Experimental Report H7 - Free Essay Example

Experimental Report Aim The objective of the laboratory report was to determine the thermal expansion coefficient of copper and other materials by measuring the relative change in length of bars of the materials as a function of temperature. Introduction The average coefficient of thermal expansion ? over a temperature interval ? T is given by ?L / L0 = ? T Where L0 is the length at some initial temperature, and ?L is the change in length corresponding to a change in temperature ? T. Therefore the thermal expansion coefficient can be determined from the slope of a graph of the relative change in length ? L/L0 versus temperature change ? T. Experimental Method 1. A small amount of water was placed in the copper kettle (there was enough water so it did not take all day for it to boil, and not too little so it would have boiled dry). The kettle was placed on the tripod with an asbestos mat and heated with the bunsen burner. 2. The rest of the apparatus was set up as shown below. kettle test material bunsen burner Figure 1. Apparatus showing the set up and items used for the experiment. 3. The dial gauge was observed to understand how to read it. The initial length Lo was marked on the apparatus, and the room temperature. 4. When the water boiled the temperature of the bar was noted with the thermocouple thermometer and the dial gauge reading was also noted. These are the initial conditions taken. 5. The bunsen burner was then turned off. 6. As the bar cooled, the temperature was noted as well as the dial gauge reading every ten degrees. From this the calculation of the change in length ? L and change in temperature ? T from the initial conditions were done. 7. A graph of ? L/Lo versus ? T was then plotted. The thermal expansion coefficient ? was then found from the slope of the graph. 8. The stop screw was undone on the non-dial gauge end of the steam jacket, (reference to Figure 1was used, and the test rod was then carefully removed). Replacing with another test rod, steps (3) to (7) were repeated. Results Material | Aluminum| Temp (? C)| Extension (mm)| ? L/Lo| ? T (? C) | 96. 6| 0. 50| 1. 515 * 10-3| 71. 6| 86. 6| 0. 4607| 1. 264 * 10-3| 61. 6| 76. 6| 0. 378| 1. 145 * 10-3| 51. 6| 66. 6| 0. 30| 0. 909 * 10-3| 41. 6| 56. 6| 0. 22| 0. 667 * 10-3| 31. 6| 46. 6| 0. 48| 0. 449 * 10-3| 21. 6| 36. 6| 0. 07| 0. 212 * 10-3| 11. 6| 29. 5| 0. 025| 0. 076 * 10-3| 4. 5| Table 1. Results from the experiment of aluminum Material | Copper| Temp (? C)| Extension (mm)| ? L/Lo| ? T (? C) | 97. 5| 0. 36| 1. 091 * 10-3| 75. 2| 87. 5| 0. 32| 0. 970 * 10-3| 65. 2| 77. 5| 0. 27| 0. 818 * 10-3| 55. 2| 67. 5| 0. 222| 0. 673 * 10-3| 45. 2| 57. 5| 0. 177| 0. 536 * 10-3| 35. 2| 47. 5| 0. 13| 0. 393 * 10-3| 25. 2| 37. 5| 0. 092| 0. 279 * 10-3| 15. 2| 30. 5| 0. 052| 0. 158 * 10-3| 8. 2| Table 2. Results from the experiment of copper Data analysis To find the coefficient of thermal expansion, the change in length is divided by the initial length. This is shown by the formula: ? L/Lo Example from the results of Aluminum: 0. 5mm / 330mm = 1. 515 x 10-3 The results are shown above in Tables 1 and 2. See appendix for graph of Thermal Expansion ? L/Lo vs. ?T Experimental Discussion Problem (1) In order to find if the expansion gaps have closed up you need to find how much the material will expand at the temperatures given. To calculate this certain data of the material is needed. The material has an expansion coefficient of 20 x 10-6 ? C Each beam member of the bridge is 50m long with a cross sectional area of 5m2. To find the thermal expansion of the beam is can be expressed linearly. ? L = L0  ? (t1   t0)  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚   (1) where ? L = change in length (m) L0  = initial length (m) ? =  linear expansion coefficient (m/m ? C) t0  = initial temperature  (? C) t1  = final temperature  (? C) ? L = 50m x (20 x 10-6 ? C) x (60 ? C – 20 ? C) ? L = 0. 04m ? L = 4cm At a temperature of 60 ? C and with the bridge members being gapped 2cm apart the expansion gaps will close up and then compression on both members will take place. The expansion gaps will close up at 30 ? C. The bridge expands linearly 1cm with a change of temperature of 10 ? C (2) To calculate the stress set up on the bridge after the expansion gaps close up it is necessary to use the elastic modulus and the stress and strain calculations. Each beam member of the bridge is 50m long with a cross sectional area of 5m2. It has an elastic modulus of 2 x 1011 Pa Strain Strain can be expressed as strain = ? L / L where strain  = (m/m) ? L  = elongation or compression (offset) of the object (m) L  = length of the object (m) Stress Stress can be expressed as stress = F / A here stress  = (N/m2) F  = force (N) A  = area of object (m2) Youngs Modulus (Tensile Modulus) Youngs modulus or Tensile modulus can be expressed as E = stress / strain = (F / A) / (? L / L) where E = Youngs modulus (N/m2) Firstly use the formula for strain = ? L / L = 0. 06m / 50m = 1. 2 x 10-3 Next Young’s modulus E = stress / strain = (F / A) / (? L / L)   Re arrange the values to get the force therefore Force = E x Strain x Area = (2 x 1011) x (1. 2 x 10-3) x 5m2 = 12 x 109 = 1. 2 GN (3) If this bridge was in the design stage, it would be recommended to increase the gap of the bridge members. This gap would need to be more than 8cm if they were not to touch. Question 1 Copper has an expansion coefficient of 1. 7 x 10-5 ? C The beam is 7m long Initial temperature is 20 ? C Final temperature is 60 ? C ? Temperature is 40 ? C Expansion of Copper pipe = 7m x (1. 7 x 10-5) x 40 ? C = 0. 00476m = 0. 476mm Polythene pipe has an expansion coefficient of 200 x 10-6 With the beam at the same length and the change in temperature being the same the expansion of the pipe would be: Expansion of Polythene pipe = 7m x (200 x 10-6) x 40 ? C = 0. 056m = 0. 6mm This shows that it expands more than the copper pipe and would therefore not provide the smallest space between the pipe and the wall. Question 2 All piping systems will expand and contract due to temperature changes. Changes in temperature tend to cause a change in dimensions of any matter. The allowance of moderate change in length of an installed piping system is needed as a consequence of a temperature change is generally benefic ial, regardless of the piping material, in that it tends to reduce and redistribute the stresses that are generated should the tendency for a dimensional change be fully restrained. The main way to avoid damage or without creating a wide cavity in the wall is with the typical expansion loop as seen in Figure 2. Expansion loops are periodically placed in long pipelines subject to wide temperature changes, such as steam lines and the Alaska oil pipeline. These loops absorb changes in pipe length and thereby mute and redistribute the large stresses that would result if pipe thermal movements had been physically restrained. Similar measures for safely absorbing thermal expansion/contraction reactions need to be taken with thermoplastic piping systems or within the case of the copper piping next to the wall. Figure 2. Expansion loop. Summary Understanding thermal expansion is very important as it can impact on every kind of material. From the experiment and the results produced is it clear that as there is an increase in temperature of a material, there will be an increase in length of the material. One particular way it can increase is linearly as shown above in the Tables 1 and 2 and also in the graph Thermal Expansion ? L/Lo vs. ?T. The expansion that is shown is due to the different expansion coefficients of each material. Some will expand more than others. From the results is it clear that aluminum has a large expansion coefficient due to the change in length in relation to temperature. There are also means that can be put in place to absorb changes in pipe length such as using expansion loops or bends. Overall the results that were produced support the expected relationships between the change of length of a material and temperature.