Chapter 3 - Mass Conservation and General Energy Equation PDF

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Fluid Mechanics: Mass Conservation and General Energy Equation

3.0 Mass Conservation Equation

and

General

Energy

The rules in fluid mechanics are the conservation of mass, conservation of momentum and conservation of energy. Since conservation of momentum will involve a moving fluid, it will be omitted in this chapter, but further explained in the next chapter: Momentum Equation and Its Application. For this chapter, only conservation of mass and conservation of energy are discussed as it is more compatible to be applied when the fluid remains static. Rather than looking into the details of origins of mass conservation equation and energy equation, this note will focus on the principles and their direct applications only. For mass conservation and energy equation are listed in Equation (3.1) and (3.2) just for information sharing 1 . The equations are widely used in the computational fluid dynamics (CFD). 2D Mass Conservation Equation for Incompressible Flow: 𝜕𝑢

+ 𝜕𝑥

𝜕𝑣

𝜕𝑦

=0

(3.1)

2D Energy Equation for Incompressible Flow: 𝜕𝑒

𝜌 ( 𝜕𝑡 +𝑢

𝜕𝑒

𝜕𝑢 2

𝜕𝑥

+𝑣

𝜇(2 ( 𝜕𝑥) +2 (

3.1

𝜕𝑒

𝜕𝑦

𝜕𝑣

) = 𝜌𝑞󰇗 + 2

) +( 𝜕𝑦

𝜕𝑢

𝜕

+ 𝜕𝑥

𝜕𝑥

(𝑘

𝜕𝑣 2 ) 𝜕𝑦

𝜕𝑇

𝜕𝑥

)+

𝜕

𝜕𝑦

(𝑘

𝜕𝑇

𝜕𝑦

) −𝑃(

𝜕𝑢

𝜕𝑥

+

𝜕𝑣

𝜕𝑦

𝜕𝑢

)+ 𝜆 ( 𝜕𝑥 +

𝜕𝑣 2 𝜕𝑦

) + (3.2)

Conservation of Mass

Conservation of Mass states that, for a closed system the mass of the system remains constant during the process. For a control volume (CV), the conservation of mass can be expressed as: 𝑚󰇗𝑖𝑛 −𝑚󰇗𝑜𝑢𝑡 =

𝑑𝑚𝐶𝑉

(3.3)

𝑑𝑡

where𝑚󰇗𝑖𝑛 is the total mass flow into the control volume, 𝑚󰇗𝑜𝑢𝑡 is the total mass flow out of the control volume, while

𝑑𝑚𝐶𝑉 𝑑𝑡

is the rate of change of mass within the CV.

For an incompressible flow in which the density of a fluid is constant over time, the mass flow rate (𝑚󰇗) can be described as: 1 Anderson J.D. (1995). Computational Fluid Dynamics: The Basics with Application, McGraw-Hill Series in Mechanical Engineering. Pg. 74.

1

Fluid Mechanics: Mass Conservation and General Energy Equation 𝑚󰇗 = 𝜌𝑣𝑎𝑣𝑒 𝐴

(3.4)

vave is the average velocity of the flow, while A is the cross sectional area of the flow. Now, if the flow is in steady state, then the rate of change of mass will become zero, to make the inlet mass flow rate equals outlet mass flow rate: 𝑑𝑚𝐶𝑉 𝑑𝑡

= 0 → 𝑚󰇗𝑖𝑛 −𝑚󰇗𝑜𝑢𝑡 = 0 → 𝑚󰇗𝑖𝑛 = 𝑚󰇗𝑜𝑢𝑡

(3.5)

For incompressible flow in which there is no change of density over time, the volumetric flow rate (𝑉󰇗 ) of the flow will be: (𝜌𝑣𝑎𝑣𝑒 𝐴)𝑖𝑛 = (𝜌𝑣𝑎𝑣𝑒 𝐴)𝑜𝑢𝑡 𝜌𝑖𝑛 = 𝜌𝑜𝑢𝑡 → 𝑣𝑖𝑛 𝐴𝑖𝑛 = 𝑣𝑜𝑢𝑡 𝐴𝑜𝑢𝑡

(3.6)

where𝑉𝑖𝑛󰇗 = 𝑣𝑖𝑛 𝐴𝑖𝑛 while𝑉𝑜𝑢𝑡󰇗 = 𝑣𝑜𝑢𝑡 𝐴𝑜𝑢𝑡 . For an arbitrary shape of CV, the Equation (3.3) is also written in differential form such that: 𝑑𝑚𝐶𝑉

= 𝑚󰇗𝑖𝑛 −𝑚󰇗𝑜𝑢𝑡 = −(𝑚󰇗𝑜𝑢𝑡 −𝑚󰇗𝑖𝑛 ) = −𝑚󰇗𝑛𝑒𝑡

𝑑(𝜌𝑉)

= −𝜌𝑣𝐴

𝑑𝑡

𝑑𝑡

𝑑 ∫ 𝜌 𝑑𝑉 𝑑𝑡 𝐶𝑉

= − ∫ 𝐶𝑆𝜌(𝑣. 𝑛󰇍 )𝑑𝐴

𝑑 ∫ 𝜌 𝑑𝑉 𝑑𝑡 𝐶𝑉

+∫ 𝐶𝑆𝜌(𝑣. 𝑛󰇍 )𝑑𝐴 = 0

(3.7)

𝑛󰇍 is the vector component which resolves the vector in a direction normal to the surface of the control volume.

𝑑 ∫ 𝜌 𝑑𝑉 𝑑𝑡 𝐶𝑉

refers to the rate of change of mass over

time within a CV, while ∫𝐶𝑆 𝜌(𝑣. 𝑛󰇍 )𝑑𝐴 refers to the net mass entering and leaving the CV across the inlet or outlet of the CV.

Examples: 1. A garden hose attached with a nozzle is used to fill a 40L bucket. The inner diameter of the hose is 2cm, and it reduces to 0.8cm at the nozzle exit. If it takes 50s to fill the bucket with water, determine: (a) the volume and mass flow rates of water through the hose, and (b) the average velocity of water at the nozzle exit.

2

Fluid Mechanics: Mass Conservation and General Energy Equation 2. A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls the air in and forces it through the resistors where it is heated. If the density of air is 1.2 kg/m3 at the inlet and 1.05kg/m3 at the exit, determine the percent increase in the velocity of air as it flows through the dryer.

Figure 3.1. Hair dryer. (Taken from: Çengel, Y., &Cimbala, J. (2014). Fluid mechanics: Fundamentals and applications (3rd ed.). McGraw-Hill Higher Education.)

3. A 1 m3 rigid tank initially contains air whose density is 1.18kg/m3. The tank is connected to a high pressure supply line through a valve. The valve is opened, and air is allowed to enter the tank until the density in the tank rises to 7.2 kg/m3. Determine the mass of air that has entered the tank. 4. Water (density = 1000 kg/m3) flows steadily across a device made up of one inlet and two exits as shown in Figure 3.2. The water flows into the device with the velocity of 10 m/s through a circular pipe with diameter 7.5 cm. The volume flow rate at the First Exit is 0.01 m3/s.

First exit

D = 7.5 cm

Device

v1 = 10 m/s Inlet

Second exit

𝑉󰇗 2 = 0.01 m3/s 𝑚󰇗3

Figure 3.2. Conservation of mass in a water device.

(a)

Calculate the velocity at the First Exit, if the First Exit is in circular shape with diameter of 10 cm.

(b)

Calculate the mass flow rate at the Second Exit (𝑚󰇗3) in kg/s.

3

Fluid Mechanics: Mass Conservation and General Energy Equation 3.2

Bernoulli Equation

The Bernoulli Equation is an approximate relation between pressure, velocity and elevation, which is valid in regions of steady, incompressible flow with negligible frictional forces. In other words, the equation implies extremely small viscous effect compared with the inertial, gravitational and pressure effects. The equation was first introduced by Daniel Bernoulli (1700-1782) in 1738, and derived by Leonhard Euler (1707-1783) in 1755. Now let us consider a fluid particle motion of a steady flow field as shown in Figure 3.3. The fluid particle is moving towards the direction of s. Direction n is perpendicular to direction of s.

(P+dP) dA ds s

n

θ PdA W

Figure 3.3. The schematic of a fluid particle motion for derivation of Bernoulli Equation.

Recall the Newton Second’s Law: ∑ 𝐹 = 𝑚𝑎

(3.8)

Taking the consideration for the s-direction force component to satisfy equation (3.8): 𝑃𝑑𝐴 −(𝑃 +𝑑𝑃)𝑑𝐴−𝑊sin 𝜃 = 𝑚𝑎

(3.9)

P is the pressure exerted on the fluid particle, A is the cross sectional area of fluid particle, W is the weight of fluid particle, as the acceleration of fluid particle, and m is the mass of fluid particle. The weight, W can be further evolved as: 𝑊 = 𝑚𝑔 = 𝜌𝑉𝑔 = 𝜌(𝑑𝐴. 𝑑𝑠)𝑔

(3.10)

Then the sin term is changed as well to a function of dz and ds, in which dz and ds is the angular component and tangential component respectively. This is because one needs dz to form up elevation component at the end of derivation and ds would come along as a package with dz. It can be formed by referring to the Figure 3.4. 4

Fluid Mechanics: Mass Conservation and General Energy Equation

ds dz θ

dx dz

θ

ds

Figure 3.4. The schematic of to form sin θ into a function of dz and ds.

Do note that dz is perpendicular to dx. Then sin θ is: sin 𝜃 =

𝑑𝑧

(3.11)

𝑑𝑠

The acceleration term, a, requires some attention. Recall from the kinematics equation2: 𝑣 2 = 𝑢 2 +2𝑎𝑠 𝑎=

𝑣 2 −𝑢 2 2𝑠

=

1 𝑑𝑉 2

(3.12)

2 𝑑𝑠

By substituting Equation (3.10) - (3.12) into Equation (3.9), 𝑃𝑑𝐴 −(𝑃 +𝑑𝑃)𝑑𝐴−𝜌(𝑑𝐴. 𝑑𝑠)𝑔 −𝑑𝑃𝑑𝐴−𝜌(𝑑𝐴. 𝑑𝑠)𝑔

𝑑𝑧

𝑑𝑧

= 𝜌(𝑑𝐴. 𝑑𝑠) 𝑑𝑠

= 𝜌(𝑑𝐴. 𝑑𝑠) 𝑑𝑠

1 𝑑𝑉 2 2 𝑑𝑠

1 𝑑𝑉 2 2 𝑑𝑠

(3.13)

By dividing Equation (3.13) with dA and ρ, and then −𝑑𝑃−𝜌(𝑑𝑠)𝑔

𝑑𝑧

𝑑𝑠

−𝑑𝑃−𝜌𝑔𝑑𝑧 = 𝜌 𝑑𝑉 2 𝑑𝑃 + 2 𝜌

= 𝜌(𝑑𝑠)

1 𝑑𝑉 2 2 𝑑𝑠

𝑑𝑉 2

2

+𝑔𝑑𝑧 = 0

(3.14)

For a steady and incompressible flow, the differential term in Equation (3.14) vanishes and it can be perceived as the mechanical energy of a flowing fluid, 𝑒𝑚𝑒𝑐ℎ . ∫

𝑑𝑃 𝜌

+∫

𝑑𝑉 2 2

+ ∫ 𝑔𝑑𝑧 = ∫ 0

(3.15)

The equation becomes the well-known Bernoulli Equation: 𝑉2 𝑃 + 2 𝜌

+𝑔𝑧 =constant (along a streamline)

2

Online Physics Hypertextbook. Retrieved from: http://physics.info/energy-kinetic/.

5

(3.16)

Fluid Mechanics: Mass Conservation and General Energy Equation Bernoulli Equation can also be written between two points (point 1 and point 2) on the same streamline: 𝑷𝟏 𝝆

+

𝑽𝟏 𝟐 𝟐

+𝒈𝒛𝟏 =

𝑷𝟐 𝝆

+

𝑽𝟐 𝟐 𝟐

+𝒈𝒛𝟐

(3.17)

The term P/ρ refers to the flow energy (pressure effect), the term v2/2 refers to the kinetic energy (inertia effect) while the gz refers to the potential energy (gravitational effect). From here, the important principle that we can deduct is that:

The energy cannot be created nor destroyed. The energy can only be transformed from one to another.

The abovementioned principle is the famous Conservation of Energy, or the First Law of Thermodynamics. Each term in Bernoulli Equation can also described as some kind of pressure. For instance, P = the static pressure 2 ρv /2 = the dynamic pressure (pressure formed due to flow kinetics) ρgz = the hydrostatic pressure (pressure formed due to depth difference) The summation of static and dynamics pressure is stagnation pressure. Although Bernoulli Equation is perceived as one of the simplest (quite effective sometimes), the equation has quite a number of weaknesses. These encompass: 1. Applicable only to steady flow. 2. Only suitable for case in which viscous effect is not significant. 3. Not applicable to the case when a moving shaft such as turbine is involved. 4. Due to its constant density, it is not applicable to compressible flow as well. 5. Heat transfer effect has been omitted.

6

Fluid Mechanics: Mass Conservation and General Energy Equation Examples: 1. Consider a river flowing towards a lake at an average velocity of 3 m/s at a rate of 500 m3/s at a location 90 m above the lake surface. Determine the total mechanical energy of the river water per unit mass and the power generation potential of the entire river at that location.

Figure 3.5. Example of calculation of mechanical energy. (Taken from: Çengel, Y., & Cimbala, J. (2014). Fluid mechanics: Fundamentals and applications (3rd ed.). McGraw-Hill Higher Education.)

