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ChemE_150A_HW_solutions

# ChemE_150A_HW_solutions - HW2 Problem 1 Solution First...

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Unformatted text preview: HW2 Problem 1 Solution: First, construct a table of the variables and arameters and their dimensions: Symbol L L/Tz H _ G _ w Efficienc of turbine 17 MLZ/T3 Next, generate a dimensional matrix of variables and dimensions: There are Z=8 total variables involved in the problem and Y=3 dimensions. So there should be Z—Y=8-3=5 dimensionless groups. We will call these dimensionless groups In 7132' 3133 3174’ 7135. The core variables, among them, must 1) contain all the dimensions (M, L, T) in the problem and 2) not form a dimensionless group themselves. We then write out expressions for the dimensionless groups using the core variables and each remaining variable/parameter: M a 1 ° 1 n= anwc = —— Lb— — =>a=0,b=0,c=0 1 p 7” 1(L3) (TM) M d 1‘ n2=pdD°wa[=](—l—?-)Li?)L=>d=0,f=0,e+1=0——>e=—1 , M g i L . . n3=pgthg[=] (13—) Lh(}—)(F)=>g=0,h+l=0, —1—2=O—>h=—1,1=—2 . k 1 M j k 1 ‘ L3 n = ’Da) = — L — — =>'=0,k+3=0, —l—1=0—>k=—3,1=—1 m o 2 ns=menw°P[=](ﬂ3) L“(l) M? =m+1=0,n—3m+2=0, —o—3=0—+m=—1,n=—5,o=—3 L T T plugging the exponents: Jtl=1’] H ”2-11—3— g Jr = 3 Da)2 Q ﬂ = 4 D300 P 7115— 3 a.- ~£J HWQ‘. Probftmék. Oi Cﬂmmﬁ 09} 59* 99> H31 02) T; main—Iain simJarH-‘a bt-Jwrcrz 4kg, ywadrl d 4LC’ Fro'fv‘hdfo 2:; _ :2 _ ha = J2. Hm ‘ H ’9 j) H H“ h, = ( IV.“ \. j) (\$9.3m) ((1014011) (Imiﬂglm) : 0 03w; QDmm :3 H’ us‘nrrt—nc" is ALA- 4L1:— b)TL»c 660?va 6"’Y’"“*‘ M‘+ b5 “Hm. GJUWHE— “gr +‘L‘ﬁ... 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W 1 _ _ W _ ._ 0 1 d/D 1 gﬁh’sﬂﬂyimﬂe__ u ‘LVﬁﬂLE1QLS_._~_ﬁ_L_i1‘£_L/EE§L—0‘3o 1-9M__. Wt “We Na "'1-_,_tég__‘mh7;..._,__.__.__._5_£77,~th. -Wz. _V*‘_\g¢ --~¢ 10“? (“W 9* lure.- j’kts Ama' ye) a sari“ rag Hindi \s gaze—”(1 gmwluqmt Problem 4 Solution The Reynolds number is n 1 79 x 111“S Pa 3 A Reynolds number greater than 2100 in a pipe normally indicates turbulent flow. (a) If the ﬂow is laminar, then - 16 - 16 - 0. 001164 f Re 13700 " The pressure dropL' 1n a 0.1--m long horizontal section ofthe pipe would be a 1n .. mm = 116;- ){2p1~-)= (0 001164) (0. 00:1) (2)(1 23 kg/m3xso m/s)-— - o 179 kPa (b) If 4000 < Re < 105 in a smooth pipe, the friction factor can be approximated by the Blasius equation ’ ' f = 0.079119%” = 0.079 C13700)'1"‘ = 0.007296 The pressure dropL' 1n a 0. l- -—m long horizontal section of the pipe would be 0 1m Mp}: f—G )(2pu‘)= (0 007296) (0. 004 mm) {231: 1 23 kg/mSXSO m/s}-— — 1.122 kPa (c) If the ﬂow is turbulent in a rough pipe (k = 0.045 mm), the friction factor can be calculated by the Colebrook equation 1 —J__ = —4' 0 logic 0')” 1.1+ 4.67 0+ Revtf + 2. 28: «1 01 (0045 mm + 4‘67 + :2 1’8 ' ng _ 41mm 1370(3‘31? ._ f = 0.01073 The pressure drop 1n a 0. 1- m- -long horizontal section of the pipe would be 0.1m Mp]: f(-5){2p113 )= (0 01078) (0 004 mm) (2X1 23 kg/n13)(50 nil/S): = 1. 657 kPa The pressure drop of 1.657 kPa in a length of 0.1 m of pipe corresponds to a change in absolute pressure (assuming p = 101 kPa) of approximately 1.657/101 = 0.016 or about 2%. 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ChemE_150A_HW_solutions - HW2 Problem 1 Solution First...

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