2.1.1.1 Enthalpy

As discussed in Section 2.2.5.4, in a flow system, the fluid specific internal energy uu appears with the quantity P_s/ρPs/ρ, which equals specific flow work as a result of pressure force. By combining these two terms, we define the static enthalpy as

hs=u+Psρ(2.5)

Note that the static enthalpy h_shs given by Equation 2.5 is a state property, which is a combination of three state properties u, P_s,u,Ps, and ρρ. Only for a flow system, P_s/ρPs/ρ can be interpreted to equal the specific flow work. For a calorically perfect gas, which has constant c_pcp, we can write h_s = c_pT_shs=cpTs where we have assumed that h_s = 0hs=0 at the reference temperature T_s = 0Ts=0.

For a perfect gas with P_s = ρRT_sPs=ρRTs as its equation of state, we can rewrite Equation 2.5 as

giving

as a simple relation between the two specific heats and gas constant RR.

When we add the specific kinetic energy associated with the bulk gas flow to the static enthalpy of the gas, we obtain the total specific enthalpy of the flow as

ht=hs+12V2(2.7)

For a calorically perfect gas, we can rewrite Equation 2.7 as

ht=cpTs+12V2=cpTs+V22cp,ht=cpTt(2.8)

which yields a simple relation between total (stagnation) temperature and total (stagnation) enthalpy.

For a flow system, it is desirable to express the first law of thermodynamics in terms of enthalpy. Accordingly, we express Equation 2.4 as follows:

δq=du+Psdv=du+dPsv−vdPs=du+Psρ−dPsρ

where we have used v=1/ρ. Thus, we finally obtain

δq=dhs−dPsρ(2.9)

For an isobaric (constant pressure) process Equation 2.9 reduces to

which shows that, for a system at constant pressure, the amount of heat transfer appears as a change in fluid static enthalpy.

2.1.2.1 Entropy

We can have an infinite number of paths connecting states 1 and 2 of a fluid system, each path with different values of heat transfer and work transfer, and the associated process could be either reversible or irreversible. According to the second law of thermodynamics, for any two reversible paths AA and BB connecting states 1 and 2, the following equality holds:

∫12δqrevTsA=∫12δqrevTsB,(2.11)

which points to the existence of a state property, which we call entropy. Thus, we can write

s2−s1=∫12δqrevTsA=∫12δqrevTsB(2.12)

which is true for any reversible process connecting states 1 and 2. The second law of thermodynamics further states that the following inequality must be true for an irreversible path, say CC, connecting states 1 and 2

∫12δqirrevTsC<s2−s1,(2.13)

which means that a part of the entropy increase along path CC occurs as a result of irreversible work transfer. Because entropy is a state property, both reversible and irreversible processes connecting states 1 and 2 will have the same change in entropy, that is, (s₂ − s₁)_A = (s₂ − s₁)_B = (s₂ − s₁)_Cs2−s1A=s2−s1B=s2−s1C. This fact gives us a means to compute entropy change between any two states. A reversible process that is also adiabatic must, therefore, be isentropic. Thus, the second law of thermodynamics provides us with an important concept of the state property called entropy, which serves to delineate the feasible flow solutions from those that are not physically possible even if they satisfy the remaining conservation equations.

Computing entropy change between two states. Let us consider a reversible process connecting two states of a fluid flow. In this case Equation 2.9 yields

δqrev=dhs−dPsρ

Replacing the reversible heat transfer δq_revδqrev in terms of entropy in the aforementioned equation by using Equation 2.12, we obtain

Tsds=dhs−dPsρ(2.14)

Because Equation 2.14 involves only state properties, it holds good for both reversible and irreversible processes connecting any two states. For a calorically perfect gas, we can write Equation 2.14 as

ds=cpdTsTs−dPsρTs=cpdTsTs−PsρTsdPsPsds=cpdTsTs−RdPsPs(2.15)

Integrating Equation 2.15 between states 1 and 2 yields

∫12ds=cp∫12dTsTs−R∫12dPsPss2−s1=cplnTs2Ts1−RlnPs2Ps1(2.16)

For an isentropic (adiabatic and reversible) process, Equation 2.16 yields

cplnTs2Ts1−RlnPs2Ps1=0(2.17)

giving

Ps2Ps1=Ts2Ts1cpR=Ts2Ts1κκ−1(2.18)

where κ = c_p/c_vκ=cp/cv. At any point in a gas flow, the total pressure and total temperature are obtained assuming an isentropic stagnation process. Accordingly, at state 1, Equation 2.18 yields

cplnTt1Ts1−RlnPt1Ps1=0cplnTs1−RlnPs1=cplnTt1−RlnPt1(2.19)

Similarly, we can write at state 2

Combining Equations 2.19 and 2.20, we obtain

cplnTs2Ts1−RlnPs2Ps1=cplnTt2Tt1−RlnPt2Pt1.(2.21)

Thus, in terms of total pressure and total temperature, Equation 2.16 becomes

s2−s1=cplnTt2Tt1−RlnPt2Pt1(2.22)

Note that when we replace all static quantities in Equation 2.16 with the corresponding total quantities, we obtain Equation 2.22. For an isentropic process, Equation 2.22 yields

Pt2Pt1=Tt2Tt1cpR=Tt2Tt1κκ−1(2.23)

If the total temperature in a process connecting states 1 and 2 remains constant; for example, in an adiabatic flow in a stationary duct with wall friction, Equation 2.22 reduces to

s2−s1=−RlnPt2Pt1.Pt2Pt1=e−s2−s1R(2.24)

Because the wall friction will always increase entropy downstream of an adiabatic pipe flow, Equation 2.24 implies that the total pressure must decrease along this flow.

2.2.6.1 Mass Conservation (Continuity Equation)

According to the law of mass conservation we must have Π_S = Π_B = 0ΠS=ΠB=0 as there can be no agent that will change the mass of a system. In this case, we have ϕ = 1ϕ=1, and Equation 2.51 reduces to

∂∂t∭CVρdV+∬CVSρV→•dA→=0,(2.53)

which simply states that, for a given control volume with multiple inlets and outlets, the imbalance between mass inflows and outflows equals the rate of accumulation or reduction of mass in the control volume.

For ϕ = 1ϕ=1, Equation 2.52 reduces to the volume integral

∭CV∂ρ∂t+∇•ρV→dV=0

whose integrand must be zero, giving

∂ρ∂t+∇•ρV→=0.(2.54)

Note that Equation 2.54, which is a partial differential equation, must be satisfied at every point in a flow field. For a steady compressible flow, the continuity equation becomes ∇•ρV→=0, and for an incompressible flow, be it steady or unsteady, we obtain ∇•V→=0, which states that the velocity field of an incompressible flow must be divergence free.

Expanding terms within the volume integral on the right-hand side of Equation 2.52, using the vector identity for the divergence term involving scalar coefficients, and regrouping them, we obtain

ΠS+ΠB=∭CVρ∂ϕ∂t+V→•∇ϕ+ϕ∂ρ∂t+∇•ρV→dV.

Using Equation 2.54, the expression within the second curly brackets in the right-hand side of the aforementioned equation becomes zero. We finally obtain a simpler form of Equation 2.52

ΠS+ΠB=∭CVρ∂ϕ∂t+V→•∇ϕdV.(2.55)

2.2.6.2 Linear Momentum Equation

The linear momentum equation for a system is grounded in the Newton’s second law of motion, which simply states that the force acting on a system equals the mass of the system times its acceleration in the direction of the force. So the forces acting on a system are the change agent for its linear momentum. Accordingly, Equation 2.52 for the integral form of the linear momentum equation in an inertial (nonaccelerating) reference frame becomes

F→S+F→B=∂∂t∭CVV→ρdV+∬CVSV→ρV→•dA→(2.56)

where ΠS=F→S and ΠB=F→B, which are surface and body forces, respectively, and ϕ=V→ is the intensive property corresponding to the linear momentum M→. Surface forces arising from the static pressure at inlets, outlets, and walls of a control volume are compressive in nature, being always normal to the surface. Wall shear stresses, which also contribute to the surface forces, are locally parallel to the surface. In this text, we consider only two kinds of body forces, one as a result of gravity (weight) and the other as a result of rotation (centrifugal force).

