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12 Pages

### ch2-4

Course: CVEN 205, Spring 2008
School: Texas A&M
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Word Count: 2401

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2 Axially 122 CHAPTER Loaded Members Stresses on Inclined Sections Problem 2.6-1 A steel bar of rectangular cross section (1.5 in. 2.0 in.) carries a tensile load P (see figure). The allowable stresses in tension and shear are 15,000 psi and 7,000 psi, respectively. Determine the maximum permissible load Pmax. Solution 2.6-1 Rectangular bar in tension 2.0 in. P P 1.5 in. 2.0 in. P P 1.5 in. Maximum shear...

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Texas A&M - CVEN - 205
134CHAPTER 2Axially Loaded MembersProblem 2.6-16 A prismatic bar is subjected to an axial force that produces a tensile stress 63 MPa and a shear stress 21 MPa on a certain inclined plane (see figure). Determine the stresses acting on all faces
Texas A&M - CVEN - 205
144CHAPTER 2Axially Loaded MembersProblem 2.7-9 A slightly tapered bar AB of rectangular cross section and length L is acted upon by a force P (see figure). The width of the bar varies uniformly from b2 at end A to b1 at end B. The thickness t
Texas A&M - CVEN - 205
160CHAPTER 2Axially Loaded MembersStress ConcentrationsThe problems for Section 2.10 are to be solved by considering the stress-concentration factors and assuming linearly elastic behavior. Problem 2.10-1 The flat bars shown in parts (a) and (
Penn State - CHE - 210
CHAPTER ELEVEN11.1 a. The peroxide mass fraction in the effluent liquid equals that in the tank contents, which is:xp =Mp MTherefore, the leakage rate of hydrogen peroxide is m1 M p / M b. Balance on mass: Accumulation = input outputEdM =
Penn State - CHE - 210
CHAPTER TWO2.1 (a)= 18144 10 9 ms . 1 wk 1 d 1 h 1 s 38.1 ft / s 0.0006214 mi 3600 s (b) = 25.98 mi / h 26.0 mi / h 3.2808 ft 1 h 3 wk 7d 24 h 3600 s 1000 ms(c)554 m 4 1d 1h d kg 24 h 60 min1 kg 108 cm 4 = 3.85 10 4 cm 4 / min g 1000 g 1
Penn State - CHE - 210
CHAPTER THREE3.1 (a) m =16 6 2 m3 1000 kg 2 10 5 2 103 2 105 kg 3 mbgb gb gd i4 106(b) m =8 oz 2s1 qt106 cm31g332 oz 1056.68 qt cmb3 10gd10 i3 1 102 g / s(c) Weight of a boxer 220 lb m 12 220 lb m 1 stone Wmax
Penn State - CHE - 210
CHAPTER FOUR4.1 a. b. Continuous, Transient Input Output = Accumulation No reactions Generation = 0, Consumption = 06.00c.dn kg kg dn kg - 3.00 = = 3.00 dt dt s s st=100 m3 1000 kg 1 s . = 333 s 1 m3 3.00 kg4.2a. b. c.Continuous, S
Penn State - CHE - 210
CHAPTER FIVE5.1Assume volume additivity Av. density (Eq. 5.1-1):1=0.400 0.600 + = 0.719 kg L 0.703 kg L 0.730 kg LAAOa.Dmmass of tank at time tA= mt + m0 m =mass of empty tankAb250 - 150gkg = 14.28 kg min bm = mass f
Penn State - CHE - 210
CHAPTER SIX6.1 a.AB: Heat liquid - -V constantBC: Evaporate liquid - -V increases, system remains at point on vapor - liquid equilibrium curve as long as some liquid is present. T = 100 o C.CD: Heat vapor - -T increases, V increases .b. Point
Penn State - CHE - 210
CHAPTER SEVEN0.80 L 35 10 4 kJ 0.30 kJ work . 1h 1 kW = 2.33 kW 2.3 kW h L 1 kJ heat 3600 s 1 k J s7.12.33 kW 10 3 W 1.341 10 -3 hp 1 kW7.21W= 312 hp 3.1 hp .All kinetic energy dissipated by friction(a) E k =mu 2 2 5500 lbm 552 mil
Penn State - CHE - 210
