Answer:
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Jello there, see explanstion for step by step solving.
A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.
Explanation:
A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.
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Answer:
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Explanation:
Answer:
a) P ≥ 22.164 Kips
b) Q = 5.4 Kips
Explanation:
GIven
W = 18 Kips
μ₁ = 0.30
μ₂ = 0.60
a) P = ?
We get F₁ and F₂ as follows:
F₁ = μ₁*W = 0.30*18 Kips = 5.4 Kips
F₂ = μ₂*Nef = 0.6*Nef
Then, we apply
∑Fy = 0 (+↑)
Nef*Cos 12º - F₂*Sin 12º = W
⇒ Nef*Cos 12º - (0.6*Nef)*Sin 12º = 18
⇒ Nef = 21.09 Kips
Wedge moves if
P ≥ F₁ + F₂*Cos 12º + Nef*Sin 12º
⇒ P ≥ 5.4 Kips + 0.6*21.09 Kips*Cos 12º + 21.09 Kips*Sin 12º
⇒ P ≥ 22.164 Kips
b) For the static equilibrium of base plate
Q = F₁ = 5.4 Kips
We can see the pic shown in order to understand the question.
Answer:
Change in entropy S = 0.061
Second law of thermodynamics is satisfied since there is an increase in entropy
Explanation:
Heat Q = 300 kW
T2 = 24°C = 297 K
T1 = 7°C = 280 K
Change in entropy =
S = Q(1/T1 - 1/T2)
= 300(1/280 - 1/297) = 0.061
There is a positive increase in entropy so the second law is satisfied.
