Experiment Is Performed To Determine The Friction Loss Biology Essay

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In this practical, experiment is performed to determine the friction loss in straight pipes with uniform cross sectional area. On the other hand, the mirror loss due to pipe fittings, like valves, joints and bends is also analyzed. Besides, the use of venture meter to measure the flow rate is learnt.

2. Abstract

In order to study the heat transfer of heat exchangers, two type flow arrangement, parallel and counter flow are constructed and different velocity of the flow rate is operated to observe the affect of the performance of a heat exchanger. The temperature data are collected during the experiment and calculation and plots based on the data are obtained. The results enable the experimenter to examine the different efficiency when operating the heat exchanger at a particular arrangement and flow rate.

3. Principles

When a fluid flows through a pipe, the fluid particles near the pipe wall have relative low velocity and the one near the canter move with relative high velocity. Because of this relative motion and the viscosity of the fluid, shear stress are produced. This viscous action causes dissipation, which is normally referred as pipe friction loss.

In order to study friction loss of flow in pipe, the Bernoulli's equation Considering the two surfaces normal to the direction of the fluid flow, an overall energy balance equation that governs the fluid flow can be derived:

Equation 1: Overall Energy Balance for incompressible fluid

where each term has the unit of Joule/kg (m2s-2).

The major head loss was contributed by the frictional effects that arose due to fluid flow and intermolecular forces of attraction, whereas the minor head loss occurred as a result of other physical obstructions to the flow. Examples are the presence of orifice plates, Venturi meters, bends, flow control valves, pipe fittings, and so on. The major and minor head loss are represented by the following equations:

Equation 2: Major head loss Equation 3: Minor head loss

The friction factor, f, in Equation 2 is largely dependent on the flow regime, and hence, dependent on the Reynolds number. In addition, for turbulent flow, f also becomes dependent on the relative roughness ε/D for rough pipes.

Equation 4: Friction factor for laminar flow

Equation 5: Colebrook's equation for friction factor in turbulent flow Equation 6: Blasius formula for friction factor for smooth pipes

For this experiment, appropriate equations from Equation 2 to 6 were chosen to combine with the overall energy balance equation, which were reduced accordingly, for evaluation of the specified quantities.

4. Equipment and Materials

A water to water turbulent flow concentric tube heat exchanger manufactured by P. A. Hilton Ltd. is used for the investigation. The heat exchanger is a double pipe type with hot water flowing through the central tube while cooling water flows through the annular space.

A heater (electrical resistance type) is used to heat up hot water and the water cools as it flows through the heat exchanger, and on leaving passes through a pump which provides the circulating head. Then, it passes the heater again to get reheat. On the other hand, cold water passes through a flow control valve and flow meter to the unit. Two direction control valves can swap the cooling water inlet/outlet of the exchanger. Therefore, the flow can be either a parallel or counter direction. The below shows the photo of the operating unit:

Photo - Water to water turbulent flow concentric tube heat exchanger

5. Procedure

The heat exchanger was connected in parallel flow configuration. The water level in the heater tank was checked at the correct level. The whole unit was turned on and the hot water flow valve was fully open. These steps were done by the lab staff. The following steps are the steps that experimenters performed during lab:

The hot water flow rate was reduced to 50g/s and the cold water 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 cooling water direction control valve was switched to "counter flow" position immediately. Then, step 2 was repeated.

Fully opened the hot water flow valve.

Adjusted the cold water to bring the mean hot water temperature, [T3 +T6]/2 to around 65°C. Then, step 2 was repeated.

The hot water flow rate was reduced to 80% of the initial value without changing the cold water flow rate.

Adjusted the heater control to bring the mean hot water temperature, [T3 +T6]/2 to around 65°C. Then, step 2 was repeated.

Repeated step 6 and 7 with hot water flow rate about 60%, 40% and 20% of the initial value.

