TUTORIAL 3 Solution (06/02/2017) Thermodynamics for Aerospace Engineers (AS1300) Temperature and Heat

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1 TUTORIAL 3 Solution (06/02/2017) Thermodynamics for Aerospace Engineers (AS1300) Temperature and Heat 1. A relation between the resistance and the temperature T of a resistance is given as R=R0[1+α(T-T0)], where R0 is the resistance measured in ohms at temperature T0 (in C) and α is a material constant. The thermometer reads temperatures of 0 C and 91 C corresponding to resistance values of and ohms respectively. What would be the resistance corresponding to a temperature of 50 C in this thermometer? In this question, the resistance value at two different temperatures for a material is provided. The relation between the resistance and the temperature which is linear in nature is also given with unknowns like R0, T0, and α. Many of you might be thinking with resistance value at 0 C and 91 C, we have only two equations but there are three unknowns in the function R=R0[1+α(T-T0)]. To make it simpler, let s write the above function as R=R0(1-αT0)+ R0αT = C1 + C2T, where C1 and C2 are constants which are unknowns. By using the resistance value at 0 C and 91 C the constants C1 and C2 can be found out. This information can be used to calculate the resistance value (R) at 50 C which is asked in the question. But we cannot find the value of R0, T0, and α with this information. So writing the two equations: = C 1 + C 2 0 [Eqn 1.1] = C 1 + C 2 91 [Eqn 1.2] Solving Eqn 1.1 and 1.2, we can obtain the constants C1 and C2 to be Ω and Ω/K respectively. So, the resistance value at 50 C is: R = C 1 + C 2 50 = = Ω We can solve this question using another approach. As this function is a linear function, we can write this using the resistance value at 0 C and 91 C as follows: R R 0 = T 0, where R0 = Ω and R91 = Ω [Eqn 1.3] R 91 R Substituting T value as 50 in Eqn 1.3, the resistance value at that temperature can be obtained. R 50 = R 0 + (R 91 R 0 ) T = ( ) = Ω

2 2. The readings TA and TB in C of 2 thermometers A and B agree well at ice and steam points. At other points, their temperature are related as TA= p + qtb + rtb 2, where p, q and r are constants. When A reads 51 C, B reads 50 C. Determine what A reads when B reads 30 C. In this question, it is provided that the temperature readings TA and TB in C of 2 thermometers A and B are related by the equation TA= p + qtb + rtb 2. At ice and steam point both thermometer show correct temperature i.e. 0 C and 100 C respectively. Along with this, when thermometer A reads 51 C, thermometer B reads 50 C. It can be observed in the above that, the equation relating TA and TB has 3 unknowns i.e. p, q, and r. To find these unknowns, we have three data points. So, these three equations can be written as follows: 0 = p + q 0 + r 0 2 [Eqn 2.1] 100 = p + q r [Eqn 2.2] 51 = p + q 50 + r 50 2 [Eqn 2.3] Solving Eqn 2.1, we obtain p = 0 and from Eqn 2.2 and 2.3, we get q = 1.04 & r = Using these values the reading of the thermometer A can be calculated when thermometer B shows a temperature of 30 C. T A = p + q T B + r T B 2 = = = C

3 3. A thermometer is constructed by arranging an aluminum rod so that its length can be measured by using a scale made of different material. The thermometer is calibrated at 0 C and 100 C and the distance between these points is divided uniformly into 100 degrees of equal length of the scale. What is the true temperature when this thermometer reads 40 C? The coefficient of linear expansion of aluminum in this temperature range is x10-4 cm/(cm.k) while that of the material of the scale is 10-5 cm/(cm.k). In this question, it has been mentioned that the expansion of aluminum rod is measured by a scale made of a material different from aluminum. So as the temperature increases both the scale and the aluminum rod will expand. The scale has been divided into 100 divisions. It is also mentioned that the scale has been calibrated at 0 C and 100 C which means that the expansion of the aluminum rod will exactly match the 0 th and 100 th reading at these temperatures respectively. Finally, it is asked to find the true temperature if the reading in the scale shows 40 C. Remember, as the scale expands with temperature and the division in the scale is fixed (100 division) the unit length on the scale will change with temperature. Let length of scale at T=273 K (0 C) is Lscale. So, the expression for the length of the scale at any temperature T is as follows: X scale T = L scale [1 + α material (T 273)] [Eqn 3.1] So, the unit length of the scale at any temperature T becomes: U scale T = L scale 100 [1 + α material(t 273)] [Eqn 3.2] Let s assume that at T=273 K (0 C) the length of the Aluminum rod is Laluminum. The edge of the rod exactly matches with the 0 th reading of the scale. At T=373 K (100 C) the length of the aluminum rod becomes: X aluminum T = L aluminum [1 + α aluminum (T 273)] [Eqn 3.3] = L aluminum [1 + α aluminum ( )] = L aluminum [ α aluminum ] So, length of the aluminum expanded at T=100 C is as follows: L 100 = L aluminum 100 α aluminum [Eqn 3.4] The above expanded length is equal to the 100 th reading of the scale at T=100 C. So the equation becomes as follows: L 100 = L aluminum 100 α aluminum = 100 U scale 100 [Eqn 3.5]

