389 Chapter 5 x Dimensional Analysis and Similarity Solution: Establish the variables and their dimensions: M fcn( R , :, P , T ) {ML2/T2} {L} {1/T} {M/LT} {1} Then n 5 and j 3, hence we expect n j arrangement, as follows: M P :R 3 5 3 2 Pi groups, capable of only one reasonable M P:T R 3 fcn(T ); if M v T , then constant Ans. See Prob. 1.56 of this Manual, for an analytical solution. 5.36 The rate of heat loss, Qloss through a window is a function of the temperature difference 'T, the surface area A, and the R resistance value of the window (in units of ft2hrqF/Btu): Qloss fcn('T, A, R). (a) Rewrite in dimensionless form. (b) If the temperature difference doubles, how does the heat loss change? Solution: First figure out the dimensions of R: {R} {T34/M}. Then note that n and j 3 dimensions, hence we expect only 4 3 one Pi group, and it is: 31 Qloss R A 'T Const , or: Qloss Const A 'T R 4 variables Ans. (a) (b) Clearly (to me), Q v 'T: if 't doubles, Qloss also doubles. Ans. (b) P5.37 The volume flow Q through an orifice plate is a function of pipe diameter D, pressure drop 'p across the orifice, fluid density U and viscosity P, and orifice diameter d. Using D, U, and 'p as repeating variables, express this relationship in dimensionless form. Solution: There are 6 variables and 3 primary dimensions (MLT), and we already know that j = 3, because the problem thoughtfully gave the repeating variables. Use the pi theorem to find the three pi’s: 31 D a U b 'p c Q ; Solve for a 32 D a U b 'p c d ; Solve for a 33 D a U b 'p c P ; Solve for a 2, b 1 / 2, c 1 / 2. Thus 1 b 0 c 0. Thus 1, b 1 / 2, c 1 / 2. Thus 31 31 31 Q U 1/ 2 D 2 'p1 / 2 d D P D U 1 / 2 'p1 / 2

Download
# P T