Fluid Mechanics: Fundamentals and Applications
Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 3, Problem 48P

Freshwater and seamier flowing in parallel horizontal pipelines are connected to each other by a double U-tube manometer, as shown in Fig. P3-48. Determine the pressure difference between the two pipelines. Take the density of seawater at that location to be ρ = 71035 kg/m 3 . Can the air column be ignored in the analysis?

Expert Solution & Answer
Check Mark
To determine

The pressure difference between two pipelines.

Answer to Problem 48P

The pressure difference is 5.39kPa. The air column can be ignored since the density of air very low.

Explanation of Solution

Given information:

The density of seawater is 1035kg/m3.

The following figure represents freshwater and seawater flowing in parallel horizontal pipelines which are connected to each other by a double U-tube manometer.

  Fluid Mechanics: Fundamentals and Applications, Chapter 3, Problem 48P

Figure 1

Write the equation for the pressure difference of sea water and fresh water.

  ΔP=PHg+PairPfwPsw   ...... (I)

Here, the pressure difference is ΔP, the pressure on mercury is PHg, the pressure on air is Pair, pressure on freshwater is Pfw and pressure on seawater is Psw.

Write the expression for pressure on air.

  Pair=ρair×g×hair   ...... (II)

Here, density of air is ρair, acceleration due to gravity is g, and height of air column is hair.

Write the expression for density of mercury.

  ρHg=SGHg×ρw   ...... (III)

Here, specific gravity of mercury is SGHg and density of water is ρw.

Write the expression for pressure on mercury.

  PHg=ρHg×g×hHg   ...... (IV)

Here, pressure on mercury is PHg, density of mercury is ρHg, acceleration due to gravity is g and height of mercury column is hHg.

Write the expression for pressure on seawater.

  Psw=ρsw×g×hsw...... (V)

Here, pressure on seawater is Psw, density of seawater is ρsw, acceleration due to gravity is g, height of seawater column is hsw.

Write the expression for pressure on seawater.

  Pfw=ρfw×g×hfw...... (VI)

Here, pressure on freshwater is Pfw, density of freshwater is ρfw, acceleration due to gravity is g, height of freshwater column is hfw.

Calculation:

Substitute 0 for ρair in Equation (II).

  Pair=0×g×hair=0

Substitute 13.6 for SGHg and 1000kg/m3 in Equation (III).

  ρHg=(13.6)(1000kg/ m 3)=13600kg/m3

Substitute 13600kg/m3 for ρHg, 9.81m/s2 for g and 10cm for hHg in Equation (IV).

  PHg=(13600kg/ m 3)(9.81m/ s 2)(10cm)=(13600kg/ m 3)(9.81m/ s 2)(10cm× 1m 100cm)=(13600kg/ m 3)(9.81m/ s 2)(0.1m)=13341.6kg/ms2

  PHg=13341.6kg/ms2( 1Pa 1 kg/ ms 2 )=13341.6Pa( 1Pa 1000kPa)=13.34kPa

Substitute 1035kg/m3 for ρsw

  9.81m/s2 for g and 30cm for hsw in Equation (V).

  Psw=(1035kg/ m 3)(9.81m/ s 2)(30cm)=(1035kg/ m 3)(9.81m/ s 2)(30cm× 1m 100cm)=(1035kg/ m 3)(9.81m/ s 2)(0.3m)=3045.0055kg/ms2

  Psw=3045.0055kg/ms2( 1Pa 1 kg/ ms 2 )=3045.0055Pa( 1Pa 1000kPa)=3.045kPa

Substitute 1000kg/m3 for ρfw, 9.81m/s2 for g and 50cm for hfw in Equation (VI).

  Pfw=(1000kg/ m 3)(9.81m/ s 2)(50cm)=(1000kg/ m 3)(9.81m/ s 2)(50cm× 1m 100cm)=(1000kg/ m 3)(9.81m/ s 2)(0.5m)=4905kg/ms2

  Pfw=4905kPa( 1Pa 1 kg/ ms 2 )=4905kPa( 1Pa 1000kPa)=4.9kPa

Substitute 13.34kPa for PHg, 0 for Pair, 4.905kPa for Pfw and 3.045kPa for Psw in Equation (I)/

  ΔP=13.34kPa+0+4.905kPa+3.045kPa=5.39kPa

Conclusion:

The pressure difference is 5.39kPa.The air column can be ignored since the density of air very low.

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Chapter 3 Solutions

Fluid Mechanics: Fundamentals and Applications

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