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3 years ago

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The velocity profile for a thin film of a Newtonian fluid that is confined between the plate and a fixed surface is defined by u

= (5y-0.5y2) mm/s where y is in mm. Determine the force P that must be applied to the plate to cause this motion. The plate has a surface area of 4000mm2 in contact with the fluid. Take µ= 0.55 N.s/m2

1 answer:

zimovet [89]3 years ago

8 0

Answer:

F = 0.0022N

Explanation:

Given:

Surface area (A) = 4,000mm² = 0.004m²

Viscosity = µ = 0.55 N.s/m²

u = (5y-0.5y²) mm/s

Assume y = 4

Computation:

F/A = µ(du/dy)

F = µA(du/dy)

F = µA[(d/dy)(5y-0.5y²)]

F = (0.55)(0.004)[(5-1(4))]

F = 0.0022N

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A strain gauge with a 5 mm gauge length gives a displacement reading of 1.25 um. Calculate the stress value given by this displa

KengaRu [80]

Answer:

stress = 50MPa

Explanation:

given data:

Length of strain guage is 5mm

displacement $\delta = 1.25 \mu m =\frac{1.25}{1000} = 0.00125 mm$

stress due to displacement in structural steel can be determined by using following relation

$E =\frac{stress}{strain}$

$stress = E \times strain$

where E is young's modulus of elasticity

E for steel is 200 GPa

$stress = 200\times 10^3 *\frac{1.25*10^{-3}}{5}$

stress = 50MPa

7 0

3 years ago

A manufacturer makes two types of drinking straws: one with a square cross-sectional shape, and the other type the typical round

Harlamova29_29 [7]

Answer:

$\frac{Q_{square}}{Q_{circle}} = 0.785$

Explanation:

given data

types of drinking straws

square cross-sectional shape
round shape

solution

we know that both perimeter of the cross section are equal

so we can say that

perimeter of square = perimeter of circle

4 × S = π × D

here S is length and D is diameter

S = $\frac{\pi D}{4}$ ....................1

and

ratio of flow rate through the square and circle is here

$\frac{Q_{square}}{Q_{circle}} = \frac{AV^2}{AV^2}$

$\frac{Q_{square}}{Q_{circle}} = \frac{S^2}{\frac{\pi D^2}{4}}$

$\frac{Q_{square}}{Q_{circle}} = \frac{(\frac{\pi D}{4})^2}{\frac{\pi D^2}{4}}$

$\frac{Q_{square}}{Q_{circle}} = \frac{\pi }{4}$

$\frac{Q_{square}}{Q_{circle}} = 0.785$

4 0

4 years ago

The diffusion coefficients for species A in metal B are given at two temperatures:

Kruka [31]

Answer:

a) 149 kJ/mol, b) 6.11*10^-11 m^2/s ,c) 2.76*10^-16 m^2/s

Explanation:

Diffusion is governed by Arrhenius equation

$D = D_0e^{\frac{-Q_d}{RT} }$

I will be using R in the equation instead of k_b as the problem asks for molar activation energy

I will be using

$R = 8.314\ J/mol*K$

and

°C + 273 = K

here, adjust your precision as neccessary

Since we got 2 difusion coefficients at 2 temperatures alredy, we can simply turn these into 2 linear equations to solve for a) and b) simply by taking logarithm

So:

$ln(6.69*10^{-17})=ln(D_0) -\frac{Q_d}{R*(1030+273)}$

and

$ln(6.56*10^{-16}) = ln(D_0) -\frac{Q_d}{R*(1290+273)}$

You might notice that these equations have the form of

$d=y-ax$

You can solve this equation system easily using calculator, and you will eventually get

$D_0 =6.11*10^{-11}\ m^2/s\\ Q_d=1.49 *10^3\ J/mol$

After you got those 2 parameters, the rest is easy, you can just plug them all including the given temperature of 1180°C into the Arrhenius equation

$6.11*10^{-11}e^{\frac{149\ 000}{8.143*(1180+273)}$

And you should get D = 2.76*10^-16 m^/s as an answer for c)

5 0

3 years ago

Determine the reactor volume (assume a CSTR activated sludge aerobic reactor at steady state) required to treat 5 MGD of domesti

12345 [234]

Answer:

1.0MG

Explanation:

to solve this problem we use this formula

S₀-S/t = ksx --- (1)

the values have been given as

concentration = S₀ = 250mg

effluent concentration = S= 10mg

value of K = 0.04L/day

x = 3000 mg

when we put these values into this equation,

250-10/t = 0.04x10x3000

240/t = 1200

we cross multiply from this stage

240 = 1200t

t = 240/1200

t = 0.2

remember the question says that 5MGD is required to be treated

so the volume would be

v = 0.2x5

= 1.0 MG

4 0

2 years ago

2. A F-22 Raptor has just climbed through an altitude of 9,874 m at 1,567 kph when a disk

BabaBlast [244]

The pressure difference across the sensor housing will be "95 kPa".

According to the question, the values are:

Altitude,

9874

Speed,

1567 kph

Pressure,

122 kPa

The temperature will be:

→ $T = 15.04-[0.00649(9874)]$

→ $= 15.04-64.082$

→ $= -49.042^{\circ} C$

now,

→ $P_o = 101.29[\frac{(-49.042+273.1)}{288.08} ]^{(5.256)}$

→ $= 27.074$

hence,

→ The pressure differential will be:

= $122-27$

= $95 \ kPa$

Thus the above solution is correct.

Learn more about pressure difference here:

brainly.com/question/15732832

3 0

3 years ago