{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

AET432-CH6-page11

AET432-CH6-page11 - Review problems 6-59 The Coutte flow...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Review problems 6-59 The Coutte flow of a fluid between two parallel plates is considered. The temperature distribution is to be sketched and determined, and the maximum temperature of the fluid, as well as the temperature of the fluid at the contact surfaces with the lower and upper plates are to be determined. Assumptions Steady operating conditions exist. Properties The viscosity and thermal conductivity of the fluid are given to be p = 08 N-s/m2 and kf= 0.145 W/m-K. The thermal conductivity of lower plate is given to be kp = 15 W/m-K. Analysis: (a) Insulation ‘ v9 fee. 9". ova-’0 9‘2'7'9'0 9 7‘0. vvv .‘Q' 202020:¢I¢$2€OI§OX¢I029202631020102033301020202.10‘ r, = 40°C k,, The sketch of temperature distribution is given in the figure. We observe from this figure that there are different slopes at the interface (y = 0) because of different conductivities (k,, > kf). The slope is zero at the upper plate 02 = L) because of adiabatic condition. (b) The general solution of the relevant differential equation is obtained as follows: u=lV———)iu—=lf— L dy L 2 2 _ 2 a; if? iLJLwCI dy kf L dy kf L2 —/-l V2 2 a T:————- +C +C Zkf L2 y 1y 2 Applying the boundary conditions: . . dT T(0)—T ””17“”? 7:7 y=0 y o p kp kal =T(C2 ‘TJ (1) 2 y=L.adiabatic dT =0————-)Cl:iZ—— dy L kf L k 2 FromEq. (1), C2 :b—fC] +Tv =b—g—V—+Tx kp ' kp L Substituting the coefficients, the temperature distribution becomes ...
View Full Document

{[ snackBarMessage ]}