B2) An automotive engine can be assumed as rectangular block with approximation size 0.4 m high, 0.6 m wide, and 0.7 m long as shown in Figure 6. The ambient air is at 1 atm and 15°C where air density (p) = 1.225 kg/m³ and kinematic viscosity (V) = 1.47x105 m²/s. Air 85km/h 15°C Engine Block Figure 6 a) Calculate the friction coefficient (C) and investigate the force acting on the bottom surface of the engine block as the car travels at the velocity of 85 km/h. Assume the flow to be turbulent (Cı = 0.074/Re5) over the entire surface because of the constant agitation of the engine block. b) An engine oil at 40°C with kinematic viscosity (v) = 2.485×104 m²/s is flowing over a long flat plate with velocity of 4 m/s. Determine the critical distance xer from the leading edge of the plate where the flow become turbulent (Rec = 5x105), and analyze the thickness of the boundary layers over a length of 2xer. Given boundary layer thickness for laminar flow is, o, = 4.91x/Re,12 and turbulent flow, ỗ., = 0.38x/Re,15. %3D

Problem 1: Streamlining From dimensional analysis, we know the drag force is FD- CopAV2/2, where Cp is a dimensionless coefficient of drag. The two space shuttles below have the same area A, the same velocity V, and are flying through the same air. Which has a smaller Co and why? (<25 words) thine e

Please solve and write out equations used. Thanks

% ParametersD = 0.1; % Diameter of the tube (m)L = 1.0; % Length of the tube bundle (m)N = 8; % Number of tubes in the bundleU = 1.0; % Inlet velocity (m/s)rho = 1.2; % Density of the fluid (kg/m^3)mu = 0.01; % Dynamic viscosity of the fluid (Pa.s) % Define the grid size and time stepdx = D/10; % Spatial step size (m)dy = L/10; % Spatial step size (m)dt = 0.01; % Time step size (s) % Calculate the number of grid points in each directionnx = ceil(D/dx) + 1;ny = ceil(L/dy) + 1; % Create the velocity matrixU_matrix = U * ones(nx, ny); % Perform the iterationsfor iter = 1:100    % Calculate the velocity gradients    dUdx = (U_matrix(:, 2:end) - U_matrix(:, 1:end-1)) / dx;    dUdy = (U_matrix(2:end, :) - U_matrix(1:end-1, :)) / dy;        % Calculate the pressure gradients    dpdx = -mu * dUdx;    dpdy = -mu * dUdy;        % Calculate the change in velocity    dU = dt * (dpdx / rho);        % Update the velocity matrix    U_matrix(:, 2:end-1) = U_matrix(:, 2:end-1) + dU;        % Apply…