Answer:
The flow speed , pressure and volume flow rate in the bathroom are 6 m/s , 3.3\times10^{5}\ Pa3.3×10
5
Pa and 0.47 L/s.
Explanation:
Given that,
Inside diameter = 2.0 cm
Pressure P= 4.0\times10^{5}\ paP=4.0×10
5
pa
Diameter of pipe = 1.0 cm
Speed at inlet = 1.5 m/s
Height = 5.0 m
(I). We need to calculate the flow speed
Using continuity equation
A_{1}v_{1}=A_{2}v_{2}A
1
v
1
=A
2
v
2
v_{2}=(\dfrac{D_{1}}{D_{2}})^2\times v_{1}v
2
=(
D
2
D
1
)
2
×v
1
Where, A₁ = area of inside
A₂ = area of
v₁ = speed at inlet
Put the value into the formula
v_{2}=(\dfrac{2.0\times10^{-2}}{1.0\times10^{-2}})^2\times1.5v
2
=(
1.0×10
−2
2.0×10
−2
)
2
×1.5
v_{2}=6 m/sv
2
=6m/s
(II). We need to calculate the pressure
Using Bernoulli equation
P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}=P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho g h_{1}P
2
+
2
1
ρv
2
2
+ρgh
2
=P
1
+
2
1
ρv
1
2
+ρgh
1
Put the value in the equation
P_{2}+\dfrac{1}{2}\times1000\times6^2+1000\times9.8\times5.0=4.0\times10^{5}+\dfrac{1}{2}\times1000\times(1.5)^2+1000\times9.8\times0P
2
+
2
1
×1000×6
2
+1000×9.8×5.0=4.0×10
5
+
2
1
×1000×(1.5)
2
+1000×9.8×0
P_{2}+18000+49000=4.0\times10^{5}+1125P
2
+18000+49000=4.0×10
5
+1125
P_{2}=4.0\times10^{5}+1125-67000P
2
=4.0×10
5
+1125−67000
P_{2}=3.3\times10^{5}\ PaP
2
=3.3×10
5
Pa
(III). We need to calculate the volume flow rate
Using formula of the volume flow rate
Q= vAQ=vA
Where, v = final velocity
A = area
Put the value into the formula
Q=6\times\pi\times(0.005)^2Q=6×π×(0.005)
2
Q=4.7\times10^{-4}\ m^3/sQ=4.7×10
−4
m
3
/s
Conversation:
Q=\dfrac{4.7\times10^{-4}\ m^3}{s}\times\dfrac{1000\ L}{1\ m^3}Q=
s
4.7×10
−4
m
3
×
1 m
3
1000 L
Q=0.47\ L/sQ=0.47 L/s
Hence, The flow speed , pressure and volume flow rate in the bathroom are 6 m/s , 3.3\times10^{5}\ Pa3.3×10
5
Pa and 0.47 L/s.yap
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