Answer to Question #156480 in Chemical Engineering for Mohammedazeem Rokad

Question

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).

Answer 1

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 D_{2 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

L_s = 0.1m and K_s = 0.7Wm^-1k^-1

Heat flux = (T₂ - T₁)/ (1/h₁ + 1/h₂ + L_s/K_s)

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

answer = 16.074Wm^-2