University:
Massachusetts Institute of Technology
Course:
16.01 | Unified Engineering I, II, III, & IV
Academic year:

2019
Views:

278

Pages:

2
Author:

Raina Ballard

Certainly, here's the analysis of the image you provided: Visual Elements: Handwritten Text: The image presents handwritten notes or calculations. Writing Style: The writing is in blue or black ink. Content: The text is organized into sections labeled "a)" and "b)", with further sub-sections for different parts of the problem. Within each section: There are equations, calculations, and some explanatory text. Textual Content: TG SOLUTIONS (WAITE) 1 OF 2 a) P_e = 125 atm = P_c T_c = 3000K = T_re (ξ_0 in combustor) (ξ_0 in combustor) ∂ξ/∂t = -γξ(1-ξ²) = -ξ(1-ξ²) ∂ξ/∂t + ξ² = ξ G_Tξ + ξ² = G_Tξ_e + ξ²_e = G_Tξ_e + 1 G_Tξ + ξ² = 1 ALSO (P_c/P_e) = T_c SINCE ADIABATIC ⇒ G_T = √2G_T_c [1 - (P_e/P_c) ^ (γ-1)/γ] = √2(1500)(3000) [1 - (1/125)^(2/7)] = 2230 m/s C_a = 2230 m/s (IN SPACE P_e → 0) b) (G_T_c)² + C_a² = 2H_0 1500(3000) + 2230² = 2H_0 H_0 = 1342 K α = γ/(γ-1) = 7/2 = 3.5 M_a = C_a / √(αRT_e) = 3.5 b) IN SPACE C_a = 3000 m/s T_e → 0, ... M_a → ∞ 2 OF 2 c) If H_TC reduced 20%, then H_re reduced 20%. (C_p T_c) (0.8) = C_p T_re + C_a²/2 C_a = √2C_p T_c (0.8)[1 - (T_re/T_c)]^(1/2) C_a,atm = √2(1500)(3000)(0.8)[1 - (1342/3000)]^(1/2) C_a,atm = 1995 m/s C_a,space = 0 P_e → 0 C_a,space = √2(1500)(3000)(0.8)[1 - 0]^(1/2) C_a,space = 2683 m/s

T9 Solutions

of 2

Description

Free up your schedule!

Take 5 seconds to unlock