It is instructive to analyze Equation 2.56 as three independent equations, one for each coordinate direction of a Cartesian coordinate system. This is possible because the forces acting along one coordinate direction will not change the momentum in the other coordinate direction, which is orthogonal. Accordingly, for the following presentation, let us only consider the xx component of Equation 2.56 in an assigned Cartesian coordinate system:

FSx+FBx=∂∂t∭CVVxρdV+∬CVSVxρV→•dA→.(2.57)

If the velocity, static pressure, and density are uniform at each inlet and outlet of a control volume, we can replace the surface integral on the right-hand of Equation 2.57 by algebraic summations. For a control volume with N_iNi inlets and N_jNj outlets having M_xMx as its total instantaneous value of xx momentum, Equation 2.57 simplifies to

FSx+FBx=∂Mx∂tCV+∑j=1j=NoutVxjṁjoutlets−∑i=1i=NinVxiṁiinlets(2.58)

where

• 
  

  V_xi≡Vxi≡ xx-momentum velocity at inlet i

  
  
• 
  

  V_xj≡Vxj≡ xx-momentum velocity at outlet j

  
  
• 
  

  ṁi≡ Mass flow rate at inlet i (ṁi=ρiAiVi)

  
  
• 
  

  ṁj≡ Mass flow rate at outlet j (ṁj=ρjAjVj)

For an inertial (nonaccelerating) control volume, Equation 2.58 reads the following:

• 
  

  (Sum of surface forces: pressure force and shear force)

  
  
  
  
  • 
    

    + (Sum of body forces: gravitational force and centrifugal force under rotation)

    
    
    
    
    • 
      

      = (Time-rate of change of xx-momentum within the control volume)

      
      
      
      
      • 
        

        + (Total xx-momentum flow rate leaving the control volume)

        
        
        
        
        • 
          

          – (Total x-momentum flow rate entering the control volume)

In Equation 2.58, V_iVi and V_jVj are the mass velocity at each inlet and outlet, respectively. They are always positive and independent of the chosen coordinate system. By contrast, V_xiVxi and V_xjVxj are the xx-momentum velocity at inlet i and outlet j, respectively. Whether they are positive or negative is determined relative to the positive direction of the chosen coordinate system. Identifying mass velocity (the velocity that produces mass flow rate) and momentum velocity (the velocity component for which the linear momentum equation is being considered) greatly simplifies the evaluation of the momentum flow rate with the correct sign at each inlet and outlet of a control volume.

To demonstrate the use of mass velocity and momentum velocity in the control volume analysis of linear momentum, let us consider a steady flow through the control volume shown in Figure 2.6. Based on the direction of each flow velocity, it is easy to see that sections 2 and 3 are inlets and sections 1, 4, and 5 are outlets. The mass velocity and xx-momentum velocity for each inlet and outlet are summarized in Table 2.1. Note that, with reference to the Cartesian coordinate system shown in the figure, the xx-momentum velocities for inlet 3 and outlet 1 are negative. For outlet 5, whereas the mass velocity V₅V5 in the yy direction is nonzero, the corresponding xx-momentum velocity is zero. With various quantities given in Table 2.1 for a steady flow through the control volume, the right-hand side of Equation 2.58 can be easily evaluated.

Figure 2.6 A control volume showing flow properties at its two inlets and three outlets.

Table 2.1 xx-Momentum Flow Rate and Stream Thrust at Each Inlet and Outlet of the Control Volume of Figure 2.6

Inlet/ Outlet	Mass Velocity	Mass Flow Rate	x-Momentum Velocity	x-Momentum Flow Rate	Stream Thrust
1 (outlet)	V1V1	ṁ1=ρ1A1V1	−V1−V1	−ṁ1V1	−ṁ1V1+Ps1A1
2 (inlet)	V2V2	ṁ2=ρ2A2V2	V2V2	ṁ2V2	ṁ2V2+Ps2A2
3 (inlet)	V3V3	ṁ3=ρ3A3V3	−V3−V3	−ṁ3V3	−ṁ3V3+Ps3A3
4 (outlet)	V4V4	ṁ4=ρ4A4V4	V4V4	ṁ4V4	ṁ4V4+Ps4A4
5 (outlet)	V5V5	ṁ5=ρ5A5V5	00	00	00

If we separate out the pressure force at each inlet and outlet of the control volume from the total surface forces F→S on the left-hand side of Equation 2.58, and combine it with the corresponding momentum flow on the right-hand side of the equation, we can express the momentum equation in terms of stream thrust in the xx direction. In Table 2.1, note that the stream thrust at each inlet and outlet has the same sign as the momentum velocity.

How to handle nonuniform flow properties at inlets and outlets and to express the linear momentum equation in a noninertial (accelerating or decelerating) reference frame attached to the control volume are discussed in comprehensive details by Sultanian (2015).

2.2.6.3 Angular Momentum Equation

Because angular momentum is expressed as H→=r→×M→, the angular momentum equation is not a new conservation law, but simply a moment of the linear momentum equation. In word form, the angular momentum equation can be expressed as

• 
  

  (Sum of the torque from surface forces: pressure force and shear force)

  
  
  
  
  • 
    

    + (Sum of the torque from body forces: gravitational force and centrifugal force)

    
    
    
    
    • 
      

      = (Time-rate of change of angular momentum within the control volume)

      
      
      
      
      • 
        

        + (Total angular momentum flow rate leaving the control volume)

        
        
        
        
        • 
          

          – (Total angular momentum flow rate entering the control volume)

In equation form, we can write the aforementioned statement in an inertial coordinate system as

Γ→S+Γ→B=∂H→∂tCV+∑j=1j=NoutH→̇joutlets−∑i=1i=NinH→̇iinlets.(2.58)

Similar to the approach presented in the previous section for the evaluation of linear momentum flow at an inlet or outlet of a control volume by using the concept of mass velocity and momentum velocity, we can evaluate each angular momentum flow rate by taking the product of mass velocity and angular momentum, which is considered positive in the counterclockwise direction and negative in the clockwise direction. Table 2.2 summarizes the flow rate of zz-component of the angular momentum for each inlet and outlet of the control volume shown in Figure 2.6, demonstrating the application of the approach proposed here.

A coordinate system rotating at constant angular velocity is considered noninertial. In this noninertial coordinate system with no linear and rotational accelerations, we can write, see Sultanian (2015), the integral angular momentum equation for a control volume as

Γ→S+Γ→B−∭CVr→×2Ω→×W→+Ω→×Ω→×r→ρdV=∬CVSr→×W→ρW→•dA⇀(2.59)

where Ω→ is the constant speed of rotation and the relative velocity W→ is the flow velocity in this coordinate system with its corresponding value in the stationary (inertial) coordinate system represented by the absolute V→. Note that expressing the linear momentum equation in this noninertial coordinate system gives rise two new terms: the Coriolis acceleration term 2Ω→×W→ and the centrifugal acceleration term Ω→×(Ω→×r→), whose moments appear in Equation 2.59.

For turbomachinery flows, including those in a gas turbine with a single axis of rotation, a cylindrical coordinate system shown in Figure 2.7 is a natural choice. In this coordinate system, the axial coordinate direction is aligned with the axis of rotation. As shown in the figure, the tangential direction is perpendicular to the meridional plane, which is formed by the axial and radial directions. For a constant rotational speed ΩΩ and steady flow, Sultanian (2015) presents a mathematically rigorous simplification of Equation 2.59 to yield

ΓS+ΓB=∬CVSrVθρW→•dA⇀.(2.60)

Figure 2.7 Cylindrical coordinate system used in turbomachinery flows.

In the derivation of Equation 2.60 from Equation 2.59, it is seen that the moment of the centrifugal force term makes no contribution and the volume integral of the moment of the Coriolis term contributes to the surface integral and converts the flow of angular momentum in the noninertial reference into that in the inertial reference frame at each inlet and outlet of the control volume. Equation 2.60 involves only the tangential velocity V_θVθ in the inertial reference frame, forming the basis for the Euler’s turbomachinery equation discussed in Section 2.2.8.

With reference to Figure 2.7, we see that the angular momentum vector at point AA in the flow about the turbomachinery axis of rotation is the cross-product of the radial vector r→ and the velocity vector V→. Because the radial velocity V_rVr is collinear with the radial vector, its contribution to angular momentum will be zero. The contribution of axial velocity V_xVx to angular momentum will be in the tangential direction and not along the axis of rotation. Thus, only the tangential velocity V_θVθ contributes to the angular momentum vector along the rotation axis, giving the angular momentum flow rate Ḣx=ṁrVθ.

For uniform properties at control volume inlets and outlets, we can replace the surface integral in Equation 2.60 by algebraic summations, giving

ΓS+ΓB=∑j=1j=NoutrjVθjṁjoutlets−∑i=1i=NinriVθiṁiinlets.(2.61)

2.2.6.4 Energy Equation

The energy equation embodies the first law of thermodynamics. According to Equation 2.1, heat transfer and work transfer are the agents responsible for change in the total energy of a system. Thus, with ϕ = e = u + V²/2 + gzϕ=e=u+V2/2+gz as the intensive property corresponding to the total energy EE, Equation 2.51 for the energy conservation becomes

Q̇−Ẇ=∂∂t∭CVρedV+∬CVSeρV→•dA→.(2.62)

In Equation 2.62, according to the convention used in Equation 2.1, the work transfer into the control volume becomes negative. Recognizing different types of work transfer into the control volume, we can write this equation as

Q̇+Ẇpressure+Ẇshear+Ẇrotation+Ẇshaft+Ẇother=∂∂t∭CVeρdV+∬CVSeρV⇀•dA⇀(2.63)

where

• 
  

  Q̇≡ Rate of heat transfer into the control volume through its control surface

  
  
• 
  

  Ẇpressure≡ Rate of work transfer into the control volume by pressure force

  
  
• 
  

  Ẇshear≡ Rate of work transfer into the control volume by shear force

  
  
• 
  

  Ẇrotation≡ Rate of Euler work transfer into the control volume by rotation

  
  
• 
  

  Ẇshaft≡ Rate of shaft work into the control volume

  
  
• 
  

  Ẇother≡ Rate of other energy transfers into the control volume: chemical energy, electrical energy, electromagnetic energy, and so on.

In a nondeformable control volume, the pressure force can do work only at its inflows and outflows. Within the control volume itself, the pressure force at any point has an equal and opposite pressure force, and does not contribute to the work transfer term Ẇpressure. At an inlet to the control volume, because V→ and A→ are in opposite direction, ρV⇀⋅dA⇀ will be negative, and the pressure force will do work on the flow entering the control volume.

According to our convention, we express the work transfer by the pressure force at inlets as

Ẇpressureinlets=−∬CVSPsρ(ρV⇀⋅dA⇀)inlets.

At control volume outlets, ρV⇀⋅dA⇀ is positive, and the pressure force opposes the flow exiting the control volume. We can thus write

Ẇpressureoutlets=−∬CVSPsρρV⇀⋅dA⇀outlets.

Combining the aforementioned two equations, we can express Ẇpressure by the following surface integral, which is true at all inlets and outlets:

Ẇpressure=−∬CVSPsρρV⇀⋅dA⇀

where P_s/ρPs/ρ is the specific flow work.

Transferring Ẇpressure to the right-hand side of Equation 2.63 yields

Q̇+Ẇshear+Ẇrotation+Ẇshaft+Ẇother=∂∂t∭CVeρdV+∬CVSu+Psρ+V22+gzρV⇀•dA⇀.(2.64)

Replacing u + P_s/ρu+Ps/ρ by the specific static enthalpy h_shs in Equation 2.64, we obtain

Q̇+Ẇshear+Ẇrotation+Ẇshaft+Ẇother=∂∂t∭CVeρdV+∬CVShs+V22+gzρV⇀•dA⇀.(2.65)

For uniform properties at control volume inlets and outlets, we can replace the surface integral in Equation 2.65 by algebraic summations, giving

Q̇+Ẇshear+Ẇrotation+Ẇshaft+Ẇother=∂E∂tCV+∑j=1j=Nouths+V22+gzjṁjoutlets−∑i=1i=Ninhs+V22+gziṁiinlets.(2.66)

Note that, in Equation 2.66, we can further combine the specific static enthalpy h_shs and specific kinetic energy V²/2V2/2 to yield the specific total enthalpy h_t = h_s + V²/2ht=hs+V2/2.

2.2.6.5 Entropy Equation

The entropy equation is founded in the second law of thermodynamics, which states that the entropy of an isolated system increases if the processes within the system are irreversible; the entropy remains constant when the processes are reversible. By introducing the concept of entropy production as a change agent, in addition to the change in entropy as a result of heat transfer, we can write Equation 2.51 for entropy as

Ṗentropy+Q̇T=∂∂t∭CVρsdV+∬CVSsρV→•dA→(2.67)

where ϕ = sϕ=s is the entropy per unit mass. In this equation, the rate of entropy production Ṗentropy accounts for entropy generation within the control volume from all irreversible processes and Q̇ is the rate of heat transfer into the control volume through its surface at temperature TT.

The unsteady integral term on the right-hand side of Equation 2.67 is the rate of storage of entropy within the control volume and should not be confused with Ṗentropy, which represents the rate of entropy production associated with the irreversibility of energy transfer into the control volume. For the case with no heat transfer, Ṗentropy equals the rate of increase in entropy stored within the control volume plus the excess of the entropy outflow rate over its inflow rate through the control-volume surface. Further note that there is no counterpart to Ṗentropy in the equations of mass, momentum, and energy whose conservation laws dictate zero production within the control volume.

In a steady flow, the unsteady integral term in Equation 2.67 vanishes, yielding Equation 2.68 in which Ṗentropy and other terms are considered time-independent.

Ṗentropy+Q̇T=∬CVSsρV⇀•dA⇀(2.68)

For uniform properties at control volume inlets and outlets, we can again replace the surface integral in Equation 2.68 by algebraic summations, giving

Ṗentropy+Q̇T=∑j=1j=Noutsjṁjoutlets−∑i=1i=Ninsiṁiinlets(2.69)

2.2.7.1 Rothalpy

The concept of rothalpy is grounded in Euler’s turbomachinery equation presented in the preceding section. Rearranging Equation 2.73 yields

which reveals that the quantity (h_t − UV_θ)ht−UVθ at any point in a rotor flow remains constant under adiabatic conditions (no heat transfer). This quantity is called rothalpy expressed as

I=ht−UVθ=hs+V22−UVθ=cpTtS−UVθ(2.75)

where both h_tht and V_θVθ are in the inertial (stationary) reference frame. Let us now convert Equation 2.75 into the rotor reference frame. If V_xVx, V_rVr, and V_θVθ are, respectively, the axial, radial, and tangential velocity components of the absolute velocity V→ at any point in the flow; and similarly, if W_xWx, W_rWr, and W_θWθ are, respectively, the axial, radial, and tangential velocity components of the corresponding relative velocity W→ at that point, we can write the following equations:

V2=Vx2+Vr2+Vθ2(2.76)

W2=Wx2+Wr2+Wθ2(2.77)

where UU is the local tangential velocity of the rotor. Substituting for V_θVθ from Equation 2.78 into Equation 2.76 and noting that W_x = V_xWx=Vx and W_r = V_rWr=Vr, we obtain

V2=Wx2+Wr2+Wθ2+2WθU+U2.(2.79)

Using Equations 2.78 and 2.79, we rewrite Equation 2.75 as

I=hs+Wx2+Wr2+Wθ2+2WθU+U22−UWθ+U,

which simplifies to

I=hs+W22−U22=htR−U22(2.80)

where h_tRhtR is the specific total enthalpy in the rotor reference frame. Equation 2.80 yields an interesting interpretation of rothalpy in a rotating system in that the change in fluid relative total enthalpy between any two radii in this system equals, under adiabatic conditions, the change in dynamic enthalpy associated with the solid-body rotation over these radii.

For a calorically perfect gas (constant c_pcp), we can also write Equation 2.80 as

I=cpTtR−U22(2.81)

where T_tRTtR is the fluid total temperature in the rotor reference frame.

2.2.7.2 An Alternate Form of the Euler’s Turbomachinery Equation

For an adiabatic flow in a rotor, the rothalpy remains constant, that is, I₁ = I₂I1=I2. Using Equation 2.80 we can write

hs1+W122−U122=hs2+W222−U222hs2−hs1=W122−W222+U222−U122.(2.82)

Equation 2.82 expresses the change in fluid static enthalpy in a rotor in terms of changes in the flow relative velocity and rotor tangential velocity. From the definition of total enthalpy, we can write its change between locations 1 and 2 as

ht2−ht1=hs2−hs1+V222−V122.(2.83)

Using Equation 2.82 to substitute for (h_s2 − h_s1)hs2−hs1 in Equation 2.83 yields the following alternate form of Euler’s turbomachinery equation earlier derived as Equation 2.73:

ht2−ht1=W122−W222+U222−U122+V222−V122(2.84)

where the terms on the right-hand side have the following physical interpretations:

• 
  

  W122−W222≡ Static enthalpy change as a result of change in rotor passage geometry

  
  
• 
  

  U222−U122≡ Static enthalpy change as a result of rotation of the rotor

  
  
• 
  

  V222−V122≡ Dynamic enthalpy change as a result of change in flow absolute velocity

2.2.7.3 Total Pressure and Total Temperature in Stator and Rotor Reference Frames

In turbomachinery, we need to consider flow properties in both stator (absolute, stationary, or inertial) and rotor (relative, rotating, or noninertial) reference frames. Because turbomachinery generally has a single axis of rotation, tangential (swirl) velocity is the only velocity component that is impacted by the change of reference frames from stator to rotor or vice versa. Both the axial and radial velocity components do not change. Further, all static properties of the fluid; for example, the static temperature and static pressure, remain the same in both stator and rotor reference frames.

Total temperature in both incompressible and compressible flows is computed by the equations:

Statorreferenceframe:TtS=Ts+V22cp(2.85)

Rotorreferenceframe:TtR=Ts+W22cp(2.86)

In an incompressible flow (constant density), the total pressure is evaluated by the equations:

Statorreferenceframe:PtS=Ps+ρV22(2.87)

Rotorreferenceframe:PtR=Ps+ρW22(2.88)

In a compressible flow, an isentropic stagnation process is assumed to obtain total properties. Accordingly, we invoke the isentropic relations to compute the total pressure using the following equations in each reference frame:

Statorreferenceframe:PtSPs=TtSTsκκ−1=1+κ−12MS2κκ−1(2.89)

Rotorreferenceframe:PtRPs=TtRTsκκ−1=1+κ−12MR2κκ−1(2.90)

Conversion from stator reference frame to rotor reference frame. Tangential velocities in the two reference frames are related by Equation 2.78, giving W_θ = V_θ − UWθ=Vθ−U. Because the fluid static temperature remains invariant in both reference frames, we can write

TtR−W22cp=TtS−V22cp.(2.91)

Because W_x = V_xWx=Vx and W_r = V_rWr=Vr, Equation 2.91 reduces to

TtR=TtS+Wθ22cp−Vθ22cp.(2.92)

Substituting W_θ = V_θ − UWθ=Vθ−U in Equation 2.92, we obtain

TtR=TtS+Vθ−U22cp−Vθ22cpTtR=TtS+UU−2Vθ22cp,(2.93)

which yields T_tR = T_tSTtR=TtS for V_θ = U/2Vθ=U/2.

