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faltersainse [42]
2 years ago
7

prove that the heat transfer at the constant pressure is given by the enthalpy change during the process​

Engineering
1 answer:
mihalych1998 [28]2 years ago
5 0

Answer:

at constant pressure, the heat flow for any process is equal to the change in the internal energy of the system plus the PV work done. Comparing the previous two equations shows that at constant pressure, the change in the enthalpy of a system is equal to the heat flow: ΔH=qp.

Explanation:

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Students are expected to respond to one of the two questions described below. Students should provide examples to clarify their
Flura [38]

Answer:

can u explain what we are answering?

Explanation:

7 0
3 years ago
What is a two stroke engine and what is a four stroke engine, please keep the definitions as simple as can be and please explain
rewona [7]

Answer:

A two-stroke engine is a type of internal combustion engine that completes a power cycle with two strokes of the piston during only one crankshaft revolution.

A four-stroke engine is an internal combustion engine in which the piston completes four separate strokes while turning the crankshaft. A stroke refers to the full travel of the piston along the cylinder, in either direction.

process for two stroke engine

This is in contrast to a "four-stroke engine", which requires four strokes of the piston to complete a power cycle during two crankshaft revolutions.In a two-stroke engine, the end of the combustion stroke and the beginning of the compression stroke happen simultaneously, with the intake and exhaust (or scavenging) functions occurring at the same time.Two-stroke engines often have a high power-to-weight ratio, power being available in a narrow range of rotational speeds called the "power band". Compared to four-stroke engines, two-stroke engines have a greatly reduced number of moving parts.

process for four stroke engine

A stroke refers to the full travel of the piston along the cylinder, in either direction. The four separate strokes are termed:

Intake: Also known as induction or suction. This stroke of the piston begins at top dead center (T.D.C.) and ends at bottom dead center (B.D.C.). In this stroke the intake valve must be in the open position while the piston pulls an air-fuel mixture into the cylinder by producing vacuum pressure into the cylinder through its downward motion. The piston is moving down as air is being sucked in by the downward motion against the piston.

Compression: This stroke begins at B.D.C, or just at the end of the suction stroke, and ends at T.D.C. In this stroke the piston compresses the air-fuel mixture in preparation for ignition during the power stroke (below). Both the intake and exhaust valves are closed during this stage.

Combustion: Also known as power or ignition. This is the start of the second revolution of the four stroke cycle. At this point the crankshaft has completed a full 360 degree revolution. While the piston is at T.D.C. (the end of the compression stroke) the compressed air-fuel mixture is ignited by a spark plug (in a gasoline engine) or by heat generated by high compression (diesel engines), forcefully returning the piston to B.D.C. This stroke produces mechanical work from the engine to turn the crankshaft.

Exhaust: Also known as outlet. During the exhaust stroke, the piston, once again, returns from B.D.C. to T.D.C. while the exhaust valve is open. This action expels the spent air-fuel mixture through the exhaust valve.

characteristics of two stroke and four stroke engines (which one gets hotter)

Since there are twice as many power strokes during the operation of a two-stroke engine as there are during the operation of a four-stroke engine, the engine tends to heat up more, and thus is likely to have a shorter life. Also, in the two-stroke engine lubricating oil must be mixed with the fuel.

6 0
2 years ago
Read 2 more answers
. An ideal vapor compression refrigeration cycle operates with a condenser pressure of 900 kPa. The temperature at the inlet to
tatuchka [14]

Answer:

Actual COP = 5.368

Maximum theoretical COP = 6.368

Explanation:

Given - An ideal vapor compression refrigeration cycle operates with a condenser pressure of 900 kPa. The temperature at the inlet to the compressor is -5oC.

To find -  If this device operates using R134a as the working fluid. Calculate the actual COP of this device as well as the maximum theoretical COP.

Proof -

Given that,

An ideal vapor compression refrigeration cycle operates with a condenser pressure of 900 kPa.

From Refrigerant 134-a Table

At T1 = -5°C

h1 = 247.505 KJ/kg

S1 = 0.93434 KJ/kg

At P2 = 900 KPa

S1 = S2

h2 = 274.679 Kj/Kg

h3 = h4 = 101.61 KJ/g

So,

Compressor work (Wc) = h2 - h1

                                       = 274.679 - 247.505

                                       = 27.174

⇒Compressor work (Wc) = 27.174 KJ/kg

Now,

Heat out (Qout) = h2 - h3

                          = 274.679 - 101.61

                          = 173.069

⇒Heat out (Qout) = 173.069 KJ/kg

Now,

Heat input (Qin) = h1 - h4

                          = 274.505 - 101.61

                          = 145.895

⇒Heat input (Qin) = 145.895 KJ/kg

So,

Actual COP at the refrigerator is -

(COP)R = (Qin)/(Wc)

            = (145.895)/ (27.174)

            = 5.368

⇒Actual COP = 5.368

Now,

Maximum theoretical COP is -

(COP) = (Qout)/(Wc)

          = (173.069)/ (27.174)

          = 6.368

⇒Maximum theoretical COP = 6.368

8 0
2 years ago
A well penetrates an unconfined aquifer. Prior to pumping, the water level (head) is 25 meters. After a long period of pumping a
GarryVolchara [31]

Answer:

Explanation:

Find attach the solution

3 0
3 years ago
Compute the number of kilo- grams of hydrogen that pass per hour through a 6-mm-thick sheet of palladium having an area of 0.25
nydimaria [60]

Answer:

The number of kilo- grams of hydrogen that pass per hour through this sheet of palladium is 4.1 * 10^{-3} \frac{kg}{h}

Explanation:

Given

x1 = 0 mm

x2 = 6 mm = 6 * 10^{-3} m

c1 = 2 kg/m^{3}

c2 = 0.4 kg/m^{3}

T = 600 °C

Area = 0.25 m^{2}

D = 1.7 * 10^{8} m^{2}/s

First equation

J = - D \frac{c1 - c2}{x1 - x2}

Second equation

J = \frac{M}{A*t}

To find the J (flux) use the First equation

J = - 1.7 * 10^{8} m^{2}/s * \frac{2 kg/m^{3}  - 0.4 kg/m^{3}}{0 - 6 * 10^{-3} } = 4.53 * 10^{-6} \frac{kg}{m^{2}s }

To find M use the Second equation

4.53 * 10^{-6} \frac{kg}{m^{2}s} = \frac{M}{0.25 m^{2} * 3600s/h}

M = 4.1 * 10^{-3} \frac{kg}{h}

4 0
2 years ago
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