Why does the compression-refrigeration cycle have a high-pressure side and a low-pressure side?

How can I find the quotient of 27 divided 91.8

gayaneshka [121]

Step 1:
Start by setting it up with the divisor 8 on the left side and the dividend 27 on the right side like this:

8 ⟌ 2 7

Step 2:
The divisor (8) goes into the first digit of the dividend (2), 0 time(s). Therefore, put 0 on top:

0
8 ⟌ 2 7

Step 3:
Multiply the divisor by the result in the previous step (8 x 0 = 0) and write that answer below the dividend.

0
8 ⟌ 2 7
0

Step 4:
Subtract the result in the previous step from the first digit of the dividend (2 - 0 = 2) and write the answer below.

0
8 ⟌ 2 7
- 0
2

Step 5:
Move down the 2nd digit of the dividend (7) like this:

0
8 ⟌ 2 7
- 0
2 7

Step 6:
The divisor (8) goes into the bottom number (27), 3 time(s). Therefore, put 3 on top:

0 3
8 ⟌ 2 7
- 0
2 7

Step 7:
Multiply the divisor by the result in the previous step (8 x 3 = 24) and write that answer at the bottom:

0 3
8 ⟌ 2 7
- 0
2 7
2 4

Step 8:
Subtract the result in the previous step from the number written above it. (27 - 24 = 3) and write the answer at the bottom.

0 3
8 ⟌ 2 7
- 0
2 7
- 2 4
3

You are done, because there are no more digits to move down from the dividend.

The answer is the top number and the remainder is the bottom number.

Therefore, the answer to 27 divided by 8 calculated using Long Division is:

3

8 0

3 years ago

If welding is being done in the vertical position, the torch should have a travel angle of?

siniylev [52]

Answer:

Between 35°– 45°

Explanation:

In the vertical position, Point the flame in the direction of travel. Keep the flame tip at the correct height above the base metal. An angle of 35°–45° should be maintained between the torch tip and the base metal. This angle may be varied up or down to heat or cool the weld pool if it is too narrow or too wide

4 0

2 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

a.

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%

b.

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%

c.

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

What is a heat engine? State its efficiency (a sketch would be appropriate).

dimulka [17.4K]

Answer:

$\eta =\dfrac{Q_1-Q_2}{Q_1}$

Explanation:

Heat engine

Heat engine is a device which produce work while consuming some amount of energy .This engine works in cycle.In other words we can say that Heat engine is a device which covert thermal or chemical energy in to mechanical energy and after that this mechanical energy can be use for producing mechanical work.

Heat engine takes heat from high temperature and rejects heat to lower temperature and producing work.

We know that efficiency given as

$\eta =\dfrac{out\ put}{in\ put}$