22. Dil of relative density 0.750 flows through the nozzle shown in Fig. 1-9 below and deflects the mercury in the U-tube gage. Determine the value of h if the pressure at A is 1.38 bar. P=10³ = 1000 g=9,81 Solution:

Structural Analysis
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Chapter2: Loads On Structures
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Can you please explain step by step how the final answer was obtained? Tried it out myself & don't know how they ended up getting the answer , please help!

22. Oil of relative density 0.750 flows through the nozzle shown in Fig. 1-9 below and deflects the
mercury in the U-tube gage. Determine the value of h if the pressure at A is 1.38 bar.
p=103=1000
g=9,81
Solution:
or, using bar units,
Another method:
0.8 m
Using convenient metres of water units,
B
Р'
9
1.38+ (0.75 x 9810)(0.8 + h) × 10¯5 = (13.57 × 9810)h × 10-5 and h =
1.14 m
*Convert Pa to bat x 10-5
B=C
Fig. 1-9
D
Pressure at B = pressure at C
pgh
pgh
105
10s (mercury)
PA +
bat
1.38 x 105
9810
C
-(oil) = PD +
Pressure head at B
_why not ()?
+ (0.8+ h)0.75=
1.38 bar
pressure head at C
13.57h and h
??
1.14 m, as before.
Transcribed Image Text:22. Oil of relative density 0.750 flows through the nozzle shown in Fig. 1-9 below and deflects the mercury in the U-tube gage. Determine the value of h if the pressure at A is 1.38 bar. p=103=1000 g=9,81 Solution: or, using bar units, Another method: 0.8 m Using convenient metres of water units, B Р' 9 1.38+ (0.75 x 9810)(0.8 + h) × 10¯5 = (13.57 × 9810)h × 10-5 and h = 1.14 m *Convert Pa to bat x 10-5 B=C Fig. 1-9 D Pressure at B = pressure at C pgh pgh 105 10s (mercury) PA + bat 1.38 x 105 9810 C -(oil) = PD + Pressure head at B _why not ()? + (0.8+ h)0.75= 1.38 bar pressure head at C 13.57h and h ?? 1.14 m, as before.
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