.370.
7Hot watercourse 1stintermediate, T4 ( °C )65.468.0Hot watercourse 2ndintermediate, T5 ( °C )62.864.
7Hot watercourse at issue, T6 ( °C )61.059.9Cold watercourse entry/exit, T7 ( °C )22.455.6Cold watercourse intermediate, T8 ( °C )38.647.4Cold watercourse intermediate, T9 ( °C )47.637.
2Cold watercourse entry/exit, T10 ( °C )52.824.6Hot H2O indicated flow, V ( L/min )33Water denseness at hot H2O recess, I? ( kg/m3 )977.5977Hot H2O existent flow rate, mh ( Kg/s )0.
05170.0516Mean hot H2O temperature, ( T3+T6 ) /2 ( °C )65.765.
3Cooling H2O flow rate, megahertz ( kg/s )0.0150.015Heat transportation from hot H2O, Qh ( W )2009.
82329.4Heat transportation to cold H2O, Qc ( W )1906.11943.7Log Mean Temp Diff, I?ln22.423.8Overall heat transportation coefficient, U ( W/m2K )3115.43398.4
Table Parallel and Counter flow
Sample Calculation for Parallel Flow:Hot H2O existent flow rate, mhActual rate ( L/min ) = indicated Flow + T6 ( °C ) x 0.
0041 – 0.0796= 3 + 61 x 0.0041 – 0.0796= 3.17053.1705 L/min = 3.
1705 L/min x ( m3 /1000L ) ten ( 1min / 60s )= 5.284 ten 10-5 m3/smh = 5.284 ten 10-5 m3/s x 977.5 kg/m3= 0.0517 kg/sHeat transportation from hot H2O, QhQh = mh x Cp x ( T3 – T6 )Qh = 0.0517 kg/s x 4.
18kJ/kg.K x ( 70.3 – 61.0 )= 2.0098 kJ/s= 2009.8 WHeat transportation to cold H2O, QcQc = megahertz x Cp x ( T7 – T10 )= 0.
015 kg/s x 4.18kJ/kg.K x |22.4 – 52.8|= 1.9061 kJ/s= 1906.1 WLog Mean Temp Diff, I?lnI?ln = [ ( T3 – T7 ) – ( T6 – T10 ) ] / ln [ ( T3 – T7 ) / ( T6 – T10 ) ]= [ ( 70.
3 – 22.4 ) – ( 61 – 52.8 ) ] / ln [ ( 70.3 – 22.4 ) / ( 61 – 52.8 ) ]= 22.4Overall heat transportation coefficient, UU = Qh / ( Am x I?ln )= 2009.
8 watt / ( 0.0288m2 x 22.4 )= 3115.4 W/m2KSample Calculation for Counter Flow:Hot H2O existent flow rate, mhActual rate ( L/min ) = indicated Flow + T6 ( °C ) x 0.0041 – 0.0796= 3 + 59.
9 x 0.0041 – 0.0796= 3.
1663.1705 L/min = 3.166 L/min x ( m3 /1000L ) ten ( 1min / 60s )= 5.2766 ten 10-5 m3/smh = 5.2766 ten 10-5 m3/s x 977 kg/m3= 0.0516 kg/sHeat transportation from hot H2O, QhQh = mh x Cp x ( T3 – T6 )Qh = 0.
0516 kg/s x 4.18kJ/kg.K x ( 70.7 – 59.
9 )= 2.3294 kJ/s= 2329.4 WHeat transportation to cold H2O, QcQc = megahertz x Cp x ( T7 – T10 )= 0.015 kg/s x 4.18kJ/kg.K x |55.
6 – 24.6|= 1.9437 kJ/s= 1943.7 WLog Mean Temp Diff, I?lnI?ln = [ ( T3 – T7 ) – ( T6 – T10 ) ] / ln [ ( T3 – T7 ) / ( T6 – T10 ) ]= [ ( 70.
