Question

The density of a fluid is given by the empirical equation rho=70.5 exp(8.27 X 10^-7 P) where rho is density (lbm/ft^3) and P is pressure (lbf/in.^2) a) What are the units of 70.5 and 8.27 x 106-7? b) Calculate the density in g/cm3 for a pressure of 9.00 x10^6 N/m^2. e) Derive a formula for rho (g/cm^3) as a function of P(N/m^2).

Answer 1

(a) The unit of 70.5 is lbm/ft^3 and the unit of 8.27×10^-7 is in^2/lbf

(b) density = 0.1206g/cm^3

(c) rho = 0.1206exp(8.27×10^-7P)

Explanation:

(a) The unit of 70.5 is the same as the unit of rho which is lbm/ft^3. The unit of 8.27×10^-7 is the inverse of the unit of P (lbf/in^2) because exp is found of a constant. Therefore, the unit of 8.27×10^-7 is in^2/lbf

(b) P = 9×10^6N/m^2

rho = 70.5exp(8.27×10^-7× 9×10^6) = 70.5exp7.443 = 70.5×1.71 = 120.6kg/m^3

rho = 120.6kg/m^3 × 1000g/1kg × 1m^3/10^6cm^3 = 0.1206g/cm^3

(c) Formula for rho (g/cm^3) as a function of P (N/m^2) is

rho = 0.1206exp(8.27×10^-7P) (the unit of 0.1206 is g/cm^3)