Chapter1_A - Chapter 1 - Section A - Mathcad Solutions 1.4...

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P 3000atm := D 0.17in := A π 4 D 2 := A 0.023in 2 = FP A := g 32.174 ft sec 2 = mass F g := mass 1000.7lb m = Ans. 1.7 P abs ρ g h P atm + = ρ 13.535 gm cm 3 := g 9.832 m s 2 := h 56.38cm := P atm 101.78kPa := P abs ρ g h P atm + := P abs 176.808kPa = Ans. 1.8 ρ 13.535 gm cm 3 := g 32.243 ft s 2 := h 25.62in := P atm 29.86in_Hg := P abs ρ g h P atm + := P abs 27.22psia = Ans. Chapter 1 - Section A - Mathcad Solutions 1.4 The equation that relates deg F to deg C is: t(F) = 1.8 t(C) + 32. Solve this equation by setting t(F) = t(C). Guess solution: t0 := Given t 1.8t 32 + = Find t () 40 = Ans. 1.5 By definition: P F A = F mass g = Note: Pressures are in gauge pressure. P 3000bar := D 4mm := A π 4 D 2 := A 12.566mm 2 = F PA := g 9.807 m s 2 = mass F g := mass 384.4kg = Ans. 1.6 By definition: P F A = F mass g = 1
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F Mars Kx := F Mars 41 0 3 × mK = g Mars F Mars mass := g Mars 0.01 kg = Ans. 1.12 Given: z P d d ρ g = and: ρ MP RT = Substituting: z P d d g = Separating variables and integrating: P sea P Denver P 1 P d 0 z Denver z Mg d = After integrating: ln P Denver P sea M g z Denver = Taking the exponential of both sides and rearranging: P Denver P sea e M g z Denver = P sea 1atm := M2 9 gm mol := g 9.8 m s 2 := 1.10 Assume the following: ρ 13.5 gm cm 3 := g 9.8 m s 2 := P 400bar := h P ρ g := h 302.3m = Ans. 1.11 The force on a spring is described by: F = K s x where K s is the spring constant. First calculate K based on the earth measurement then g Mars based on spring measurement on Mars.
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Chapter1_A - Chapter 1 - Section A - Mathcad Solutions 1.4...

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