Answer to Question #154974 in Chemical Engineering for Candy

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) Shear rate is the velocity of the moving plate divided by the distance between the plates. ... According to Newton's Law, shear stress is viscosity times shear rate. Therefore, the viscosity (eta) is shear stress divided by shear rate. Only Newtonian liquids can be described by this simple relation.

b) 20000 /0.77 = 25 974,026 L.

25 974,026/3600= 0.3 L/s.

c) 1.8 x 150/0.75= 360 cm/s.

d) 0.998 x 0.337 = 0,336326.

e) 20 .