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morpeh [17]
3 years ago
8

Write the official Kelvin-Plank statement of the Second Law as well as a simplified version of it that is more easily grasped. W

rite the official Clausius statement of the Second Law as well as a simplified version of it that is more easily grasped.
Engineering
2 answers:
slava [35]3 years ago
8 0

Answer:

Kelvin-Plank statement of the second law:

" It is not possible to construct a heat engine which operates cyclically which in effect absorbs energy from a thermal reservoir (single) in the form of heat and deliver in output an equivalent or net work work"

Simplified form:

It is impossible to device a heat engine with  100% thermal efficiency i.e, it cannot transfer heat from a thermal reservoir to net work done i.e, some amount of heat will be lost.

Clausius Statement of the second law:

" It is impossible to design a device in a way that when operating in a cycle , it will transfer heat from a cooler body to a hotter body only without any other effect"

Simplified form:

It is not possible to design a device which can transfer heat from low to high temperatures without the requirement of any external work as input. With the help of external work, heat, in the system can be transferred from low to high temperatures.

marshall27 [118]3 years ago
3 0

Answer:

Explanation:

<u>The official statement of Kelvin-Plank :</u>

''It is impossible for a heat engine to produce net work in a complete cycle if it exchanges heat only with bodies at a single fixed temperature''.

Two reversible adiabatic paths can not intersect each other which violates the kelvin-planks statement.The machine which violate Kelvin-plank statement is called PMM2.

<u>Simplified form:</u>

We can say that it is impossible to covert all heat energy in to work with out rejecting heat in to the surrounding when operating in a cycle.

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The outer surface of an engine is situated in a place where oil leakage can occur. When leaked oil comes in contact with ahot su
VladimirAG [237]

Answer:

TBC thickness of 4 mm is insufficient to prevent fire hazard

Explanation:

Given:

- Temperature of hot-fluid inner surface T_i = 333°C

- The convection coefficient hot-fluid h_i = 7 W/m^2K

- The thermal conductivity of engine cover k_1 = 14 W/mK

- The thickness of engine cover L_1 = 0.01 m

- The thermal conductivity of TBC layer k_2 = 1.1 W/mK  ... (Typing error)

- The thickness of TBC layer L_2 = 0.004 m

- Temperature of ambient air outer surface T_o = 69°C

- The convection coefficient ambient air h_o = 7 W/m^2K

Find:

Would a TBC layer of 4 mm thickness be sufficient to keep the engine cover surface below autoignition temperature of 200°C to prevent fire hazard?

Solution:

- We will use thermal circuit analogy for the 1-D problem and steady state conduction with no heat generation in the cover or TBC layer.

 The temperature at each medium interface and the Thermal resistance for each medium is given in the attachment schematic and circuit analogy.

 - We will calculate the total heat flux for the entire system q:

                       q = ( T_i - T_o ) / R_total

- R_total is the equivalent thermal resistance of the entire circuit. Since all resistances are in series we have:

                       R_total = 1 / h_i + L_1 / k_1 + L_2 / k_2 + 1 / h_o

- Plug in the values and compute:

                       R_total = 1 / 7 + 0.01 / 14 + 0.004 / 1.1 + 1 / 7

                       R_total = 0.2900649351 T-m^2 / W

- Calculate the Total heat flux q:

                       q = ( 333 - 69 ) / 0.2900649351

                       q = 910.141 W / m^2

- Just like the total current in a circuit remains same, the total heat flux remains same. We will use the total heat flux q to calculate the temperature of outer engine surface T_2 as follows:

                      q = ( T_i - T_2 ) / R_i2

Where,

                      R_i2 = 1 / h_i + L_1 / k_1

                      R_i2 = 1 / 7 + 0.01 / 14 = 0.14357 T-m^2 / W

Hence,

                      ( T_i - T_2 ) = q*R_i2

                        T_2 = T_i - q*R_i2

Plug the values in:

                        T_2 = 333 - 910.141*0.14357

                        T_2 = 202.33°C

- The outer surface of the engine cover has a temperature above T_ignition = 200°C. Hence, the TBC thickness of 4 mm is insufficient to prevent fire hazard

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