For an isentropic compressible flow, we can use Equation 2.93 to obtain the pressure ratio as

PtRPtS=TtRTtSκκ−1=1+UU−2Vθ2cpTtSκκ−1PtR=PtS1+UU−2Vθ2cpTtSκκ−1.(2.94)

Conversion from rotor reference frame to stator reference frame. Using Equation 2.78 to substitute for V_θVθ in Equation 2.93, and rearranging terms, we obtain

TtS=TtR−UU−2Wθ−2U22cp=TtR−UU+2Wθ22cp(2.95)

PtSPtR=TtSTtRκκ−1=1−UU+2Wθ2cpTtRκκ−1,(2.96)

which uses the fact that both the static pressure and static temperature, being fluid state properties, are independent of the reference frame. Equation 2.96 thus gives

PtS=PtR1−UU+2Wθ2cpTtRκκ−1.(2.97)

An easier approach to first converting total temperatures, and then total pressures, between stator and rotor reference frames is by equating the expressions of rothalpy in these reference frames, as given by Equations 2.75 (I = c_pT_tS − UV_θI=cpTtS−UVθ) and 2.81 (I = c_pT_tR − U²/2I=cpTtR−U2/2) and converting tangential velocities in these reference frames by using Equation 2.78.

2.2.8.1 Isentropic Flow

Isentropic gas flows are hard to find in a real word, including gas turbines. Nevertheless, these flows serve as ideal reference flows in many design applications. From Equation 2.17, we obtain the following relation between the pressure ratio and temperature ratio at two points of an isentropic compressible flow:

Ps2Ps1=Ts2Ts1cpR=Ts2Ts1κκ−1(2.98)

Because the stagnation process in a compressible flow is assumed isentropic, a practical application of Equation 2.98 is to compute the maximum local stagnation pressure using the equation

PtPs=TtTscpR=TtTsκκ−1(2.99)

where the total temperature is computed as T_t = T_s + 0.5V²/c_pTt=Ts+0.5V2/cp. It is often convenient to express compressible flow equations in terms of Mach number. These equations for an isentropic compressible flow are summarized in Table 2.3.

Table 2.3 Isentropic Compressible Flow Equations

TtTs=1+κ−12M2
PtPs=TtTscpR=TtTsκκ−1=1+κ−12M2κκ−1		ρtρs=TtTs1κ−1=1+κ−12M21κ−1
With starred quantities corresponding to the sonic condition (M = 1M=1):
TtTs∗=κ+12	PtPs∗=κ+12κκ−1	ρtρs∗=κ+121κ−1
AA∗=1M2+κ−1M2κ+1κ+1κ−1

2.2.8.2 Mass Flow Functions

In an internal flow, mass flow rate is simply calculated as ṁ=ρAV, where the velocity VV is normal to the flow area AA. In steady state, regardless of changes in density, area, and velocity, the mass flow rate remains constant at each section. Further, for an incompressible flow and subsonic compressible flow in a duct, the exit static pressure must always equal the ambient static pressure, which acts as the discharge pressure boundary condition. In our approach in this textbook, we use only the properties at a section to compute mass flow rate at that section. For the flow to occur in either direction at a section, the total pressure at the section must be greater than the static pressure. Discounting the change in entropy as a result of heat transfer, an internal flow at any section occurs in the direction of increasing entropy.

In a compressible flow, the mass flow functions provide a convenient means of computing mass flow rate at a section without explicitly using the fluid density, which does not remain constant in such a flow. This approach is widely used by design engineers in the gas turbine industry. Starting with the equation ṁ=ρAV and using the isentropic flow equations along with the equation of state (P_s/ρ = RT_sPs/ρ=RTs), we can easily express the mass flow rate at a section as

ṁ=AF̂fsPsRTt=AFfsPsTt(2.100)

where F̂fs is the dimensionless static-pressure mass flow function given by

F̂fs=Mκ1+κ−12M2(2.101)

and F_fsFfs, having the dimensions of 1/R, is the static-pressure mass flow function given by

Ffs=MκR1+κ−12M2.(2.102)

Note that, for a calorically perfect gas, both F̂fs and F_fsFfs are functions of Mach number only.

Equation 2.100 shows that, for a given Mach number, the mass flow rate at a section is proportional to the static pressure and inversely proportional to the square root of the total temperature. Both the static pressure and total temperature are assumed to be uniform over that section.

Using P_t/P_sPt/Ps in terms of Mach number from Table 2.3, we can replace P_sPs in Equation 2.100 by P_tPt and alternatively express the mass flow rate as

ṁ=AF̂ftPtRTt=AFftPtTt(2.103)

where F̂ft is the dimensionless total-pressure mass flow function given by

F̂ft=Mκ1+κ−12M2κ+1κ−1(2.104)

and F_ftFft, having the dimensions of R, is the total-pressure mass flow function given by

Fft=MκR1+κ−12M2κ+1κ−1.(2.105)

For a calorically perfect gas both F̂ft and F_ftFft are functions of Mach number only.

Equation 2.103 shows that, for a given Mach number, the mass flow rate at a section is proportional to the total pressure and inversely proportional to the square root of the total temperature. Both the total pressure and total temperature are assumed to be uniform over that section.

Figure 2.9 shows the variations of mass flow functions F̂ft and F̂fs with Mach number for κ = 1.4κ=1.4. As shown in the figure, F̂ft increases as Mach number increases for M < 1M<1 (subsonic flow) and decreases as Mach number increases for M>1M>1 (supersonic flow), and it has a maximum value of 0.6847 at M = 1M=1(sonic flow), which corresponds to the choked flow conditions at the flow area. For example, for the ambient air at 20^°C20°C and 1.013 bar1.013bar, Equation 2.103 computes 239.2 kg/s239.2kg/s as the maximum possible mass flow rate through an area of 1.0 m²1.0m2. Furthermore, the figure shows that each value of Mach number yields a single value of mass flow function F̂ft, but each value of flow function F̂ft corresponds to two values (except at M = 1M=1) of Mach number – one subsonic and the other supersonic. While the calculation of F̂ft by Equation 2.103 for a given value of Mach number is direct, we need to use an iterative method (e.g., “Goal Seek” in MS Excel) to compute subsonic and supersonic Mach numbers for a given value of F̂ft.

Figure 2.9 Variations of mass flow functions F̂ft and F̂fs with Mach number for κ = 1.4κ=1.4.

As shown in Figure 2.9, unlike F̂ft, F̂fs increases monotonically with Mach number without exhibiting a maximum value at M = 1M=1, implying one-to-one correspondence between F̂fs and MM. Note that the ratio F̂fs/F̂ft equals the total-to-static pressure ratio P_t/P_sPt/Ps, which increases monotonically with Mach number, more rapidly for M>1M>1. For a given value of F̂fs, we present here a direct method to calculate the corresponding Mach number.

Finding Mach number for a given static-pressure mass flow function. Squaring both sides of Equation 2.101, we obtain

F̂fs2=M2κ1+κ−12M2=κM2+0.5κκ−1M4

0.5κκ−1M4+κM2−F̂fs2=0,

which is a quadratic equation in M²M2 with the positive root given by

M2=−κ+κ2+2κκ−1F̂fs2κκ−1M=−κ+κ2+2κκ−1F̂fs2κκ−1.(2.106)

Use of isentropic flow tables to compute mass flow functions. Not all isentropic flow tables list mass flow functions; however, they all list quantities like pressure ratio P_t/P_sPt/Ps and area ratio A/A^∗A/A∗ against Mach number. For an isentropic flow, we know that the total pressure and total temperature remain constant in a variable-area duct. In a duct flow, we can equate the mass flow rate at a give section of area AA to that at the sonic throat area A^∗A∗(imaginary if it is outside the duct), giving

ṁ=AF̂ftPtRTt=A∗F̂ft∗PtRTtF̂ft=F̂ft∗AA∗=0.6847AA∗(2.107)

and

F̂fs=F̂ftPtPs.(2.108)

Knowing the Mach number at a section, we can look up P_t/P_sPt/Ps and A/A^∗A/A∗ in isentropic flow tables; then use Equation 2.107 to compute F̂ft, followed by Equation 2.108 to compute F̂fs at that section.

2.2.8.3 Impulse Functions

As discussed in Section 2.2.5, the concept of stream thrust is grounded in the linear momentum equation. The change in stream thrust (the total stream thrust at outflows minus the total stream thrust at inflows) across a duct represents the total thrust on a duct as a result of fluid flow.

Using impulse functions, we can express the stream thrust at any section of a compressible duct flow as

where I_fsIfs is the static-pressure impulse function given by

and I_ftIft is the total-pressure impulse function given by

Ift=1+κM21+κ−12M2κκ−1.(2.111)

Note that the ratio I_fs/I_ftIfs/Ift equals the total-to-static pressure ratio P_t/P_sPt/Ps. For a given gas flow, both I_fsIfs and I_ftIft are functions of Mach number only.

Figure 2.10 shows the monotonically increasing parabolic variation of I_fsIfs with Mach number for κ = 1.4κ=1.4. As shown in the figure, I_fsIfs increases with MM for M < 1M<1, has a maximum value of 1.2679 at M = 1M=1, and decreases with MM for M>1M>1.

Figure 2.10 Variations of impulse functions I_ftIft and I_fsIfs with Mach number for κ = 1.4κ=1.4.

2.2.8.4 Normal Shock Function

When a supersonic compressible flow needs to adjust to a sudden increase in downstream static pressure, it undergoes a normal shock, which is a very thin nonequilibrium region featuring an abrupt increase in entropy, while its total temperature and mass flow rate remain constant. The flow upstream of a normal shock is always supersonic and the downstream flow always subsonic. Thus, the static temperature across a normal shock increases, and the flow velocity decreases with simultaneous increase in density to satisfy continuity.

Let us consider a constant-area control volume simulating a normal shock under the boundary conditions of no heat transfer and surface force. In this case, therefore, the mass flow rate, the stream thrust, and the total temperature over the control volume all remain constant. Thus, the ratio of mass flow rate to stream thrust at the inlet (upstream of the normal shock) and outlet (downstream of the normal shock) of our control volume must be equal. Evaluating this ratio using Equations 2.103 and 2.109 yields

ṁST=AF̂ftPtRTtPtAIft=F̂ftIft1RTt.

The aforementioned equation reveals that the ratio F̂ft/Ift must remain constant across a normal shock. We call this ratio, which is the ratio of total-pressure mass flow function to the total-pressure impulse function, the normal shock function NN expressed as the following function of MM and κκ:

N=F̂ftIft=F̂fsIfs=M1+κM2κ1+κ−12M2(2.112)

For M = 1M=1, Equation 2.112 yields

N∗1κ=κ21+κ,(2.113)

which for κ = 1.4κ=1.4 equals 0.540. For M→∞M→∞, Equation 2.112 yields N_∞N∞ as

N∞∞κ=κ−12κ.(2.114)

For κ = 1.4κ=1.4 we obtain N_∞ = 0.378N∞=0.378. It is interesting to note that the corresponding value of the subsonic Mach number is also equal to 0.378. This can be easily demonstrated by substituting M=κ−1/2κ in Equation 2.112 and obtaining N=κ−1/2κ after some simplification of the right-hand of the equation. From this we can conclude that, even for the strongest normal shock, the downstream subsonic Mach number will not be lower than κ−1/2κ.

The behavior of the normal shock function NN as a function of Mach number is depicted in Figure 2.11 for κ = 1.4κ=1.4. Like the total-pressure mass flow function F̂ft and the total-pressure impulse function I_ftIft, the figure shows that NN also increases with MM for M < 1M<1, reaches a maximum value of N^∗ = 0.540N∗=0.540 at M = 1M=1, and decreases with MM for M>1M>1.

Figure 2.11 Variation of normal shock function NN with Mach number for κ = 1.4κ=1.4.

In addition to simplifying the computations of compressible flows without the explicit use of density, the mass flow functions, impulse functions, and normal shock function presented in the foregoing sections are very helpful in explaining many seemingly nonintuitive behaviors these flows, as we will see in the following sections.

2.2.8.5 Isentropic Flow with Area Change

For the differential control volumes for mass conservation, force-momentum balance, and energy conservation shown in Figure 2.12, Sultanian (2015) presents a detailed derivation of the following differential equation for flow velocity change under the influence of area change, friction, heat transfer, and rotation:

1−M2dVV=−dAA−Mθ2drr+κM2ρAV2δFsh+δQ̇ṁcpTs(2.115)

where the rotational Mach number Mθ=rΩ/κRTs. Being nonlinear, Equation 2.115 generally requires a numerical solution when all effects are presents, as in a serpentine passage used in internal cooling of gas turbine blades (with rotation) and vanes (without rotation).

Figure 2.12 Differential control volumes: (a) mass, (b) linear momentum, and (c) energy.

To develop a good intuitive understanding of a compressible duct flow, let us consider one effect at a time. For an isentropic compressible flow in a variable-area duct, Equation 2.115 reduces to

1−M2dVV=−dAA,(2.116)

which clearly shows that, for subsonic (M < 1M<1) flows with the coefficient (1 − M²1−M2) being positive, the flow velocity increases as the duct area decreases and vice versa. For supersonic (M>1M>1) flows; however, the coefficient (1 − M²1−M2) being negative, the flow velocity increases with duct area and vice versa. As discussed earlier, a supersonic C-D nozzle used in rocket engines features this counterintuitive compressible flow behavior.

2.2.8.6 Isentropic Flow with Rotation

For an isentropic flow in a constant-area duct rotating around an axis other than along the flow direction, Equation 2.115 reduces to

1−M2dVV=−Mθ2drr.(2.117)

Note that the duct rotation essentially effect the flow in two ways; first, as a body force (centrifugal force) in the momentum equation, and second, as rotational work transfer in the energy equation. For the velocity change in the flow, both these effects are combined into the term on the right-hand side of Equation 2.117.

Because Mθ2 is positive definite, Equation 2.117 reveals that, for a subsonic flow, the velocity will increase when the flow is radially inward (dr < 0dr<0) and decrease when the flow is radially outward (dr>0dr>0). When the flow is supersonic in a constant-area rotating duct flow, it will exhibit the opposite behavior.

2.2.8.7 Nonisentropic Flow with Friction: Fanno Flow

The nonisentropic compressible flow with friction in a constant-area duct with no rotation and no heat transfer is known as Fanno flow. In this case, Equation 2.115 reduces to

1−M2dVV=κM2ρAV2δFsh.(2.118)

Because the right-hand side of Equation 2.118 is always positive, we infer from this equation that the friction will always accelerate a subsonic flow and decelerate a supersonic flow. The fact that the friction accelerates a subsonic flow in a constant-area duct is counterintuitive to our experience on the effects of friction, which is known to slow things down.

A simple physics-based reasoning behind why the flow accelerates in a subsonic Fanno flow, in which the total temperature, flow area, and mass flow rate remain constant, runs as follows. For the flow to overcome wall shear force as a result of friction, which always opposes the flow, we need a net pressure force in the flow direction. This entails that the static pressure and density (from the equation of state) decrease downstream. Because the duct area remains constant, the flow velocity must increase to maintain the constant mass flow rate.

Another way to explain the flow behavior in a Fanno flow is through the continuity equation expressed in terms of the total-pressure mass flow function, Equation 2.103. The effect of friction in an adiabatic flow is always to increase entropy and decrease total pressure in the flow direction. Because the mass flow rate at each section of a steady internal flow remains constant, Equation 2.103 dictates that the downstream total-pressure mass flow function (F̂ft) must increase in the flow direction. According to the variation of F̂ft with Mach number shown in Figure 2.9, it is clear that the downstream Mach number should move closer toward the sonic condition (M = 1M=1). This explains why in a Fanno flow the downstream Mach number increases if the flow is subsonic and decreases if it is supersonic. This reasoning can also be extended to conclude that the exit Mach number of a Fanno flow is limited to M = 1M=1(frictional choking) for both subsonic and supersonic flows at the inlet.

Table 2.4 summarizes a set of equations to evaluate changes in various properties in a Fanno flow. Detailed derivations of these equations are given by Sultanian (2015).

Table 2.4 Fanno Flow Equations

TsTs∗=κ+12+κ−1M2	ρρ∗=1M2+κ−1M2κ+1	PsPs∗=1Mκ+12+κ−1M2
PtPt∗=1M2+κ−1M2κ+1κ+12κ−1		PiPi∗=Ps1+κM2Ps∗κ+1
fLmaxDh=κ+12κlnκ+1M22+κ−1M2+1κ1M2−1

2.2.8.8 Nonisentropic Flow with Heat Transfer: Rayleigh Flow

The nonisentropic compressible flow in a nonrotating, constant-area, frictionless duct with heat transfer (both heating and cooling of the flow) is known as Rayleigh flow. Because friction is always present in real duct flows, Rayleigh flows are seldom found in engineering applications. Nevertheless, it is instructive to study these flows to develop a better understanding of the effects of heat transfer. The following reduced form of Equation 2.115 shows that the heating will increase flow velocity for a subsonic Rayleigh flow (M < 1M<1) and will decrease it for a supersonic Rayleigh flow (M>1M>1). In addition, this equation also shows that cooling the flow will have opposite trends in velocity changes in the flow direction.

1−M2dVV=δQ̇ṁcpTs.(2.119)

In a Rayleigh flow, in addition to the area and mass flow rate, the stream thrust (zero shear force from the wall) remains constant at each section. Combining Equations 2.103 and 2.109, we can express the stream thrust as

ST=ṁRTtN.(2.120)

If the upstream and downstream sections of a Rayleigh flow are designated by subscripts 1 and 2, respectively, we have S_T1 = S_T2ST1=ST2, which according to Equation 2.120 yields

N2N1=Tt2Tt1.

For heating, we have T_t2>T_t1Tt2>Tt1, which means N₂>N₁N2>N1. According to the variation of NN with Mach number shown in Figure 2.11, we further conclude that M₂M2 should be closer to M = 1M=1 than to M₁M1. In other words, if the upstream flow is subsonic, the Mach number will increase downstream. By contrast, if the flow is supersonic, the Mach number will decrease downstream. Thus, heating in a Rayleigh flow raises a subsonic Mach number and lowers a supersonic Mach number toward the sonic condition (thermal choking).

From the constancy of stream thrust between any two sections of a Rayleigh flow, we can write

Pt1Ift1=Pt2Ift2Pt2Pt1=Ift1Ift2.(2.121)

Based on the dependence of I_ftIft on Mach number shown in Figure 2.10, because M₂M2 is on the same side as M₁M1 and closer to M = 1M=1, we can infer that I_ft1 < I_ft2Ift1<Ift2. Thus, Equation 2.121 yields P_t2 < P_t1Pt2<Pt1 for both subsonic and supersonic Rayleigh flows. This result explains another nonintuitive behavior of a Rayleigh flow in that the increase in entropy as a result of heating leads to decrease in total pressure in the flow direction. Conversely, cooling in a Rayleigh flow increases the total pressure downstream.

