water is flowing in a fire house with velocity of 10 m/s and a pressure of 200 000 Pa. At the nozzle, the pressure decreases to atmospheric pressure (101 300), there no change in height. Use the Bernoulli equation equation to calculate the velocity of the water exiting the nozzle. The density of water is1000 kgm-3 and gravity is 9.8 ms-2.
Answers
Answer:
Let's write the Bernoulli equation:
\dfrac{1}{2}\rho v_1^2+\rho gh_1+p_1=\dfrac{1}{2}\rho v_2^2+\rho gh_2+p_2.
2
1
ρv
1
2
+ρgh
1
+p
1
=
2
1
ρv
2
2
+ρgh
2
+p
2
.
Since there is no change in height (h_1=h_2h
1
=h
2
) the equation takes the next form:
\dfrac{1}{2}\rho v_1^2+p_1=\dfrac{1}{2}\rho v_2^2+p_2,
2
1
ρv
1
2
+p
1
=
2
1
ρv
2
2
+p
2
,
v_2=\sqrt{\dfrac{\dfrac{1}{2}\rho v_1^2+p_1-p_2}{\dfrac{1}{2}\rho}},v
2
=
2
1
ρ
2
1
ρv
1
2
+p
1
−p
2
,
v_2=\sqrt{\dfrac{\dfrac{1}{2}\cdot1000\ \dfrac{kg}{m^3}\cdot(1.0\ \dfrac{m}{s})^2+2\cdot10^5\ Pa-1.013\cdot10^5\ Pa}{\dfrac{1}{2}\cdot1000\ \dfrac{kg}{m^3}}}=14.0\ \dfrac{m}{s}.v
2
=
2
1
⋅1000
m
3
kg
2
1
⋅1000
m
3
kg
⋅(1.0
s
m
)
2
+2⋅10
5
Pa−1.013⋅10
5
Pa
=14.0
s
m
.
Answer:
Hope it helps u...
Explanation:
