Lecture 6 - 2nd Law Introduction Flashcards
The second law of thermodynamics helps us know …
the direction the energy is flowing in a process and the efficiency of the process
If you throw a ball, you are doing work by …
changing the potential energy to kinetic energy
If you throw a ball and the temperature of the ball doesn’t change, is there a change in internal energy?
No
Work done ON a system is positive/negative
Negative
Work done throwing a ball raises the kinetic energy of the ball. So, why does the ball slow down after throwing?
Due to air drag reducing the kinetic energy of the ball
In the first law of thermodynamics, Q is …
heat transfer
Energy equation for an incompressible stationary fluid
change in internal energy of the fluid (specific heat * density * change in temperature with respect to time) = thermal conduction within fluid + viscous dissipation (shear velocity)
Heat diffusion equation
(density*specific volume * change in temperature with respect to time)/k = d^2T/dx^2 + d^2T/dy^2 + d^2T/dz^2 + q/k
Drag force causes a fluid to churn, contributing to the viscous dissipation, which results in a temperature change from a change in thermal energy.
True
An increase in the random kinetic energy of a system does not ensure that work will be done.
True
Direction of energy transfer is determined by the equation of the first law.
FALSE ; the second law tells us the direction of energy flow
Q-W = ∆U + ∆KE + ∆PE does not influence the direction of energy flow in the real world
The first law shows us that work can easily be converted into other forms of energy (KE, PE, Q, ∆U).
True
What is a heat engine?
A cyclical machine that receives heat from a high temperature source and converts part of the heat into work and rejects the other half of the heat to a heat sink.
Describe a heat engine.
HEAT SOURCE (Temp. high)
↓ Q(in) = Q(high) ** Q(high) = high heat
HEAT ENGINE → W (net, out)
↓ Q(out) = Q(low) ** Q(low) = low heat
SINK (Temp. low)
Describe the Rankine Cycle.
↓Q(in)
BOILER (Caldera) → TURBINE (Turbina) → W(out)
↑ ↓
W(in) → PUMP (Bomba) ← CONDENSER (Condensador)
↓ Q(out; rejected)
Boiler
Turbine
Condenser
Pump
Caldera
Turbina
Condensador
Bomba
The Rankine Cycle is an example of a heat engine.
True
How do we determine the net work out of a heat engine?
W(net, out) = W(out) - W(in) = Q(in) - Q(out)
Symbol for thermal efficiency
ɳ
Thermal efficiency, ɳ =
ɳ = net work out/total heat in
ɳ = W(net, out)/Q(total, in)
ɳ = 1 - Q(out)/Q(in)
If net work out = Q(in) - Q(out), we can write the thermal efficiency, ɳ as …
ɳ = [Q(in) - Q(out)]/Q(in, total) = 1 - Q(out)/Q(in)
In terms of a heat engine, thermal efficiency, ɳ, can be written as …
ɳ = 1 - Q(low)/Q(high)
Thermal efficiency is bounded between …
0 & 1
0 < ɳ < 1
What is the Kelvin - Planck statement?
A statement that says that it is impossible for a device that operates on a cycle to receive heat from a single source (reservoir) and produce a net amount of work.
The Kelvin - Planck statement means …
that a cyclical device that gets its heat from one source needs to have a sink to eject the residual or waste heat in order to produce net work
Refrigerators and heat pumps are the opposite of a heat engine in that …
refrigerators and heat pumps transfer heat from a low temperature source to a high temperature source
Describe a heat pump
HEAT PUMP ← W(in)
Describe a heat pump
SOURCE (T, high) ↑ Q(high) HEAT PUMP ← W(net, in) ↑ Q(low) Sink (T, low)
Describe the cycle associated with evacuating a rigid tank.
↑ Q(high)
CONDENSER (Condensador)
↓ ↑
EXPANSION COMPRESSOR ← W(net, in)
VALVE (Compresor)
(válvula de ↑
expansión) → EVAPORATOR (Evaporador)
↑ Q(low)
SINK (T, low)
The cycle associated with evacuating a rigid tank has to do with the uniform process theorem. Describe what happens in this cycle.
A compressor goes into a tank (condenser) that is hotter than the environment and therefore rejects heat into the environment. An expansion valve is opened very quickly and expands and gets cold. When the valve gets cold, it absorbs heat from the environment.
Air comes out of a compressed tank, and water and ice form outside of the tank. This is due to the temperature in the tank being lower than the temperature outside the tank, and the air around the valve being cooled as a result.
A compressor goes into a tank (condenser) that is hotter than the environment and therefore rejects heat into the environment. An expansion valve is opened very quickly and expands and gets cold. When the valve gets cold, it absorbs heat from the environment (evaporator).
If the heat sink is the outside, we have a …
heat pump
An AC unit where you can go from cooling to heating is an example of a …
heat pump
If a heat sink is food, we have a …
refrigerator
Describe a refrigeration cycle
↑ Q(high)
CONDENSER (Condensador)
↓ ↑
EXPANSION COMPRESSOR ← W(net, in)
VALVE (Compresor)
(válvula de ↑
expansión) → EVAPORATOR (Evaporador)
↑ Q(low)
FOOD
Fluid flows out of a compressor and into a …
condenser
Expansion valves work by ….
lowering the pressure in a vessel
In a heat pump or refrigerator, the sink is …
the location (source) where the heat was coming from
In a heat pump or refrigeration cycle, the net work in can be found by …
W(net, in) = Q(high) - Q(low)
The equation W(net, in) = Q(high) - Q(low) tells us …
how well a heat engine is working
We we talk about thermal efficiency of a heat pump or refrigerator, we are talking about …
the coefficient of performance, COP
The coefficient of performance, COP, is found by …
COP = Desired Output/Required Input
The coefficient of performance, COP, of a refrigerator is equal to …
COPR = Q(low)/W(net, in) = Desired Output/Required Input = Q(low)/[Q(high)-Q(low)]
For the coefficient of performance, COP, of a refrigerator, Q(low) and W(net, in) are …
Desired Output & Required Input respectively
The coefficient of performance, COP, of a heat pump is equal to …
COPP = Q(high)/W(net, in) = Q(high)/[Q(high)-Q(low)]
The typical range of values for the coefficient of performance, COP, of a heat pump is …
2 - 3, depending on the working fluid and design of the system
↑ Q(high)
CONDENSER (Condensador)
↓ ↑
EXPANSION COMPRESSOR ← W(net, in)
VALVE (Compresor)
(válvula de ↑
expansión) → EVAPORATOR (Evaporador)
↑ Q(low)
SINK (T, low)
Label the steps from 1 to 4.
↑ Q(high)
CONDENSER (Condensador)
3 ↓ ↑ 2
EXPANSION COMPRESSOR ← W(net, in)
VALVE (Compresor)
(válvula de 4 ↑ 1
expansión) → EVAPORATOR (Evaporador)
↑ Q(low)
SINK (T, low)
A throttling valve is a constant enthalpy process.
True
Draw the T-v diagram of a refrigeration/heat pump process within the Andrews curve, ∩
(EXP. VALVE) Q(high)
3 ₒ _________↑____________/ₒ 2 (CONDENSER)
/ ↘ ↑ ← W(in)
4 ₒ_____________________ₒ 1 (COMPRESSOR)
/ (EVAPORATOR) ↑ Q(low)
In the compressor process of a refrigeration/ heat pump cycle, we …
increase in temperature
(EXP. VALVE) Q(high)
3 ₒ _________↑____________/ₒ 2 (CONDENSER)
/ ↘ ↑ ← W(in)
4 ₒ_____________________ₒ 1 (COMPRESSOR)
/ (EVAPORATOR) ↑ Q(low)
In state 1 we ….
In state 2, we …
In state 4 we …
increase in temperature
are at a higher temperature than our environment and therefore reject heat to the environment
are at a lower temperature than our surroundings and are able to absorb heat
(EXP. VALVE) Q(high)
3 ₒ _________↑____________/ₒ 2 (CONDENSER)
/ ↘ ↑ ← W(in)
4 ₒ_____________________ₒ 1 (COMPRESSOR)
/ (EVAPORATOR) ↑ Q(low)
In the case of a heat pump, at step 2, where we would be rejecting heat, we …
heat the fluid
What is the Clausius statement?
No device can transfer heat from a cold body to a hot body without leaving an effect on the surroundings.
What is does the Clausius statement mean?
We do work when transferring heat from a cold body to a hot body; work is the effect on the surroundings.
compressor
condenser
expansion/throttling valve
evaporator
compresor
condensador
válvula de expansión / estrangulamiento
evaporador
