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nata0808 [166]
3 years ago
14

Use the map to complete the table by selecting the boxes that best describe how ocean currents affect the

Engineering
1 answer:
DerKrebs [107]3 years ago
4 0

Western coast is frequented by the cold californian current which reduces the temperature.

Southern Alaskan coast is frequented by Warm Alaskan current which increases the temeprature of the coast.

Eastern coast is frequented by Warm Gulf Stream which increases the temperature of the coast.

Explanation:

Oceanic current (Gulf stream) increases the temperature of the coastal region along the Eastern Coast of the US mainland.  Gulf stream represents the system of warm oceanic current. This current causes excessive hot conditions in the US mainland.

West Coast of the US is frequented by Cold Californian current and warm Alaskan current. Hence, the Californian current decreases the temperature of the western coast of the US and cause atmospheric stability in the region.

Southern Alaska is frequented by the warm Alaskan current which raises the temperature of the southern Alaskan coast.  

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A company buys a machine for $12,000, which it agrees to pay for in five equal annual payments, beginning one year after the dat
Yuki888 [10]

Answer:

$7,778.35

Explanation:

At year 3, the final payment of the remaining balance is equal to the present worth P of the last three payments.

First, calculate the uniform payments A:

A = 12000(A/P, 4%, 5)

= 12000(0.2246) = 2695.2  (from the calculator)

Then take the last three payments as its own cash flow.

To calculate the new P:

P = 2695.2 + 2695.2(P/A, 4%, 2) = 2695.2 + 2695.2(1.886) = 7778.35

Therefore, the final payment is $7,778.35

4 0
3 years ago
A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the
raketka [301]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Intertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy = ?

(b) rotor acceleration = ?

(c) change in torque angle = ?

(c) rotor speed = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

(c) change in torque angle (in elec deg) = 6.75 elec deg

(c) change in torque angle (in rmp/s) = 28.12 rpm/s

(c) rotor speed = 1505.62 rpm

Explanation:

(a) Find the stored energy in the rotor at synchronous speed.

The stored energy is given by

E = G \times H

Where G represents complex rated power and H is the inertia constant of turbo-generator.

E = 100 \times 8 \\\\E = 800 \: MJ

(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.

The rotor acceleration is given by

$ P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}  $

Where M is given by

$ M = \frac{E}{180 \times f} $

$ M = \frac{800}{180 \times 50} $

M = 0.0889 \: MJ \cdot s/ elec \: \: deg

So, the rotor acceleration is

$ P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}  $

$  30 = 0.0889 \frac{d^2 \delta}{dt^2}  $

$   \frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}  $

$   \frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2 $

(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.

The change in torque angle is given by

$ \Delta  \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2 $

Where t is given by

1 \: cycle = 1/f = 1/50 \\\\10 \: cycles = 10/50 = 0.2  \\\\t = 0.2 \: sec

So,

$ \Delta  \delta = \frac{1}{2} \cdot 337.46 \cdot (0.2)^2 $

$ \Delta  \delta = 6.75 \: elec \: deg

The change in torque in rpm/s is given by

$ \Delta  \delta = \frac{337.46 \cdot 60}{2 \cdot 360\circ  }   $

$ \Delta  \delta =28.12 \: \: rpm/s $

The rotor speed in revolutions per minute at the end of this period (10 cycles) is given by

$ Rotor \: speed = \frac{120 \cdot f}{P}  + (\Delta  \delta)\cdot t  $

Where P is the number of poles of the turbo-generator.

$ Rotor \: speed = \frac{120 \cdot 50}{4}  + (28.12)\cdot 0.2  $

$ Rotor \: speed = 1500  + 5.62  $

$ Rotor \: speed = 1505.62 \:\: rpm

4 0
3 years ago
Given a books table with columns named book_title and year_released, the following SQL statement will select the number of books
lord [1]

Question Continuation

True or False?

Answer:

False

Explanation:

Given

Columns: book_title and year_released

SQL statement: SELECT book_title, year_released FROM books GROUP BY year_released

Expected Result: To select the number of books released each year

Although the GROUP BY used in the SQL statement is meant to group rows with same values into summary rows but it's not user properly because the GROUP BY clause, as used here needs the COUNT aggregate functions to group the result-set by one or more columns.

8 0
3 years ago
What limits the practical realization of higher efficiencies in the Otto cycle?
AlekseyPX

Answer:

We know that all petrol engines are works on Otto cycle.Otto cycle have four process out of four two are constant volume process and others two are isentropic processes.

There are lots of limitations for practical Otto cycle these are as follows

1.In practical cycle heat can not add at constant volume.

2.In practical cycle there is a gap between combustion of two set of fuel.

3.Lots of heat is dissipated by cylinders.

4.Valve opening and closing is not a sudden process it requires some time.

5.There is a limitations of cylinder material ,it means that temperature of cycle can not rise after a specified limit of material.

Due to these above limitations practical cycles have low efficiency as compare to ideal cycle.

5 0
3 years ago
When a starter motor begins to turn it produces a high…. *
Georgia [21]

Answer:

The motor produces a high Torque when it begins to start.

8 0
2 years ago
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