CHAPTER EIGHT8.1 a.U (T ) = 25.96T + 0.02134T 2 J / molU (0 o C) = 0 J / mol U (100 o C) = 2809 J / mol Tref = 0 o C (since U(0 o C) = 0)b. We can never know the true internal energy. U (100 o C) is just the change from U (0 o C) to c.U (100 o
Penn State - CHE - 210
CHAPTER NINE9.14 NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) H ro = -904.7 kJ / mola.When 4 g-moles of NH3(g) and 5 g-moles of O2(g) at 25C and 1 atm react to form 4 g-moles of NO(g) and 6 g-moles of water vapor at 25C and 1 atm, the change in enth
Penn State - CHE - 210
CHAPTER TEN10.1 b. Assume no combustionn 1 (mol gas),T1 (C) x 1 (mol CH4 /mol) x 2 (mol C2 H6 /mol) 1 x 1 x 2 (mol C3 H8 /mol) n 2 (mol air), T2 (C) n 3 (mol), 200C y 1 (mol CH /mol) 4 y 2 (mol C2 H6 /mol) y 3 (mol C3 H8 /mol) 1 y 1 y 2 y 3 (m
Virginia Tech - CHE - 2114
CHAPTER TWO2.1 (a)= 18144 10 9 ms . 1 wk 1 d 1 h 1 s 38.1 ft / s 0.0006214 mi 3600 s (b) = 25.98 mi / h 26.0 mi / h 3.2808 ft 1 h 3 wk 7d 24 h 3600 s 1000 ms(c)554 m 4 1d 1h d kg 24 h 60 min1 kg 108 cm 4 = 3.85 10 4 cm 4 / min g 1000 g 1
Virginia Tech - CHE - 2114
CHAPTER THREE3.1 (a) m =16 6 2 m3 1000 kg 2 10 5 2 103 2 105 kg 3 mbgb gb gd i4 106(b) m =8 oz 2s1 qt106 cm31g332 oz 1056.68 qt cmb3 10gd10 i3 1 102 g / s(c) Weight of a boxer 220 lb m 12 220 lb m 1 stone Wmax
Virginia Tech - CHE - 2114
CHAPTER FOUR4.1 a. b. Continuous, Transient Input Output = Accumulation No reactions Generation = 0, Consumption = 06.00c.dn kg kg dn kg - 3.00 = = 3.00 dt dt s s st=100 m3 1000 kg 1 s . = 333 s 1 m3 3.00 kg4.2a. b. c.Continuous, S
Virginia Tech - CHE - 2114
CHAPTER SIX6.1 a.AB: Heat liquid - -V constantBC: Evaporate liquid - -V increases, system remains at point on vapor - liquid equilibrium curve as long as some liquid is present. T = 100 o C.CD: Heat vapor - -T increases, V increases .b. Point
Virginia Tech - CHE - 2114
CHAPTER FIVE5.1Assume volume additivity Av. density (Eq. 5.1-1):1=0.400 0.600 + = 0.719 kg L 0.703 kg L 0.730 kg LAAOa.Dmmass of tank at time tA= mt + m0 m =mass of empty tankAb250 - 150gkg = 14.28 kg min bm = mass f
Virginia Tech - CHE - 2114
CHAPTER SEVEN0.80 L 35 10 4 kJ 0.30 kJ work . 1h 1 kW = 2.33 kW 2.3 kW h L 1 kJ heat 3600 s 1 k J s7.12.33 kW 10 3 W 1.341 10 -3 hp 1 kW7.21W= 312 hp 3.1 hp .All kinetic energy dissipated by friction(a) E k =mu 2 2 5500 lbm 552 mil
Virginia Tech - CHE - 2114
CHAPTER EIGHT8.1 a.U (T ) = 25.96T + 0.02134T 2 J / molU (0 o C) = 0 J / mol U (100 o C) = 2809 J / mol Tref = 0 o C (since U(0 o C) = 0)b. We can never know the true internal energy. U (100 o C) is just the change from U (0 o C) to c.U (100 o
Virginia Tech - CHE - 2114
CHAPTER NINE9.14 NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) H ro = -904.7 kJ / mola.When 4 g-moles of NH3(g) and 5 g-moles of O2(g) at 25C and 1 atm react to form 4 g-moles of NO(g) and 6 g-moles of water vapor at 25C and 1 atm, the change in enth
Virginia Tech - CHE - 2114
CHAPTER TEN10.1 b. Assume no combustionn 1 (mol gas),T1 (C) x 1 (mol CH4 /mol) x 2 (mol C2 H6 /mol) 1 x 1 x 2 (mol C3 H8 /mol) n 2 (mol air), T2 (C) n 3 (mol), 200C y 1 (mol CH /mol) 4 y 2 (mol C2 H6 /mol) y 3 (mol C3 H8 /mol) 1 y 1 y 2 y 3 (m
Virginia Tech - CHE - 2114
CHAPTER ELEVEN11.1 a. The peroxide mass fraction in the effluent liquid equals that in the tank contents, which is:xp =Mp MTherefore, the leakage rate of hydrogen peroxide is m1 M p / M b. Balance on mass: Accumulation = input outputEdM =
Cornell - ECE - 2100
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