A subsonic Rayleigh flow exhibits yet another nonintuitive feature. In the Mach number range 1/κ<M<1, the static temperature decreases with heating and increases with cooling. For the case of heating in this Mach number range, although the total temperature increases, the flow accelerates very rapidly to satisfy the continuity equation with concomitant reduction in the static temperature. The opposite happens for the case of cooling.

Table 2.5 summarizes a set of equations to evaluate changes in various properties in a Rayleigh flow. Detailed derivations of these equations are given by Sultanian (2015).

Table 2.5 Rayleigh Flow Equations

TsTs∗=M2κ+12+κM22	ρρ∗=1M21+κM2κ+1	PsPs∗=κ+11+κM2
PtPt∗=κ+11+κM22+κ−1M2κ+1κκ−1	TtTt∗=M2κ+11+κM222+κ−1M2κ+1

2.2.8.9 Nonisentropic Flow: Normal Shock

The normal shock, which occurs only in a supersonic compressible flow, is a compression wave normal to the flow direction. With no area-change, friction, heat transfer, and rotation, the right-hand side of Equation 2.115 vanishes, yielding

1−M2dVV=0.(2.122)

A trivial solution of Equation 2.122 corresponds to dV = 0dV=0, which implies equal velocities upstream and downstream of a normal shock. The second solution corresponds to abrupt changes in flow properties across the normal shock. These changes are primarily driven by the nonequilibrium and nonisentropic processes, allowing the flow to adjust itself to a higher downstream static pressure boundary condition by becoming subsonic. Because the changes of properties across a normal shock are discontinuous, the governing differential equations cannot be integrated across it. However, we can use a control volume analysis of the governing conservation laws over a normal shock to develop simple algebraic equations, summarized in Table 2.6, which relate flow properties before and after the shock.

Table 2.6 Normal Shock Equations

M22=2+κ−1M122κM12−κ−1	Ps2Ps1=2κM12−κ−1κ+1	ρ2ρ1=κ+1M122+κ−1M12
Ts2Ts1=2κM12−κ−1κ+12+κ−1M12κ+1M12
Pt2Pt1=κ+12κM12−κ−11κ−1κ+1M122+κ−1M12κκ−1
Pt2Ps1=2κM12−κ−1κ+1−1κ−1κ+12M12κκ−1

As we noted in Section 2.2.8.4, the normal shock function NN remains constant across a normal shock. Figure 2.11 shows that, for a given value of NN above the asymptotic value of 0.378, we obtain two values of MM, one subsonic and the other supersonic. Let us go through a simple physical reasoning to show that the Mach number upstream of a normal shock must be supersonic. Let the sections upstream and downstream of a normal shock be denoted by subscripts 1 and 2, respectively. As discussed earlier, the flow across a normal shock is adiabatic (T_t1 = T_t2Tt1=Tt2). For entropy to increase across a normal shock, Equation 2.22 yields P_t2 < P_t1Pt2<Pt1. Because the static pressure increases across a normal shock (compression wave), we obtain P_t2/P_s2 < P_t1/P_s1Pt2/Ps2<Pt1/Ps1, which from isentropic relations leads to the result M₂ < M₁M2<M1. Thus, for NN remaining constant across a normal shock, we conclude that the flow before a normal shock must be supersonic, resulting in a subsonic flow after the shock.

2.2.9.1 Inertial Reference Frame: Cartesian Coordinates

In this coordinate system, the velocity V→ (also called absolute velocity) has components V_x, V_y,Vx,Vy, and V_zVz in x, y,x,y, and zz coordinate directions, respectively. Equation 2.215 in each coordinate direction and the continuity equation are written as follows:

• 
  

  xx-coordinate direction:

ρ∂Vx∂t+Vx∂Vx∂x+Vy∂Vx∂y+Vz∂Vx∂z=−∂Ps∂x+Fx+μ∂2Vx∂x2+∂2Vx∂y2+∂2Vx∂z2(2.127)

• 
  

  yy-coordinate direction:

ρ∂Vy∂t+Vx∂Vy∂x+Vy∂Vy∂y+Vz∂Vy∂z=−∂Ps∂y+Fy+μ∂2Vy∂x2+∂2Vy∂y2+∂2Vy∂z2(2.128)

• 
  

  zz-coordinate direction:

ρ∂Vz∂t+Vx∂Vz∂x+Vy∂Vz∂y+Vz∂Vz∂z=−∂Ps∂z+Fz+μ∂2Vz∂x2+∂2Vz∂y2+∂2Vz∂z2(2.129)

• 
  

  Continuity equation:

  
  
  ∂Vx∂x+∂Vy∂y+∂Vz∂z=0(2.130)
  
  

2.2.9.2 Inertial Reference Frame: Cylindrical Coordinates

In this coordinate system, the velocity V→ (also called absolute velocity) has components V_r, V_θ,Vr,Vθ, and V_xVx in r, θ,r,θ, and xx(axial) coordinate directions respectively, as shown in Figure 2.13a. Equation 2.215 in each coordinate direction and the continuity equation are written as follows:

• 
  

  rr-coordinate direction:

ρ∂Vr∂t+Vr∂Vr∂r+Vθr∂Vr∂θ+Vx∂Vr∂x−Vθ2r=−∂Ps∂r+Fr+μ∂2Vr∂r2+1r∂Vr∂r+1r2∂2Vr∂θ2−2r2∂Vθ∂θ+∂2Vr∂x2−Vrr2(2.131)

• 
  

  θθ-coordinate direction:

ρ∂Vθ∂t+Vr∂Vθ∂r+Vθr∂Vθ∂θ+Vx∂Vθ∂x+VrVθr=−1r∂Ps∂θ+Fθ+μ∂2Vθ∂r2+1r∂Vθ∂r+1r2∂2Vr∂θ2+2r2∂Vr∂θ+∂2Vθ∂x2−Vθr2(2.132)

• 
  

  xx-coordinate direction:

ρ∂Vx∂t+Vr∂Vx∂r+Vθr∂Vx∂θ+Vx∂Vx∂x=−∂Ps∂x+Fx+μ∂2Vx∂r2+1r∂Vx∂r+1r2∂2Vx∂θ2+∂2Vx∂x2(2.133)

• 
  

  Continuity equation:

  
  
  ∂Vr∂r+Vrr+1r∂Vθ∂θ+∂Vx∂x=0(2.134)
  
  

Figure 2.13 (a) Inertial cylindrical coordinate system and (b) rotating (noninertial) cylindrical coordinate system.

Note that the last term −Vθ2/r within brackets on the left-hand side of Equation 2.131 is the centripetal acceleration (negative of the centrifugal acceleration), which results from the flow tangential velocity. For V_r <  < V_θ, V_x <  < V_θ, F_r = 0,Vr<<Vθ,Vx<<Vθ,Fr=0, and away from the viscosity dominated region, Equation 2.131 reduces to ∂Ps/∂r=ρVθ2/r, which is known as the radial equilibrium equation for a rotating flow.

2.2.9.3 Noninertial Reference Frame: Cylindrical Coordinates with Constant Rotation

Let us now consider a cylindrical coordinate system with its coordinate axes rotating with constant Ω→ along an arbitrary axis of rotation, as shown in Figure 2.13b for constant rotation around the xx axis. In this coordinate system, which is noninertial, the flow velocity W→ (also called relative velocity) has components W_r, W_θ,Wr,Wθ, and W_xWx in r, θ,r,θ, and xx(axial) coordinate directions, respectively. The relative velocity W→ is related to the absolute velocity V→ by the following equation:

V→=W→+Ω→×r→(2.135)

where r→ is the position vector in the rotating reference frame.

Centrifugal and Coriolis forces. Using Equation 2.135, the N–S equations (Equation 2.125) in absolute velocity can be easily converted (see Greitzer, Tan, and Graf (2004) for details) into the corresponding equations in relative velocity of the rotating coordinate system

ρDW→Dt+ρΩ→×Ω→×r→+2ρΩ→×W→=ρ∂W→∂t+W→•∇W→+ρΩ→×Ω→×r→+2ρΩ→×W→=−∇Ps+F→+μ∇2W→.(2.136)

Transferring ρΩ→×Ω→×r→ and 2ρΩ→×W→ to the right-hand side of Equation 2.136, we obtain

ρDW→Dt=ρ∂W→∂t+W→•∇W→=−∇Ps−ρΩ→×Ω→×r→−2ρΩ→×W→+F→+μ∇2W→.(2.137)

Note that the left-hand side of Equation 2.137, representing the substantial derivative of the relative velocity W→, is identical in form to the left-hand side of Equation 2.125, which uses the absolute velocity V→. Further note that the pressure forces, viscous forces, and F→ on the right-hand side of Equation 2.137 also take the same form as in Equation 2.125.

On the right-hand side of Equation 2.137, ρΩ→×(Ω→×r→) and 2ρΩ→×W→ are the centrifugal force and Coriolis force per unit volume, respectively. Because these forces have no resemblance to the forces that appear in this equation, they are sometimes referred to as the “fictitious forces” for an observer in the inertial reference frame. These forces, however, are real for an observer in the rotating (noninertial) reference frame. They simply arise from the kinematics of going from rotating reference frame, which is noninertial, to an inertial reference frame in which the Newton’s laws of motion are valid. Note that the vector that results from ρΩ→×(Ω→×r→) is co-linear with r→ and points toward the origin (centripetal). Any relative velocity component in the direction of Ω→ will have no contribution to the Coriolis force (2ρΩ→×W→). Only the relative velocities in a plane normal to the direction of Ω→ will contribute to this force.

