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anastassius [24]
2 years ago
10

Every two years or at recommendation by manufacturer.

Engineering
1 answer:
9966 [12]2 years ago
4 0

Answer:

Manufacturer’s Recommendations means the instructions, procedures, and recommendations which are issued by the manufacturer of any equipment used at the Facility relating to the operation, maintenance, or repair of such equipment, and any revisions or updates thereto from time to time issued by the manufacturer.

Manufacturer’s Recommendations means the instructions, procedures and recommendations which are issued by any manufacturer of the Equipment relating to the operation, maintenance and repair of the Equipment and any revisions to such instructions, procedures and recommendations agreed to by any manufacturer of the Equipment and which are valid at the time such operation, repair and maintenance is being carried out.

Manufacturer’s Recommendations means the written instructions, procedures and recommendations which are issued by the original equipment manufacturer of any plant or equipment used at the Utility Plant relating to the operation, maintenance and repair of such plant or equipment and any revisions thereto issued by the manufacturer, which are valid and applicable at the time such operation, maintenance or repair is undertaken. Notwithstanding the above, Manufacturer’s Recommendations shall not include any instructions, procedures or recommendations of a manufacturer of any plant or equipment that the Owner and the Operator have agreed in writing to exclude from this definition or have agreed in writing should not be followed.

Explanation:

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A sheet of steel 3-mm thick has nitrogen atomospheres on both sides at 900 C and is permitted to achieve a steady-state di usion
kati45 [8]

Answer:

X_B = 1.8 \times 10^{-3} m = 1.8 mm

Explanation:

Given data:

Diffusion constant for nitrogen is = 1.85\times 10^{-10} m^2/s

Diffusion flux = 1.0\times 10^{-7} kg/m^2-s

concentration of nitrogen at high presuure = 2 kg/m^3

location on which nitrogen  concentration is 0.5 kg/m^3   ......?

from fick's first law

J = D \frac{C_A C_B}{X_A X_B}

Take C_A as point  on which nitrogen concentration is 2 kg/m^3

x_B = X_A + D\frac{C_A -C_B}{J}

Assume X_A is zero at the surface

X_B = 0 + ( 12\times 10^{-11} ) \frac{2-0.5}{1\times 10^{-7}}

X_B = 1.8 \times 10^{-3} m = 1.8 mm

4 0
3 years ago
In order to impress your neighbors and improve your vision in traffic jams, you decide to mount a cylindrical periscope 2.0 m hi
kondaur [170]
Follow @richard.gbe on Instagram for the answer
5 0
3 years ago
Write short notes on: (any four) a) Suspended ground floor b) Soil exploration c) Baulking of sand d) Bearing capacity of soil e
vredina [299]

Answer:

a) A suspended floor is a ground floor with a void underneath the structure. The floor can be formed in various ways, using timber joists, precast concrete panels, block and beam system or cast in-situ with reinforced concrete. However, the floor structure is supported by external and internal walls.

b) Soil exploration consists of determining the profile of the natural soil deposits at the site, taking the soil samples and determining the engineering properties of soils using laboratory tests as well as in-situ testing methods

c) Bulking in sand Occurs When dry sand interacts with the atmospheric moisture. Presence of moisture content forms a thin layer around sand particles. This layer generates the force which makes particles to move aside to each other. This results in the increase of the volume of sand.

d) In a nutshell, bearing capacity is the capacity of soil to support the loads that are applied to the ground above. It depends primarily on the type of soil, its shear strength and its density. It also depends on the depth of embedment of the load – the deeper it is founded, the greater the bearing capacity.

Explanation:

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6 0
3 years ago
Base course aggregate has a target dry density of 119.7 Ib/cu ft in place. It will be laid down and compacted in a rectangular s
djyliett [7]

Answer:

total weight of aggregate =  5627528 lbs = 2814 tons  

Explanation:

we get  here volume of space to be filled with aggregate that is

volume = 2000 × 48 × 0.5

volume = 48000 ft³

now filling space with aggregate of the density that is

density = 0.95 × 119.7

density = 113.72 lb/ft³

and dry weight of this aggregate is

dry weight = 48000 × 113.72

dry weight = 5458320 lbs

we consider here percent moisture is by weigh

so weight of moisture in aggregate will be

weight of moisture = 0.031 × 5458320

weight of moisture = 169208 lbs

so here total weight of aggregate is

total weight of aggregate = 5458320 + 169208

total weight of aggregate =  5627528 lbs = 2814 tons  

3 0
3 years ago
Car B is traveling a distance dd ahead of car A. Both cars are traveling at 60 ft/s when the driver of B suddenly applies the br
vagabundo [1.1K]

Answer:

Explanation:

Using the kinematics equation v = v_o + a_ct to determine the velocity of car B.

where;

v_o = initial velocity

a_c = constant deceleration

Assuming the constant deceleration is = -12 ft/s^2

Also, the kinematic equation that relates to the distance with the time is:

S = d + v_ot + \dfrac{1}{2}at^2

Then:

v_B = 60-12t

The distance traveled by car B in the given time (t) is expressed as:

S_B = d + 60 t - \dfrac{1}{2}(12t^2)

For car A, the needed time (t) to come to rest is:

v_A = 60 - 18(t-0.75)

Also, the distance traveled by car A in the given time (t) is expressed as:

S_A = 60  * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2

Relating both velocities:

v_B = v_A

60-12t = 60 - 18(t-0.75)

60-12t =73.5 - 18t

60- 73.5 = - 18t+ 12t

-13.5 =-6t

t = 2.25 s

At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars

i.e.

S_B = S_A

d + 60 t - \dfrac{1}{2}(12t^2) = 60  * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2

d + 60 (2.25) - \dfrac{1}{2}(12*(2.25)^2) = 60  * 0.75 +60((2.25)-0.75) -\dfrac{1}{2}*18*((2.25)-0.750)^2

d + 104.625 = 114.75

d = 114.75 - 104.625

d = 10.125 ft

3 0
3 years ago
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