Question

4. In a cylindrical pipe of uniform diameter D, the pressure loss per unit length (pa/m) due to viscous effects can be computed using the Darcy-Weisbach equation: Use a first order error analysis to determine the error in the pressure loss for a fluid of with density ρ=876±3 kg/m3 flowing in a pipe of diameter D 0.45 ±0.001 m with a mean velocity of u 10t0.1 m/s. The Darcy-Weisbach factor is gives as fp 0.064 ves as JD

Answer 1

Pressure loss per unit length, Delta P_L = delta_0 delta u^2/2 D = g (p, u, D) (say) Now, Delta g = (Partialdifferential g/Partialdifferential p) Delta p + (Partialdifferential g/Partialdifferential u) Delta u + (Partialdifferential g/Partialdifferential D) Delta D Here, Delta g rightarrow error in g, we have to calculate. Delta p rightarrow error in p, = 3 kg/m^3 D u rightarrow error in u = 0.1 m/s Delta D rightarrow error in D = 0.001 m Partialdifferential g/Partialdifferential p = f_D u^2/2D, Partialdifferential g/Partialdifferential u = f_D p u/D, Partialdifferential g/Partialdifferential D = -f_D p u^2/2 D^2 Putting, u = 10 m/s, p = 876 kg/m^3, D = 0.45 m and f_D = 0.064, we have Partialdifferential g/Partialdifferential p = f_D 10^2/2 times 0.45 = 111.11 f_D Partialdifferential g/Partialdifferential u = f_D times 876 times 10/0.45 = 19466.67 f_D Partialdifferential g/Partialdifferential D = -f_D times 10^2 times 876/2 times (0.45)^2 = -216296.296 F_D So, Delta g

$\\ \text{So, the error in pressure loss for the fluid is } \Delta(\Delta P_L) = 132 \text{ Pa/m}\\ \text{ and the pressure loss is }\Delta P_L = 62293.33\pm 132 \text{ Pa/m}$