Let us now see how we can express the centrifugal force ρΩ→×(Ω→×r→) as the gradient of a scalar quantity, just as the conservative gravitation force can be expressed as the gradient of a potential function. Without loss of generality, let the xx axis in our rotating coordinate system coincide with the rotational velocity vector Ω→. Then the radius of rotation will be measured off this axis, and (Ω→×r→) can be written as (Ω→×rêr), where êr is the unit vector in the radial direction, because the component of the displacement vector r→ along the direction of Ω→ will have no contribution to the cross product. Thus, we obtain ρΩ→×(Ω→×r→)=ρΩ→×(Ω→×rêr). Using the vector identity A→×(B→×C⇀)=B→(A→•C⇀)−C⇀(A→•B→), we can write ρΩ→×(Ω→×r→)=ρΩ→×(Ω→×rêr)=−ρrΩ2êr, which can be further expressed as ∇(−ρ r²Ω²/2)∇−ρr2Ω2/2. Using this result in Equation 2.137 and defining a reduced static pressure P˜s=Ps−ρr2Ω2/2, we can alternatively write this equation as

ρDW→Dt=ρ∂W→∂t+W→•∇W→=−∇P˜s−2ρΩ→×W→+F→+μ∇2W→(2.138)

which reveals that the net effect of the centrifugal force as a result of rotation of the coordinate system is to alter the static pressure field to the extent needed to balance this centrifugal force.

A cylindrical coordinate system rotating with constant ΩΩ about the xx axis, which could be thought of as the gas turbine axis of rotation, is shown in Figure 2.13b. For this case, Equation 2.137 in each coordinate direction and the continuity equation can be written as follows:

rr-coordinate direction:

ρ∂Wr∂t+Wr∂Wr∂r+Wθr∂Wr∂θ+Wx∂Wr∂x−Wθ2r=−∂Ps∂r+ρΩ2r+2ρΩWθ+Fr+μ∂2Wr∂r2+1r∂Wr∂r+1r2∂2Wr∂θ2−2r2∂Wθ∂θ+∂2Wr∂x2−Wrr2(2.139)

θθ-coordinate direction:

ρ∂Wθ∂t+Wr∂Wθ∂r+Wθr∂Wθ∂θ+Wx∂Wθ∂x+WrWθr=−1r∂Ps∂θ−2ρΩWr+Fθ+μ∂2Wθ∂r2+1r∂Wθ∂r+1r2∂2Wr∂θ2+2r2∂Wr∂θ+∂2Wθ∂x2−Wθr2(2.140)

xx-coordinate direction:

ρ∂Wx∂t+Wr∂Wx∂r+Wθr∂Wx∂θ+Wx∂Wx∂x=−∂Ps∂x+Fx+μ∂2Wx∂r2+1r∂Wx∂r+1r2∂2Wx∂θ2+∂2Wx∂x2(2.141)

Continuity equation:

∂Wr∂r+Wrr+1r∂Wθ∂θ+∂Wx∂x=0(2.142)

In the radial direction, according to Equation 2.139, we have two centrifugal force terms ρWθ2/r and ρΩ²rρΩ2r, the first results from the relative tangential velocity W_θWθ of the flow, and the second from the fact that the coordinate system is rotating with constant ΩΩ. Note that the Coriolis force has two components. The component 2ρΩ W_θ2ρΩWθ is in the radial direction and (−2ρΩ W_r−2ρΩWr) in the tangential direction (Equation 2.139).

The centrifugal and Coriolis forces are real forces in a rotating noninertial reference frame. We often experience centrifugal forces while driving on a curve or when a car skids on an icy road with a sharp turn. If you are standing on a counterclockwise rotating platform, such as a merry-go-round in a theme park, you will experience a centrifugal force trying to push you radially outward. If the platform is slippery or icy, you will soon slip to the outer edge of the platform, but not exactly along the radial direction from where you started. As you are slipping radially outward, you will gradually turn to your right as a result of the Coriolis force.

Consider a steady, two-dimensional, incompressible, inviscid flow from a source located at the origin (r = 0r=0). For an observer in an inertial reference frame, the flow will have radial streamlines emanating from the origin with the absolute radial velocity V_rVr, decaying inversely with radius to satisfy continuity. The static pressure gradient ∂P_s/∂r∂Ps/∂r in this flow will equal (−ρV_r∂V_r/∂r)−ρVr∂Vr/∂r. This pressure gradient is independent of the reference frame. The kinematics of this source flow will appear different to an observer in a reference frame rotating counterclockwise at an angular velocity ΩΩ. This observer will see the flow having two velocities in the plane, one radial W_r = V_rWr=Vr and the other tangential W_θ =  − rΩWθ=−rΩ, spiraling away from the origin at lower radii and assuming circular streamlines for V_r <  < W_θVr<<Wθ. Because the pressure gradient at each radial location also equals (−ρW_r∂W_r/∂r)−ρWr∂Wr/∂r, we find that the additional centrifugal force term due to the tangential velocity (W_θWθ), ρWθ2/r=ρrΩ2, together with the centrifugal force ρrΩ²ρrΩ2 as a result of coordinate system rotation cancels the Coriolis force (2ρΩW_θ =  − 2ρrΩ²2ρΩWθ=−2ρrΩ2).

The existence of the Coriolis and centripetal accelerations as a result of coordinate system rotation is easily demonstrated for a vortex flow with circular streamlines and circular isobars. For a streamline of radius rr with velocity V_θVθ in this flow, the centripetal acceleration Vθ2/r in the inertial reference frame is driven by the static pressure gradient ∂P_s/∂r∂Ps/∂r; recall the local equilibrium equation ∂Ps/∂r=ρVθ2/r. Substituting V_θ = W_θ + rΩVθ=Wθ+rΩ for the rotating reference frame, we obtain Vθ2/r=Wθ2/r+2ΩWθ+rΩ2, which shows that the centripetal acceleration Wθ2/r in the rotating reference frame is not sufficient to yield the required static pressure gradient, which is independent of the reference frame. The Coriolis acceleration 2ΩW_θ2ΩWθ and the centripetal acceleration rΩ²rΩ2 as a result of coordinate system rotation are needed to make up for the difference.

To develop further insight into the centrifugal and Coriolis forces, let us consider a frictionless water sprinkler shown in Figure 2.14. Only one sprinkler arm is shown flowing, the other is blanked off. Each arm is made of a straight pipe in the radial direction and a curved pipe in the form of a circular arc, having the radius of curvature equal to that of the circle traversed by the sprinkler. For the sprinkler, the maximum angular velocity is given by Ω = W_θ/RΩ=Wθ/R, where W_θWθ is the constant relative tangential velocity in the curved pipe. The radial pipe has constant relative radial flow velocity W_rWr and has no tangential and axial velocities.

Figure 2.14 A sprinkler system featuring centrifugal and Coriolis forces.

In the blanked-off arm of the sprinkler, water is in a rigid-body rotation. Because all relative velocity components are zero in this arm, there are no Coriolis forces. However, the centrifugal force in this arm corresponds to ρ Ω²rρΩ2r, increasing linearly with radius. As a result, we have uniform static pressure at each section of the straight pipe, but radially increasing static pressure distribution in each section of the curved pipe, as shown in Figure 2.14 for sections A and B.

In the rotating sprinkler arm that is flowing, the straight pipe features a radial velocity W_rWr and will generate a uniform Coriolis force (−2ρΩW_r−2ρΩWr) in the negative tangential direction, causing a static pressure gradient in the direction shown for section C in Figure 2.14. The flow features in the curved pipe with flow are quite interesting. In this rotating pipe, we have no radial static pressure gradient at any section, as shown for section D in Figure 2.14. How do we explain this apparently nonintuitive flow behavior? The answer lies in finding the centrifugal and Coriolis forces in this pipe. Because the absolute flow velocity in this pipe is zero (V_θ = RΩ − W_θ = 0Vθ=RΩ−Wθ=0), the centrifugal force ρVθ2/R=0, resulting in a radially uniform static pressure distribution. Now, let us look at the centrifugal and Coriolis forces in the rotating reference frame attached to the sprinkler. The centrifugal force from W_θWθ equals ρWθ2/R and that from ΩΩ equals ρΩ²RρΩ2R. Because Ω = W_θ/RΩ=Wθ/R, both these forces are equal, giving 2ρΩ²R2ρΩ2R as the total centrifugal force. The Coriolis force in this curve pipe can be computed as 2ρΩ(−W_θ) =  − 2ρΩ²R2ρΩ−Wθ=−2ρΩ2R, which is opposite and equal to the centrifugal force, yielding a net zero force, which we concluded from using the absolute velocity (inertial reference frame).

2.2.9.4 Rotating Couette Flow

In this section, we demonstrate a procedure to obtain exact solutions of the Navier-Stokes equations governing a steady incompressible laminar flow between concentric cylinders of radii R₁R1 and R₂R2, as shown in Figure 2.15. Among other applications, like CFD validation, the solutions offer a basis for measuring fluid viscosity by a device. In this case, the inner cylinder is rotating at a constant angular velocity Ω₁Ω1, and the outer cylinder with Ω₂Ω2, with no imposed pressure gradient and zero flow in the axial direction (V_x = 0Vx=0).

