Answer to Question #156480 in Chemical Engineering for Mohammedazeem Rokad

Question #156480

a)  Write down the equation for the relationship between the shear stress and the shear rate for Newtonian fluid defining the variables and parameters used.


b) Oil with a density of 770 kg m−3 flows through a pipe with a diameter of 30 cm at a rate of 20,000 tonnes per day. Calculate the velocity of the oil in the pipe.

 

c)  A liquid flows in a hosepipe of internal diameter 1.8 cm with a velocity of 150 cm s-1. Calculate the velocity of the water in the nozzle at the end of the pipe which has a diameter of 0.75 cm.

 

d) Water at 20°C and 1 atm has a density = 998.0 kg/m3. Calculate the density at 35°C to 3 significant figures if the coefficient of volume

∆𝜌𝜌⁄𝜌𝜌

expansion of the water is given by 𝛽𝛽 = − = 0.337x10-3 K-1.

∆𝑇𝑇

 

e) A brick wall with 0.1 m thickness (k = 0.7 W m−1K−1) is exposed to a cold wind at 270 K through a convection heat transfer coefficient of

                          40 W m−2K−1. On the other side is calm air at 330K, with a convection

heat transfer coefficient of 10 W m−2K−1. Calculate the rate of heat transfer per unit area (heat flux).


1
Expert's answer
2021-01-20T04:56:52-0500

a) The equat1ion that shows relationship between shear stress and shear rate is; 

Answer = "\\tau" = "\\eta" x "\\gamma" 11


where; "\\tau" = shrear stress

"\\gamma" = share rate

and "\\eta" = viscosity


b) D =770Kg/m3

d = 30cm

From the flow rate equation; Q = Av where A = crossection Area and v = velocity

A = "\\pi" r2 = 3.142 x 0.152 = 0.7855m2

And v = Q/ A

Hence = 20000/0.7855

Answer= 25461.49



c) Velocity ( V1) = 150cm/s

Diameter D1 = 1.8cm

Diameter D2 1= 0.75

1

continuity equation states that; A1 V1 = A2V2

A1 = 3.142 X 0.92 = 2.545

A2 = 3.142 X 0.3752 = 0.4418

Therefore V2 = A1V1/A2

V2 = (2.545 x 150)/ 0.4418

Answer= 831.82cm/s


d) Density of walter at 20degress and 1atm is 998kg/ m3

what about at 35dgress

change in density = (0.337 x 10-3/K x 998Kg/m2) x (350 - 200) = 5.044


Answer = Density = 1003Kg/m3


e) T1 = 270 , h1 = 40Wm-1k-1 and T2 = 330, h2 = 10Wm-1k-1

Ls = 0.1m and Ks = 0.7Wm-1k-1


Heat flux = (T2 - T1)/ (1/h1 + 1/h2 + Ls/Ks)

= (330 - 270)/ (0.025 + 0.1 + o.1429)


answer = 16.074Wm-2



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS