Question

Question 3 (10 marks) What is the permeability of a vertical sand pack of length 1 m and radius 10 cm, displayed in the following Figure, where the top is exposed to atmospheric pressure and fluid density and viscosity are 1000 kg/m3 and 10 cP respectivelyThe flow rate of fluid is 1 cc/s. Show your calculation - Liquid P= 1 atm

Answer 1

Liquid and Land D length are the diameta of vertical qand t (in meter) Po are inet and SPi and outlet AAD Pressures in pascal at 3 pon po Zso h=0 and not respective. > sil, o are density (kg/m3), m (given) viscosity (kg|m-p) '80 and flow rate (m3 18) latm = 101325 w a Pa serpeuvely 80 Pia 101325 pal - 0 h=L at 23 point Pi= 1 atm (give to gravity = 9.81 ml8ec².. L= 1m given) I= 1000 kg/m3 (given) g=accleration due Po= SOL Po - 1000*9.81 * Po=9810 pa] - ② D = 28 ; where 8= radius of tube 8= 10cm (m) po ) 2* 10 = 20 cm Algo lcm = 10-2 m. 80 D=20*10-2 m D = 0.2 m) A= crops - pectional Area of tube (m²) A = ĘDZ A = I (0.22 A = 0.0314 m2 - M= viscosity of fluid (kg/m-sec). M = 10cp (given). Also lep= 10-3 kg/m pec- 80 u= 10cp = 10x10 3 Kg 1m-sec. Tu= 0.01 kg/m-8ec-6.

S = flowrate of fluid (m318 ed. S = Icclsec. 1cc = 10-6 m3 so s= 10-6 malrec.] - Defining; k=permebiality of pand pack (m2). using there the redation; 8 = k A (Pi-Po) - A UL Pugging the value of Pi, po, A, ll, s from equation 0 ② 0 0 0 in equation ®; we get - 10-6 = K*0.0314 (101325 - 9810) 0.01 * 1. 3) 10-8 0.0314 gisis) =k. = 3.48 * 10-12 m2 = ex. TK = 3.48 * 10-12 m2t Answer dari alm2 = to 1013 darce 9.87 80 3.48*10-12 m²= 3.48 * 10-12 * 1013 9.87 I k = 3.525 darcy. K - darcy Ansur.