Question

please explain how to do it!

3) The possibility of a typo in Excel increases as the formula to be entered gets more complex. Sometimes it is advisable to break the problem into steps and obtain answers in which you have confidence. Then you can attempt to code the problem in a single formula, Gas-sparging is a method used in chemical engineering to remove impurities in liquids, in which the cleaning gas is bubbled through the liquid in a tank. Often the tank of liquid is stirred (think a giant eggbeater) to increase the rate of impurity removal. An engineer working with a gas-sparging system needs to compute power Pc required for tank mixing using the equation: 438 / 01152, 1900) logie --1926) Wd where P, is the power required if there is no sparging, d is the impeller (mixer) diameter, Dy is the tank diameter, N is the rotational speed of the impeller, g is the gravitational acceleration constant, v is the kinematic viscosity of the liquid, and is the volumetric flow rate of the gas (Katoh et al., 2015). Make a worksheet similar to the one below, in which you compute the four terms on the right-hand-side of the equation separately in E2:E5. They are combined with the-192 constant in E6, and Pais calculated in E7. You should check each term with a hand calculator (or phone). Then in B9 enter a single formula to compute P. Cells 89 and E7 should match. 0.004750 1 Gas-sparge system 2 POW 1d (m) OT) 5 N (1/5) 6 m /s) (d/OT (d'NIV ( / Nd .38 .115 196/d/or 0 0.436494 .031841 ROS 8.9E-07 9.81 0.00416 Computed PG (W) 8 m3/s) 9 Computed PG (W)

Answer 1

Solution:

Formula given is,

Log₁₀ (P_g/P_o) = -192 (d/D_r)^4.38 * (d²N/v)^0.335 * (dN²/g)^{^}(1.96*d/D_r) * (Q/Nd³)

P_o (W) ₌ Power required if there is no sparing = 794

d (m) = impeller diameter = 0.36

D_r (m) = Tank diameter = 1.22

N (1/s) = Rotational speed of the impeller = 2.8

g (m²/s) = The gravitational acceleration constant = 9.81

v (m/s²) = Kinetic viscosity of liquid = 8.93 E-07

Q (m³/s) = The volumetric flow rate of gas = 0.00416

Let solve the right hand side terms one by one

d/D_r = 0.36/1.22 = 0.2951

(d/D_r)^4.38 = 0.2951 ^4.38 = 0.00476

d²N/v = 0.36² * 2.8 / (8.93 * E^-7)

                       = 0.36288 / (8.93 * 2.72^-7) Note : the value of e is 2.72 or else you can use

                                                                                          Scientific calculator)

                        = 0.36288 / (8.93 * 0.00091)

                        = 0.36288 / 0.00814

                        = 44.5799

(d²N/v)^0.335 = 44.5799 ^ 0.335

                        = 3.56

dN²/g            = 0.36 * 2.8² / 9.81

                        = 0.2877

1.96*d/D_r    = 1.96 * 0.36/1.22

                        = 0.57836

(dN²/g)^{^}(1.96*d/D_r) = 0.2877^0.57836

                                        = 0.48648

Q/Nd³ = 0.00416/ (2.8*0.36^3)

             = 0.00416/ (2.8*0.046656)

             = 0.00416/ 0.1306368

             = 0.031844

-192 (d/D_r)^4.38 * (d²N/v)^0.335 * (dN²/g)^{^}(1.96*d/D_r) * (Q/Nd³)

                                        = -192 *0.00476 * 3.56 * 0.48648 * 0.031844

                                        = -0.0504

LHS = RHS

Therefore,

Log₁₀ (P_g/P_o) = -0.0504

Log₁₀ (P_g/794) = -0.0504

According to logarithmic rule,

This equation can be rewritten as,

P_g/794 = 10^(-0.0504)

P_g/794 = 0.890430

P_g = 0.890430 * 794

P_g = 707.001 W