Question

Water enters a house from a ground like with an inside diameter of 2cm at an absolute pressure of 4.0×10^5Pa. A pipe that is 1cm leads to the second floor bathroom 5m above the ground floor. the flow speed at the ground pipe is 1.5m/s.
find the flow speed. ​

find the pressure.

find the volume flow rate in the bathroom.

Answer 1

Answer:

The flow speed , pressure and volume flow rate in the bathroom are  6 m/s , 3.3\times10^{5}\ Pa and 0.47 L/s.

Explanation:

Answer 2

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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