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Rufina [12.5K]
3 years ago
9

What forces are not present in space

Engineering
1 answer:
ss7ja [257]3 years ago
7 0

Answer:

C.) Weight and distance I believe

Explanation:

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(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5
bezimeni [28]

Answer:

a.) 1.453MW/m2,  b.)  2,477,933.33 BTU/hr  c.) 22,733.33 BTU/hr  d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

                        50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, Q = \frac{109*0.5*100}{0.0075} = 726,666.667W

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, Q=\frac{1*0.5*100}{0.0075} = 6,666.67W

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, Q=\frac{109*0.5*100}{0.015} =363,333.33W

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0
2 years ago
When engineers design solutions to problems (bridge a river, support a building, etc.) there are always certain conditions which
zheka24 [161]

Explanation:

what is your main question

6 0
2 years ago
A thin rim with a mean diameter of 1.2 m cross-section of 15 mm x 200 mm is subjected to an internal pressure of 10 MPa and rota
Soloha48 [4]

Answer:

The net centrifugal force over the rim is 30000N, the radial stress is 397887 Pa and the total change in diameter is 4.98 mm.

Explanation:

Lets first calculate the force in the rim due to the centrifugal force. For doing this we may assume that the centrifugal force is constant along with thick because of the thin thick.

Fc = m.ω^2/R

Where m is the mass, w the angular speed and R the mean radius.  The mass is computing by the rim density and its volume:

m=p.V

m=p*(A*R)

Where A is the cross-sectional area in meters:

m=((0.015m*0.200m)*0.6m)*(7800 kg/m^3)=28.08 kg

The angular speed in rad/s is:

ω = 800r/m . 1m/60s = 133.33 r/s

Thus the centrifugal force is:

Fc = (28.08 kg)*(133.33 rad/s)^2*(0.6m) = 299505N = 30000N

Note that the calculating value is the net contribution to the whole rim but the centrifugal force is distributed along the rim's external area:

fc = Fc / (2π .R .b)

Where b is rim's with equal to 200mm :

fc = 300000 N / (2π*0.6m*0.2m) = 397887 N/m^2

The centrifugal force can be taken as internal pressure:

Pfc = 397887 N/m^2 = 3978787 Pa

As both pressures act expanding the rim it can be summed:

Pt=Pi+Pfc

Pt = 10MPa+397887Pa= 10000000Pa+397887Pa= 10397887Pa

Then for a thinner thick the stress is calculated by:

Pt*d =2σr*t

Take into account that the stress σr is over the radial direction. Then solving for o and by replacing the total pressure:

σr = Pt.d/(2*t)

σr = 10397887 Pa / (2*0.015m*0.2m) = 415915480Pa = 415MPa

We know that the radial specific deformation ε is:

σr = E / εr

εr = σr / E

For a young modulus of 200GPa:

εr = 415MPa / 200GPa

εr = 415MPa / 200000MPa=0.002075

By definition the specific deformation is written in terms of the total change in the radius:

εr = Δr / R

Δr = R / εr =0.002075 * 1.2 m = 0.00249m

As we need the change in diameter:

Δd = 2Δr =0.00498m= 4.98mm [/tex]

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3 years ago
A heat pump with refrigerant-134a as the working fluid is used to keep aspace at 25°C by absorbing heat from geothermal water th
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Answer:12

Explanation:

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2 years ago
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Two grenades, A and B, are thrown horizontally with different speeds from the top of a cliff 70 m high. The speed of A is 2.50 m
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Explanation:

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