University:
Massachusetts Institute of Technology
Course:
16.50 | Introduction to Propulsion Systems
Academic year:

2023
Views:

321

Pages:

6
Author:

ilikemyicons0y5

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 related to propulsion systems, specifically the analysis of a two-position nozzle. Equations: Numerous mathematical equations are included, describing various properties of the nozzle and the flow through it, such as characteristic velocity, mass flow rate, Mach number, and area ratio. Values: Numerical values are plugged into the equations to obtain specific results. Textual Content: Homework 3: Two-Position Nozzle a) c* = √(RT_c/γ) ............................................. (1) Γ = √γ (2/(γ+1))^((γ+1)/(2(γ-1))) ............................. (2) Γ = √1.2 (2/2.2)^(2.2/(2(1.2-1))) = 0.6485 R = 8.314/mol-K = 461.9 J/mol-K ............................. (3) Solving for c* and ṁ: c* = √(461.9*3300)/0.6485 = 1903.7 m/s ṁ = ρ_c A_t/c* = 30*1.013E5*0.1/1903.7 = 372.5 kg/s For the inner bell: P_e1/P_c = 0.4 Therefore, the exit Mach number is given by: 0.4 = (1 + (γ-1)/2 M_e1^2)^(-γ/(γ-1)) M_e1 = 3.695 The area ratio is then: A_e1/A_t = 1/M_e1 ((1 + (γ-1)/2 M_e1^2)/((γ+1)/2))^((γ+1)/(2(γ-1))) A_e1/A_t = 18.234 A_e1 = 1.8234 m² ............................. (4) The vacuum thrust is: F_v1 = ṁu_e1 + P_e1 A_e1 ............................. (5) u_e1 = √(2γ/(γ-1)RT_c[1 - (P_e1/P_c)^((γ-1)/γ)]) ............................. (6) u_e1 = √(2*1.2/(1.2-1)*461.9*3300[1 - (0.4)^(0.2/1.2)]) = 3241 m/s On the ground: F_0 = F_v1 - P_a0 A_e1 F_v1 = 1.2842e6 N, A_e1 = 1.8234 m² F_0 = 1.2842e6 - 1.013e5 * 1.8234 = 1.0995e6 N b) m dv/dt = F - mg = F_v - P_a A_e - mg (8) dm = -ṁdt (9) dv = -g dt - ṁdt (10) Putting equations together: dv = -g dt - ṁdt - A_e p_a/m dt (12) Integrating between t = 0 and t = t_1 (with F_v = F_v1 and A_e = A_e1): v_1 = -g t_1 - ṁ ln(m_0/m_1) - A_e1 ∫₀^t_1 (p_a(t)/m(t)) dt (13) Integrating between t = t_1 and t = t_b (with F_v = F_v2 and A_e = A_e2): v_b - v_1 = -g(t_b - t_1) - ṁ ln(m_1/m_b) - A_e2 ∫_t_1^t_b (p_a(t)/m(t)) dt (14) Adding equations (13) and (14): v_b = -g t_b - ṁ ln(m_0/m_1) - ṁ ln(m_1/m_b) - g t_1 - A_e1 ∫₀^t_1 (p_a(t)/m(t)) dt - A_e2 ∫_t_1^t_b (p_a(t)/m(t)) dt (15) Here, of course: m_1 = m_0 - ṁt_1 (16) So, t_1 appears in m_1 and in the limits of integration. To optimize, set ∂v_b/∂t_1 = 0, using ∂m_1/∂t_1 = -ṁ: F_v1 - F_v2 + (A_e2 - A_e1) p_a(t_1) = 0 F_v(t_1) = (F_v1 A_e2 - F_v2 A_e1) / (A_e2 - A_e1) (for optimum t_1) Just before the transition: F_1(t_1 - ε) = F_v1 - (F_v2 - F_v1) / (A_e2 - A_e1) A_e1 = (F_v1 A_e2 - F_v2 A_e1) / (A_e2 - A_e1) (17)

16.50 Homework 3 (Solutions)

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