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Bimetallic Strip Worksheet Answers

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Bimetallic Strip Worksheet Answers Bimetallic strip worksheet answers Continue The bimetallic strip consists of two different materials with different expansion ratios that are related to each other. For example, for brass and steel, linear expansion ratios: brass: 19 x 10-6/C steel: 11 x 10-6/C When this bimetallic band is heated, brass expands more than steel and strip curves with brass on the outside. If the band is cooled, it curves with steel on the outside. Bimetallic strips are used as switches in thermostats. Copper and zinc strips are the same length of 20 cm at 20 degrees Celsius (a) What will be the difference in the length of the strips at 100 degrees Celsius? b) The stripes were chained to each other at 20 degrees Celsius and formed the so-called bimetallic strip. Suppose it bends into an arc when heated. Determine which metal will be on the outside of the arc and what will be the arc radius at 100 degrees Celsius. The thickness of each band is 1 mm. t0 - 20 degrees Celsius, in which both strips are equally long l0 and 20 cm, the length of both bands is 0.20 m at t0 t. difference in length of both bands at t d 1 mm and thickness of 1'10-3 m of each band r ? The radius of the curved bimetallic strip From tables No 30-10- 6K-1 ratio of linear thermal zinc expansion NoCu 17'10-6K-1 ratio of linear thermal expansion of copper substances expands when we raise their temperature. Different substances expand in different ways, so there will be a difference in the length of the two bands. As long as the strips are firmly chained to each other so that they can't move along each other, they will bend. A band that expands less will be on the inside, so in our case it will be copper. Because the thickness of the stripes is constant, the two curves differ in length. The difference in the length of both arcs should correspond to the difference in length caused by the different thermal expansion of the two metals. First, we express the length of both strips at temperature t. \[l_{Cu}=l_0\,(1+\alpha_{Cu}\Delta t)\] \[l_{Zn}=l_0\,(1+\alpha_{Zn}\Delta t)\] We subtract one length from another: \[\Delta l=l_{Zn}-l_{Cu}\] \[\Delta l=l_0\,(1+\alpha_{Zn}\Delta t)-l_0\,(1+\alpha_{Cu}\Delta t)\] \[\Delta l=l_0\,(\alpha_{Zn}-\alpha_{Cu})\Delta t\] \[\Delta l=l_0\,(\alpha_{Zn}-\alpha_{Cu})\,\left(t-t_0\right)\] Finally, we substitute for numerical values: \[\Delta l=l_0\,(\alpha_{Zn}-\alpha_{Cu})\,\left(t-t_0\right)\] \[\Delta l=0.2\cdot\,(30\cdot{10^{-6}}-17\cdot{10^{-6}})\cdot\,\left(100-20\right)\,\mathrm{m}\] \[\Delta l=2.08\cdot{10^{-4}}\ , mathrm'm0.208, mathrmmm, as the strips are firmly attached to each other, they will bend during heating. Consider that the length of the strip corresponding to the increased temperature is in the Strip. We've already expressed the length of both bands partially a), but we can also express this through φ and desired radius r. From there we can determine the radius of r. Note: When the temperature rises, the thickness of the bands also increases. However, this effect has very little effect on the result, so you can ignore it. We will use the image to calculate the radius of the arc. Because the strips are firmly attached to each other, they will bend during heating. Keep in mind that the strip length corresponding to the elevated temperature is in the middle of the strip. The radius of the smaller arc to r. The radius of the larger arc is R: Rfracd_1 d_2 {2}, where d1 and d2 are increased thicknesses of both bands. The length of the circular arc can be defined as the product of the radius and angle of the φ (in the radian). l_ Cur'varphi (l_ n (r'fracd_ n{2}frac d_ Cu{2}) We subtract the length of one arc from another and understand that their difference should be equal to the difference in the length of the bands that we expressed in part a) of the task. Delta ll_n-l_Cu (r'fracd_ n{2}fracd_ Cu{2})varphi-r'varphi'frac{1} {2} (d_ d_ lengths and increased thickness at high temperature l_0 (alpha_ -alpha_ Delta t'frac{1}{2}'d (1alpha_ CuDelta t)d (1'alpha_ 'n'Delta t)l_0 ,alpha_ yen alpha_ Ku{1}{2})Delta t'frak{1}{2}d2 (alpha_ (Kua alpha_)t) : Warfi 2l_o (alpha_)-alpha_ (Kua)) Delta t22 (alpha_ Cualpha_)Delta t) For the length of the heated copper band it corresponds to the reality of l_o (1alpha_ CuDelta t φ) : Frak l_o (1alpha_ Delta t) alpha_ alpha_ 2l_o l_o (1 alpha_ Delta talpha_) alpha_ Delta t) Ryo Frak (alpha_ Ku Delta t) 2 (alpha_ Kue alpha_) Delta etc.2 (Alpha_ ing-alpha_) Delta etc. we replace these values : Frak (117'cdot10'-6'cdot{80})2 (30'cdot-10'-6 17-kdot 10-10-10-2 80'2'cdot (30'cdot-10'-6'-1 7'cdot-10'-6')) mathrm-m'dot)96, mathrmcm: If we look more closely at both expressions in brackets as a result of the faction, it is obvious that we can omit terms that contain heat coefficients No. The first expression in brackets is associated with a change in the length of one band, and the second expression in brackets corresponds to a change in the thickness of the strip. On the contrary, we cannot make any omission of expression brackets in the denominator, because both terms are of similar magnitude. This expression is associated with a difference in the linear expansion of both metals. This will give us a simpler attitude: the dot tailcoats (1 euro...) (2 euros) No 2 (alpha_---Ku alpha_) Delta i.e. d'frak de (alpha_.- alpha_Ku) (Delta- t'10-3' (30'cdot-10'-1-16'-17'c 10-6)) kdot{80}, matrm-dot 0.9615, matemarmamamamamam Difference in the length of two bands will be about 0.2 mm, and the bimetallic band will bend into an arc with a radius of about 96 cm. This article needs additional quotes to verify. Please help improve this article by adding quotes to reliable sources. Non-sources of materials can be challenged and removed. Find sources: Bimetallic stripe - News newspaper book scientist JSTOR (February 2012) (Learn how and when to remove this pattern message) A bimetallic strip diagram showing how the difference in thermal expansion in the two metals leads to a much larger lateral shift of the strip of the bimetallic coil from the thermometer reacts to the heat from the lighter, by unwinding and then the backup. The bimetallic band is used to convert temperature changes into mechanical displacement. The band consists of two strips of different metals that expand at different speeds as they heat up, usually steel and copper, and in some cases steel and brass. Different extensions cause the flat band to bend to one side when heated, and in the opposite direction if it cools below the initial temperature. A metal with a higher thermal expansion factor is on the outside of the curve when the band heats up and on the inside when cooled. The invention of the bimetallic band is usually attributed to John Harrison, an eighteenth-century watchmaker who did so for his third marine chronometer (H3) of 1759 to compensate for temperature changes in the balance sheet in the spring. Harrison's invention is recognized at a memorial to him at Westminster Abbey, England. This effect is used in a number of mechanical and electrical devices. The characteristics of the band consists of two strips of different metals that expand at different speeds as they heat up, usually steel and copper, and in some cases steel and brass. The stripes are connected to each other along their entire length, riveting, rationing or welding. Different extensions cause the flat band to bend to one side when heated, and in the opposite direction if it cools below the initial temperature. A metal with a higher thermal expansion factor is on the outside of the curve when the band heats up and on the inside when cooled. The side-shifting band is much larger than a slight extension of length in either of the two metals. In some apps bimetallic strips are used in a flat form. In others, it is wrapped in a coil for compactness. The long length of the spiral version gives an improved sensitivity. The curvature of the bimetall beam can be described by the following equation: No. 6 E 1 E 2 (h 1 and h 2) h 1 h 2 ε E 1 2 h 1 4 4 4 E 1 E 2 h 1 3 h 2 x 6 E 1 E 2 h 1 1 2 2 2 2 2 h 2 x 4 E 1 E 2 h 2 3 h e E 2 h 2 2 2 4 display kappa Frak (6E_{1}E_{2} h_{1})h_{2}) h_{1}h_{2} Epsilon E_{1} {2}h_{1} {4} 4E_{1}E_{2}h_{1} {3}h_{2} 6E_{1}E_{2}h_{1} {2}h_{2} {2} 4E_{1}E_{2}h_{2}-{3}h_{1}-E_{2}-{2}h_{2}-{4} where no 1/R display kappa 1/R and R R display is a curvature radius, E 1 displaystyle E_{1} and h 1 displaystyle h_{1} is a module and height (thickness) of material 1 and E 2 displaystyle E_{2} and h 2 displaystyle h_{2} is a module and height (thickness) of two material.
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