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Ratling [72]
3 years ago
5

List the main activities of exploration??

Engineering
1 answer:
Trava [24]3 years ago
3 0

Answer:  Exploration includes plethora of activities and depend upon the kind  of exploration a person is doing. But most include some of the basic activities like research , investigation, planning and execution.

Suppose we want to explore new petroleum sites then we would have to start with studying the geography of that area, then according to our research we will analyse the hot spots or the sector where probability of finding of oil field is highest, post that appropriate man power is skilled professionals, tools and machinery will be brought at the site so that execution can take place.

You might be interested in
A slug travels 3 centimeters in 3 seconds. A snail travels 6 centimeters in 6 seconds. Both travel at constant speeds. Mai says,
Irina-Kira [14]

Answer:

i dont agree with mai because they were both going 1cm per second

Explanation:

3÷3=1

6÷6=1

they both are difrent numbers but equal the same thing

8 0
4 years ago
Precast double-tee units are generally used as structural floors when:________.
TiliK225 [7]

Answer:

<em>d. floor spans exceed that which is economical with site-cast or hollow-core slabs and exceed 60 feet.</em>

<em></em>

Explanation:

<em> Double tee slabs are prestressed concrete castings usually about 100 m to 150 m long</em>. <em>They are usually used in the construction industry in structures requiring a long floor span (or roof span) exceeding 60 feet in length, that is required to bear additional load (additional load carrying capability</em>). Double tee slabs are usually employed in construction because they shorten construction time. reduce congestion at the construction site, and also reduce construction cost.

<em>They are the required choice in construction when the cost of site casting or hollow-core slabs is no longer economical due to the length of the structure exceeding 60 feet.</em>

8 0
3 years ago
Air exits a compressor operating at steady-state, steady-flow conditions at 150 oC, 825 kPa, with a velocity of 10 m/s through a
ioda

Answer:

a) Qe = 0.01963 m^3 / s , mass flow rate m^ = 0.1334 kg/s

b) Inlet cross sectional area = Ai = 0.11217 m^2 , Qi = 0.11217 m^3 / s    

Explanation:

Given:-

- The compressor exit conditions are given as follows:

                  Pressure ( Pe ) = 825 KPa

                  Temperature ( Te ) = 150°C

                  Velocity ( Ve ) = 10 m/s

                  Diameter ( de ) = 5.0 cm

Solution:-

- Define inlet parameters:

                  Pressure = Pi = 100 KPa

                  Temperature = Ti = 20.0

                  Velocity = Vi = 1.0 m/s

                  Area = Ai

- From definition the volumetric flow rate at outlet ( Qe ) is determined by the following equation:

                   Qe = Ae*Ve

Where,

           Ae: The exit cross sectional area

                   Ae = π*de^2 / 4

Therefore,

                  Qe = Ve*π*de^2 / 4

                  Qe = 10*π*0.05^2 / 4

                  Qe = 0.01963 m^3 / s

 

- To determine the mass flow rate ( m^ ) through the compressor we need to determine the density of air at exit using exit conditions.

- We will assume air to be an ideal gas. Thus using the ideal gas state equation we have:

                   Pe / ρe = R*Te  

Where,

           Te: The absolute temperature at exit

           ρe: The density of air at exit

           R: the specific gas constant for air = 0.287 KJ /kg.K

             

                ρe = Pe / (R*Te)

                ρe = 825 / (0.287*( 273 + 150 ) )

                ρe = 6.79566 kg/m^3

- The mass flow rate ( m^ ) is given:

               m^ = ρe*Qe

                     = ( 6.79566 )*( 0.01963 )

                     = 0.1334 kg/s

- We will use the "continuity equation " for steady state flow inside the compressor i.e mass flow rate remains constant:

              m^ = ρe*Ae*Ve = ρi*Ai*Vi

- Density of air at inlet using inlet conditions. Again, using the ideal gas state equation:

               Pi / ρi = R*Ti  

Where,

           Ti: The absolute temperature at inlet

           ρi: The density of air at inlet

           R: the specific gas constant for air = 0.287 KJ /kg.K

             

                ρi = Pi / (R*Ti)

                ρi = 100 / (0.287*( 273 + 20 ) )

                ρi = 1.18918 kg/m^3

Using continuity expression:

               Ai = m^ / ρi*Vi

               Ai = 0.1334 / 1.18918*1

               Ai = 0.11217 m^2          

- From definition the volumetric flow rate at inlet ( Qi ) is determined by the following equation:

                   Qi = Ai*Vi

Where,

           Ai: The inlet cross sectional area

                  Qi = 0.11217*1

                  Qi = 0.11217 m^3 / s    

- The equations that will help us with required plots are:

Inlet cross section area ( Ai )

                Ai = m^ / ρi*Vi  

                Ai = 0.1334 / 1.18918*Vi

                Ai ( V ) = 0.11217 / Vi   .... Eq 1

Inlet flow rate ( Qi ):

                Qi = 0.11217 m^3 / s ... constant  Eq 2

               

6 0
3 years ago
4.In a hydroelectric power plant, 100 m3/s of water flows from an elevation of 120 m to a turbine, where electric power is gener
Dmitrij [34]

Answer:

The rate of irreversible loss will be "55.22 MW".

Explanation:

The given values are:

Elevation,

h = 120 m

Flow of water,

Q = 100 m³/s

Efficiency,

= 80%

i.e,

= 0.8

Efficiency turbine,

= 50 MW

Now,

Without any loss,

The power generated by turbine will be:

⇒ P=\delta gQh

On substituting the values, we get

⇒     =1000\times 9.8\times 100\times 120

⇒     =117.72 \ MW

Power generated in actual will be:

= \frac{50}{0.8}

= 62.5 \ MW

Hence,

Throughout the piping system,

The rate of irreversible loss is:

= Power \ generated \ by \ turbine-Power \ generated  \ in \ actual

= 117.72-62.5

= 55.22 \ MW

4 0
3 years ago
Exercise 19
musickatia [10]
Uhh idk this can u mabye look it up
6 0
4 years ago
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