Water is flowing through a horizontal pipe of
non-uniform cross-section. The speed of water is
30 cm s-1
at a place where pressure is 10 cm (of
water). Calculate the speed of water at the other place
where the pressure is half of that at the first place.
Answers
Answer:
Which is,
P + \frac {\rho v^2} {2} +\rho hg = constantP+
2
ρv
2
+ρhg=constant
Where,
P = pressure
\rho = density of the liquidρ=densityoftheliquid
v = velocity
h = height
g = acceleration due to gravity
Since it is a horizontal pipe, the height (potential energy part becomes zero)
Now the equation can be changed as below,
but we know that, pressure (P) = ⍴hg
\frac {\rho v^2} {2} +\rho hg = constant
2
ρv
2
+ρhg=constant
\frac {v^2} {2} + hg = constant
2
v
2
+hg=constant
substituting the given values in the above equation,
\Rightarrow h_1g + \frac {1} {2} v_1^2 = h_2 g + \frac {1} {2} v^2_2⇒h
1
g+
2
1
v
1
2
=h
2
g+
2
1
v
2
2
\Rightarrow 10\times10 + \frac {1} {2} \times(30)^2 = 5\times10 + \frac {1} {2} \times (v_2^2)⇒10×10+
2
1
×(30)
2
=5×10+
2
1
×(v
2
2
)
\Rightarrow 100 + 450 = 50 + (\frac {v_2^2} {2})⇒100+450=50+(
2
v
2
2
)
\Rightarrow 500 = \frac {v_2^2} {2}⇒500=
2
v
2
2
\Rightarrow 1000 = v_2^2⇒1000=v
2
2
\Rightarrow 31.62 \frac { cm }{ s } = v_2⇒31.62
s
cm
=v
2
Therefore, speed of water = 31.62 \frac { cm }{ s }=31.62
s
cm
"
Explanation:
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Water is flowing through a horizontal pipe of non-uniform cross-section. the speed of water is 30 cm/s at a place where pressure is 10 cm (of water). calculate the speed of water at the other place where the pressure is half of that at the first place
one year ago
Answers : (2)
576-1253_Screenshot_2019-08-15-04-26-17-671_com.android.chrome.pngone year ago
"We can solve this question by using Bernoulli's theorem.
Which is,
Where,
P = pressure
v = velocity
h = height
g = acceleration due to gravity
Since it is a horizontal pipe, the height (potential energy part becomes zero)
Now the equation can be changed as below,
but we know that, pressure (P) = ⍴hg
substituting the given values in the above equation,