Chemical Engineering - Fluid Mechanics - Discussion
Discussion Forum : Fluid Mechanics - Section 1 (Q.No. 47)
47.
A Bingham fluid of viscosity μ = 10 Pa.s and yield stress, τ0 = 10 KPa, is shared between flat parallel plates separated by a distance of 10-3 m. The top plate is moving with a velocity of 1 m/s. The shear stress on the plate is
Discussion:
8 comments Page 1 of 1.
Rahul said:
1 decade ago
Since yield stress is 10kPa. It will be addition to the stress we calculated from shear stress formula.
Total shear stress = from formula + yield stress.
= 10 + 10.
= 20kPa.
Total shear stress = from formula + yield stress.
= 10 + 10.
= 20kPa.
(2)
Pongli said:
9 years ago
τ = σ0+ μ* dv/dy.
dv/dy = (1-0)/10^(-3)= 10^(3) s^(-1) since the lower plate is constant.
Ï„ = 10kPa + 10kPa = 20kPa.
dv/dy = (1-0)/10^(-3)= 10^(3) s^(-1) since the lower plate is constant.
Ï„ = 10kPa + 10kPa = 20kPa.
(5)
RAJ MUSALE said:
1 decade ago
Calculated stress:
t = dynamic viscosity*du/dy.
So t = 10000 pa.
Total stress = 20 kPa.
t = dynamic viscosity*du/dy.
So t = 10000 pa.
Total stress = 20 kPa.
(1)
Ramsajan said:
4 years ago
Dynamic viscosity 10 pa.s
10*10^(-3)=10^{-2) kpa = 10kpa. + 10k.pa = 20kpa.
10*10^(-3)=10^{-2) kpa = 10kpa. + 10k.pa = 20kpa.
(4)
Ramesh Vishwakarma said:
9 years ago
T = t¢+ (du/dy) * n.
So, T = 10 + 10 = 20 kPa.
So, T = 10 + 10 = 20 kPa.
Devendra said:
2 years ago
(10*(10^3)) + (10*(1/10)) = 20kpa.
(1)
Arif said:
1 decade ago
T=T¢+(viscosity)*du/dy.
(1)
Rana said:
5 years ago
Thanks all for explaining.
(2)
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