Figure 2.15 Couette flow between two concentric, rotating cylinders.

The solutions developed here are based on the following assumptions for the flow being: (i) axisymmetric (∂/∂θ = 0∂/∂θ=0), (ii) fully developed (∂/∂x = 0∂/∂x=0), and (iii) with zero body force as a result of gravity.

Continuity equation in cylindrical coordinates. The continuity equation, Equation 2.134, in the inertial cylindrical polar coordinates is expressed as

∂Vr∂r+Vrr+1r∂Vθ∂θ+∂Vx∂x=0

1r∂rVr∂r+1r∂Vθ∂θ+∂Vx∂x=0

Using V_x = ∂V_θ/∂θ = 0Vx=∂Vθ/∂θ=0 from the assumptions, this equation reduces to

∂Vr∂r+Vrr=∂rVr∂r=0

whose integration yields rV_r = CrVr=C in rr direction, where CC is the integration constant. Because V_r = 0Vr=0 at the inner cylinder wall at r = R₁r=R1, it must be zero everywhere in the flow, giving C = 0C=0.

Reduced Navier-Stokes Equations in cylindrical coordinates. With V_x = 0Vx=0, the xx-momentum equation, Equation 2.133, is identically satisfied. Under the assumptions made in this problem and V_r = 0Vr=0 from the continuity equation, the rr-momentum equation, Equation 2.131, reduces to

dPsdr=ρVθ2r,(2.143)

which governs the radial distribution of the static pressure between two cylinders.

The θθ-momentum equation, Equation 2.132, reduces to

d2Vθdr2+1rdVθdr−Vθr2=0,

which can be alternatively written as

ddr1rddrrVθ=0.(2.144)

Equation 2.144 is a homogeneous second-order ordinary differential equation and requires two boundary conditions for a unique solution. These two boundary conditions in this case correspond to no-slip condition at the walls of the inner and outer cylinders, namely, V_θ = R₁Ω₁Vθ=R1Ω1@r = R₁r=R1 and V_θ = R₂Ω₂Vθ=R2Ω2@r = R₂r=R2.

Tangential velocity distribution. Integrating Equation 2.144 once, we obtain

1rddrrVθ=C˜

ddrrVθ=rC˜.

Integrating this equation once more yields

rVθ=r2C˜/2+C2=C1r2+C2

where C1=C˜/2. Thus, the general solution for the distribution of V_θVθ between the cylinders is given by

where C₁C1 and C₂C2 are the integration constants, which are to be determined from the boundary conditions at the inner and outer cylinders, yielding the following two equation:

Solving the aforementioned two simultaneous algebraic equations yields:

C1=R22Ω2−R12Ω1R22−R12(2.146)

and

C2=R12R22Ω1−Ω2R22−R12(2.147)

Substituting Equations 2.146 and 2.147 into Equation 2.145, we finally obtain

Vθ=1R22−R12R22Ω2−R12Ω1r+R12R22rΩ1−Ω2(2.148)

Static pressure distribution. Substituting for V_θVθ from Equation 2.145 into Equation 2.143, we obtain

dPsdr=ρC12r+2C1C2r+C22r3,

which upon integration and substitution for C₁C1 and C₂C2 from Equations 2.146 and 2.147, respectively, finally yields

Ps−Ps1=ρR22−R122R22Ω2−R12Ω12r2−R122+2R12R22R22Ω2−R12Ω1Ω1−Ω2lnrR1−R14R242Ω1−Ω221r2−1r12(2.149)

where P_s1Ps1 is the static pressure at the surface of the inner cylinder.

Torques on inner and outer cylinders. The torque on each cylinder, per unit length in the axial direction, can be computed as the product of the wall shear stress, cylinder circumference, and radius. Knowing the distribution of velocity V_θVθ from Equation 2.148, we can evaluate the tangential shear stress distribution in the flow between the two cylinders by the equation

τrθ=μrddrVθr

Substituting for V_θVθ yields

τrθ=2μR12R22Ω2−Ω1r2R22−R12(2.150)

For r = R₁r=R1, Equation 2.150 yields the torque on the inner cylinder

Γ1=4πμR12R22Ω2−Ω1R22−R12.

Similarly, for r = R₂r=R2, Equation 2.150 yields the torque on the outer cylinder

Γ2=−4πμR12R22Ω2−Ω1R22−R12.

Note that the torques Γ₁Γ1 and Γ₂Γ2 on the two cylinders are equal and opposite, implying zero net torque acting on the fluid; as a result, its angular momentum remains constant.

2.2.9.5 Taylor-Proudman Theorem

Taylor-Proudman theorem embodies the behavior of a steady, incompressible, laminar flow that is dominated by rotation (Coriolis forces). To understand this theorem, let us consider the Navier-Stokes equation, Equation 2.138. When we use the velocity scale W_refWref and length scale L_refLref to nondimensionalize this equation, two dimensionless numbers emerge. The first is called the Rossby number Ro = W_ref/ΩL_refRo=Wref/ΩLref and the second, the Ekman number Ek=μ/ρΩLref2. While the Rossby number is a measure of the inertia force associated with the relative velocity in comparison to the Coriolis forces, the Ekman number is the ratio of viscous forces to Coriolis forces. For a steady incompressible flow outside a wall boundary layer and with Ro <  < 1Ro<<1 and Ek <  < 1Ek<<1, that is, the Coriolis forces dominate over the inertia and viscous forces, Equation 2.138 reduces to

2ρΩ→×W→=−∇P˜s.(2.151)

The flows that are governed by Equation 2.151 are known as geostrophic flows in which the modified (by including the potential function associated with the centrifugal forces as a result of the rotation of the coordinate axes) static pressure gradients are balanced by the Coriolis forces.

Taking the curl of both sides of Equation 2.151 and noting that the curl of the gradient of a scalar function is identically zero, we obtain

∇×(Ω→×W→)=0.(2.152)

Using the vector identity ∇×(A→×B→)=(B→•∇)A→−B→(∇•A→)−(A→•∇)B→+A→(∇•B→) and invoking the continuity (∇•W→=0) and the fact that Ω→ is constant, we finally obtain from Equation 2.152

(Ω→•∇)W→=0.(2.153)

Assuming the direction of Ω→ as the axis of rotation along the xx direction, Equation 2.153 yields ∂W→/∂x=0, which indicates that all three components of the relative velocity do not change in the direction parallel to the axis of rotation – the statement of the Taylor-Proudman theorem. In a flow system with solid boundaries perpendicular to the axis of rotation, we have W_x = 0Wx=0 at some specified value of xx. The Taylor-Proudman theorem implies that this flow system will remain entirely two-dimensional in planes perpendicular to the axis of rotation.

Taylor columns. Taylor columns are a unique and interesting manifestation of the Taylor-Proudman theorem. These columns occur when there is relative motion between an obstacle and fluid in a rotating system dominated by rotation. To illustrate a Taylor column, let us consider a rotating cylinder filled with liquid. We have a short cylinder attached to the bottom surface of the cylinder. If you rotate this system at some angular velocity ΩΩ, eventually the liquid will attain solid-body rotation with all the solid surfaces. Now, if we reduce the cylinder angular velocity slightly, the liquid will continue to rotate at the old angular velocity and develop a relative velocity (Ω_liquid>Ω_cylinderΩliquid>Ωcylinder) such that the short cylinder at the bottom of the rotating cylinder is rendered into cross-flow with the surrounding fluid, as shown in Figure 2.16.

Figure 2.16 Taylor column.

In the flow system shown in Figure 2.16, as a result of strong rotation, there is no flow on the top of the short cylinder. According to the Taylor-Proudman theorem, the velocity components will not change in the direction of the axis of rotation, the xx direction in this case. As a result, a solid column of liquid develops on the top of the short cylinder, as if the cylinder is extended up to the free surface of the liquid in the rotating cylinder. In planes perpendicular to the xx axis, we have identical two-dimensional velocity distributions, as shown in the figure.

2.2.9.6 Ekman Boundary Layers

Rotor surfaces associated with rotor disks and rotor cavities abound in gas turbines. Boundary layers formed on these surfaces are generally known as Ekman layers. These layers feature some unique characteristics not found in a typical boundary layer such as the one developed on a flat plate under zero pressure gradient where wall shear stress reduces momentum in the flow direction, resulting in a continuously (growing) thickening boundary layer. In contrast, the Ekman layer does not entrain fluid from the region outside the boundary layer and consequently maintains a constant thickness. For low Rossby numbers, where the Ekman boundary layer is sandwiched between the geostrophic flow and the rotating wall, we obtain a set of linear boundary layer equations with closed-form analytical solutions for the distributions of velocities in the boundary layer. These solutions, systematically developed here, provide further insight into an Ekman layer.

Let us consider a rotating disk whose axis of rotation is along the xx-axis. For an axisymmetric (∂/∂θ = 0∂/∂θ=0), steady, incompressible, laminar flow with constant viscosity, no body forces, and Ro <  < 1Ro<<1, the Navier-Stokes equations (Equations 2.139, 2.140, and 2.141) reduce to:

rr-momentum equation:

−2ρΩWθ=−∂P˜s∂r+μ∂∂r1r∂∂rrWr+μ∂2Wr∂x2(2.154)

θθ-momentum equation: