Civil Engineering - Hydraulics - Discussion

Sure, let me explain in more detail.

The pressure exerted by a fluid increases with depth. In this case, we have an oil tank with a specific gravity of 0.8 and an atmospheric pressure of 0.1 kg/cm2 on the surface. To find the pressure at a depth of 2.5 m, we can use the formula:

Pressure = Atmospheric pressure + (Density of fluid x Acceleration due to gravity x Depth).

Since the specific gravity of oil is less than 1, we can use the formula:
Pressure = Atmospheric pressure + (Specific gravity x Atmospheric pressure x Depth).

Plugging in the values, we get:
Pressure = 0.1 kg/cm2 + (0.8 x 0.1 kg/cm2 x 2.5 m)
Simplifying the equation, we find that the pressure at a depth of 2.5 m is equivalent to the pressure exerted by 3 m of water.

Given, Atmospheric pressure P, oil = 0.1kg/cm² = 10000N/m².

{How I converted 0.1kg/cm² to N/m².
For this, we should know.
1kg = 10N & 1m = 100Cm.
So, (0.1x10)N÷(100-¹x100-¹)m².
= 10000N/m²}.

Density of oil = Specific gravity of oil x Density of water.
=>0.8x1000 = 800kg/m³.

The pressure at 2.5 m depth of oil surface will be:

P = Poil+{Density of oil (p)x Acc due to gravity(g) x depth of oil surface(h)}.
P = 10000N/m²+(800x10x2.5)N/m²,
P = 30000N/m² or Pa.

Now, converting to water head:

P = {Density of water(p) x Acc due to gravity(g) x depth of water(h)}
=>30000N/m²=1000x10xh
h = 3 meters of the water column.

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 3m.
P-atm = 0.1kg/cm2=0.1*10000=1000kg/m2=1000*9.81=9810 N/m2=9810/1000=9.81 KN/m2.
P-oil => p=0.8*1000 of watr=800kg/m3.

As P=pgh = 800*9.81*2.5=19620 N/m2=196209/1000 = 19.62 KN/m2.

P-total = P-atm + P-oil.
P-total = 19.62 + 9.81 = 29.43 KN/m2.
P in-term of water =>pg of water = 9.81KN/m3 so 29.43/9.81 = 3m height of water.

Specific gravity oil = 0.8.

Density of oil=0.8*1000=800 kg/m3
Atmospheric pressure at surface of oil=0.1 kg/cm2=0.1*10000*10=20,000 pascal
Pressure of oil at depth of 2.5 m = 800 * 2.5 * 10 = 10,000 pascal.

Total pressure = 10,000 + 20,000 = 30000 pascal.

In terms of water depth pressure = density of water * depth of water * gravity.

The Depth of water = pressure/density * gravity = 30,000/(10000*10) = 3 meter.

P(total pressure)= pressure of oil at 2.5m + p(ATM)
= 0.8X1000X9.81X2.5 + 0.1X9.81X10^4 (kg is converted into Newton by multiplying 9.81 & cm^2 is also converted into m^2).
= 29430 N/m^2.

Now,
The total pressure = pressure term of water.
29430 = 1000X9.81XH,
or, H = 29430/9810,
H = 3 m.
Hence, option C. 3 m of water is correct.

We know that;

1kg=9.81N and 1m=100cm.
let atmosppheric pressure at top surface be; P1=denssity*g*h=0.1kg/cm2=0.1*9.81*100*100=9810N/m2.
let pressure at heigt of 2.5m be p2=0.8*1000*9.81*2.5=19620N/M2.
now total pressureP =p1+p2=9810+19620=29430N/m2.
NOW pressure in terms of water =P/p1=29430/9810=3m so option C is correct.

First Patm = 0.1kg/sq/cm = 1000 kg/sq.m.
Now Poil = S(oil) * ρ(water) *h(oil) => 0.8 * 1000 * 2.5 = 2,000 kg/sq.m.

So, P(total at bottom) = 2,000 + 1,000 = 3,000 kg/sq.m.

Now h(pressure in terms of head of water) = P(total)/ρ(water) => 3,000 kg/sq.m/1,000 kg/cu.m = 3 m of water.

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.

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.

Given, S = 0.8, Patm = 0.1 kg/cm2 = 10000 N/m2.
Density of Oil= S x ρ(w)= 0.8x1000= 800 kg/m3.
Pressure at 5m below surface.

P = Patm+ ρ(oil)*g*h = 10000 + (800 * 10 * 2.5)= 30000 N/m2.
Now for water head.
P=ρ(w)*g*h.
30000=100*10*h,
h = 3m of water.