Answer:
Explanation:
Mean temperature is given by
Tmean = (Ti + T∞)/2
Tmean = 107.5⁰C
Tmean = 107.5 + 273 = 380.5K
Properties of air at mean temperature
v = 24.2689 × 10⁻⁶m²/s
α = 35.024 × 10⁻⁶m²/s
= 221.6 × 10⁻⁷N.s/m²
= 0.0323 W/m.K
Cp = 1012 J/kg.K
Pr = v/α = 24.2689 × 10⁻⁶/35.024 × 10⁻⁶
= 0.693
Reynold's number, Re
Pv = 4m/πD²
where Re = (Pv * D)/
Substituting for Pv
Re = 4m/(πD)
= (4 x 0.003)/( π × 6 ×10⁻³ × 221.6 × 10⁻⁷)
= 28728.3
Since Re > 2000, the flow is turbulent
For turbulent flows, Use
Dittus - Doeltr correlation with n = 0.03
Nu = 0.023Re⁰⁸Pr⁰³ = (h₁D)/k
(h₁ × 0.006)/0.0323 = 0.023(28728.3)⁰⁸(0.693)⁰³
(h₁ × 0.006)/0.0323 = 75.962
h₁ = (75.962 × 0.0323)/0.006
h₁ = 408.93 W/m².K
Answer:
<em>No, the velocity profile does not change in the flow direction.</em>
Explanation:
In a fluid flow in a circular pipe, the boundary layer thickness increases in the direction of flow, until it reaches the center of the pipe, and fill the whole pipe. If the density, and other properties of the fluid does not change either by heating or cooling of the pipe, <em>then the velocity profile downstream becomes fully developed, and constant, and does not change in the direction of flow.</em>
Answer:
The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero
Explanation:
The expression for the maximum shear stress is given:
Where
σx = stress in vertical plane = 20 ksi
σy = stress in horizontal plane = -30 ksi
τM = 32 ksi
Replacing:
Solving for τxy:
τxy = ±19.98 ksi
The principal stress is:
Where
σp1 = 20 ksi
σp2 = -30 ksi
(equation 1)
equation 2
Solving both equations:
σp1 = 27 ksi
σp2 = -37 ksi
The shear stress on the vertical plane is zero
Answer:
The young lady was his daughter.The shoemaker was frightened when he saw that she wants to sit near him and took his knife to frighten her and leave him alone to do his work
Explanation:
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