Question
Ytrria Stabilized Zirconia (YSZ) is a very important ceramic because it provides outstanding strength and toughness over a wide temperature range--up to a very high
Ytrria Stabilized Zirconia ("YSZ") is a very important ceramic because it provides outstanding strength and toughness over a wide temperature range--up to a very high melting temperature of 2700 degrees Celsius! Consequently, one application is as a coating on turbine blades for high performance aircraft. Of course you never want to reach that melting temperature, because the mechancial properties degrade dramatically as it is approached.
But, engineering is always about trade-offs, and optimal engine performance is achieved when the blade surface is at the highest possible temperature. For this to work without the engine melting, the inside of the blade must remain relatively cool. This naturally establishes a thermal gradient across the YSZ coating, and correspondingly a thermal stress. If that stress surpasses the failure stress, the coating will fail catastrophically. That's really bad for airplane engines!
Consider that the core temperature for the inside of a turbine blade, which is necessarily made out of a different materia, is designed to be no hotter than 500 degrees C. Otherwise it will melt (no noke, this is a real world example, where the internal temperature is helped to stay low using active cooling through internal channels with forced air).
Thus, what is the peak temperature possible at the outer blade surface (in the combustion chamber) before the YSZ coating fails due to compressive thermal stresses? Provide this peak temperature in degrees Celsius.
Possibly useful details: YSZ has a thermal expansion coefficient of 10 E-6/Kelvin, an elastic Young's modulus of 210GPa, and a compressive fracture strength of 2.4 GPa.
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