Isentropic efficiency

Given steam entering a turbine at $3$ MPa and $350$ degrees Celsius and leaving at $10$ kPa, the energy balance is equally content with a machine that produces $980$ kJ per kilogram and one that produces nothing at all.

The gap the energy balance leaves

The steady-flow energy equation says $w = h_{1} - h_{2}$ for an adiabatic turbine, and that is a true statement about any turbine. It is also a statement with two unknowns in it. Nothing in the First Law fixes $h_{2}$ , so nothing in the First Law fixes the work. Feed the equation an exhaust enthalpy of $2136$ kJ kg⁻¹ and the turbine is excellent; feed it $3116$ kJ kg⁻¹ and the steam has passed through unchanged, producing nothing, while conserving energy perfectly.

The same hole appears everywhere. A compressor that raises air from $100$ to $800$ kPa may deliver it at $534$ K or at $700$ K, and the energy balance simply reports a larger work input in the second case without complaint. A nozzle may reach $449$ m s⁻¹ or $400$ m s⁻¹ from the same inlet state. In each case the missing statement is not about energy but about direction, and the previous course established what supplies it.

The entropy balance for a control volume

Entropy is not conserved. It is transferred with heat, carried by mass, and generated by irreversibility, and the balance says so term by term. For a control volume,

\frac{d S_{c v}}{d t} = \sum \frac{\dot{Q}}{T_{b}} + \sum_{i n} \dot{m} s - \sum_{o u t} \dot{m} s + {\dot{S}}_{g e n}

where $T_{b}$ is the temperature of the boundary where the heat crosses, and ${\dot{S}}_{g e n} \geq 0$ always, equalling zero only for a reversible process. The second and third terms are new relative to the closed-system version, and they are simple: a kilogram crossing a port takes its entropy with it, exactly as it takes its enthalpy.

Under steady operation the left side vanishes, and for one inlet and one outlet the balance rearranges to

{\dot{S}}_{g e n} = \dot{m} (s_{2} - s_{1}) - \sum \frac{\dot{Q}}{T_{b}} \geq 0

Now make the device adiabatic, which as the earlier lesson argued is a good description of a turbine, a compressor, a nozzle or a valve. The heat term goes and what is left is stark:

s_{2} \geq s_{1}

An adiabatic steady-flow device cannot lower the entropy of the fluid passing through it. Not "usually does not": cannot. Every real one raises it, and the amount by which it does is ${\dot{S}}_{g e n} / \dot{m}$ , a direct measure of how badly the device is behaving.

The ideal device is the isentropic one

The inequality has a boundary, and the boundary is the benchmark we were missing. The best conceivable adiabatic device is the one that generates no entropy at all, so it operates at $s_{2} = s_{1}$ and is called isentropic. Isentropic means adiabatic and reversible together, and neither half alone is enough: a throttling valve is adiabatic and hopeless.

The isentropic device is not achievable, and that does not weaken it. It is a fixed, computable target defined by the inlet state and the exit pressure alone, which is exactly what a benchmark has to be. Given $3$ MPa and $350$ degrees Celsius at inlet and $10$ kPa at exhaust, the isentropic exit state is fully determined: $s_{2} = s_{1} = 6.7450$ kJ kg⁻¹ K⁻¹, which at $10$ kPa where $s_{f} = 0.6492$ and $s_{f g} = 7.4996$ gives a quality

x_{2 s} = \frac{6.7450 - 0.6492}{7.4996} = 0.8128

and hence $h_{2 s} = 191.81 + 0.8128 \times 2392.1 = 2136.2$ kJ kg⁻¹. The ideal work is $3116.1 - 2136.2 = 980.0$ kJ kg⁻¹, and no adiabatic turbine between those states can beat it.

For an ideal gas with constant specific heats the isentropic exit state comes from a formula rather than a table. Starting from $d s = c_{p} d T / T - R d p / p$ and setting $d s = 0$ ,

\frac{T_{2}}{T_{1}} = {(\frac{p_{2}}{p_{1}})}^{(k - 1) / k}

with $k = c_{p} / c_{v}$ , which for air is $1.400$ and gives an exponent of $0.2857$ . This one relation carries most of the gas turbine work later in the course.

Reversible work, and why pumping liquid is cheap

Before scoring real machines, one result explains a fact that has been asserted twice already. For a reversible steady-flow process, the work per kilogram is

w_{r e v} = - \int_{1}^{2} v d p

ignoring velocity and height changes. The derivation is two lines. The property relation $T d s = d h - v d p$ holds for any substance between any two equilibrium states. For a reversible process $δ q = T d s$ , and the energy equation in differential form is $δ q - δ w = d h$ . Substitute: $δ w = T d s - d h = T d s - (T d s + v d p) = - v d p$ . Integrate.

Read it as an instruction. The work of a steady-flow machine is set by the specific volume of whatever is being pushed against the pressure change. To spend little, arrange for $v$ to be small, which means compress liquids and expand vapours, never the reverse.

The numbers are brutal. Water at $10$ kPa has $v = 0.001010$ m³ kg⁻¹ and is nearly incompressible, so the integral is just $v (p_{2} - p_{1})$ . Raising it to $15$ MPa costs $0.001010 \times 14990 = 15.1$ kJ kg⁻¹. Saturated steam at the same $10$ kPa has $v = 14.67$ m³ kg⁻¹, fourteen thousand times larger, and compressing it over the same range would cost of the order of the entire output of the turbine. This is why every vapour power cycle condenses its working fluid completely before raising its pressure, and it is the single most consequential line in this course.

Example. A boiler feed pump raises water from saturated liquid at $10$ kPa, where $v = 0.001010$ m³ kg⁻¹, to $15$ MPa. Its isentropic efficiency is $0.85$ . Find the ideal and actual work per kilogram, and the power required at $462$ kg s⁻¹.

Treating the water as incompressible, $w_{s} = v (p_{2} - p_{1}) = 0.001010 \times (15000 - 10) = 15.14$ kJ kg⁻¹. A pump's efficiency is ideal work over actual, since work is being paid for, so $w_{a} = 15.14 / 0.85 = 17.81$ kJ kg⁻¹. The power is $462 \times 17.81 = 8230$ kW, or $8.2$ MW, out of a station producing several hundred.

Now you. A smaller plant pumps the same water from $10$ kPa to $5$ MPa with a pump of isentropic efficiency $0.80$ . Find the ideal and actual work per kilogram.

Answer

$w_{s} = 0.001010 \times 4990 = 5.04$ kJ kg⁻¹ and $w_{a} = 5.04 / 0.80 = 6.30$ kJ kg⁻¹.

Turbine isentropic efficiency

Now the definition. For a turbine, work is the product, so the efficiency is what you got over what you could have got, both taken between the same inlet state and the same exit pressure:

η_{T} = \frac{w_{a}}{w_{s}} = \frac{h_{1} - h_{2 a}}{h_{1} - h_{2 s}}

Large steam turbines reach $0.88$ to $0.92$ , gas turbines $0.85$ to $0.90$ , small machines much less. The definition has a subtlety worth stating: $h_{2 a}$ and $h_{2 s}$ are at the same pressure but not the same state. The real turbine ends at a higher entropy, and therefore at a higher enthalpy, than the ideal one.

Example. The turbine above, taking steam at $3$ MPa and $350$ degrees Celsius and exhausting to $10$ kPa, has $η_{T} = 0.87$ . Find the actual work and the actual exit state.

The ideal work was computed above as $980.0$ kJ kg⁻¹, so the actual work is $0.87 \times 980.0 = 852.6$ kJ kg⁻¹ and the actual exit enthalpy is $h_{2 a} = 3116.1 - 852.6 = 2263.5$ kJ kg⁻¹. That is still inside the dome at $10$ kPa, at a quality of $(2263.5 - 191.81) / 2392.1 = 0.866$ , against $0.813$ for the ideal machine. The real turbine exhausts drier than the ideal one, because the losses reappear as internal energy in the fluid and re-evaporate some of the moisture. The entropy generated is $s_{2 a} - s_{1} = 7.144 - 6.745 = 0.399$ kJ kg⁻¹ K⁻¹, and a later lesson will convert that number into wasted joules.

Now you. A worn machine on the same duty manages $η_{T} = 0.80$ . Find its actual work, exit enthalpy and exit quality.

Answer

$w_{a} = 0.80 \times 980.0 = 784.0$ kJ kg⁻¹, so $h_{2 a} = 3116.1 - 784.0 = 2332.1$ kJ kg⁻¹ and $x = (2332.1 - 191.81) / 2392.1 = 0.895$ .

Compressors, pumps and nozzles

For a device that consumes work the ratio is inverted, so that the efficiency is again a number below one:

η_{C} = \frac{w_{s}}{w_{a}} = \frac{h_{2 s} - h_{1}}{h_{2 a} - h_{1}}

Typical values run from $0.75$ for a small machine to $0.90$ for a large axial compressor. Pumps are defined identically and reach $0.70$ to $0.90$ .

Example. A compressor takes air at $100$ kPa and $295$ K and delivers it at $800$ kPa. Its isentropic efficiency is $0.82$ . With $c_{p} = 1.005$ kJ kg⁻¹ K⁻¹ and $k = 1.400$ , find the ideal work, the actual work and the delivery temperature.

The isentropic exit temperature is $T_{2 s} = 295 \times 8^{0.2857} = 295 \times 1.8115 = 534.4$ K, so $w_{s} = c_{p} (T_{2 s} - T_{1}) = 1.005 \times 239.4 = 240.6$ kJ kg⁻¹. The actual work is $w_{a} = 240.6 / 0.82 = 293.4$ kJ kg⁻¹, and since the air really does absorb all of it, $T_{2 a} = 295 + 293.4 / 1.005 = 587$ K, or $314$ degrees Celsius. Fifty-two kelvin of that is pure inefficiency, and it has to be got rid of downstream.

Now you. A cheaper machine on the same duty has $η_{C} = 0.75$ . Find its actual work and delivery temperature.

Answer

$w_{a} = 240.6 / 0.75 = 320.8$ kJ kg⁻¹ and $T_{2 a} = 295 + 320.8 / 1.005 = 614$ K, which is $341$ degrees Celsius.

A nozzle produces neither work nor heat, so its efficiency is written in kinetic energy: $η_{N} = V_{2 a}^{2} / V_{2 s}^{2}$ , the ratio of actual to ideal exit kinetic energy. Well-designed nozzles are extremely good, $0.90$ to $0.97$ , because the flow is accelerating and accelerating flows do not separate from walls. The air nozzle of an earlier lesson, ideally reaching $449$ m s⁻¹, manages $449 \sqrt{0.95} = 438$ m s⁻¹ at $η_{N} = 0.95$ .

Diffusers are the mirror image and are much worse, often $0.70$ to $0.85$ , because there the flow is decelerating against a rising pressure and will separate at the smallest excuse. Anyone who has designed a duct knows that expansions are hard and contractions are easy, and this is the thermodynamic statement of it.

What the benchmark does not say

Three honest limits, because the number is used more casually than it deserves.

First, isentropic efficiencies are not comparable across devices. A turbine at $0.87$ and a compressor at $0.87$ generate different amounts of entropy on the same duty, because the two definitions divide by different things. Comparing them properly needs a common currency, which is what the exergy lesson supplies.

Second, a multistage turbine's overall isentropic efficiency exceeds the efficiency of its individual stages, which sounds impossible and is not. Each stage's losses reheat the fluid, and the reheated fluid arriving at the next stage has slightly more enthalpy available to drop, since the constant-pressure lines on an enthalpy-entropy diagram diverge as entropy rises. Some of each stage's loss is therefore recovered downstream. The effect, called the reheat factor, is worth two to four points on a large machine, and it means a stage efficiency quoted as an overall one is being flattered.

Third, the benchmark is silent on devices whose ideal work is zero. A throttling valve has $w_{s} = 0$ and $w_{a} = 0$ , so the definition gives nothing at all, yet the valve is thoroughly irreversible: the R-134a valve of an earlier lesson generates $0.0174$ kJ kg⁻¹ K⁻¹ every time a kilogram goes through it. Isentropic efficiency measures how well a device does the job it was built for. It cannot notice a device whose job is to waste something.

With efficiencies for the individual machines in hand, the pieces can finally be chained together: the exhaust of one device becomes the inlet of the next, and the last one feeds back into the first. That closed chain is a cycle, and it is where this course starts producing power.