7 – 55.6 ) – ( 59.9 – 24.6 ) ] / ln [ ( 70.7 – 55.
6 ) / ( 59.9 – 24.6 ) ]= 23.
8Overall heat transportation coefficient, UU = Qh / ( Am x I?ln )= 2329.4 watt / ( 0.0288m2 x 23.8 )= 3398.4 W/m2KTrial12345Metal wall at recess, T1 ( °C )65.
8Metal wall at issue, T2 ( °C )56.455.950.352.046.
6Hot watercourse at recess, T3 ( °C )67.168.068.269.673.0Hot watercourse 1stintermediate, T4 ( °C )66.966.
566.667.568.9Hot watercourse 2ndintermediate, T5 ( °C )65.
065.164.464.663.8Hot watercourse at issue, T6 ( °C )62.
6Cold watercourse entry/exit, T7 ( °C )58.157.957.555.953.0Cold watercourse intermediate, T8 ( °C )51.
444.9Cold watercourse intermediate, T9 ( °C )43.342.
1Cold watercourse entry/exit, T10 ( °C )30.230.030.330.430.0Hot H2O indicated flow rate, V ( L/min )8.256.
604.953.301.65Hot H2O existent flow rate, mh ( Kg/s )0.13750.11050.
08350.05650.0294Cooling H2O flow rate, megahertz ( kg/s )0.
0210.0210.0210.0210.021Water denseness at hot H2O recess, I? ( kg/m3 )979978.7978.4977.
7975Mean hot H2O temperature, ( T3+T6 ) /2 ( °C )65.065.264.965.165.
3Linear speed at interior tubing, V ( m/s )2.86252.30111.73801.17660.
6122Reynolds No. at average hot H2O temperature, Re51554.341430.
931307.521184.111023.3Heat transportation from hot H2O, Qh ( W )2471.42586.62303.
62125.51892.5Heat transportation to cold H2O, Qc ( W )2449.13326.
92387.62238.42018.9Log Mean Temp Diff, I?ln H3.
19963.74005.74445.87698.9662Log Mean Temp Diff, I?ln degree Celsius15.107715.
389813.303414.997014.6178Surface heat transportation Coef. at interior tubing, hh ( W/m2K )29594.226498.115364.613857.
48087.2Surface heat transportation Coef. at outer tubing, hc ( W/m2K )5229.25133.
45789.54814.74455.3Log Mean Temp Diff, I?ln overall18.33619.13119.
19220.87423.596Overall heat transportation coefficient, U ‘ ( W/m2K ) neglect thermic opposition of the metal wall4680.14694.54167.73535.
62784.9Overall heat transportation coefficient, U ( W/m2K )4444.04300.34205.03573.22872.
Table – Consequence of fluid speed on the surface transportation coefficients ( Counter Flow )
Sample computation for Test 1:Hot H2O existent flow rate, mhActual rate ( L/min ) = indicated Flow + T6 ( °C ) x 0.0041 – 0.0796= 8.25 + 62.8 x 0.0041 – 0.0796= 8.42798.
4279 L/min = 8.4279 L/min x ( m3 /1000L ) ten ( 1min / 60s )= 1.4046 ten 10-4 m3/smh = 1.4046 ten 10-4 m3/s x 979 kg/m3= 0.1375 kg/sLinear speed at interior tubing, Vmh = V x Ah x I? , I? value based on the hot H2O average temperature.V = mh / ( Ah x I? )= 0.1375 kg/s / ( 0.0261m2 x 980.
3 kg/ M3 )= 0.005374 m/sReynolds No. at average hot H2O temperature, ReRe = ( 5 x Ah x I? ) / I? , I? and I? value based on the hot H2O average temperature.= ( 0.