2. A piezometer and a pitot tube are tapped into a horizontal water pipe to measure the static and stagnation pressure, as shown in Figure 3.6. Determine the velocity at the centre of the pipe.

Figure 3.6. Piezometer and pitot tube. (Taken from: Çengel, Y., & Cimbala, J. (2014). Fluid mechanics: Fundamentals and applications (3rd ed.). McGraw-Hill Higher Education.)

3. A Pitot-static probe is used to measure the velocity of an aircraft flying at 3000 m. If the differential pressure reading is 3 kPa, determine the velocity of the aircraft. Assume the density of the air is 1 kg/m3. 4. A pressurized tank of water has a 10 cm diameter orifice at the bottom, where water discharges to the atmosphere. The water level is 3 m above the outlet. The absolute tank air pressure above the water level is 300 kPa while the atmospheric pressure is 100 kPa. Neglecting the frictional effects, determine the initial discharge rate of water from the tank. 5. Air flows steadily from a tank, through a hoze of diameter D = 0.05 m and exits to the atmosphere from a nozzle of diameter d = 0.02 m as shown in Figure 3.7. The pressure in the tank remains constant at 3.0 kPa (gage

7

Fluid Mechanics: Mass Conservation and General Energy Equation pressure). Determine the volumetric flow rate and pressure in the hose. Assume the air density is 1.26 kg/m3. D

d

Air

Tank

Hose

Nozzle

Figure 3.7. Flow across the nozzle from a tank.

3.3

General Energy Equation

According to the Conservation of Energy Principle or the First Law of Thermodynamics, energy can neither be created nor destroyed during a process, and it can only change forms. The energy content of a closed system can be changed by two mechanisms: heat transfer (Q) and work transfer (W). Heat transfer (Q) is a process of energy transmission from a higher temperature region to lower temperature region. Heat can be transferred by conduction, convection and radiation. Work transfer (W) is the energy transmission due to external forces and devices. Some of the devices produce power (engines and turbines) while some consume energy (compressors, pumps and mixers). For the current course, the formulation on heat and work transfer is omitted. The focus will be on the consolidation of energy principle through applications. Consider a typical power plant energy generation system which involves pump and turbine. By modifying the Bernoulli Equation as in Equation (3.17), the general energy equation for power plant energy becomes: 𝑚󰇗 (

𝑃1

𝜌

+

𝑉1 2 2

+𝑔𝑧1 )+𝑊 󰇗 𝑝𝑢𝑚𝑝 = 𝑚󰇗 (

𝑃2 𝜌

+

𝑉2 2 2

+𝑔𝑧2 )+𝑊 󰇗 𝑡𝑢𝑟𝑏𝑖𝑛𝑒 +𝐸󰇗𝑚𝑒𝑐ℎ,𝑙𝑜𝑠𝑠

(3.18)

where 𝑚󰇗 is mass flow rate, 𝑊󰇗 𝑝𝑢𝑚𝑝 is shaft power input due to pump’s shaft, 𝑊󰇗𝑡𝑢𝑟𝑏𝑖𝑛𝑒 is the shaft output through turbine shaft and 𝐸󰇗𝑚𝑒𝑐ℎ,𝑙𝑜𝑠𝑠 is the total mechanical power loss in the whole system due to piping friction. To transform Equation (3.17) from the “work” form to “heads” form, which is its most common equation formation, 𝑊󰇗𝑝𝑢𝑚𝑝 , 𝑊󰇗𝑡𝑢𝑟𝑏𝑖𝑛𝑒 and 𝐸󰇗𝑚𝑒𝑐ℎ,𝑙𝑜𝑠𝑠 are related with their respective heads as: ℎ𝑝𝑢𝑚𝑝 =

𝑤𝑝𝑢𝑚𝑝

ℎ𝑡𝑢𝑟𝑏𝑖𝑛𝑒 =

=

𝑔

𝑤𝑡𝑢𝑟𝑏𝑖𝑛𝑒 𝑔

𝑊󰇗𝑝𝑢𝑚𝑝 𝑚󰇗 𝑔

=

=

𝜂𝑝𝑢𝑚𝑝 𝑊𝑝𝑢𝑚𝑝

𝑊󰇗 𝑡𝑢𝑟𝑏𝑖𝑛𝑒 𝑚󰇗 𝑔

(3.19)

𝑚󰇗 𝑔

=

𝑊𝑡𝑢𝑟𝑏𝑖𝑛𝑒

𝜂𝑡𝑢𝑟𝑏𝑖𝑛𝑒 𝑚 󰇗 𝑔

8

(3.20)

