531_PartUniversity Physics Solution

531_PartUniversity Physics Solution - Temperature and Heat...

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Temperature and Heat 17-17 (b) The brass bar is given as &heavy± and the wires are given as &fine,± so it may be assumed that the stress in the bar due to the fine wires does not affect the amount by which the bar expands due to the temperature increase. This means that L Δ is not zero, but is the amount brass 0 L T α Δ that the brass expands, and so 1 0 51 8 steel brass steel ( ) (20 10 Pa)(2.0 10 (C ) 1.2 10 (C ) )(120 C ) 1.92 10 Pa. F YT A αα −− =− Δ = × × ° × ° = × ° EVALUATE: The length of the brass bar increases more than the length of the steel wires. The wires remain taut and are under tension when the temperature of the system is raised above 20²C. 17.90. IDENTIFY: Apply the equation derived in part (a) of Problem 17.89 to the steel and aluminum sections. The sum of the L Δ values of the two sections must be zero. SET UP: For steel, 10 20 10 Pa Y and 1.2 10 (C ) . ° For aluminum, 10 7.0 10 Pa Y and 2.4 10 (C ) . ° EXECUTE: In deriving Eq.(17.12), it was assumed that 0; L Δ = if this is not the case when there are both thermal and tensile stresses, Eq. (17.12) becomes 0 . F LL T AY ⎛⎞ Δ= Δ+ ⎜⎟ ⎝⎠ (See Problem 17.89.) For the situation in this problem, there are two length changes which must sum to zero, and so Eq.(17.12) may be extended to two materials a and b in the form 0a a 0b b ab 0. FF LT AY AY + = Note that in the above, , and TF A Δ are the same for the two rods. Solving for the stress /, FA a0 a b0 b oa a 0b b . (( ) ( )) FL L T AL Y L Y + = −Δ + 8 10 10 (1.2 10 (C ) )(0.350 m) (2.4 10 (C ) )(0.250 m) (60.0 C ) 1.2 10 Pa. ((0.350 m 20 10 Pa) (0.250 m 7 10 Pa)) F A ×° + = × ×+ × EVALUATE: / F A is negative and the stress is compressive. If the steel rod was considered alone and its length was held fixed, the stress would be 8 steel steel 1.4 10 Pa. = × For the aluminum rod alone the stress would be 8 aluminum aluminum 1.0 10 Pa. = × The stress for the combined rod is the average of these two values. 17.91. (a) IDENTIFY and SET UP: The diameter of the ring undergoes linear expansion (increases with T ) just like a solid steel disk of the same diameter as the hole in the ring. Heat the ring to make its diameter equal to 2.5020 in. EXECUTE: 0 L Δ so 0 0.0020 in. 66.7 C (2.5000 in.)(1.2 10 (C ) ) L T L Δ = = ° 0 20.0 C 66.7 C 87 C TT T =+ Δ= °+ ° (b) IDENTIFY and SET UP: Apply the linear expansion equation to the diameter of the brass shaft and to the diameter of the hole in the steel ring. EXECUTE: 0 (1 ) L Δ Want sb (steel) (brass) = for the same T Δ for both materials: 0s s 0b b ) ) L TL T + so 0s 0s s b L L L T = + Δ 0s 0s s b 2.5020 in. 2.5000 in. (2.5000 in.)(1.2 10 (C ) ) (2.5050 in.)(2.0 10 (C ) ) T = −× ° × ° 55 0.0020 C 100 C 3.00 10 5.00 10 T ° = ° ×−× 0 20.0 C 100 C 80 C Δ= °− ° = −° EVALUATE: Both diameters decrease when the temperature is lowered but the diameter of the brass shaft decreases more since b s ; > bs | | | | 0.0020 in. Δ− 17.92. IDENTIFY: Follow the derivation of Eq.(17.12). SET UP: For steel, the bulk modulus is 11 1.6 10 Pa B and the volume expansion coefficient is 3.0 10 K . β EXECUTE: (a) The change in volume due to the temperature increase is , VT Δ and the change in volume due to the pressure increase is .
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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531_PartUniversity Physics Solution - Temperature and Heat...

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