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**Unformatted text preview: **Manometer equation: P1+P1id1+Pfih=P2+Pzid2
gc gc gc Both ﬂuid 1 and ﬂuid 2 are air.. Phs = P0 + pf ih = 750+972 =1722-mmHg
‘ g 0
Insert this result into the expression for the other manometer. z + d2 760 - mmHg
10.33-mH20 750 - mmHg +1208 -mmHg = 1722 - mmHg + 236~mmHg 10.33-mH20 _ z+0.lO-m
760-mmHg z=3.1-m ChE 501 Fall 2016 HW Assignment 4 Solution 1. Perform the following conversions. atm 23.4 - psi
14.696 - psi 23.4 psi to atm = 1.59 - atm =1.02x105 -Pa . 5 _
768 mm—Hg to Pa —768 mmHg ‘W 760 - mmHg 2. Perform the following conversions. 538 OC to °R 5380C+273=811K
0
—811K1°8 R=1.46><1030R : — "
K
0 0
85Kto°F 85 K 1'8KR=1530R 1530 R — 459 = —3060 F ch> 3. If the atmospheric pressure i.ind the pressure 15.5 feet below the surfaceofalake. 'H’ ‘3' WQOWWVSA‘ _- -__ -
<1 1' l \ Q ‘1“ a §\ <15 56mm, ‘ﬁ‘ﬁ’ma “EVH‘V N 3% V.) 0101m® 15-5'ft-H20 60-mmHg _ _ , P = P0 + ng / gc = 755mmHg +—— = 1.10x.103,-mmH§ “55le um 40‘? u K1305“ 5.338%?” 0\ Rm N3 \05 Q
We t + m . ’6‘ 9 “M a
Wm\ g 335%; B \Lg‘“ 4. The ends of a mercury manometer are attached to the bottom of a horizontal p1pe .
through which water is ﬂowing, with one end on each side of an orifice meter. The
manometer reading is 12.5 mm. Find the pressure difference betweeln the two sides of . . Jslwd m
the oriﬁce.
mm H. WWW ends
Use the general manometer equation. Ck ‘@ g \\ U29 6
1? Q.
g Q .
P1+P1—d1+Pf—g—h=P2+P2—g‘d2 (m gc gc gc ...

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- Spring '14
- DBarkey
- Chemical Engineering, pH