6. Results

The results of the experiment are tabulated below in Table 1 and Table 2:

Test

Parallel Flow

Counter Flow

Metal wall at inlet, T1 (°C)

57.7

67.7

Metal wall at exit, T2 (°C)

57.3

45.8

Hot stream at inlet, T3 (°C)

70.3

70.7

Hot stream 1stintermediate, T4 (°C)

65.4

68.0

Hot stream 2ndintermediate, T5 (°C)

62.8

64.7

Hot stream at exit, T6 (°C)

61.0

59.9

Cold stream entry/exit, T7 (°C)

22.4

55.6

Cold stream intermediate, T8 (°C)

38.6

47.4

Cold stream intermediate, T9 (°C)

47.6

37.2

Cold stream entry/exit, T10 (°C)

52.8

24.6

Hot water indicated flow, V (L/min)

3

3

Water density at hot water inlet, ρ (kg/m3)

977.5

977

Hot water actual flow rate, mh(Kg/s)

0.0517

0.0516

Mean hot water temperature, (T3+T6)/2 (°C)

65.7

65.3

Cooling water flow rate, mc(kg/s)

0.015

0.015

Heat transfer from hot water, Qh(watt)

2009.8

2329.4

Heat transfer to cold water, Qc(watt)

1906.1

1943.7

Log Mean Temp Diff, θln

22.4

23.8

Overall heat transfer coefficient, U (W/m2K)

3115.4

3398.4

Table Parallel and Counter flow

Sample Calculation for Parallel Flow:

Hot water actual flow rate, mh

Actual rate (L/min) = indicated Flow + T6 (°C) x 0.0041 - 0.0796

= 3 + 61 x 0.0041 - 0.0796

= 3.1705

3.1705 L/min = 3.1705 L/min x (m3 /1000L) x (1min / 60s)

= 5.284 x 10-5 m3/s

mh = 5.284 x 10-5 m3/s x 977.5 kg/m3

= 0.0517 kg/s

Heat transfer from hot water, Qh

Qh = 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 watt

Heat transfer to cold water, Qc

Qc = mc 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 watt

Log Mean Temp Diff, θln

θln = [(T3 - T7) - (T6 - T10)] / ln[(T3 - T7) / (T6 - T10)]

= [(70.3 - 22.4) - (61 - 52.8)] / ln[(70.3 - 22.4) / (61 - 52.8)]

= 22.4

Overall heat transfer coefficient, U

U = Qh / (Am x θln)

= 2009.8 watt / (0.0288m2 x 22.4)

= 3115.4 W/m2K

Sample Calculation for Counter Flow:

Hot water actual flow rate, mh

Actual rate (L/min) = indicated Flow + T6 (°C) x 0.0041 - 0.0796

= 3 + 59.9 x 0.0041 - 0.0796

= 3.166

3.1705 L/min = 3.166 L/min x (m3 /1000L) x (1min / 60s)

= 5.2766 x 10-5 m3/s

mh = 5.2766 x 10-5 m3/s x 977 kg/m3

= 0.0516 kg/s

Heat transfer from hot water, Qh

Qh = 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 watt

Heat transfer to cold water, Qc

Qc = mc 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 watt

Log Mean Temp Diff, θln

θln = [(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)]

= 23.8

Overall heat transfer coefficient, U

U = Qh / (Am x θln)

= 2329.4 watt / (0.0288m2 x 23.8)

= 3398.4 W/m2K

Test

1

2

3

4

5

Metal wall at inlet, T1 (°C)

65.8

66.1

65.8

65.8

65.8

Metal wall at exit, T2 (°C)

56.4

55.9

50.3

52.0

46.6

Hot stream at inlet, T3 (°C)

67.1

68.0

68.2

69.6

73.0

Hot stream 1stintermediate, T4 (°C)

66.9

66.5

66.6

67.5

68.9

Hot stream 2ndintermediate, T5 (°C)

65.0

65.1

64.4

64.6

63.8

Hot stream at exit, T6 (°C)

62.8

62.4

61.6

60.6

57.6

Cold stream entry/exit, T7 (°C)

58.1

57.9

57.5

55.9

53.0

Cold stream intermediate, T8 (°C)

51.3

50.6

49.6

48.4

44.9

Cold stream intermediate, T9 (°C)

43.3

42.4

41.9

40.6

38.1

Cold stream entry/exit, T10 (°C)

30.2

30.0

30.3

30.4

30.0

Hot water indicated flow rate, V (L/min)

8.25

6.60

4.95

3.30

1.65

Hot water actual flow rate, mh(Kg/s)

0.1375

0.1105

0.0835

0.0565

0.0294

Cooling water flow rate, mc(kg/s)

0.021

0.021

0.021

0.021

0.021

Water density at hot water inlet, ρ (kg/m3)

