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Rus_ich [418]
3 years ago
15

An aluminum oxide component must not fail when a tensile stress of 12.5 MPa is applied. Determine the maximum allowable surface

crack length if the surface energy of aluminum oxide is 0.90 J/m2. The modulus of elasticity of this material is 393 GPa.
Engineering
1 answer:
aivan3 [116]3 years ago
4 0

Answer:

1.44 mm

Explanation:

Compute the maximum allowable surface crack length using

C=\frac {2E\gamma}{\pi \sigma_c^{2}} where E is the modulus  of elasticity, \gamma is surface energy and \sigma_c is tensile stress

Substituting the given values

C=\frac {2\times 393\times 10^{9}\times 0.9}{\pi\times (16\times 10^{6})^{2}= 0.001441103 m\approx 1.44mm

The maximum allowable surface crack is 1.44 mm

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The steam requirements of a manufacturing facility are being met by a boiler whose rated heat input is 5.5 x 106 Btu/h. The comb
Veronika [31]

Answer:

Explanation:uhhhhhh all?

7 0
3 years ago
What happens to the amperage draw of a condensing unit on a split AC system if the liquid line is restricted
cupoosta [38]

Answer:

The amperage draw of the condensing unit will be low.

Explanation:

A condensing unit is made up of a compressor and condenser, while an evaporating unit is made up of an evaporator coil.

A split AC system is a type of air conditioner system that has a condensing unit which is placed separately from the evaporative coil unit. Then the two units are connected to each other via a copper tube containing refrigerants.

The liquid line connects the condenser to the evaporator, and if this liquid line is restricted, the amp consumed by the condensing unit will be low.

6 0
3 years ago
Consider an 8-car caravan, where the propagation speed is 100 km/hour, each car takes 1 minute to pass a toll both. The caravan
melamori03 [73]

Answer:

A. 36 minutes

B. 120 minutes

C.

i. 144 minutes

ii. 984 minutes

D. Car 1 is 1.67km ahead of Cat 2 when Car 2 passed the toll B.

E. 98.33km

Explanation

A.

Given

dAb = 10km

dBc = 10km

Propagation Speed = 100km/hr

Delay time = 1 minute

Numbers of cars = 8

Number of tolls = 3

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 10km + 10km = 20km

So, Propagation delay = 20km/100km/hr

Propagation delay = 0.2 hour

                               

Translation delay = delay time* numbers of tolls * numbers of cars

Transitional delay = 1 * 3 * 8

Transitional delay = 24 minutes

Total End delay = 24 minutes + 0.2 hours

= 24 minutes + 0.2 * 60 minutes

= 24 minutes + 12 minutes

= 36 minutes

B.

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 10km + 10km = 20km

So, Propagation delay = 20km/100km/hr

Propagation delay = 0.2 hour

                               

Translation delay = delay time* numbers of tolls ------ Cars traveling separately

Transitional delay = 1 * 3

Transitional delay = 3 minutes

Total End delay for one car = 3 minutes + 0.2 hours

= 3 minutes + 0.2 * 60 minutes

= 3 minutes + 12 minutes

= 15 minutes

Total End delay for 8 cars = 8 * 15 = 120 minutes

C.

Given

dAb = 100km

dBc = 100km

Propagation Speed = 100km/hr

Delay time = 1 minute

Numbers of cars = 8

Number of tolls = 3

i. Cars travelling together

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 100km + 100km = 200km

So, Propagation delay = 200km/100km/hr

Propagation delay = 2 hours

                               

Translation delay = delay time* numbers of tolls * numbers of cars

Transitional delay = 1 * 3 * 8

Transitional delay = 24 minutes

Total End delay = 24 minutes + 2 hours

= 24 minutes + 2 * 60 minutes

= 24 minutes + 120 minutes

= 144 minutes

ii. Cars travelling separately

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 100km + 100km = 200km

So, Propagation delay = 200km/100km/hr

Propagation delay = 2 hours

                               

Translation delay = delay time* numbers of tolls ------ Cars traveling separately

Transitional delay = 1 * 3

Transitional delay = 3 minutes

Total End delay for one car = 3 minutes + 2 hours

= 3 minutes + 2 * 60 minutes

= 3 minutes + 120 minutes

= 123 minutes

Total End delay for 8 cars = 8 * 123 = 984 minutes

D.

Distance = 100km

Time = 1 min/car

Car 1 is 1 minute ahead of car 2 --- at toll A and B

If car 1 leaves toll B after 10 minutes then cat 2 leaves after 11 minutes

Time delay = 11 - 10 = 1 minute

Distance = time * speed

= 1 minute * 100km/hr

= 1 hr/60 * 100 km/hr

= 100/60

= 1.67km

E.

Given

Distance = 100km

Distance behind = 1.67

Maximum value of dBc = 100km - 1.67km = 98.33km

The maximum distance that can be reached is 98.33km

7 0
3 years ago
A 10-mm-diameter Brinell hardness indenter produced an indentation 1.55 mm in diameter in a steel alloy when a load of 500 kg wa
BigorU [14]

Answer:

HB = 3.22

Explanation:

The formula to calculate the Brinell Hardness is given as follows:

HB = \frac{2P}{\pi D\sqrt{D^{2}- d^{2} } }

where,

HB = Brinell Hardness = ?

P = Applied Load in kg = 500 kg

D = Diameter of Indenter in mm = 10 mm

d = Diameter of the indentation in mm = 1.55 mm

Therefore, using these values, we get:

HB = \frac{(2)(500)}{\pi (10)\sqrt{10^{2}- 1.55^{2} } }

<u>HB = 3.22 </u>

4 0
3 years ago
An equal-tangent vertical curve is to be constructed between grades of -2% (initial) and 1% (final). The PVI is at station 110 0
Mila [183]

Answer:

The curve length (<em>L</em>) will be = 1218 ft

The elevations and stations for PVC and PVI

a. station of PVC = 103 + 91.00

b. station of PVI = 116 + 09.00

c. elevation of PVC = 432.18ft

d. elevation of PVI = 426.09ft

Explanation:

First calculate for the length (<em>L</em>)

To calculate the length, use the formula of "elevation at any point".

where, elevation at any point = 424.5.

and ∴ PVC Elevation = (420 + 0.01L)

Then, calculate for Station of PVC and PVI and elevation of PVC and PVI

4 0
3 years ago
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