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navik [9.2K]
3 years ago
6

An experiment to determine the convection coefficient associated with airflow over the surface of a thick stainless steel castin

g involves insertion of thermocouples in the casting at distances of 10 and 20 mm from the surface along a hypothetical line normal to the surface. The steel has a thermal conductivity of 15 W/m-K. If the thermocouples measure tempe
Engineering
1 answer:
Maksim231197 [3]3 years ago
4 0

Answer:

h = 375 KW/m^2K

Explanation:

Given:

Thermo-couple distances: L_1 = 10 mm , L_2 = 20 mm

steel thermal conductivity k = 15 W / mK

Thermo-couple temperature measurements: T_1 = 50 C , T_2 = 40 C

Air Temp T_∞ = 100 C

Assuming there are no other energy sources, energy balance equation is:

                                               E_in = E_out

                                        q"_cond = q"_conv

Since, its a case 1-D steady state conduction, the total heat transfer rate can be found from Fourier's Law for surfaces 1 and 2

q"_cond = k * (T_1 - T_2) / (L_2 - L_1) = 15 * (50 - 40) / (0.02 - 0.01)

=15KW/m^2

Assuming SS is solid, temperature at the surface exposed to air will be 60 C since its gradient is linear in the case of conduction, and there are two temperatures given in the problem. Convection coefficient can be found from Newton's Law of cooling:

q"_conv = h * ( T_∞  - T_s ) ----> h = q"_conv / ( T_∞  - T_s )

                                                   h = 15000 W / (100 - 60 ) C = 375 KW/m^2K

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A 35kg block of mass is subjected to forces F1=100N and F2=75N at agive angle thetha= 20° and 35° respectively.find the distance
Talja [164]

Answer:

21 m

Explanation:

Since F₁ = 100 N and acts at an angle of 20° to the horizontal, it has horizontal component F₁' = 100cos20° = 93.97 N and vertical component F₁" = 100sin20° = 34.2 N.

Also, F₂ = 75 N and acts at an angle of -35° to the horizontal, it has horizontal component F₂' = 75cos(-35°) = 75cos35° = 61.44 N and vertical component F₂" = 75sin(-35°) = -75sin35° = -43.02 N

The resultant horizontal force F₃' = F₁' + F₂' = 93.97 N + 61.44 N = 155.41 N

The resultant vertical force F₃" = F₁" + F₂" = 34.2 N - 43.02 N = -8.82 N

If f is the frictional force on the block, the net horizontal force on the block is F = F₃' - f.

Since f = μN where μ = coefficient of kinetic friction = 0.4 and N = normal force on the block.

For the block to be in contact with the surface, the vertical forces on the block must balance.

Since the normal force, N must equal the resultant vertical force F₃" and the weight, W = mg of the object for a zero net vertical force,

N = mg + F₃" (since both the weight and the resultant vertical force act downwards)

N = mg + F₃"

Since m = mass of block = 35 kg and g = acceleration due to gravity = 9.8 m/s² and F₃" = 8.82 N

So,

N = mg + F₃"

N = 35 kg × 9.8 m/s² + 8.82 N

N = 343 N + 8.82 N

N = 351.82 N

So, the net horizontal force F = F₃' - f.

F = 155.41 N - 0.4 × 351.82 N

F = 155.41 N - 140.728 N

F = 14.682 N

Since F = ma, where a = acceleration of block,

a = F/m = 14.682 N/35 kg = 0.42 m/s²

To find the distance the block moved, x we use the equation

x = ut + 1/2at² where u = initial speed of block = 0 m/s, t = time = 10 s and a = acceleration of block = 0.42 m/s²

Substituting the values of the variables into the equation, we have

x = ut + 1/2at²

x = 0 m/s × 10 s + 1/2 × 0.42 m/s² × (10 s)²

x = 0 m + 1/2 × 0.42 m/s² × 100 s²

x = 0.21 m/s² × 100 s²

x = 21 m

So, the distance moved by the block is 21 m.

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2 years ago
A Class A fire extingisher is for use on general combustibles such as:​
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Answer:

used for ordinary combustibles, such as wood, paper, some plastics, and textiles. This class of fire requires the heat-absorbing effects of water or the coating effects of certain dry chemicals.

Explanation:

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3 years ago
Technician a says that if a tapered roller bearing is adjusted to loose
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The technician is true
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Which option should the engineers focus on as they develop the train in the following scenario?
pav-90 [236]

Answer:

  Engineers can design a train with a regenerative braking system

Explanation:

Assuming the point of the question is that the engineers want to focus on using energy efficiently when starting and stopping, they would likely want to consider a regenerative braking system. Such a system can store energy during braking so that it can be used during starting, reducing the amount of energy that must be supplied by an outside power source.

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3 years ago
A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte
Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

a(t)=5(1-\frac{t}{15})

a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})

Integrating both sides we get

\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m

Thus the remaining 125 meters will be covered with a constant speed of

v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s

in time equalling t_{2}=\frac{125}{37.5}=3.33seconds

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0
2 years ago
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