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3 years ago

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Two plates are separated by a 1/4 in space. The lower plate is stationary; the upper plate moves at 10 ft/s. Oil (viscosity of 2

.415 lb/ft-s), which fills the space between the plates, has the same velocity as the plates at the surface of contact. The variation in velocity of the oil is linear.What is the shear stress(in units of lbf/in2) in the oil?

1 answer:

Montano1993 [528]3 years ago

6 0

Answer:

τ = 0.25 lbf/in²

Explanation:

given that the oil viscosity, μ = 2.415 lb/ft-s

gap between plates = 1/4 inches = 1/4*12 = 1/48 ft

recall from newtons law of viscosity;

shear stress τ = μ du/dy =

τ = (2.415 lb/ft-s) (10 ft/s)/(1/48) ft

τ = 1159.2 lb/ft-s²

we know that, 1 slug = 32.174 lb

lb = 1/32.174 slug

∴ τ = 1159.2/32.174 slug/ft-s² = 36 slug/ft-s²

τ = 36 slug/ft-s²

multiply both the numerator and denominator by ft, this gives

τ = 36 slug-ft/ft²-s²

τ = 36 lbf/ft² where 1 slug-ft/s² = 1lbf

since 1 ft = 12 inch = 1 ft² = 12² in² = 144 in²

∴ τ = 36/144 lbf /in² = 0.25 lbf/in²

τ = 0.25 lbf/in²

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Explanation:

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4 years ago

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Calculate KI for a rectangular bar containing an edge crack loaded in three-point bending where P=35.0 kN, W=50.8 mm, B=25 mm, a

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Answer:

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Explanation:

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Width of bar W = 50.8 mm

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Ratio of crack length to width α = a/W = 0.2

solution

we get here KI for a rectangular bar that is express as

$K_{I} = \frac{6P}{BW}Y\sqrt{\pi a}$ ................................1

here Y is the geometrical function

so

Y = $\frac{1.12+\alpha (3.43\alpha -1.89)}{1-0.55\alpha}$

Y = $\frac{1.12+0.2(3.43\times 0.2-1.89)}{1-0.55\times 0.2}$

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so put here value in equation 1

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$K_{I} = 165354.33\times 10^{3}\times 0.9878\times 0.0319$

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3 years ago

The heat flux through a 1-mm thick layer of skin is 1.05 x 104 W/m2. The temperature at the inside surface is 37°C and the tempe

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Answer:

a) Thermal conductivity of skin: $k_{skin}=1.5W/mK$

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Explanation:

a)

$k=\frac{QL}{A(T_{2}-T_{1})}$

Where: $k$ is thermal conductivity of a material, $\frac{Q}{A}$ is heat flux through a material, $L$ is the thickness of the material, $T_{1}$ is the temperature on the first side and $T_{2}$ is the temperature on the second side

$k_{skin}=\frac{QL}{A(T_{2}-T_{1})}$

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$k_{skin}=1.5W/mK$

b)

$k_{insulation}=\frac{k_{skin}}{2}$

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$\frac{Q}{A}=\frac{k(T_{2}-T_{1})}{L}$

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$\frac{1.5*(37-T_{interface})}{0.001}=\frac{0.75*(T_{interface}-30)}{0.002}$

$55500-1500T_{interface}=375T_{interface}-11250$

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$T_{interface}=35.6\°C$

$\frac{Q}{A}=\frac{k_{skin}(T_{skin}-T_{interface})}{L_{skin}}$

$\frac{Q}{A}=\frac{1.5*(37-35.6)}{0.001}$

$\frac{Q}{A}=2100W/m^2$

3 0

3 years ago