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larisa86 [58]
3 years ago
8

The flexural strength or MOR of a ceramic is 310 MPa. A block of the ceramic, which is 20 mm wide, 15 mm high, and 300 mm long,

is supported between two rods 150 mm apart. Determine the force required to fracture the material, assuming no plastic deformation occurs.

Engineering
1 answer:
Alja [10]3 years ago
3 0

Answer:

F=6200\ \text{N}\\

Explanation:

In this problem you need to define the force that acts upon a beam in a 3 point bending problem. I put a picture of the problem taken from Wikipedia:

In this problem the flexural strength is defined with the following formula:

\sigma=\cfrac{3FL}{2bd^2}

where F is the force applied, L the length between the two rods, b the width of the ceramic block and d it's height.

The force is then defined as:

F=\cfrac{2\sigma bd^2}{3L}=6200\ \text{N}

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aliina [53]

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Explanation:

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. Carly's Catering provides meals for parties and special events. In Chapter 2, you wrote an application that prompts the user f
klio [65]

Answer:

Explanation:

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6 0
3 years ago
Give two methods on how powder is produced in powder metallurgy.
Katen [24]

Answer:

Explanation:

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Molten metal is forced through a small orifice and is shatter by a jet of compressed air,inert gas .

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7 0
3 years ago
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Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2/s is being discharged by an 8-mm-diameter, 42-m-long horiz
sladkih [1.3K]

Answer:

The flow rate of oil through the pipe is 1.513E-7 m³/s.

Explanation:

Given

Density, ρ = 850 kg/m³

Kinematic viscosity, v = 0.00062 m²/s

Diameter, d = 8-mm = 0.008m

Length of horizontal pipe, L = 42-m

Height, h = 4-m.

We'll solve the flow rate of oil through the pipe by using Hagen-Poiseuille equation.

This is given as

∆P = (128μLQ)/πD⁴

Where ∆P = Rate of change of pressure

μ = Dynamic Viscosity

Q = Flow rate of oil through the pipe.

First, we need to determine the dynamic viscosity and the rate of change in pressure

Dynamic Viscosity, μ = Density (ρ) * Kinematic viscosity (v)

μ = 850 kg/m³ * 0.00062 m²/s

μ = 0.527kg/ms

Then, we calculate the rate of change of pressure.

Assuming that the velocity through the pipe is so small;

∆P = Pressure at the bottom of the tank

∆P = Density (ρ) * Acceleration of gravity (g) * Height (h)

Taking g = 9.8m/s²

∆P = 850kg/m³ x 9.8m/s² x 4m

∆P = 33320N/m²

Recall that Hagen-Poiseuille equation.

∆P = (128μLQ)/πD⁴ --- Make Q the subject of formula

Q = (πD⁴P)/(128μL)

By substituton;

Q = (π * 0.008⁴ * 33320)/(128 * 0.527 * 42)

Q = 0.00000015133693643099

Q = 1.513E-7 m³/s.

Hence, the flow rate of oil through the pipe is 1.513E-7 m³/s.

8 0
2 years ago
A wastewater treatment plant discharges 1.0 m3/s of effluent having an ultimate BOD of 40.0 mg/ L into a stream flowingat 10.0 m
kondaur [170]

Answer:

a) 6.4  mg/l

b) 5.6 mg/l

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Given data:

effluent Discharge Q_w = 1.0 m^3.s

Ultimate BOD L_w = 40 mg/l

Discharge of stream Q_r = 10 m^3.s

Stream ultimate BOD L_r = 3  mg/l

a) Ultimate BOD of mixture= \frac{Q_w l_w + Q_r L_r}{Q_w + Q_r}

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b) utlimate BOD at 10,000 m downstream

t =\frac{distance}{speed} = \frac{10,000}{\frac{Q_r +Q+w}{55}} \times \frac{hr}{3600} \times  \frac{day}{24 hr}

putting Q_r + Q_w = 1+ 10 = 11 m^3/s

t = 0.578  days

we know

L_t = L_o e^{-kt}

L_t = 6.4 \times e^{-0.22 \times 0.578}

L_t = 5.6 mg/l

7 0
2 years ago
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