979

978.7

978.4

977.7

975

Mean hot water temperature, (T3+T6)/2 (°C)

65.0

65.2

64.9

65.1

65.3

Linear velocity at inner tube, v (m/s)

2.8625

2.3011

1.7380

1.1766

0.6122

Reynolds No. at mean hot water temperature, Re

51554.3

41430.9

31307.5

21184.1

11023.3

Heat transfer from hot water, Qh(watt)

2471.4

2586.6

2303.6

2125.5

1892.5

Heat transfer to cold water, Qc(watt)

2449.1

3326.9

2387.6

2238.4

2018.9

Log Mean Temp Diff, θln h

3.1996

3.7400

5.7444

5.8769

8.9662

Log Mean Temp Diff, θln c

15.1077

15.3898

13.3034

14.9970

14.6178

Surface heat transfer Coef. at inner tube, hh (W/m2K)

29594.2

26498.1

15364.6

13857.4

8087.2

Surface heat transfer Coef. at outer tube, hc (W/m2K)

5229.2

5133.4

5789.5

4814.7

4455.3

Log Mean Temp Diff, θln overall

18.336

19.131

19.192

20.874

23.596

Overall heat transfer coefficient, U' (W/m2K) neglect thermal resistance of the metal wall

4680.1

4694.5

4167.7

3535.6

2784.9

Overall heat transfer coefficient, U (W/m2K)

4444.0

4300.3

4205.0

3573.2

2872.7

Table - Effect of fluid velocity on the surface transfer coefficients (Counter Flow)

Sample calculation for Test 1:

Hot water actual flow rate, mh

Actual rate (L/min) = indicated Flow + T6 (°C) x 0.0041 - 0.0796

= 8.25 + 62.8 x 0.0041 - 0.0796

= 8.4279

8.4279 L/min = 8.4279 L/min x (m3 /1000L) x (1min / 60s)

= 1.4046 x 10-4 m3/s

mh = 1.4046 x 10-4 m3/s x 979 kg/m3

= 0.1375 kg/s

Linear velocity at inner tube, v

mh = v x Ah x ρ, ρ value based on the hot water mean temperature.

v = mh / (Ah x ρ)

= 0.1375 kg/s / (0.0261m2 x 980.3 kg/ m3)

= 0.005374 m/s

Reynolds No. at mean hot water temperature, Re

Re = (v x Ah x ρ) / μ, ρ and μ value based on the hot water mean temperature.

= (0.005374 m/s x 0.0261m2 x 980.3 kg/ m3) / (430 x 10-6 Ns/m)

= 319.76

Surface heat transfer Coef. at inner tube, hh

hh = Qh / (Ah x θln 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 watt

θln 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)]

= 3.1996

hh = Qh / (Ah x θln h)

= 2471.4 watt / (0.0261m2 x 3.1996)

= 29594.2 watt/ m2 K

Surface heat transfer Coef. at outer tube, hc

hc = Qc / (Ac x θln c)

Qc = mc 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 watt

θln 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)]

= 15.1077

hc = Qc / (Ac x θln c)

= 2449.1 watt / (0.031m2 x 15.1077)

= 5229.2 watt/ m2 K

Overall heat transfer coefficient, U

U = Qh / (Am θln overall)

θln 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)]

= 18.336

U = Qh / (Am θln overall)

= 2471.4 watt / (0.0288m2 x 18.3)

= 4680.1W/m2K

Overall heat transfer coefficient, U´ (after neglecting the thermal resistance 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 K

Use result shown in Table 1, the temperature plot distribution of the metal wall, the hot and cold streams is illustrated below in Graph 1 and 2:

Graph - Temperature distribution of Metal Wall, Hot and Cold streams in Parallel flow

Graph - Temperature distribution of Metal Wall, Hot and Cold streams in Counter flow

Use result shown in Table 2, the surface heat transfer coefficient inside and outside the tube and overall heat transfer coefficient against the tube water velocity is plotted below:

Graph - Surface heat transfer coefficient inside and outside the tube and overall heat transfer coef. Against the tube water velocity

7. Discussion

The parallel and counter flow arrangement and different velocity on counter flow are discussed. Firstly, for parallel flow pattern, refer to Graph 1, the temperature difference (temperature of hot stream subtract temperature of cold stream) is very large at beginning about 48 °C, then it gradually decrease to 5 °C. The parallel flow is quite efficient to bring hot and cold stream to an almost same temperature, but the outlet temperature of the cold stream never exceeds the lowest temperature of the hot stream. Moreover, the large temperature difference might cause thermal expansion which is very disadvantage for the materials used to construct the heat exchanger, might cause thermal failure.

