Civil Engineering - Hydraulics - Discussion
Discussion Forum : Hydraulics - Section 1 (Q.No. 5)
5.
If the atmospheric pressure on the surface of an oil tank (sp. gr. 0.8) is 0.1 kg/cm2, the pressure at a depth of 2.5 m, is
Discussion:
42 comments Page 1 of 5.
Kanishk said:
1 decade ago
How come the answer is C?
Initial Pressure in terms of ht. of water column= 1000/(1000*10)=0.1 m --- (i).
Pressure @ 2.5 m depth in oil tank = 2.5*800*10.
Representing above pressure as ht of water column, the answer comes out to be 2.0 m --- (ii).
(i) + (ii) = 2.1m.
Initial Pressure in terms of ht. of water column= 1000/(1000*10)=0.1 m --- (i).
Pressure @ 2.5 m depth in oil tank = 2.5*800*10.
Representing above pressure as ht of water column, the answer comes out to be 2.0 m --- (ii).
(i) + (ii) = 2.1m.
Nikku said:
1 decade ago
Pressure head in terms of water = 2.5/0.8 = 3.1m.
(1)
Hitesh kumar said:
1 decade ago
Total pressure = ((.1*10^4)/1000)+(.8*2.5) = 3.
KUMAR said:
1 decade ago
Initial pressure is 1000 kg/m^2.
Pressure at 2.5 m depth is 19620 kg/m^2.
Total pressure at a depth 2.5 m is = 19620+1000 = 20620 kg/m^2.
Depth in terms of water = 20620/(1000*9.81) = 2.1 m.
Pressure at 2.5 m depth is 19620 kg/m^2.
Total pressure at a depth 2.5 m is = 19620+1000 = 20620 kg/m^2.
Depth in terms of water = 20620/(1000*9.81) = 2.1 m.
Ruben said:
1 decade ago
From Basic Hydro static principle (FM text book by Frank m White).
P = Pa-Gamma (Z2-Z1).
Where Pa = Atmospheric pressure in n/m2 Z1 = 0 (Reference surface) and Z2 = negative downwards.
= (0.98 n/m2-7848(-2.5)).
= 19620.8 pa.
H = (19620.8/(1000*9.81)) = 2 m of water.
P = Pa-Gamma (Z2-Z1).
Where Pa = Atmospheric pressure in n/m2 Z1 = 0 (Reference surface) and Z2 = negative downwards.
= (0.98 n/m2-7848(-2.5)).
= 19620.8 pa.
H = (19620.8/(1000*9.81)) = 2 m of water.
Nikita patil said:
1 decade ago
How do you calculate the pressure at the given height? Please explain me.
Manpraveen said:
1 decade ago
P = $gh.
Pressure for oil = 0.8*1000*9.81*2.5 = 19620 n/m^2.
Pressure for water = 1000*9.81*h = 19620.
H1 = 19620/9810 = 2 m.
H2 = atmospheric head = p/$g.
H2 = 1000*9.81/1000*9.81= 1 m.
Total head = H1+H2 = 2+1 = 3 m (Answer).
Pressure for oil = 0.8*1000*9.81*2.5 = 19620 n/m^2.
Pressure for water = 1000*9.81*h = 19620.
H1 = 19620/9810 = 2 m.
H2 = atmospheric head = p/$g.
H2 = 1000*9.81/1000*9.81= 1 m.
Total head = H1+H2 = 2+1 = 3 m (Answer).
MA MURTAZA said:
1 decade ago
p*ATM = 0.1 kg/cm*2 = 1000 kg/m*2 = 9810 N/m*2.
p*oil = 19620 N/m*2 (after calculation).
Total pressure = 29430 N/m*2.
Hence, h*water = 29430/9810 = 3 m of water.
p*oil = 19620 N/m*2 (after calculation).
Total pressure = 29430 N/m*2.
Hence, h*water = 29430/9810 = 3 m of water.
Abdilahi khalif said:
1 decade ago
1000/(1000+2.5*0.8) = 3.
Rizgar Karim said:
1 decade ago
Patm = 0.1kg/cm2 = 0.1*10000*9.81/1000 = 9.81 KN/m2.
Poil = 0.8*9.81*2.5 = 19.62 KN/m2.
Ptotal = Patm + Poil.
Ptotal = 19.62 + 9.81 = 29.43 KN/m2.
P in-term of water = 29.43/9.81 = 3m of water.
Poil = 0.8*9.81*2.5 = 19.62 KN/m2.
Ptotal = Patm + Poil.
Ptotal = 19.62 + 9.81 = 29.43 KN/m2.
P in-term of water = 29.43/9.81 = 3m of water.
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