Mechanical Engineering - Hydraulics and Fluid Mechanics - Discussion
Discussion Forum : Hydraulics and Fluid Mechanics - Section 2 (Q.No. 15)
15.
A water tank contains 1.3 m deep water. The pressure exerted by the water per metre length of the tank is
Discussion:
25 comments Page 2 of 3.
Arun kumar said:
8 years ago
P=ρgh^2/2
ρ=1000
g=9.81
h=1.3
Than put,
P=1000*9.81*1.3*1.3/2=8289.45N=8.289KN.
ρ=1000
g=9.81
h=1.3
Than put,
P=1000*9.81*1.3*1.3/2=8289.45N=8.289KN.
(1)
Waseem iqbal said:
8 years ago
Hydraulic force P= rAh'.
Where h' = depth of center of gravity = h/2 in this case.
So P = r (L*h) h/2 = 9.81 * (1.3)^2 /2. ( L cancel out for per meter length).
Where h' = depth of center of gravity = h/2 in this case.
So P = r (L*h) h/2 = 9.81 * (1.3)^2 /2. ( L cancel out for per meter length).
Ram said:
8 years ago
Thank you all for the explanation.
Vikaah Choudhary said:
8 years ago
Good explanation, thanks @Suman.
Vaibhav said:
8 years ago
Can anyone explain me how that area came?
Narpat said:
8 years ago
Here, w = &rho * g.
Gupta said:
8 years ago
Thanks for the explanation.
Manish kumar said:
9 years ago
We know that, pressure= £gh ,where £= density.
Now, pressure exerted by the water (force) = pressure* area = £gh* wh/2 (pre dia is triangular =(1/2)*base*h),
So, pressure exerted per unit lenght(pressure/w) = £gh* h/2 = (1000*9.81*1.3*1.3)/2 = 8.29 kN.
Now, pressure exerted by the water (force) = pressure* area = £gh* wh/2 (pre dia is triangular =(1/2)*base*h),
So, pressure exerted per unit lenght(pressure/w) = £gh* h/2 = (1000*9.81*1.3*1.3)/2 = 8.29 kN.
Vinayak said:
9 years ago
Thank you for giving the solution.
Amit said:
9 years ago
Thanks for giving the correct explanation.
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