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 1 of 3.
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.
YATHEESH KUMAR D B said:
2 years ago
Let's consider a triangle and it has a height of 1.3 m and the pressure occurring at its bottom is p = 1.3 * 9.81 * 1000 = 12.753kPa.
Then, the triangle can be written as = 0.5 * B * H.
=0.5 * 12.753 * 1.3.
= 8.289kN/m.
Then, the triangle can be written as = 0.5 * B * H.
=0.5 * 12.753 * 1.3.
= 8.289kN/m.
(2)
Naresh said:
8 years ago
Yes right @Arun Kumar.
Here it is not mentioned that tank is a triangular or rectangular or as square tank.
so the pressure at a particular point is given by p=wh^2/2 and density of water is given by pgh.
Here it is not mentioned that tank is a triangular or rectangular or as square tank.
so the pressure at a particular point is given by p=wh^2/2 and density of water is given by pgh.
SUMAN BISWAS said:
1 decade ago
We know pressure exerted per unit length = ωH2/2.
So, the pressure of water per metre length = (9. 81x103x1.32) /2.
= 8.29 KN.
So, the pressure of water per metre length = (9. 81x103x1.32) /2.
= 8.29 KN.
Anomie said:
7 years ago
@All.
The pressure cannot have the unit kiloNewton. And the Pressure can only be exerted on an area there is nothing called as the pressure exerted on a unit length.
The pressure cannot have the unit kiloNewton. And the Pressure can only be exerted on an area there is nothing called as the pressure exerted on a unit length.
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).
Jsh said:
8 years ago
Pressure =density *gravity *cg of projected area.
Cg lies at half of height.
Force = pressure * area.
Area = height * length.
Force =ρg(h/2)*h*l.
F=8.2.
Cg lies at half of height.
Force = pressure * area.
Area = height * length.
Force =ρg(h/2)*h*l.
F=8.2.
(1)
Ravindra said:
9 years ago
Pressure diagram is triangular. Base is equal to wh and height h so pressure becomes 1/2 base * height. So pressure becomes wh*h/2.
Sushil Dumka said:
8 years ago
Yes, the answer is right,
8.29 kN.
Pressure diagram becomes triangular and area of triangular is 1/2 wh^2.
8.29 kN.
Pressure diagram becomes triangular and area of triangular is 1/2 wh^2.
(1)
Aman kumar said:
8 years ago
Is it pressure exerted per unit length or pressure force exerted per unit length? Please tell me.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers