--- title: "Thermodynamic laws and efficiencies of turbomachines" subject: "Turbo Machines" chapter: 1 key: "semester_7/turbo_machines:laws-of-thermodynamics-efficiencies" source: https://www.bctengineer.com/q/semester_7/turbo_machines/laws-of-thermodynamics-efficiencies site: BCT Engineering --- # Thermodynamic laws and efficiencies of turbomachines Turbo Machines · Chapter 1: Introduction - Asked in: year not given, 4+4 marks - Video: [Turbines, Compressors, and Pumps - ISENTROPIC EFFICIENCY in 8 Minutes!](https://www.youtube.com/watch?v=K2A5gdROVZE) — Less Boring Lectures - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/turbo_machines/laws-of-thermodynamics-efficiencies - Whole chapter: https://www.bctengineer.com/md/semester_7/turbo_machines/chapter-1 ## Question 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. ## 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. ## Answer #### First law (steady flow energy equation) For one inlet and one outlet, neglecting potential energy, per unit mass: $$ q - w = (h_2 - h_1) + \frac{C_2^2 - C_1^2}{2} $$ For an adiabatic machine $q = 0$. Using stagnation enthalpy $h_0 = h + C^2/2$: - Turbine: $w = h_{01} - h_{02}$ - Compressor: $w_{in} = h_{02} - h_{01}$ For a perfect gas, $h_0 = c_p T_0$, so $w = c_p (T_{01} - T_{02})$ for the turbine. #### Second law For an adiabatic machine $s_2 \ge s_1$. The ideal (reversible) machine is isentropic, $s_2 = s_1$. Irreversibility (friction, shock, tip leakage) raises the entropy, so the turbine delivers less work and the compressor needs more work than the isentropic value. ```text 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 $$ \eta_t = \frac{h_{01} - h_{02}}{h_{01} - h_{02s}} \approx \frac{T_{01} - T_{02}}{T_{01} - T_{02s}}, \qquad \eta_c = \frac{h_{02s} - h_{01}}{h_{02} - h_{01}} \approx \frac{T_{02s} - T_{01}}{T_{02} - T_{01}} $$ #### Polytropic (small-stage) efficiency It is the isentropic efficiency of an infinitesimal stage, assumed constant through the machine. For a compressor with pressure ratio $r$ and $\gamma$: $$ \eta_{p} = \frac{\frac{\gamma - 1}{\gamma}\ln r}{\ln (T_{02}/T_{01})}, \qquad \eta_c = \frac{r^{(\gamma-1)/\gamma} - 1}{r^{(\gamma-1)/(\gamma \eta_p)} - 1} $$ For a turbine expanding by ratio $r$: $$ \eta_t = \frac{1 - r^{-\eta_p (\gamma-1)/\gamma}}{1 - r^{-(\gamma-1)/\gamma}} $$ For the same $\eta_p$, the compressor $\eta_c$ is less than $\eta_p$ (preheat effect: the losses in each stage heat the air entering the next stage, which then needs more work), while the turbine $\eta_t$ is greater than $\eta_p$ (reheat effect: the losses in each stage are partly recovered as work in the following stages). The difference grows with the pressure ratio. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For an adiabatic turbine, work per unit mass equals: A. h02 - h01 B. h01 - h02, the fall in stagnation enthalpy C. h01 + h02 D. Heat supplied q Answer: B. With q = 0, the steady-flow equation gives w = h01 - h02 using stagnation enthalpy. (Key point: First law for adiabatic turbine) 2. Because of irreversibility in an adiabatic machine, what happens? A. Turbine gives more work and compressor needs less B. Both machines are unaffected C. Entropy falls through the machine D. Turbine delivers less work and compressor needs more work than isentropic Answer: D. Friction and shock raise entropy (s2 >= s1), so the turbine delivers less and the compressor absorbs more than the isentropic value. (Key point: Second law effect) 3. The isentropic efficiency of a compressor is: A. Isentropic work input divided by actual work input B. Actual work divided by isentropic work C. Exit temperature divided by inlet temperature D. Pressure ratio divided by temperature ratio Answer: A. Compressor efficiency is (h02s - h01)/(h02 - h01), the ideal over the actual work input. (Key point: Isentropic efficiency definition) 4. For the same polytropic efficiency, how do isentropic efficiencies compare with it? A. Compressor higher, turbine lower B. Compressor lower, turbine higher C. Both are higher D. Both equal it Answer: B. Reheating makes compressor eta_c less than eta_p, while the reheat factor makes turbine eta_t greater than eta_p. 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