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) 6cp := 50 lb ft 3 := P 13.5 psi := D 2 2in := D 2 D 1 := 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 P 1 4- ( ) 0.5 := V 2 9.6 m s = V 2 31.485 ft s = Q V 2 D 2 2 4 := Q 0.687 ft 3 s = Q 34.34 lb s = Vpipe 4 Q D 1 2 := Vpipe 7.77 ft s = Vpipe 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 := N re 6.508 10 4 = The assumed value for C v is justified. Variables with SI units for above: P 93079.2235 N m 2 = 900 kg m 3 := V 2 9.6 m s = D 2 0.0508 m = 6 10 3- kg m s = Part b) The permanent pressure loss: P final = (1- 2 )( 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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