005374 m/s x 0.0261m2 ten 980.3 kg/ M3 ) / ( 430 x 10-6 Ns/m )= 319.76Surface heat transportation Coef. at interior tubing, hhhh = Qh / ( Ah x I?ln H )Qh = mh x Cp x ( T3 – T6 )= 0.1375 kg/s x 4.18 kJ/kg.
K x ( 67.1 – 62.8 )= 2.4714 kJ/s= 2471.4 WI?ln H = [ ( T3 – T1 ) – ( T6 – T2 ) ] / ln [ ( T3 – T1 ) / ( T6 – T2 ) ]= [ ( 67.1 – 65.8 ) – ( 62.
8 – 56.4 ) ] / ln [ ( 67.1 – 65.8 ) / ( 62.8 – 56.
4 ) ]= 3.1996hh = Qh / ( Ah x I?ln H )= 2471.4 watt / ( 0.0261m2 x 3.
1996 )= 29594.2 watt/ M2 KSurface heat transportation Coef. at outer tubing, hchc = Qc / ( Ac x I?ln degree Celsius )Qc = megahertz x Cp x ( T7 – T10 )= 0.021 kg/s x 4.18 kJ/kg.
K x ( 58.1 – 30.2 )= 2.4491 kJ/s= 2449.1 WI?ln H = [ ( T1 – T7 ) – ( T2 – T10 ) ] / ln [ ( T1 – T7 ) / ( T2 – T10 ) ]= [ ( 65.
8 – 58.1 ) – ( 56.4 – 30.2 ) ] / ln [ ( 65.
8- 68.1 ) / ( 56.4 – 30.2 ) ]= 15.1077hc = Qc / ( Ac x I?ln degree Celsius )= 2449.1 watt / ( 0.031m2 x 15.
1077 )= 5229.2 watt/ M2 KOverall heat transportation coefficient, UU = Qh / ( Am I?ln overall )I?ln overall = [ ( T3 – T7 ) – ( T6 – T10 ) ] / ln [ ( T3 – T7 ) / ( T6 – T10 ) ]= [ ( 67.1 – 58.
1 ) – ( 62.8 – 30.2 ) ] / ln [ ( 67.1 – 58.1 ) / ( 62.
8 – 30.2 ) ]= 18.336U = Qh / ( Am I?ln overall )= 2471.4 watt / ( 0.
0288m2 x 18.3 )= 4680.1W/m2KOverall heat transportation coefficient, UA? ( after pretermiting the thermic opposition of the mental wall )U = 1 / ( 1/ hh + 1/ hc )= 1 / ( 1/ 29594.2 watt/ M2 K + 1/ 5229.2 watt/ M2 K )= 4444.0 watt/ M2 KUse consequence shown in Table 1, the temperature secret plan distribution of the metal wall, the hot and cold watercourse is illustrated below in Graph 1 and 2:
Graph – Temperature distribution of Metal Wall, Hot and Cold watercourse in Parallel flow
Graph – Temperature distribution of Metal Wall, Hot and Cold watercourse in Counter flow
Use consequence shown in Table 2, the surface heat transportation coefficient inside and outside the tubing and overall heat transportation coefficient against the tubing H2O speed is plotted below:
Graph – Surface heat transportation coefficient inside and outside the tubing and overall heat transportation coef.
Against the tubing H2O speed
The analogue and counter flow agreement and different speed on counter flow are discussed. First, for parallel flow form, refer to Graph 1, the temperature difference ( temperature of hot watercourse subtract temperature of cold watercourse ) is really big at get downing approximately 48 °C, so it bit by bit decrease to 5 °C.
The parallel flow is rather efficient to convey hot and cold watercourse to an about same temperature, but the outlet temperature of the cold watercourse ne’er exceeds the lowest temperature of the hot watercourse. Furthermore, the big temperature difference might do thermic enlargement which is really disadvantage for the stuffs used to build the heat money changer, might do thermic failure.Following, for the counter flow, refer Graph 2, the temperature difference starting is 15 °C, at the stoping is 35 °C, the temperature difference is bit by bit increased.