4 = 100 L scale 100 [1 + α material 100] = L scale [1 + α material 100] From above we can obtain Lscale as a function of Laluminum. L scale = 100 L aluminum α aluminum [1+α material 100] [Eqn 3.6] To find the true temperature T when the scale reads 40 C can be found by equating the length of the 40 th reading of the scale and the expanded length of aluminum rod at the true temperature T. L T = L aluminum (T 273) α aluminum = 40 U scale T [Eqn 3.7] = 40 L scale 100 [1 + α material (T 273)] Writing Lscale in terms of Laluminum by borrowing the relationship from Eqn 3.6, the equation becomes as follows: (T 273) L aluminum α aluminum [Eqn 3.8] = L aluminum α aluminum [1+α material 100] 100 [1 + α material (T 273)] The above equation becomes as follows: (T 273) (1 40 α material ) = 40 [1+α material 100] [Eqn 3.9] So, the true temperature can be solved from the Eqn 3.9 as follows: T = T = 40 [1+α material 100] (1 40 α material ) 40 [1+α material 100] (1 40 α material ), where temperature is in K [Eqn 3.10], where temperature is in C [Eqn 3.11] So, the true temperature is obtained by inserting the value of α material = 10 5 K 1 in the Eqn The true temperature is C

5 4. Two mercury-in-glass thermometers are made of identical materials and are accurately calibrated at 0 C and 100 C. One has a tube of constant diameter, while the other has a tube of conical bore, 10 % greater in diameter at 100 C than at 0 C. Both thermometers have the length between 0 and 100 subdivided uniformly. What will be the straight bore thermometer read in a place where the conical bore thermometer reads 50 C. The volume of mercury in the tube at T 0 C, V T, is given by V T = V 0 [1 + β(t T 0 )] where V 0 is the volume of mercury at T 0 = 0 0 C, β is the coefficient of volume expansion of mercury. The volume of mercury expands linearly with temperature. For the straight bore thermometer with constant diameter, the height of the mercury column varies linearly with temperature. Therefore, at 50 0 C, mercury will be at a height half way between that of 0 0 C and C (see Fig. 1a). This is not true in the case of a conical bore since its cross sectional area is not a constant. So, if the mercury level is at half the length in the conical thermometer, the actual temperature is different from 50 0 C although we the thermometer reads 50 0 C. Let T be the true temperature when mercury rises half the length of the conical tube (the apparent temperature being 50 0 C). The volume filled up by mercury is ACDB, which is less than half of the mercury volume at C, i.e., volume AEFB (see Fig. 1b). Let EA and FB be extended to meet at G. Let l represent the length of thermometers and l the vertical height of the cone ABG, as shown in the figure. We can see that l l + l = d 1.1d Hence, l = 10l. Also, l l + l 2 = d CD Therefore CD = 1.05d. Now V = V 0. β. 100 V 0 T = V 0 βt We can arrive at V 0 T = T V or Volume ACDB Volume AEFB = T 100 Solve this equation to obtain T = C.

6 Fig It is desired to melt aluminum (melting point 660 C) with solid state specific heat capacity 0.9 kj/kg K and latent heat 390 kj/kg. The initial temperature of solid aluminum is 15 C and the final temperature of molten aluminum is 700 C. The specific heat capacity of molten metal is 1.11 kj/kg K and density is 2400 kg/m 3.Find how much metal can be melted to the required state per hour in a 2.17 MW furnace. Solution: Heat required per kg of aluminum= heating solid to melting temperature+ latent heat of fusion + heating liquid metal to final temperature = 0.9(660-15) ( ) = kj Mass of aluminum that can be melted per hour = (2.17x1000x3600) kj/h / kj/kg = kg/hr.

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