Next, for the counter flow, refer Graph 2, the temperature difference starting is 15 °C, at the ending is 35 °C, the temperature difference is gradually increased. Compare to parallel flow, the temperature difference is not extreme large at the beginning. Therefore, the temperature difference is more generally distributed for counter flow. It will reduce the thermal expansion problem in the heat exchanger tubing. From the graph, the outlet temperature of the cold stream is not limited by the lowest temperature of the hot stream as shown in parallel flow. The outlet temperature of the cold stream can exceed the lowest temperature of the hot stream in the counter flow arrangement. The outlet temperature of cold stream in counter flow is higher then in parallel flow configuration.

From the results calculated in Table 1, heat transfer from hot water, Qh and heat transfer to cold water, Qc values for counter flow are higher than parallel flow. The counter flow is more efficient for the heat transfer. The counter flow configuration, the heat transfer between hotter part and colder part are at two ends. The temperature difference is quite uniform, unlike parallel flow, the temperature difference is kept decreased. Therefore, the counter flow is more efficient for heat transfer as it has a larger temperature difference throughout the whole distribution.

The table below show the overall heat transfers and the difference:

Linear velocity at inner tube, v (m/s)

2.8625

2.3011

1.7380

1.1766

0.6122

Overall heat transfer coefficient, U' (W/m2K) neglect thermal resistance of the metal wall

4680.1

4694.5

4167.7

3535.6

2784.9

Overall heat transfer coefficient, U (W/m2K)

4444.0

4300.3

4205.0

3573.2

2872.7

The difference

236.1

394.2

-37.3

-37.6

-87.8

Table - Difference between overall heat transfers

The difference calculated based on U' subtract U, from the Table 3, as the velocity increases, the difference is creasing. Therefore, at higher velocity of the hot water stream, the thermal resistance of the metal wall is no longer can be neglected.

The second part of the experiment, refer to Graph 3, when the hot water flow rate increases, surface heat transfer coefficient, hh at inner tube increases significantly. The hh is the highest at the maximum flow rate. Therefore, the heat transfer is more efficient at higher flow rate. The higher flow rate results in a higher linear velocity of the stream and the increase the Reynolds number. The flow region might end in turbulent region where the layers of the fluid mix, convection enhanced the heat transfer. On the other hand, surface heat transfer coefficient, hc at outer tube increases slightly then decrease back to nearly the starting value. Then, the value of the hc does not change much with increasing of hot water flow rate. Since the cold water stream is kept constant in the experiment, therefore, the hc does not vary much as it mainly depends on the flow rates. The overall heat transfer coefficient, U is increasingly slightly as hot water flow rate increases. Hence, increase the hot water flow rate is able to enhance the performance of the heat exchanger.

In this experiment, the data obtained based on the steady state temperature shown on the indicator. Therefore, the results obtained might not absolutely correct as the temperature might not be stabilized when collecting data. Especially, for second part of the experiment, change of the hot water flow rate need to adjust the heater control to bring the mean hot water temperature back to the original value. It is not easy to control the heater as the temperature keeps changing when hot water flow rate changes. As a result, human error might occur when performing the experiment. To overcome the possible errors, the solutions can be:

Repeat the experiment, and take average of both reading and proceed to the calculation. This can reduce the error if the temperature taken down is not steady yet.

The heater control can be programmed and bring the mean hot water temperature to the original automatically. This can reduce human error when controlling the temperature.

8. Conclusion

In conclusion, the parallel and counter flow configurations have their own advantage and disadvantages. For parallel flow, it is very good to bring two streams to a nearly same temperature, but the temperature difference is very large at front end, might cause thermal stress to the heat exchanger. For counter flow, it is more efficient as the temperature difference distribution is uniform ensure a constant heat transfer and the cold water outlet temperature is not limited by the hot water lowest temperature. For the hot water flow rate changes, the higher the flow rate, the higher the performance of the heat transfer.

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