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borishaifa [10]
2 years ago
7

When engineers design solutions to problems (bridge a river, support a building, etc.) there are always certain conditions which

they must work under. For example, there may be a limit to funds, space, labor, or technology; or there may be environmental restrictions to observe. These conditions are MOST GENERALLY referred to as design
Engineering
1 answer:
zheka24 [161]2 years ago
6 0

Explanation:

what is your main question

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Conclude from the scenario below which type of documentation Holly should use, and explain why this would be the best choice. Ho
NARA [144]

Answer:

Okay

Explanation:

7 0
3 years ago
An object of irregular shape has a characteristic length of � = 1.00 [�] and is maintained at a uniform surface temperature of �
Pani-rosa [81]

The correct question;

An object of irregular shape has a characteristic length of L = 1 m and is maintained at a uniform surface temperature of Ts = 400 K. When placed in atmospheric air at a temperature of Tinfinity = 300 K and moving with a velocity of V = 100 m/s, the average heat flux from the surface to the air is 20,000 W/m² If a second object of the same shape, but with a characteristic length of L = 5 m, is maintained at a surface temperature of Ts = 400 K and is placed in atmospheric air at Too = 300 K, what will the value of the average convection coefficient be if the air velocity is V = 20 m/s?

Answer:

h'_2 = 40 W/K.m²

Explanation:

We are given;

L1 = 1m

L2 = 5m

T_s = 400 K

T_(∞) = 300 K

V = 100 m/s

q = 20,000 W/m²

Both objects have the same shape and density and thus their reynolds number will be the same.

So,

Re_L1 = Re_L2

Thus, V1•L1/v1 = V2•L2/v2

Hence,

(h'_1•L1)/k1 = (h'_2•L2)/k2

Where h'_1 and h'_2 are convection coefficients

Since k1 = k2, thus, we now have;

h'_2 = (h'_1(L1/L2)) = [q/(T_s - T_(∞))]• (L1/L2)

Thus,

h'_2 = [20,000/(400 - 300)]•(1/5)

h'_2 = 40 W/K.m²

5 0
3 years ago
How many parts (screws and bolts included) does the average car have?
Leokris [45]
A single car has about 30,000 parts, counting every part down to the smallest screws
5 0
3 years ago
Read 2 more answers
In a major human artery with an internal diameter of 5mm, the flow of blood, averaged over the cardiac cycle is 5cm3·s−1. The ar
antiseptic1488 [7]

9514 1404 393

Answer:

  see attached

Explanation:

Assuming flow is uniform across the cross section of the artery, the mass flow rate is the product of the volumetric flow rate and the density.

  (5 cm³/s)(1.06 g/cm³) = 5.3 g/s

If we assume the blood splits evenly at the bifurcation, then the downstream mass flow rate in each artery is half that:

  (5.3 g/s)/2 = 2.65 g/s

__

The average velocity will be the ratio of volumetric flow rate to area. Upstream, that is ...

  (5 cm³/s)/(π(0.25 cm)²) ≈ 25.5 cm/s

Downstream, we have half the volumetric flow and a smaller area.

  (2.5 cm³/s)/(π(0.15 cm)²) ≈ 35.4 cm/s

7 0
3 years ago
The pump of a water distribution system is powered by a 6-kW electric motor whose efficiency is 95 percent. The water flow rate
Sonja [21]

Answer:

a) Mechanical efficiency (\varepsilon)=63.15%  b) Temperature rise= 0.028ºC

Explanation:

For the item a) you have to define the mechanical power introduced (Wmec) to the system and the power transferred to the water (Pw).

The power input (electric motor) is equal to the motor power multiplied by the efficiency. Thus, Wmec=0.95*6kW=5.7 kW.

Then, the power transferred (Pw) to the fluid is equal to the flow rate (Q) multiplied by the pressure jump \Delta P. So P_W = Q*\Delta P=0.018m^3/s * 200x10^3 Pa=3600W.

The efficiency is defined as the ratio between the output energy and the input energy. Then, the mechanical efficiency is \varepsilon=3.6kW/5.7kW=0.6315=63.15\%

For the b) item you have to consider that the inefficiency goes to the fluid as heat. So it is necessary to use the equation of the heat capacity but in a "flux" way. Calling <em>H</em> to the heat transfered to the fluid, the specif heat of the water and \rho the density of the water:

[tex]H=(5.7-3.6) kW=\rho*Q*c*\Delta T=1000kg/m^3*0.018m^3/s*4186J/(kg \ºC)*\Delta T[/tex]

Finally, the temperature rise is:

\Delta T=2100/75348 \ºC=0.028 \ºC

7 0
3 years ago
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