2.1.1 The First Law of Thermodynamics

For a closed system, the first law of thermodynamics simply states that the change in its total energy dEdE going from state 1 to state 2 results from heat transfer δQδQ and work transfer δWδW, the former is considered positive if it occurs into the system from the surroundings and the later is considered positive when the system does work on the surroundings. In equation form we can write

dE=δQ−δW=δQ−PsdV(2.1)

where the total stored energy consists of the internal energy UU, kinetic energy KEKE = mV²/2mV2/2, and potential energy PEPE (in the gravitational force field: PE = mghPE=mgh), that is,

E=U+12mV2+mgh(2.2)

For a system doing work on the surrounding, ΔV is positive. Dividing Equation 2.2 by mass mm yields

de=δq+δw=δq−Psdv(2.3)

where the specific total energy e=u+12V2+gh, uu being the specific internal energy. Neglecting changes in the kinetic and potential energy, Equation 2.3 becomes

du=δq−Psdv(2.4)

When the gas volume does not change during its state change (isochoric process), Equation 2.4 reduces to du = δqdu=δq, indicating that any change in internal energy in this process is entirely as a result of heat transfer. For a calorically perfect gas, which has constant cv, we can write u = c_vT_su=cvTs where we have assumed that u = 0u=0 at the reference static temperature T_s = 0Ts=0.

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

h_s = u + RT_s

hs=u+RTs

c_pT_s = c_vT_s + RT_s

cpTs=cvTs+RTs

giving

c_p = c_v + R

cp=cv+R(2.6)

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

δq = dh_s

δq=dhs,(2.10)

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 The Second Law of Thermodynamics

The first law of thermodynamics embodies the energy conservation principle in that the energy is neither created nor destroyed; it may only be converted from one form to another. Empirical evidence shows that, while all of the work energy can be completely converted into heat energy (internal energy), the reverse is not true. The practical restriction placed on an energy conversion process is governed by the second law of thermodynamics with its focus on irreversibility, which entails conversion of organized motion to random molecular motion through the dissipation of mean flow energy by viscosity. As a result of the presence of friction in practical applications, a reversible work transfer is merely an idealization.

One of the tenets of the second law of thermodynamics is that heat transfer always occurs from higher temperature to lower temperature. This fact, for example, is built into the Fourier’s law of heat conduction governed by the equation q̇x=−kdT/dx, where the negative sign ensures that the one-dimensional heat conduction in the xx direction takes place in the direction of negative temperature gradient.

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

c_p ln T_s2 − R ln P_s2 = c_p ln T_t2 − R ln P_t2

cplnTs2−RlnPs2=cplnTt2−RlnPt2(2.20)

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.

2.1.3 Thermodynamic Cycles

Figure 2.1a depicts the operation of a simple gas turbine on an open cycle in which both the compressor and turbine are mounted on the same shaft. In this cycle, the ambient air enters the compressor, which increases both air pressure and temperature from the specific work transfer w_CwC. The compressed air temperature is further increased in the combustor through fuel addition. Hot gases (products of combustion), treated here to have properties same as air (assumed to be a perfect gas), enters the turbine, which converts part of the energy of these gases into the specific work output w_TwT.

Figure 2.1 Key operational components of a simple gas turbine: (a) open cycle and (b) closed cycle with a fictitious heat exchanger.

Figure 2.1b uses a fictitious heat exchanger to cool the hot exhaust gases back to the ambient air conditions, making the gas turbine operation a closed cycle, at least thermally. From the steady flow energy equation (derived from the first law of thermodynamics) in this closed cycle we can write

Q̇in−Q̇out=ẆT−ẆC(2.25)

The thermal efficiency of an open-cycle gas turbine shown in Figure 2.1a is defined as the ratio of the net power output to the heat rate (rate of energy transfer from the combustion of fuel in the combustor), giving

ηth=ẆnetQ̇in=ẆT−ẆCQ̇in(2.26)

which also equals the cycle efficiency η_cycleηcycle for the closed-cycle gas turbine shown in Figure 2.1b.

2.1.4 Isentropic Efficiency

The Brayton cycle of Figure 2.2 maps the operation of the gas turbine engine shown in Figure 2.1 on the T − ST−S diagram. In this cycle, an ideal compressor operates along the isentropic process 1–2i, while the real compressor works along 1–2 with increase in flow entropy as a result of unavoidable friction and other irreversible effects present in the compressor. For the same pressure ratio π_C = P_t2/P_t1πC=Pt2/Pt1, the actual compressor work input along 1–2 will be higher than the corresponding isentropic (ideal) work input along 1–2i. Accordingly, we define the compressor isentropic efficiency as follows:

ηCi=ẆCiẆC=Tt2i−Tt1Tt2−Tt1(2.27)

which further yields

Tt2Tt1=1+Tt2i/Tt1−1ηCi(2.28)

Figure 2.2 Brayton cycle of a gas turbine on T − ST−S diagram.

Using Equation 2.23 in Equation 2.28 results in

Tt2Tt1=1+πCκ−1κ−1ηCi(2.29)

With no drop in total pressure in the combustor (P_t2 = P_t3Pt2=Pt3), the turbine inlet point 3 lies on the constant-pressure line established by the compressor exit pressure. Let us assume that the expansion in the turbine during work extraction either along the isentropic line 3–4i or the actual line 3–4 with higher entropy occurs to the same total pressure P_t4Pt4, which is equal to the compressor inlet pressure P_t1Pt1. Because point 4 lies to the right of 4i, the actual work output of the turbine along the process line 3–4 is lower than that along the isentropic (ideal) process line 3–4i. Accordingly, we define the turbine isentropic efficiency as follows:

ηTi=ẆTẆTi=Tt3−Tt4Tt3−Tt4i,(2.30)

which further yields

Tt3Tt4=Tt3/Tt4i1−ηTiTt3/Tt4i+ηTi.(2.31)

Substitution of Equation 2.23 in Equation 2.31 gives

Tt3Tt4=πTκ−1κ1−ηTiπTκ−1κ+ηTi(2.32)

where π_T = P_t3/P_t4πT=Pt3/Pt4 is the pressure ratio across the turbine and equals the pressure ratio π_C = P_t2/P_t1πC=Pt2/Pt1 across the compressor.

With reference to the gas turbine cycle shown in Figure 2.2, we can express the cycle efficiency or the thermal efficiency in terms of total temperatures as follows:

ηth=ηcycle=ẆnetQ̇in=ẆT−ẆCQ̇in=Tt3−Tt4−Tt2−Tt1Tt3−Tt2ηth=ηcycle=1−Tt4−Tt1Tt3−Tt2ηth=ηcycle=1−1/Tt3/Tt4−1/Tt3/Tt11−Tt2/Tt1/Tt3/Tt1.(2.33)

In Equation 2.33, the temperature ratio T_t2/T_t1Tt2/Tt1 across the compressor is given by Equation 2.28 and the temperature ratio T_t3/T_t4Tt3/Tt4 across the turbine by Equation 2.32. The equation involves an additional parameter T_t3/T_t1Tt3/Tt1, which is the ratio of turbine inlet temperature to compressor inlet temperature. For an ideal Brayton cycle with η_Ci = η_Ti = 1.0ηCi=ηTi=1.0, which for π_C = π_TπC=πT implies that T_t2/T_t1 = T_t3/T_t4Tt2/Tt1=Tt3/Tt4, Equation 2.33 simplifies to

ηBrayton=1−1Tt2/Tt1=1−1πκκ−1(2.34)

where π = π_C = π_Tπ=πC=πT. It is interestingly to note from Equation 2.34 that η_BraytonηBrayton is independent of the parameter T_t3/T_t1Tt3/Tt1.

Both Equations 2.33 and 2.34 are plotted in Figure 2.3, showing variation of gas turbine cycle efficiency with pressure ratio π = π_C = π_Tπ=πC=πT. The ideal Brayton cycle efficiency η_BraytonηBrayton with η_Ci = η_Ti = 1.0ηCi=ηTi=1.0 shown in the figure increases monotonically with pressure ratio, initially rather sharply for π < 10π<10 and then gradually reaching a value of 0.673 for π = 50π=50. The trend of cycle efficiency variation with pressure ratio with nonideal compressor and turbine depends upon the ratio of turbine inlet temperature to compressor inlet temperature as shown in the figure for η_Ci = η_Ti = 0.9ηCi=ηTi=0.9. For T_t3/T_t1 = 4Tt3/Tt1=4, cycle efficiency first increases with pressure ratio, reaches a maximum value of 0.380 at π = 18π=18, and then decreases to η_cycle = 0.191ηcycle=0.191 at π = 50π=50. The curve for T_t3/T_t1 = 5Tt3/Tt1=5 exhibits a shallower maximum that corresponds to η_cycle = 0.447ηcycle=0.447 at π = 33π=33, decreasing to η_cycle = 0.433ηcycle=0.433 at π = 50π=50. The figure also shows a well known result that, for a given pressure ratio and compressor inlet temperature, the cycle efficiency increases with turbine inlet temperature.

Figure 2.3 Variation of gas turbine cycle efficiency with pressure ratio.

Optimum pressure ratio for maximum net specific work output. With reference to Figure 2.2, we can express the net specific work output (turbine specific work output minus compressor specific work input, neglecting the fuel mass addition in the combustor) as

Because T_t2/T_t1 = T_t3/T_t4Tt2/Tt1=Tt3/Tt4, we obtain T_t4/T_t1 = T_t3/T_t2Tt4/Tt1=Tt3/Tt2. We now rewrite Equation 2.35 as

wnet=cpTt3−Tt2+Tt1−Tt3Tt2Tt1

which for an optimum value of w_netwnet with respect to T_t2Tt2 yields

dwnetdTt2=cp−1+Tt3Tt1Tt22=0Tt2=Tt3Tt1(2.36)

Equation 2.36 further leads to

πoptimum=Pt2Pt1optimum=Tt2Tt1κ−1κ=Tt3Tt1κ−12κ(2.37)

which, for a given ratio of turbine inlet temperature to compressor inlet temperature, yields the optimum pressure ratio across the compressor needed for maximum net specific work output of the gas turbine. Note that, in the aforementioned derivation, we have assumed a single value of κκ, which is typically close to 1.4 for the compressor and 1.33 for the turbine.

2.1.5 Polytropic Efficiency

In a gas turbine, the isentropic efficiencies of compressor and turbine depend upon their operating conditions. For example, if two identical compressors operate in series with equal pressure ratio across each, their isentropic efficiency will be equal. This isentropic efficiency, however, will be higher than if the same compressor were to operate over the higher combined pressure ratio. This results from the fact that the increase in entropy as a result of friction in the upstream compressor requires more work input in the downstream one, generally known as the preheat effect. If we divide the compression process across a compressor or the expansion process across a turbine into a large number of very small, consecutive compression or expansion processes, respectively, then the isentropic efficiency across each is called the polytropic efficiency, or small-stage efficiency, which may be assumed constant for the machine, reflecting its state-of-the-art design engineering.

With reference to Figure 2.2, we assume that the index of polytropic compression along 1–2 is nn, which yields

Tt2Tt1=Pt2Pt1n−1n=πCn−1n.(2.37)

Note that, for n = κn=κ, this equation yields the isentropic compression along 1–2i.

We define the compressor polytropic efficiency as

ηpC=dTtisentropicdTtactual=dTt/TtisentropicdTt/Ttactual(2.38)

Integrating Equation 2.38 between the compressor inlet and exit yields

ηpC∫12dTt/Ttactual=∫12idTt/TtisentropicTt2Tt1ηpC=Tt2iTt1πCn−1ηpCn=πCκ−1κn−1n=κ−1ηpCκ.(2.39)

To establish a relation between the compressor isentropic efficiency and its polytropic efficiency, we write

ηCi=Tt2i−Tt1Tt2−Tt1=Tt2i/Tt1−1Tt2/Tt1−1

ηCi=πCκ−1κ−1πCn−1n−1

which with the substitution of Equation 2.39 yields

ηCi=πCκ−1κ−1πCκ−1κηPC−1.(2.40)

Equation 2.40 expresses compressor isentropic efficiency in terms of its polytropic efficiency and pressure ratio. Alternatively, to express compressor polytropic efficiency in terms of its isentropic efficiency and pressure ratio, we can rewrite Equation 2.40 as

ηpC=lnπCκ−1κln1+πCκ−1κ−1ηCi.(2.41)

Again, with reference to Figure 2.2, we assume that the index of polytropic expansion in the turbine along 3–4 is nn, which yields

Tt3Tt4=Pt3Pt4n−1n=πTn−1n(2.42)

Note that for n = κn=κ, the aforementioned equation yields the isentropic expansion along 3–4i.

We define the turbine polytropic efficiency as

ηpT=dTtactualdTtisentropic=dTt/TtactualdTt/Ttisentropic.(2.43)

Integrating Equation 2.43 across the turbine yields

ηpT∫34idTt/Ttisentropic=∫34dTt/TtactualTt3Tt4=Tt3Tt4iηpTπTn−1n=πTκ−1ηpTκn−1n=κ−1ηpTκ.(2.44)

To establish a relation between the turbine isentropic efficiency and its polytropic efficiency, we write

ηTi=Tt3−Tt4Tt3−Tt4i=1−Tt4/Tt31−Tt4i/Tt3

ηTi=1−1/πTn−1n1−1/πTκ−1κ

which using Equation 2.44 yields

ηTi=1−1/πTηpTκ−1κ1−1/πTκ−1κ.(2.45)

Equation 2.45 expresses turbine isentropic efficiency in terms its polytropic efficiency and pressure ratio. Alternatively, to express turbine polytropic efficiency in terms of its isentropic efficiency and pressure ratio, we rewrite Equation 2.45 as

ηpT=ln1−ηTi1−1/πTκ−1κκ−1κln1/πT.(2.46)

Equations 2.40 and 2.45 are plotted in Figure 2.4 for η_pC = η_pT = 0.9ηpC=ηpT=0.9. In this figure, we have assumed κ = 1.4κ=1.4 for the compressor and κ = 1.33κ=1.33 for the turbine. For the pressure ratio very close to 1.0, the compressor and the turbine each has equal isentropic and polytropic efficiencies. The figure further shows that, for the compressor, the isentropic efficiency decreases with pressure ratio and, for the turbine, it increases with the pressure ratio. The rate of efficiency decrease for the compressor, however, is higher than the corresponding rate of efficiency increase for the turbine.

Figure 2.4 Variation of compressor and turbine isentropic efficiencies with pressure ratio for a constant value of their polytropic efficiency.

2.2.1 Secondary Flows

In a gas turbine operation, flows that directly partake in the energy conversion along the turbine and compressor flow path (core flow) are generally referred to as the primary flows. The flows that are used for cooling and sealing of gas turbine components are often called secondary flows. These flows should not be confused with the secondary flows, which are normal to the main flow direction, found in many internal flows. For example, a turbulent flow in a noncircular duct features secondary flows that have velocities in the plane of the duct cross-section, which is perpendicular to the main flow direction. These are driven by the anisotropy in turbulence intensity and gradients in normal turbulent stresses and are called secondary flow of the second kind. Note that these secondary flows do not exist in a constant-viscosity laminar flow in a noncircular duct or the turbulent flow in a circular duct.

Because the local gradient in static pressure is the common primary mechanism to generate the velocity field in a flow, when a uniform flow enters a bend in a circular duct it exits with secondary flows. In the bend region, as a result of the centrifugal force, the static pressure at the outer radius of curvature is higher than at the inner radius of curvature. This radial pressure gradient in the pipe cross-section generates secondary flows that we observe at the exit plane. These are known as the secondary flows of the first kind.

2.2.2 Vorticity and Vortex

Vorticity is a kinematic vector property of a flow field and equals the curl of its velocity vector (ζ→=∇×V→). At any point in the flow, the vorticity represents the local rigid-body rotation of fluid particles and equals twice the rotation vector (ζ→=2ω→). A vortex, on the other hand, refers to the bulk circular motion of the entire flow field. Streamlines in a vortex flow are essentially concentric circles, and so are the isobars (lines of constant static pressure). The local pressure gradient along the radius of curvature is given by the local equilibrium equation dPs/dr=ρVθ2/r. This characteristic of a vortex flow is so different from a nonrotating flow where the static pressure typically varies along a streamline, as dictated by the Bernoulli equation for an ideal incompressible flow.

A forced vortex features bulk rigid-body rotation with constant angular velocity (in radians per second) for all streamlines. As a result, the outer streamline features higher tangential, or swirl, velocity than the inner ones. In a free vortex, which is assumed to be free from any external torque, the angular momentum (the product of radius and tangential velocity), not the angular velocity, remains constant everywhere in the flow. In this vortex, the outer streamline has lower tangential velocity than the inner one. As a result of the presence of centrifugal force, all vortex motions feature positive radial gradient in the static pressure distribution. Lugt (1995) presents an excellent discussion of vortex flows in both nature and technology.

2.2.3 Sudden Expansion Pipe Flow With and Without Swirl

In gas turbine design applications, flows often feature changes in pipe diameter, going from a smaller-diameter pipe to a larger-diameter pipe (sudden expansion) or from a larger-diameter pipe to a smaller-diameter pipe (sudden contraction). Such flows, with their undesirable excess loss in total pressure, generally become necessary as a result of spatial constraints.

Following a sudden expansion, shown in Figure 2.5a for the case without swirl, the flow enters the larger pipe in the form of a circular jet whose shear layer grows both inward toward the pipe axis and outward toward the pipe wall. While the inward growth of the shear layer occurs at the expense of the potential (inviscid) core flow, which decays downstream until the shear layers merge at the pipe axis, its outward growth happens through the process of entrainment until it finally reattaches to the pipe wall. The fluid entrained into this shear layer is continuously replenished from downstream through a favorable reverse pressure gradient. This gives rise to a primary recirculation region such that the flow in its forward branch is driven by the central jet momentum, and in its reverse branch, by an adverse pressure gradient. Sultanian, Neitzel, and Metzger (1986) present accurate CFD predictions of this flow field using an advanced algebraic stress transport model of turbulence. They also demonstrate that the standard high-Reynolds number κ − εκ−ε turbulence model significantly under-predicts the experimentally found reattachment lengths.

Figure 2.5 (a) Sudden expansion pipe flow without swirl, (b) Sudden expansion pipe flow with swirl, and (c) Sudden contraction pipe flow without swirl.

Figure 2.5b shows the main features of a swirling flow in a sudden pipe expansion. The wall-bounded primary recirculation zone in this flow field is found to shrink with the increasing incoming swirl velocity. When the swirl exceeds a critical value (greater than 45-degree swirl angle), an on-axis recirculation region appears. Sultanian (1984) discusses at length the overwhelming complexity of this flow field from a computational viewpoint. The major difficulties associated with predicting this three-dimensional flow field lies in accurately modeling its highly anisotropic turbulence field.

A distinguishing feature of a critically swirling flow in a sudden pipe expansion is the region of on-axis recirculation caused by vortex breakdown. Such a feature is widely used in modern gas turbine combustors as an aerodynamic flame holder, which in early designs corresponded to the wake region of a bluff body, to ignite the fresh charge entering the combustor. Here is a simple explanation behind the vortex breakdown in this flow field. The swirl in a free or forced vortex causes positive radial pressure gradient, that is, the static pressure always increases radially outward – the higher the swirl, the higher is this pressure gradient. Near the sudden expansion, as a result of high swirl, the static pressure at the pipe axis is much lower than the section-average static pressure. Further downstream, swirl decays as a result of wall shear stress, and the static pressure distribution tends to become more uniform over the pipe cross section. As a result, the static pressure at the pipe axis becomes higher downstream than upstream, causing on-axis flow reversal.

2.2.4 Sudden Contraction Pipe Flow: Vena Contracta

Key features of a sudden contraction pipe flow are shown in Figure 2.5c. When the flow from the large pipe of area A₁A1 abruptly enters the smaller pipe of area A₂ < A₁A2<A1, the outer streamline detaches from the pipe wall near the area change, creating a short region of corner recirculation and the minimum flow cross-section, called the vena contracta, in the downstream pipe. At the vena contracta, having area A_c < A₂Ac<A2 and surrounded by the primary recirculation zone in the downstream pipe, the flow has the maximum velocity and minimum static pressure. Downstream of the vena contracta, the flow behaves very similar to the one in a sudden pipe expansion, discussed in Section 2.2.2, expanding from A_cAc to A₂A2. Experimental data in a sudden contraction pipe flow indicate that A_c/A₂Ac/A2 is a function of A₁/A₂A1/A2. For example, A_c/A₂ = 0.68Ac/A2=0.68 for A₁/A₂ = 2A1/A2=2. It is worth noting here that, for given A₁/A₂A1/A2, A_cAc does not depend on the total-to-static pressure ratio at the vena contracta for incompressible flows but increases slightly for compressible flows.

2.2.5 Stream Thrust and Impulse Pressure

The concept of stream thrust is grounded in the linear momentum equation. The stream thrust at any section of an internal flow, be it incompressible or compressible, is defined as the sum of the pressure force and inertia force (momentum flow rate). In any particular direction, the stream thrust thus represents the total force of the flow stream – we need this much force in the opposite direction to bring the flow velocity to zero (stagnation). For example, the force needed to hold a plate in place to an incoming fluid jet equals the stream thrust associated with the jet, and not the jet area times the jet stagnation pressure at the plate. As another example, let us consider a constant-area duct with zero wall friction, say a stream tube. The stream thrust associated with an incoming flow with nonuniform velocity distribution will remain constant throughout the duct, although its static pressure and total pressure at each section will change downstream. If the velocity profile is uniform at the duct exit, its static pressure will be higher and total pressure lower than their respective values at the duct inlet.

Given by the equation

ST=PsA+ṁV(2.47)

the stream thrust is a vector quantity having the same direction as the momentum velocity (see Section 2.2.5) in the chosen coordinate system. Use of stream thrust at inlets and outlets of a control volume for the momentum equation in each coordinate direction makes it analogous to the continuity and energy equations, which involve only scalar quantities. The net change in stream thrust summed over all inflows and outflows of a control volume must equal all body and surface forces (except the pressure forces at inlets and outlets) acting on the control volume.

Impulse pressure equals the stream thrust per unit area. It is given by the equation

which is valid for both incompressible and compressible flows. The aforementioned equation should not be confused with the equation P_t = P_s + ρV²/2Pt=Ps+ρV2/2, which is used to compute total pressure in an incompressible flow and approximately in a compressible flow at low Mach numbers (M ≤ 0.3M≤0.3). Note that the second term on the right-hand side of Equation 2.48 equals twice the dynamic pressure of an incompressible flow.

2.2.6 Control Volume Analyses of Conservation Laws

Control volume analyses of the conservation laws of mass, momentum, energy, and entropy form the core foundation for the mathematical modeling of all fluid flows and their quantitative predictions for practical applications. In Newtonian mechanics, conservation laws are written in an inertial reference frame for a control system of identified mass – the Lagrangian viewpoint. If Φ_CSΦCS stands for the mass, linear momentum, energy, or entropy of a control system with the corresponding intensive (per unit mass) properties represented by ϕϕ, the substantial, total, or material derivative of Φ_CSΦCS in a flow field, where the control system instantaneously assumes the local flow properties, can be written as

DΦCSDt=ΠS+ΠB.(2.49)

In Equation 2.49, the terms on the right-hand side are the change agents, Π_SΠS working on the surface of the control system and Π_BΠB working on the body of the control system. Their net effect is to cause DΦ_CS/DtDΦCS/Dt on the left-hand side of the equation. Note that Φ_CSΦCS is the instantaneous value of the extensive property ΦΦ for the entire control system.

In most engineering applications, a control volume analysis consistent with the Eulerian viewpoint is used. Facilitated by the Reynolds transport theorem, see Sultanian (2015), the left-hand side of Equation 2.49 can be replaced by

DΦCSDt=∂∂t∭CVρϕdV+∬CVSϕρV→•dA→(2.50)

in which the first term on the right-hand side represents the time rate of change of ΦΦ within the control volume and the second term represents the net outflow (total outflow minus total inflow) of ΦΦ through the control volume surface.

Combining Equations 2.49 and 2.50, we obtain

ΠS+ΠB=∂∂t∭CVρϕdV+∬CVSϕρV→•dA→.(2.51)

Invoking Gauss divergence theorem, we can replace the surface integral in Equation 2.51 by volume integral, giving

ΠS+ΠB=∭CV∂ρϕ∂t+∇•ρϕV→dV(2.52)

where we have assumed that ∂V/∂t=0, allowing us to pull ∂/∂t∂/∂t inside the volume integral. Note that Equation 2.51 is convenient to use for an integral control volume analysis, and Equation 2.52 forms the basis for a point-wise differential control volume analysis.

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.

Table 2.2 z-Component Angular Momentum Flow Rates at Inlets and Outlets of the Control Volume of Figure 2.6

Inlet / Outlet	Mass velocity	Specific z-component angular momentum	z-component angular momentum flow rate
1 (outlet)	V1V1	a1V1a1V1	a1V1(ρ1A1V1)a1V1ρ1A1V1
2 (inlet)	V2V2	a2V2a2V2	a2V2(ρ2A2V2)a2V2ρ2A2V2
3 (inlet)	V3V3	−a3V3−a3V3	−a3V3(ρ3A3V3)−a3V3ρ3A3V3
4 (outlet)	V4V4	−a4V4−a4V4	−a4V4(ρ4A4V4)−a4V4ρ4A4V4
5 (outlet)	V5V5	b5V5b5V5	b5V5(ρ5A5V5)b5V5ρ5A5V5

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 Euler’s Turbomachinery Equation

Euler’s turbomachinery equation is widely used in all types of turbomachinery. This equation, derived from the angular momentum equation discussed in Section 2.2.5.3, determines power transfer between the fluid and the rotor blades. In pumps, fans, and compressors, the power transfer occurs into the fluid to increase its outflow rate of angular momentum over the inflow rate; in turbines, the power transfer occurs to the rotor from the fluid to decrease its outflow rate of angular momentum over the inflow rate. The power transfer to or from the fluid is simply the product of the torque and rotor angular velocity in radians per second.

Let’s consider a steady adiabatic flow in a rotating passage between two blades of a mixed axial-radial-flow turbomachinery, as shown in Figure 2.8. Each of the velocity vectors, V→1 at the inlet 1 and V→2 at outlet 2, has components in axial, radial, and tangential directions. The meridional velocity, which is the resultant of the axial and the radial velocities, is the mass velocity in the rotating passage flow at sections 1 and 2. Let V_θ1Vθ1 be the tangential velocity at inlet 1 and V_θ2Vθ2 at outlet 2. For a constant mass flow rate ṁ through the rotating passage, the angular momentum equation, Equation 2.61, yields

Γ=ṁr2Vθ2−r1Vθ1(2.70)

Figure 2.8 Flow through an axial-radial-flow turbomachinery passage between adjacent blades.

The aerodynamic power transfer, which is the rate of work transfer as a result of the aerodynamic torque acting on the fluid control volume, is given by

Ẇ=ΓΩ=ṁr2Vθ2−r1Vθ1Ω=ṁU2Vθ2−U1Vθ1(2.71)

where U₁U1 and U₂U2 are rotor tangential velocities at inlet 1 and outlet 2, respectively. Using steady flow energy equation in terms of total enthalpy at sections 1 and 2, we can also write

Ẇ=ṁht2−ht1.(2.72)

Combining Equations 2.71 and 2.72 yields

Equation (2.73) is known as the Euler’s turbomachinery equation. This equation simply states that, under adiabatic conditions, the change in specific total enthalpy of the fluid flow between any two sections of a rotor equals the difference in the products of the rotor tangential velocity and flow tangential velocity at these sections. This equation further reveals that for turbines, where the work is done by the fluid, we have h_t2 < h_t1ht2<ht1 and U₂V_θ2 < U₁V_θ1U2Vθ2<U1Vθ1; and for compressors, fans, and pumps, where the work is done on the fluid, we have h_t2>h_t1ht2>ht1 and U₂V_θ2>U₁V_θ1U2Vθ2>U1Vθ1.

For axial flow machines, where r₁≈r₂r1≈r2 and U₁≈U₂U1≈U2, Equation 2.73 reveals that the change in total enthalpy is entirely as a result of the change in the flow tangential velocity, that is, Δht=U¯ΔVθ, requiring blades with camber (bow). By contrast, for radial machines (radial turbines and centrifugal compressors), the change in total enthalpy results largely as a result of the change in rotor tangential velocity as a result of the change in radius, that is, Δht=ΔUV¯θ.

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

h_t1 − U₁V_θ1 = h_t2 − U₂V_θ2

ht1−U1Vθ1=ht2−U2Vθ2(2.74)

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)

V_θ = W_θ + U

Vθ=Wθ+U(2.78)

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 Compressible Flow with Area Change, Friction, Heat Transfer, Rotation, and Normal Shocks

Compressible flows are ubiquitous in gas turbines. The key parameter that characterizes a compressible flow is its Mach number (M = V/CM=V/C), which is the ratio of the flow velocity to speed of sound (C=κRTs). For M ≤ 0.3M≤0.3, the gas flow may be treated as incompressible for most practical engineering calculations using a constant density, which is obtained using the perfect gas law P_s/ρ = RT_sPs/ρ=RTs. With several examples from gas turbine design applications, Sultanian (2015) presents comprehensive details of compressible flows with area change, friction, heat transfer, rotation, normal, and oblique shocks. In our presentation of internal compressible flows in this section, all the flow properties are uniform to be uniform at each section.

Let us briefly reflect on the fundamental difference between a compressible flow and an incompressible flow with constant density. The total energy of a flow, as measured by its total temperature, is the sum of the fluid static temperature, which is the average kinetic energy associated with the gas molecules, and the dynamic temperature based on the flow kinetic energy. In an incompressible flow, there is negligible coupling between the external flow energy and its internal one. For example, when water flows through a converging-diverging (C-D) nozzle with constant entropy (constant total pressure and total temperature), the throat section with its minimum area features the maximum flow velocity, or the flow kinetic energy, with negligible decrease in its static temperature, a measure of the internal energy of the flow. Note that the specific heat of water is more than four times that of air. Further, in an incompressible flow, total pressure (P_t = P_s + ρV²/2Pt=Ps+ρV2/2) may be interpreted as the total mechanical energy per unit volume of the flow. The constancy of total pressure in an ideal incompressible flow leads to the famous Bernoulli equation with its tenet of “high velocity, low pressure” at any point in the flow.

A compressible flow, by contrast, exhibits quite different characteristics, particularly at high Mach numbers. For an isentropic airflow in a choked C-D nozzle with exit pressure ratio (ratio of total-to-static pressure) of more than two, both the static pressure and static temperature continuously decrease in the flow direction with continuous increase in Mach number. In the diverging section with increasing flow area, Mach number increases mainly as a result of the increase in the flow velocity needed to maintain a constant mass flow rate through the nozzle under significantly decreasing density. This behavior is somewhat counterintuitive based on our understanding of an incompressible internal flow where, for a given mass flow rate, the flow velocity always decreases with increasing area.

Note that the square of Mach number indicates the ratio of the external energy to internal energy of the flow. This provides a useful physical insight into Mach number. When the airflow expands in a C-D nozzle, increasing Mach number entails increase in the external flow energy and decrease in the internal one, the total flow energy remaining constant (constant total temperature). In other words, the change in Mach number is associated of the exchange between the internal flow energy and external flow energy. This two-way coupling between the internal and external flow energies in a compressible flow, also known as the compressibility effect, makes it fundamentally different from an incompressible flow. Note that the compressibility effect increases with Mach number. At low Mach numbers (M ≤ 0.3M≤0.3), as stated before, the compressibility effect may be neglected, and the compressible flow may be assumed to behave like an incompressible flow. On the other hand, an incompressible flow cannot be modeled to behave like a compressible flow. Based on this discussion, one should therefore avoid the use of Bernoulli equation or its extended form, also called the mechanical energy equation, in a compressible flow.

The phenomenon of choking is again unique to a compressible flow. At any section of fixed area, when the Mach number is unity (M = 1M=1), the flow is considered to be choked at that section, meaning the downstream changes in the pressure boundary condition will have no effect on the mass flow rate through the section. Note, however, that this maximum mass flow rate can still be changed by changing the upstream boundary conditions where the flow is subsonic (M < 1M<1). Choking has a simple physical interpretation. Because small pressure changes in a gas travel at the speed of sound in all directions, for M = 1M=1 at a section, the flow velocity equals the local speed of sound and prevents any sound wave to travel upstream of this section. As a result, the lowering of the C-D nozzle exit static pressure, after it is choked at the throat, will not flow more regardless of how much we decrease this pressure as long as the upstream total pressure and total temperature remain constant.

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

S_T = P_iA = P_sAI_fs = P_tAI_ft

ST=PiA=PsAIfs=PtAIft(2.109)

where I_fsIfs is the static-pressure impulse function given by

I_fs = (1 + κM²)

Ifs=1+κM2(2.110)

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.

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