Skip to main content

Thermodynamic laws and efficiencies of turbomachines

Short answer

First law: turbine w=Δh0, compressor w=Δh0. Second law: entropy increase reduces output, raises input. Isentropic efficiency η=actual/isentropic Δh. Polytropic efficiency η_p is infinitesimal-stage efficiency; for perfect gas, η_c=(r^[(γ-1)/γ]-1)/(r^[(γ-1)/(γ η_p)]-1), η_t=(1-r^[-η_p(γ-1)/γ])/(1-r^[-(γ-1)/γ]). Preheat: η_c<η_p; reheat: η_t>η_p.

Apply the first and second laws of thermodynamics to an adiabatic turbine and an adiabatic compressor. Define the isentropic (adiabatic) efficiency and the small-stage (polytropic) efficiency of each and show how they are related for a perfect gas.

First law (steady flow energy equation)

For one inlet and one outlet, neglecting potential energy, per unit mass:

q−w=(h2−h1)+C22−C122q - w = (h_2 - h_1) + \frac{C_2^2 - C_1^2}{2}

For an adiabatic machine q=0q = 0. Using stagnation enthalpy h0=h+C2/2h_0 = h + C^2/2:

  • Turbine: w=h01−h02w = h_{01} - h_{02}
  • Compressor: win=h02−h01w_{in} = h_{02} - h_{01}

For a perfect gas, h0=cpT0h_0 = c_p T_0, so w=cp(T01−T02)w = c_p (T_{01} - T_{02}) for the turbine.

Second law

For an adiabatic machine s2≥s1s_2 \ge s_1. The ideal (reversible) machine is isentropic, s2=s1s_2 = s_1. Irreversibility (friction, shock, tip leakage) raises the entropy, so the turbine delivers less work and the compressor needs more work than the isentropic value.

 T                            T          2 (actual)
 |  1                         |    2s   /
 |  |\                        |    |   /
 |  | \                       |    |  /
 |  |  \                      |    | /
 |  |   2 (actual)            |    |/
 |  2s                        |    1
 +------------ s              +------------ s
    turbine                      compressor

In both cases 2s lies vertically above or below 1 (same entropy), and the actual state 2 lies to its right at a higher temperature.

Isentropic efficiencies

ηt=h01−h02h01−h02s≈T01−T02T01−T02s,ηc=h02s−h01h02−h01≈T02s−T01T02−T01\eta_t = \frac{h_{01} - h_{02}}{h_{01} - h_{02s}} \approx \frac{T_{01} - T_{02}}{T_{01} - T_{02s}}, \qquad \eta_c = \frac{h_{02s} - h_{01}}{h_{02} - h_{01}} \approx \frac{T_{02s} - T_{01}}{T_{02} - T_{01}}

Polytropic (small-stage) efficiency

It is the isentropic efficiency of an infinitesimal stage, assumed constant through the machine. For a compressor with pressure ratio rr and γ\gamma:

ηp=γ−1γln⁡rln⁡(T02/T01),ηc=r(γ−1)/γ−1r(γ−1)/(γηp)−1\eta_{p} = \frac{\frac{\gamma - 1}{\gamma}\ln r}{\ln (T_{02}/T_{01})}, \qquad \eta_c = \frac{r^{(\gamma-1)/\gamma} - 1}{r^{(\gamma-1)/(\gamma \eta_p)} - 1}

For a turbine expanding by ratio rr:

ηt=1−r−ηp(γ−1)/γ1−r−(γ−1)/γ\eta_t = \frac{1 - r^{-\eta_p (\gamma-1)/\gamma}}{1 - r^{-(\gamma-1)/\gamma}}

For the same ηp\eta_p, the compressor ηc\eta_c is less than ηp\eta_p (preheat effect: the losses in each stage heat the air entering the next stage, which then needs more work), while the turbine ηt\eta_t is greater than ηp\eta_p (reheat effect: the losses in each stage are partly recovered as work in the following stages). The difference grows with the pressure ratio.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Turbo Machines chapter 1 questions →