CHE218_Exam2Key

# CHE218_Exam2Key - kJ 10 3 J:= kPa 10 3 Pa ⋅:= cp 10 2...

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Unformatted text preview: kJ 10 3 J := kPa 10 3 Pa ⋅ := cp 10 2- poise ⋅ := Problem 1 Given/Directly Calculated D 1 4.026in := from A.2, page 598) μ 6cp := ρ 50 lb ft 3 ⋅ := Δ P 13.5 psi ⋅ := D 2 2in := β D 2 D 1 := β 0.497 = Part a) Assuming a discharge coefficient: (This will need to be checked later) Using Equation 5.18 (page 149) C v .61 := V 2 C v 2 Δ P ρ 1 β 4- ( ) ⋅ 0.5 ⋅ := V 2 9.6 m s = V 2 31.485 ft s = Q V 2 π D 2 2 ⋅ 4 ⋅ := Q 0.687 ft 3 s = Q ρ ⋅ 34.34 lb s = Vpipe 4 Q ⋅ π D 1 2 ⋅ := Vpipe 7.77 ft s = Vpipe β 2 31.485 ft s = To justify the Cv, we need to calculate the Reynolds number and check Figure 5.14, p.153 N re V 2 D 2 ⋅ ρ ⋅ μ := N re 6.508 10 4 × = The assumed value for C v is justified. Variables with SI units for above: Δ P 93079.2235 N m 2 = ρ 900 kg m 3 ⋅ := V 2 9.6 m s = D 2 0.0508 m = μ 6 10 3- × kg m s ⋅ = Part b) The permanent pressure loss: Δ P final = (1- β 2 )(Δ P). If the flow rate doubles, the pressure drop...
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CHE218_Exam2Key - kJ 10 3 J:= kPa 10 3 Pa ⋅:= cp 10 2...

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