A 5.7 diameter horizontal pipe gradually narrows to 3.6 cm. The the water flows through this pipe at certain rate, the gauge pressure in these two sections is 32.5 kPa and 24.0 kPa, respectively. What is the volume of rate of flow?

Answer 1

The flow rate is #=2.88m^3s^-1#

Apply Bernoulli's Principle

#P_1+1/2rhov_1^2+rhogz_1=P_2+1/2rhov_2^2+rhogz_2#

Since the Pipe is horizontal, #z_1=z_2#

So,

#P_1+1/2rhov_1^2=P_2+1/2rhov_2^2#

The flow rate is constant

#Q=A_1*v_1=A_2*v_2#

Where #v_1# and #v_2# are the velocities of water in the pipe.

Where #A_1, A_2# are the cross sectional areas of the pipe

#A_1=pid_1^2/4=pi*(5.7*10^-2)^2/4#

#A_2=pid_2^2/4=pi*(3.6*10^-2)^2/4#

Therefore,

#v_1=A_2/A_1*v_2#

#v_1=(pi*(3.6*10^-2)^2/4)/(pi*(5.7*10^-2)^2/4)*v_2#

#v_1=(3.6/5.7)^2v_2#

#v_1=0.4v_2#

The pressures are

#P_1=32.5kPa#

#P_2=24kPa#

And the density of water is #rho=1000kgm^-3#

Therefore,

#32.5*10^3+1/2*1000*0.4v_2=24*10^3+1/2*1000*v_2#

#v_2(1/2*1000-1/2*1000*0.4)=(32.5-24)*10^3#

#v_2*1/2*1000*0.6=8.5*1000#

#v_2=(2*8.5)/0.6=28.33ms^-1#

Finally,

The flow rate is #Q=A_2v_2=pi*(3.6*10^-2)^2/4*28.33#

#=2.88m^3s^-1#

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Answer 2

To calculate the volume flow rate of water through the pipe, you can use the principle of conservation of mass, which states that the mass flow rate at any point in the pipe is constant. Using the formula:

[ A_1 \times v_1 = A_2 \times v_2 ]

where:

  • ( A_1 ) and ( A_2 ) are the cross-sectional areas of the pipe at sections 1 and 2, respectively,
  • ( v_1 ) and ( v_2 ) are the velocities of the water at sections 1 and 2, respectively.

Since the pipe is horizontal, the height difference between the two sections does not affect the pressure, and the velocities at the two sections can be considered the same.

The cross-sectional areas of the pipe at sections 1 and 2 can be calculated using the formula for the area of a circle:

[ A = \pi r^2 ]

Once you have the cross-sectional areas, you can rearrange the equation to solve for the velocity (( v_1 = v_2 )) and then use the formula for the volume flow rate (( Q )):

[ Q = A_1 \times v_1 ]

Given the gauge pressures at sections 1 and 2, you can calculate the absolute pressures (( P_1 ) and ( P_2 )) using the equation:

[ P_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}} ]

where ( P_{\text{atm}} ) is the atmospheric pressure, which is approximately 101.3 kPa.

Using Bernoulli's equation, you can relate the pressures to the velocities:

[ P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant} ]

Since the pipe is horizontal and there is no change in height, the equation simplifies to:

[ P + \frac{1}{2}\rho v^2 = \text{constant} ]

Solve for the velocities at sections 1 and 2 using the given gauge pressures, and then calculate the volume flow rate using the formula mentioned earlier.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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