--- title: "Stagnation properties and Mach number relations" subject: "Turbo Machines" chapter: 4 key: "semester_7/turbo_machines:stagnation-temperature-pressure-relations" source: https://www.bctengineer.com/q/semester_7/turbo_machines/stagnation-temperature-pressure-relations site: BCT Engineering --- # Stagnation properties and Mach number relations Turbo Machines · Chapter 4: Gas Turbine Nozzles - Asked in: year not given, 5 marks - Video: [Explained: Stagnation Relations](https://www.youtube.com/watch?v=NYgScbe9gQw) — JoshTheEngineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/turbo_machines/stagnation-temperature-pressure-relations - Whole chapter: https://www.bctengineer.com/md/semester_7/turbo_machines/chapter-4 ## Question Define stagnation (total) temperature and pressure. Derive the relations $T_0/T$ and $p_0/p$ in terms of Mach number for a perfect gas, and explain why they are used in gas turbine analysis. ## Short answer Stagnation temperature and pressure are the values when the flow is adiabatically and isentropically brought to rest. T0/T = 1 + ((gamma-1)/2) Ma^2 and p0/p = (1 + ((gamma-1)/2) Ma^2)^(gamma/(gamma-1)). They are used because total conditions are independent of velocity, providing a common reference for engine station analysis. ## Answer The **stagnation (total) state** is the state reached when a flowing fluid is brought to rest adiabatically (and isentropically for the pressure). The static properties $T$, $p$ are those measured moving with the fluid. #### Stagnation temperature From the steady flow energy equation for adiabatic deceleration to rest: $$ h_0 = h + \frac{C^2}{2} \;\Rightarrow\; c_p T_0 = c_p T + \frac{C^2}{2} $$ $$ T_0 = T + \frac{C^2}{2c_p} $$ Using $c_p = \dfrac{\gamma R}{\gamma - 1}$ and $a^2 = \gamma R T$, so $C^2 = Ma^2 \gamma R T$: $$ \boxed{\frac{T_0}{T} = 1 + \frac{\gamma - 1}{2}Ma^2} $$ #### Stagnation pressure For isentropic deceleration, $p_0/p = (T_0/T)^{\gamma/(\gamma-1)}$: $$ \boxed{\frac{p_0}{p} = \left(1 + \frac{\gamma - 1}{2}Ma^2\right)^{\gamma/(\gamma-1)}} $$ At $Ma = 1$ (critical state): $T^*/T_0 = 2/(\gamma + 1) = 0.833$ and $p^*/p_0 = 0.528$ for air ($\gamma = 1.4$). #### Points to note - In adiabatic flow without work, $T_0$ stays **constant** along a nozzle even with friction; $p_0$ **falls** because of friction (entropy rise). - In a compressor or turbine, $T_0$ changes by the work done, so $w = c_p \Delta T_0$. - For small Mach number, $p_0 \approx p + \tfrac12\rho C^2$. #### Why used 1. The kinetic energy is large in gas turbines ($Ma$ of 0.3 to 1 or more), so static and total values differ greatly. 2. A thermocouple or pitot tube measures nearly total values, and performance (efficiency, pressure ratio) is defined on a total basis, e.g. $\eta_{tt}$. 3. Total conditions are independent of the flow velocity, so they give a common reference at each engine station (inlet, compressor exit, turbine exit). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The stagnation state is the state reached when the flowing fluid is A. heated at constant pressure B. brought to rest adiabatically (isentropically for pressure) C. expanded to zero pressure D. cooled to ambient temperature Answer: B. Total properties are those obtained by bringing the fluid to rest adiabatically, and isentropically for pressure. (Key point: Stagnation state definition) 2. Which combination gives T0/T = 1 + (gamma-1)/2 Ma^2? A. Momentum equation with p = rho R T B. Continuity with C = Ma a C. Energy equation cp T0 = cp T + C^2/2 with cp = gamma R/(gamma-1) and a^2 = gamma R T D. Entropy equation with ds = 0 only Answer: C. Dividing T0 = T + C^2/(2cp) by T and substituting cp and C^2 = Ma^2 gamma R T gives the relation. (Key point: Derivation of T0/T) 3. How is p0/p obtained from T0/T? A. p0/p = (T0/T)^(gamma/(gamma-1)) from isentropic deceleration B. p0/p = T0/T C. p0/p = (T0/T)^(gamma-1) D. p0/p = 1/(T0/T) Answer: A. For isentropic flow p and T are related by the exponent gamma/(gamma-1). (Key point: Stagnation pressure relation) 4. In an adiabatic nozzle with friction, what happens to T0 and p0? A. Both stay constant B. T0 falls, p0 constant C. T0 stays constant, p0 falls D. T0 rises, p0 falls Answer: C. No heat or work means T0 is constant, but friction raises entropy and so reduces the total pressure. (Key point: Friction in nozzle) 5. Why are stagnation values used in gas turbine analysis? A. Gas velocities are always negligible B. Total values equal static values at high Mach number C. Static values cannot be defined D. Kinetic energy is large, probes measure near-total values, and total values give a common reference at each station Answer: D. At Ma of 0.3 to 1 or more, static and total values differ greatly, and efficiencies such as eta_tt are defined on a total basis. 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