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4 years ago

List the four main propulsion systems.

Engineering

1 answer:

natali 33 [55]4 years ago

3 0

Answer:

1.Diesel propulsion systems

2.Gas turbine propulsion

3.Water jet propulsion

4.Fuel cell propulsion

Explanation:

A propulsion system is a machine that produces a thrust needed to push an object forward. An example is a system used in airplanes and rockets. Antimatter is so far identified to be the most efficient propulsion system.

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a ________________ is the flat, low-lying area adjacent to a stream channel that receives overbank flow

Umnica [9.8K]

Answer:

flood plain

Explanation:

it should be called a flood plain which makes sense because it’s not a deposition.

5 0

2 years ago

An electric water heater held at 120° F is kept in a 60°F room. When purchased, its insulation is equivalent to R-5. An owner pu

Trava [24]

Answer:

Total saving would be of 36.917 $\yr

Explanation:

Given Data:

$T_{heater} = 120$ Degree F

$T_{room} = 60$ Degree F

A = 30 ft^2

$\eta = 100$ %

Heat loss before previous final value $= \frac{A \Delta T}{R}$

$=\frac{30\times *(120-60)}{5}$

= 360 Btu/hr

Heat loss after new value $= \frac{30\times \times (120-60)}{15} = 120 Btu/hr$

saving would be $= 360 - 120 Btu/hr \times kw hr/ 3412 Btu\times 24 hr/day \times 365 day/year$

= 616.1782 kw hr/yr

$cost = 616.1782 \times 0.06$ $

= 36.917 $\yr

6 0

4 years ago

What is referred to as "Pyroelectric" materials?

aliina [53]

Answer and Explanation:

Pyroelectric material

Pyroelectric materials have special property of generating potential difference (although it is very less ) when these material are treated with heat or when celled down.

The potential difference generated is for very less time

The generation of potential difference is due to change in position of atoms after heating or cooling.

5 0

3 years ago

Which type of snap ring plier expands the snap ring for removal?

4vir4ik [10]

Answer:

c I think so haven't took engineering in ages

4 0

3 years ago

A 25 kVA transformer has an iron loss of 200 W and a full-load copper loss of 350 W. Calculate the transformer efficiency for th

Oksanka [162]

Answer:

a. 97.32 percent

b. 97.92 percent

c. 95.91 percent

Explanation:

given parameters

apparent power, S = 25 kVA

Iron Loss, Piron = 200 W

copper Loss at full load, P cufl = 350 W

let the transformer efficiency at full load be E

let the real power be P out

at full load and 0.8 power factor

P out = Scos∅

= 25 x 10³ x 0.8

=20000 W or 20 kW

efficiency at full load is

E = (P out)/(P out + P iron + P cufl) x 100%

= (20000)/(20000 + 200 + 350) x 100%

= 97.32%

at 70% of full load at unity power factor of 1

Pout = 70% x Scos∅

= 0.7 x 25 x 10³ x 1

= 17,500W or 17.5kW

copper loss at 70% full load

P cu0.7fl = (70%)² x 350

= (0.7)² x 350

= 171.5 W

Iron loss remain the same, P iron = 200 W

Efficiency of transformer at 70% of full load

E = (P out0.7)/(P out0.7 + P iron + P cufl0.7) x 100%

= (17500)/(17500 + 200 + 171.5) x 100%

= 97.92%

at 40% of full load at a power factor of 0.6

P out = 40% x Scos∅

= 0.4 x 25 x 10³ x 0.6

= 6000 W or 6 kW

copper loss at 40% full load

P cu0.7fl = (40%)² x 350

= (0.4)² x 350

= 56 W

Iron loss remain the same, P iron = 200 W

Efficiency of transformer at 40% of full load

E = (P out0.4)/(P out0.4 + P iron + P cu0.4) x 100%

= (6000)/(6000 + 200 + 350) x 100%

= 95.91%

6 0

3 years ago