Compare to parallel flow, the temperature difference is non utmost big at the beginning. Therefore, the temperature difference is more by and large distributed for counter flow. It will cut down the thermic enlargement job in the heat money changer tubing. From the graph, the mercantile establishment temperature of the cold watercourse is non limited by the lowest temperature of the hot watercourse as shown in parallel flow. The outlet temperature of the cold watercourse can transcend the lowest temperature of the hot watercourse in the counter flow agreement.
The outlet temperature of cold watercourse in counter flow is higher so in parallel flow constellation.From the consequences calculated in Table 1, heat transportation from hot H2O, Qh and heat transportation to cold H2O, Qc values for counter flow are higher than parallel flow. The counter flow is more efficient for the heat transportation. The counter flow constellation, the heat transportation between hotter portion and colder portion are at two terminals. The temperature difference is rather unvarying, unlike parallel flow, the temperature difference is kept reduced.
Therefore, the counter flow is more efficient for heat transportation as it has a larger temperature difference throughout the whole distribution.The tabular array below show the overall heat transportations and the difference:Linear speed at interior tubing, V ( m/s )2.86252.30111.73801.17660.6122Overall heat transportation coefficient, U ‘ ( W/m2K ) neglect thermic opposition of the metal wall4680.
14694.54167.73535.62784.9Overall heat transportation coefficient, U ( W/m2K )4444.04300.34205.03573.
Table – Difference between overall heat transportations
The difference calculated based on U ‘ subtract U, from the Table 3, as the speed additions, the difference is wrinkling. Therefore, at higher speed of the hot H2O watercourse, the thermic opposition of the metal wall is no longer can be neglected.
The 2nd portion of the experiment, refer to Graph 3, when the hot H2O flow rate additions, surface heat transportation coefficient, hh at interior tubing additions significantly. The hh is the highest at the maximal flow rate. Therefore, the heat transportation is more efficient at higher flow rate. The higher flow rate consequences in a higher additive speed of the watercourse and the addition the Reynolds figure. The flow part might stop in disruptive part where the beds of the fluid mix, convection enhanced the heat transportation. On the other manus, surface heat transportation coefficient, hc at outer tubing increases somewhat so diminish back to about the get downing value.
Then, the value of the hc does non alter much with increasing of hot H2O flow rate. Since the cold H2O watercourse is kept changeless in the experiment, hence, the hc does non vary much as it chiefly depends on the flow rates. The overall heat transportation coefficient, U is progressively somewhat as hot H2O flow rate additions. Hence, increase the hot H2O flow rate is able to heighten the public presentation of the heat money changer.
In this experiment, the informations obtained based on the steady province temperature shown on the index. Therefore, the consequences obtained might non perfectly right as the temperature might non be stabilized when roll uping informations. Particularly, for 2nd portion of the experiment, alteration of the hot H2O flow rate need to set the warmer control to convey the average hot H2O temperature back to the original value. It is non easy to command the warmer as the temperature keeps altering when hot H2O flow rate alterations. As a consequence, human mistake might happen when executing the experiment.
To get the better of the possible mistakes, the solutions can be:Repeat the experiment, and take norm of both reading and continue to the computation. This can cut down the mistake if the temperature taken down is non steady yet.The warmer control can be programmed and convey the average hot H2O temperature to the original automatically. This can cut down human mistake when commanding the temperature.
In decision, the analogue and counter flow constellations have their ain advantage and disadvantages. For parallel flow, it is really good to convey two watercourses to a about same temperature, but the temperature difference is really big at front terminal, might do thermic emphasis to the heat money changer.
For counter flow, it is more efficient as the temperature difference distribution is unvarying guarantee a changeless heat transportation and the cold H2O mercantile establishment temperature is non limited by the hot H2O lowest temperature. For the hot H2O flow rate alterations, the higher the flow rate, the higher the public presentation of the heat transportation.