Question

6. The cardiac systems of mammals have some nice scaling properties. The external work of the heart, Ew is dependent on the blood pressure and the stroke volume and has dimensions of Joules. The metabolic turnover rate, EMTR has dimensions of joules/sec/kg. The mass of an animal is M (kg) and the heartrate is fn (usually in 1/min). Use the Buckingham Pi theorem to derive a realtionship between these 4 quantities. Recall that work (measured in joules, J ) has the dimensions of force times distance. Once you have done this, apply your dimensionless ratios to the data for the three mammals: Man (M =70 Kg, Ew = 1 joule, EMTR = 1.17J/sec/kg, fn =70/min), Dog(M =20 Kg. Ew = 0.44 joule, Emtr = 1.6J/sec/kg, fn =90/min), and Guinea pig, (M =0.7 Kg. Ew = 0.0097 joule, EMTR - 3.7J/sec/kg, fn =240/min). Caution: you will have to convert minutes to seconds!

Answer 1

Variables Dimensions EMTR M2²T -2 L²T-3 In terms of dimensions, M= CM] [M] =M) tu= [T-1] EMTR = [22] = 1 [L]= [2²7-3 EMIR [T3] JEMTR [132]] or [2] = JEMTR tn 3 Dimension of Ew = M2²7-2 o Eu (M-1 2-2 T +2 ] must be a dimensionlees ,مل constanta → Ew [M-1 L-2 p +2] =k Ew (my? (Jense huis 9 (4:02 EK [ Ewote ak м. Емтр.

x = Ewo fu м. Епте . Ew = 1 ; M = 70 kg : Emte = 1.17 I I see I kg; fu= 70 /min = 7/2 1 sec. kman = (15) (7 secol) (70 kg) (1.17 I see-kg-1) = 0.014 Dog > Ew = 0.44J; M = 20kg; Emte = 1.651 sec / kg; Eu = 90 /min = 3 , sec Koog = (0.44 J) ( 3 sec ) 0.02 (20kg) (1. 6J/ sec /kg) Guinea Pig Ew = 0.0097 J; M = 0.7 kg; Emir = 3.7 JI see 4 Isec be = 240lmin = k G.p. - = 0.015 (0.0097 J) (4, sec) (0.7 kg) (3.7 J sec" kg-1)