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**The study of heat energy through random systems**

Thermodynamics The study of heat energy through random systems

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**Laws of Thermodynamics**

0th Law: If Ta=Tb and Tb=Tc then Ta=Tc 1st Law: ΔE=W+Q (Conservation of Energy) The increase in thermal energy of a system is determined by the heat of the system and the work added to it. The idea behind the Heat Engine

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**Efficiency Wnet=Qnet=Qhot-Qcold or Eff=Wnet/Qhot=1-(Qcold/Qhot)**

Ex: A steam engine absorbs 1.98 x105J and expels 1.49 x105J in each cycle. Assume that all the remaining energy is used to do work. a. What is the engine’s efficiency? b. How much work is done in each cycle?

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a. Eff=0.247 b. W=4.9 x104J

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**3rd Law: Temperature and Entropy are absolute scales.**

2nd Law: A system can be defined by its entropy. In a closed system entropy tends to increase. Entropy: A measure of the disorder (randomness) of the system 3rd Law: Temperature and Entropy are absolute scales. At some point no temperature or entropy exists. T→0 K and S→S0

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Heat Engines Engines allow heat energy to be transformed into work or mechanical energy. Work or Energy Heat increases no big deal. Friction does this. Heat Work or Energy is a big deal. Heat is easy to move around. You could just bring heat wherever you needed work done and “Boom!” you wouldn’t have to do the work, a machine could. Work or Energy Heat decreases is also a big deal. Making food cold preserves it and allows it to be moved readily. Less spoilage means less disease.

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**Any heat engine works on the same properties.**

A hot reservoir is the source of the energy. Both words mean something. Hot means that there is plenty of heat energy and reservoir means that if heat is removed the temperature doesn’t drop much. There is also a need for a cold reservoir. Again, both words mean something. Cold because it is at a lower temperature than the hot reservoir and reservoir because it must be large enough that you can dump heat into it without appreciably raising the temperature.

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**Heat transfer (or flow) is the answer.**

What happens if we put a hot and cold reservoir in contact? Thermal Equilibrium is not the answer! Heat transfer (or flow) is the answer. Remember that these are reservoirs so it would take a long time for them to come into thermal equilibrium. This is great, but we don’t get any work out of it. We need to “steal” some of the energy leaving the hot reservoir and make it do work for us.

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Heat Engine Stealing some energy to do work High Temp Low Temp ΔE Q W

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Types of Heat Engine Steam Engine

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**Internal Combustion Engine**

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**Unfortunately we don’t get an even trade!**

We lose energy to randomness/Entropy This is the 2nd Law of Thermodynamics Automobile engines are only about 15% efficient. That means for every 100J of heat energy, 15J worth of work is done on the piston and 85J of heat are discarded. Still, this is the source of energy for most of our transportation.

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**If a steam engine takes in 2. 254 x 104 kJ of heat and gives up 1**

If a steam engine takes in x 104 kJ of heat and gives up x 104 kJ of heat to the exhaust, what is the engines efficiency?

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**We can go backwards! This is a refrigerator or air conditioner**

High Temp Low Temp ΔE Q W

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**This requires Energy/Work**

This is why your refrigerator must be plugged in. It is constantly dumping heat into your kitchen Due to the 2nd Law of Thermodynamics more heat is dumped than is removed If you left the refrigerator door open you would heat up the kitchen

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Perfect Heat Engine Qcold Qhot ΔE W

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Qcold Qhot ΔE W

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**Efficiency Wnet=Qnet=Qhot-Qcold or Eff=Wnet/Qhot=(1-Qcold/Qhot)**

Ex: A steam engine absorbs 1.98 x105J and expels 1.49 x105J in each cycle. Assume that all the remaining energy is used to do work. a. What is the engine’s efficiency? b. How much work is done in each cycle?

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a. Eff=0.247 b. W=4.9 x104J

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More 2nd Law Entropy S=entropy Q=Heat (Joules) T=Temperature (In Kelvin) Entropy in a system must increase or at least stay the same!!!!!!!!!!!!

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An engine has a hot reservoir at 1000 K and uses the atmosphere at 300 K as the cold reservoir. You take 2500 J from the hot reservoir to do 1900 J of work. A. How much heat goes into the atmosphere? B. Is this engine possible? (Does the entropy increase?)

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**Wnet=Qnet=Qhot-Qcold Qcold=Qhot-Wnet=2500J -1900J =600J**

Entropy Engine is not possible. What is the maximum amount of work we can take out? How much work is done?

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**Carnot Engine Maximum efficiency**

In theory could run backwards All temperatures in Kelvin What is the efficiency of an ideal steam engine with steam at 685 K and exhaust at 298 K? What is Qcold ?

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**Carnot Engine Maximum efficiency**

In theory could run backwards All temperatures in Kelvin What is the efficiency of an ideal steam engine with steam at 685 K and exhaust at 298 K? What is Qhot if Qcold is 450J?

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