HW1_Soln - Problem N0. 1 1 a 62.331?- - 32.17—7 ’ 1450a...

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Unformatted text preview: Problem N0. 1 1 a 62.331?- - 32.17—7 ’ 1450a ‘ it s: i 7 P:P $17: A, = z 4 —4.43MP 2 I+ 1 15p51g+ 321711) .144” 15+627 6 2psrg a lbt 52 ft2 This is a higher pressure than exists in any municipal water systenn so in any such tall building the water, taken from the municipal water system must be pumped to the top. Furtheimore, one cannot tolerate such a high pressure in the drinking fountains on the ground floor; they would put out people's eyes Tall buildings have several zones in their internal water system, with suitable pressures in each, and storage tanks at the top of each zone. P 1 ad Problem No. 3 Payload : ~ pm); pgas 2 pm Wei—170— 41 Here the students are confronted with the fact that the weight, in kg, is not an SI unit. One must make the distinction between kgf and kgm to solve here, finding kg k 2 Okrf 9.81k p...=1.20—'—-g1:1~ 0 2’ ‘ g131m:1.20—o.0477=1.152 3 E“ m £(2Om)3_9'8]_ kgfs‘ 111 6 s3 f .. 1.2 0 0 Tm. : 7;“, 3e =293.15KL—O-] =305.3K: 32.14 C:89.9 F “ ‘ pgas 1.152 8 and C are at the same level in “the same liquid, mercury; the»:wa I hence we equate the pressures at B and C m 793]” gage. Pressure at B : pressure at C m + 14271 (for water) : pa + 1012(f01‘mercury) 20.x + 62.4(12.00~10.00) = O + (13.57 X 62.4)(1213571000) Solving, pi = 2115 psi‘ and p3 : 2115/1414 : 14.7 psi gage. An alternate solution using pressure heads in feet of water usually requires less arithmetic, as follows: Pressure head at B = pressure head at C p.\/w "e 2.00 ft water : 265 X 13.57 ft water Solving, 1) \/'LL‘ : 34.0 ft water and [Ill : (62.4X 34.0j/144 : 14.7 psi gage, as before. Pr /%7/“' (3’4: 5‘3): 5”” 3Q} ‘/ __ 46’] {$54) 9 M)[ 22 %)(7’””° 79 n a A (w 3" 5' 31,2 VQ/ém 117.95" [4: /,0,20 I; a. e 4,? :9 i515 ‘ Ff}? J 7C [/47“ 62 ? L”(?22'§5)/m2;) /‘/‘“L v 927—} 7 2 /9—7 M "” /7C $4 3,102 9);,“ Ibgf’” 1L4: - mm; M?” 277: ' jg: : ,2, 7/2 7C?“ X" IL1 A a”, (339!“ A—7/2 / f 0'0 #0 T} "Jr: 09? $1—+ Lazajck ' /'/05/ as o- X701; 2 «(35 ‘I = mfg“ Pf 5‘; fl" Vac = Pa + 5 h j a J; F P t 2 L1 [79 "f 6’ #7 152.5: / W‘lyri ( R?) V { fr‘ ‘ A mg; 11; Mqu .9. 62.71%: 1% M 7.7 4r” .93 fM/ _ 63,5, 5/ flag : 2%.? mg»; (22-? f5; (5;) ‘ 5+ "1 L 5:7? r r a g 1 a2; ,4 : 17757“; ...
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This homework help was uploaded on 04/17/2008 for the course CHE 353 taught by Professor Poehl during the Fall '07 term at University of Texas.

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HW1_Soln - Problem N0. 1 1 a 62.331?- - 32.17—7 ’ 1450a...

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