Fluid Mechanics: Mass Conservation and General Energy Equation ℎ𝐿 =

𝑒𝑚𝑒𝑐ℎ,𝑙𝑜𝑠𝑠 𝑔

=

𝐸󰇗𝑚𝑒𝑐ℎ,𝑙𝑜𝑠𝑠

(3.21)

𝑚󰇗 𝑔

The symbol η denotes efficiency. ℎ𝑝𝑢𝑚𝑝 is the useful head delivered to the fluid by the pump, ℎ𝑡𝑢𝑟𝑏𝑖𝑛𝑒 is the extracted head removed from the fluid by the turbine, while ℎ𝐿 is the irreversible head loss between 1 and 2 due to all components of the piping system other than turbine and pump. Wpump is lower than 𝑚󰇗𝑔ℎ𝑝𝑢𝑚𝑝 with a factor of 𝜂𝑝𝑢𝑚𝑝 due to the irreversible losses in pump, while Wturbine is larger than 𝑚󰇗𝑔ℎ𝑡𝑢𝑟𝑏𝑖𝑛𝑒 with a factor of 𝜂𝑡𝑢𝑟𝑏𝑖𝑛𝑒 due to the irreversible losses in turbine.

The irreversible head loss will be further discussed in Analysis of Pipe Flow. Divide Equation (3.18) with 𝑚󰇗 𝑔 and substitute Equation (3.19) – (3.21), then Equation (3.18) will become: 𝑉1 2 𝑃1 + 𝜌𝑔 2𝑔

+𝑧1 +ℎ𝑝𝑢𝑚𝑝, =

𝑉2 2 𝑃2 + 𝜌𝑔 2𝑔

+𝑧2 +ℎ𝑡𝑢𝑟𝑏𝑖𝑛𝑒 +ℎ𝐿

(3.22)

Gaspard Coriolis (1792-1843) discovered that the theoretical kinetic energy and actual kinetic energy is not the same, and the problem can be rectified by multiplying the kinetic energy terms with kinetic energy correction factor, α. The correction factor varies from case to case. α = 2.0 for fully developed laminar flow, and α = 1.04 to 1.11 in fully developed turbulent flow. Since in most of the real world cases, only turbulent flow is involved, the correction factor is ignored as it is near to unity. Therefore, for a steady incompressible flow, the general energy equation will become: 𝑷𝟏

+𝜶𝟏 𝝆𝒈

𝑽𝟏 𝟐

+𝒛𝟏 +𝒉𝒑𝒖𝒎𝒑,𝒖 = 𝟐𝒈

𝑽𝟐 𝟐 𝑷𝟐 +𝜶 𝟐 𝝆𝒈 𝟐𝒈

+𝒛𝟐 +𝒉𝒕𝒖𝒓𝒃𝒊𝒏𝒆,𝒖 +𝒉𝑳

(3.23)

Examples: 1. The pump of a water distribution system is powered by a 15 kW electric motor whose efficiency is 90%. The water flow rate through the pump is 50 L/s. The diameters of the inlet and outlet pipes are the same, and the elevation difference across the pump is negligible. If the absolute pressures at the inlet and outlet of the pump are measured to be 100 kPa and 300 kPa, respectively, determine： (a) the mechanical efficiency of the pump and (b) the temperature rise of water as it flows through the pump due to mechanical inefficiencies.

9

Fluid Mechanics: Mass Conservation and General Energy Equation

Figure 3.8. Example of water distribution system. (Taken from: Çengel, Y., & Cimbala, J. (2014). Fluid mechanics: Fundamentals and applications (3rd ed.). McGraw-Hill Higher Education.)

2. Water is being pumped from a large lake to a reservoir 25 m above a rate of 25 L/s by a 10 kW (shaft pump). If the irreversible head loss of the piping system is 7 m, determine the mechanical efficiency of the pump. 3. In a hydroelectric power plant, 100 m3/s of water flows from an elevation of 120 m to a turbine, where electric power is generated. The total irreversible head loss in the piping system from point 1 to point 2 (excluding the turbine unit) is determined to be 35 m. If the overall efficiency of the turbinegenerator is 80%. Estimate the electric power output.

Figure 3.9. Example of electric power plant. (Taken from: Çengel, Y., & Cimbala, J. (2014). Fluid mechanics: Fundamentals and applications (3rd ed.). McGraw-Hill Higher Education.) 4. A submersible pump with a shaft power of 5 kW and an efficiency of 72% is used to pump water from a lake to pool through a constant diameter pipe, to a tank 25 m above the lake surface. If the irreversible head loss in the piping system is 4 m, determine the discharge rate of water and the pressure difference across the pump.

10...