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Volgvan
3 years ago
11

30 points have a good day

Engineering
1 answer:
lakkis [162]3 years ago
8 0

Answer:

thank you :)

Explanation:

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Three bars each made of different materials are connected together and placed between two walls when the temperature is 12 oC. D
slega [8]

Answer:

F = 9.11 x 10³ N = 9.11 KN

Explanation:

The areas, lengths, young's modulus, and coefficient of linear thermal expansion are given in the diagram. First we find the equivalent change in length due to temperature change:

ΔL = (ΔL)steel + (ΔL)brass + (ΔL)Copper

ΔL = (∝s)(Ls)(ΔT) + (∝b)(Lb)(ΔT) + (∝c)(Lc)(ΔT)

where,

ΔL = Equivalent Change in Length = ?

ΔT = Change in Temperature = 25°C - 12°C = 13°C

Ls = Length of Steel Segment = 300 mm = 0.3 m

Lb = Length of Brass Segment = 200 mm = 0.2 m

Lc = Length of Copper Segment = 100 mm = 0.1 m

Therefore,

ΔL = (12 x 10⁻⁶ °C⁻¹)(0.3 m)(13 °C) + (21 x 10⁻⁶ °C⁻¹)(0.2 m)(13 °C) + (17 x 10⁻⁶ °C⁻¹)(0.1 m)(13 °C)

ΔL = 46.8 x 10⁻⁶ m + 54.6 x 10⁻⁶ m + 22.1 x 10⁻⁶ m

ΔL = 123.5 x 10⁻⁶ m   ----------------------- equation (1)

Now, we calculate this deflection in terms of an applied force (F):

ΔL = (F)(Ls)/(Es)(As) + (F)(Lb)/(Eb)(Ab) + (F)(Lc)/(Ec)(Ac)

ΔL = (F)(0.3 m)/(200 x 10⁹ Pa)(200 x 10⁻⁶ m²) + (F)(0.2 m)/(100 x 10⁹ Pa)(450 x 10⁻⁶ m²) + (F)(0.1 m)/(120 x 10⁹ Pa)(515 x 10⁻⁶ m²)

ΔL = F(7.5 x 10⁻⁹ m/N + 4.44 x 10⁻⁹ m/N + 1.61 x 10⁻⁹ m/N)

ΔL = F(13.55 x 10⁻⁹ m/N)   --------------------- equation (1)

Comparing equation (1) and equation (2):

123.5 x 10⁻⁶ m = F(13.55 x 10⁻⁹ m/N)

F = (123.5 x 10⁻⁶ m)/(13.55 x 10⁻⁹ m/N)

<u>F = 9.11 x 10³ N = 9.11 KN</u>

6 0
2 years ago
An 80-percent-efficient pump with a power input of 20 hp is pumping water from a lake to a nearby pool at a rate of 1.5 ft3/s th
photoshop1234 [79]

Answer:

mechanical power used to overcome frictional effects in piping is 2.37 hp

Explanation:

given data

efficient pump = 80%

power input = 20 hp

rate = 1.5 ft³/s

free surface = 80 ft

solution

we use mechanical pumping power delivered to water is

{W_{u}}= \eta  {W_{pump}}  .............1

put here value

{W_{u}}  = (0.80)(20)

{W_{u}} = 16 hp

and

now we get change in the total mechanical energy of water is equal to the change in its potential energy

\Delta{E_{mech}} = {m} \Delta pe   ..............2

\Delta {E_{mech}} = {m} g \Delta z  

and that can be express as

\Delta {E_{mech}} = \rho Q g \Delta z     ..................3

so

\Delta {E_{mech}} = (62.4lbm/ft^3)(1.5ft^3/s)(32.2ft/s^2)(80ft)[\frac{1lbf}{32.2lbm\cdot ft/s^2}][\frac{1hp}{550lbf \cdot ft/s}]      ......4

solve it we get

\Delta {E_{mech}} = 13.614 hp

so here

due to frictional effects, mechanical power lost in piping

we get here

{W_{frict}} = {W_{u}}-\Delta {E_{mech}}  

put here value

{W_{frict}} = 16 -13.614

{W_{frict}} = 2.37  hp

so mechanical power used to overcome frictional effects in piping is 2.37 hp

4 0
2 years ago
I need help with this I dont know the word ​
DiKsa [7]

Would it be Unit?

...............

5 0
3 years ago
Read 2 more answers
Consider the following grooves, each of width W, that have been machined from a solid block of material. (a) For each case obtai
kogti [31]

Answer:A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

Explanation:

8 0
3 years ago
A copper-nickel alloy of composition 60 wt% Ni-40 wt% Cu is slowly heated from a temperature of 1250°C (2280 °F). (a) At what te
makkiz [27]

Answer:

a. The very first liquid process, when heated from 1250 degree Celsius, is expected to form at the temperature by which the vertical line crosses the phase boundary (a -(a + L)) which is about <em>1310 degree Celsius. </em>

b. The structure of that first liquid is identified by the intersection with ((a+ L)-L) phase boundary; <em>47wt %of Ni</em> is of a tie line formed across the (a+ L) phase area <em>at 1310 degrees.</em>

c. To find the alloy's full melting, it is determined that the intersection of the same vertical line at 60 wt percent Ni with (a -(a+L)) phase boundary is around <em>1350 degrees.</em>

c. The structure of the last remaining solid before full melting correlates to the intersection with the phase boundary (a -(a + L), of the tie line built at 1350 degrees across the (a + L) phase area, <em>being 72wt % of Ni.</em>

4 0
3 years ago
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