Question

EXAMPLE 4.3 A major objective in advancing gas turbine engine technologies is to increase the temperature limit associated with operation of the gas turbine blades. This limit determines the permissible turbine gas inlet temperature, which, in turn, strongly influences overall system performance. In addition to fabricating turbine blades from special, high-temperature, high-strength superal- loys, it is common to use internal cooling by machining flow channels within the blades and routing air through the channels. We wish to assess the effect of such a scheme by approximat- ing the blade as a rectangular solid in which rectangular channels are machined. The blade, which has a thermal conductivity of k- 25 W/m-K, is 6 mm thick, and each channel hasa 2 mm × 6 mm rectangular cross section, with a 4-mm spacing between adjoining channels. Air channel Combustion gases 2 mm 6 mm 6 mm 4 mm CombustionTo Turbine blade, k gases

I need a Matlab code for this problem with solution using matrixes please, thanks in advance.

Answer 1

clear r = zeros (20,29) ; Initial condition % Sst all material to 20C ambient temp CRE y- 1:20; for x 1:20; sad end h11000000:0:0:01 at= .1; 11me increment , Seconds) p-9000; %LDsnadty of copper, kg/m^ 3) C具一380: %13x satis heat of Copper, J/kg *K) 虹 500; (Solar heat flux, w/m^ 2) TL 50; %(Temperature of the left side of the block, Cel31us) T sam-400; % (Temperature of steam inside the duct, Celsius) 1 = 0.005; % 0.005 Meter K = 400; % (Thermal conductivity of Coppy.Х/m*c) Alpha-K/LD"Cp) ; (Thermal diffusivity Bi-(h1): ell )=det (T) ;I tprintt'Enter Time (3n' A-t/dt: i/tep increment s1ze 동in3tant team at center for y 6:15; ez -6:15 end: for whole matrix % BOUNDRY CONDITIONS Node 1 -Top left node T (1,1)-T (1,1)(1-4Eo) +2 Eo (T (1,2) T (2,1) (2/K) TLt 1/K) % Node 21 T(1,20} = -Top right node T(1, 20) * (1-4*Fg)+2*Fo"(T ( 1,19) T(2.201+ 虹+1/K); + Left nodes - Left Bide BC Right nodes -Right Bur face BC (Insulated Bection (no gr)) for x 20; tRE y-3:19; 2*T (y, x-1)) end: end: % Bottom left Node - in3ulated with left BC T120, 1) =T(20, 1) * (1-4*Fg) +2*Fo"(T ( 19, 1) + τ(20,2) + (2 TL) /K) Bottom right Node - in ulated with right BC 7/20,20) 7/20, 20)* (1-4"Eg) 2*E9" (T (19,20) T(20,19)); · + + Top nodes calcs-Top Bur face (Heat lux addition) tor y=1; for x-2:19: end: end: 용 Calculating the in between nodal temperatures at each deltaitime) İRE y _ 2:19; 兔inside left slab and right 31ab 船Ex- [2:5, 16:19] 1.x) +1(y+1 , x)); end: central nodes for x = 6:19陪center slab portions above and below duct fot y[2:5,16:191 end; sad % Bottom Nodes - insulated section (no gl) for y = 20; for x = 2:19; 2*T (y-1, x) ) ; end: