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2 years ago

For potassium chloride (KCl), calculate the radius ratio to show why the compound has the cesium chloride structure.

Engineering

1 answer:

valentina_108 [34]2 years ago

8 0

Explanation:

The radius for potassium ion, ${r^+}$ = 152 pm

The radium for the chloride ion, ${r^-}$ = 167 pm

$\frac {r^+}{r^-}=\frac {152}{167}=0.91018$

Cesium chloride has cubic unit cell and for the ionic compounds to have such lattice, the condition is:

${\frac {r^+}{r^-}}\geq 0.732$

<u>Thus, for potassium chloride, the ratio is 0.91018 and it is greater than 0.732. It exists in the cesium chloride structure.</u>

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2 years ago

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2 years ago

A cylindrical part of diameter d is loaded by an axial force P. This causes a stress of P/A, where A = πd2/4. If the load is kno

Afina-wow [57]

Answer:

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Explanation:

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2 years ago

Same rule: If both players spend the same number of coins, player 2 gains 1 coin. Off-by-one rule: If the players do not spend t

Galina-37 [17]

Answer:

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Explanation:

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2 years ago

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

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q = ( 333 - 69 ) / 0.2900649351

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q = ( T_i - T_2 ) / R_i2

Where,

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3 0

2 years ago