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Flura [38]
1 year ago
6

Steam at 75 kPa and 8 percent quality is contained in a spring-loaded piston–cylinder device, as shown in Figure, with an initia

l volume of 2 m3 . Steam is now heated until its volume is 5 m3 and its pressure is 225 kPa. Determine the heat transferred to and the work produced by the steam during this process.
​
Engineering
1 answer:
Rashid [163]1 year ago
6 0

The heat transferred to and the work produced by the steam during this process  is 13781.618 kJ/kg

<h3>​How to calcultae the heat?</h3>

The Net Change in Enthalpy will be:

= m ( h2 - h1 ) = 11.216 ( 1755.405 - 566.78 ) = 13331.618 kJ/kg

Work Done (Area Under PV curve) = 1/2 x (P1 + P2) x ( V1 - V2)

= 1/2 x ( 75 + 225) x (5 - 2)

W = 450 KJ

From the First Law of Thermodynamics, Q = U + W

So, Heat Transfer = Change in Internal Energy + Work Done

= 13331.618 + 450

Q = 13781.618 kJ/kg

Learn more about heat on:

brainly.com/question/13439286

#SP1

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ENERGIA
VMariaS [17]

nswer:

i dont understand french

97

Explanation:

4 0
2 years ago
The air contained in a room loses heat to the surroundings at a rate of 60 kJ/min while work is supplied to the room by computer
charle [14.2K]

Answer:

The net amount of energy change of the air in the room during a 10-min period is 120 KJ.

Explanation:

Given that

Heat loss from room (Q)= 60 KJ/min

Work supplied to the room(W) = 1.2 KW = 1.2 KJ/s

We know that  1 W = 1 J/s

Sign convention for heat and work:

1. If heat is added to the system then it is taken as positive and if heat is rejected from the system then it is taken as negative.

2. If work is done by the system then it is taken as positive and if work is done on the system then it is taken as negative.

So

Q = -60 KJ/min

In 10 min Q = -600 KJ

W = -1.2 KJ/s

We know that

1 min = 60 s

10 min = 600 s

So   W = -1.2 x 600 KJ

W = -720 KJ

WE know that ,first law of thermodynamics

Q = W + ΔU

-600  =  - 720 + ΔU

ΔU = 120 KJ

The net amount of energy change of the air in the room during a 10-min period is 120 KJ.

4 0
3 years ago
Read 2 more answers
Select the best answer to the questo
Norma-Jean [14]

Answer:

C

Explanation:

7 0
3 years ago
Read 2 more answers
4. The instant the ignition switch is turned to the start position,
geniusboy [140]

Answer:

D. Both pull-in and hold-in windings are energized.

Explanation:

The instant the ignition switch is turned to the start position, "Both pull-in and hold-in windings are energized." This is because the moment the ignition switch is turned to the start position, voltage passes through to the S terminal of the solenoid.

The hold-in winding is attached to the case of the solenoid. Similarly, the pull-in winding is also attached to the starter motor. Thereby, the current will move across both windings by getting energized to generate a strong magnetic field.

4 0
2 years ago
A three-point bending test is performed on a glass specimen having a rectangular cross section of height d = 5.4 mm (0.21 in.) a
Fudgin [204]

Answer:

5.21e-2mm

Explanation:

Please see attachment

8 0
3 years ago
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