Lecture 4 - 1st Law Open System Flashcards
Forms of heat transfer
conduction
convection
radiation
Conduction is where we have …
heat transfer going through a solid, and the heat transfer itself is driven by a temperature gradient within the solid.
We express one dimensional conduction using Fourier’s Law, which states …
Q̇ = -kAdT/dx
where k is the thermal conductivity of heat
A = the cross sectional area that the heat is flowing through
dT/dx = the temperature gradient
Q̇ = heat transfer rate
In ______________ we have heat transfer between a solid and a fluid.
convective heat transfer
In convection, we have …
a flowing fluid carrying thermal energy within the fluid itself from the boundary of the system
The rate at which heat is transferred in convection is equal to …
Q̇ = hA(Ts - T∞)
where Ts is the surface temperature of the object that heat is entering or leaving
T∞ is the free stream temperature of the fluid flowing past the object
h = convective heat transfer coefficient for the particular case being studied
A = the cross sectional area that the heat is flowing through
Q̇ = heat transfer rate
In radiation heat transfer,
heat transfers without physical contact
Radiation is expressed as
Q̇ = εσA (Ts^4 - Tsurr^4)
Q̇ = heat transfer rate ε = emissivity coefficient σ = Boltzmann constant A = area Ts = surface temperature of the object expressed to the power of 4 (in Kelvin) Tsurr = temperature of the surroundings raised to the power of 4 (in Kelvin)
Forms of work
Boundary work Electrical work Gravitational work Shaft work Spring work
Work crossing the boundary itself by heat transfer is called …
Boundary work
In electrical work, heat is transferred through wires and there is a voltage difference. All of this is doing _____.
work
Electrical work is equal to ..
W = V*I*∆t ∆t = change inn time I = current flow V = voltage
Boundary work is expressed as
W = 1∫2 Pdv ∫P = change in pressure at which the boundary is moving into with respect to the volume dv = change in volume
In gravitational work, work is done …
through a potential field (gravitational) and a differential in position
Gravitational work is expressed as …
W = mg(h2-h1) = ∆Potential Energy
Acceleration work is when we …
change the velocity of the system
Acceleration work is equal to …
W = 1/2m(v2^2 - v1^2) = ∆kinetic energy of system
Shaft work is when …
you have a shaft going across the control boundary and the shaft is rotating at a given torque, thus doing work on the system
Propellers create _________ work and ____.
shaft work and heat
Shaft work
trabajo de eje
Propeller in spanish
hélice
el propulsor
Shaft work is expressed as …
W = pi*2*n*τ n = number of revolutions τ = torque exerted on shaft
In spring work, we do work by …
extending a spring
Spring work is expressed as …
W = 1/2 * k * (x2^2 - x1^2)
k = spring constant ∆x = change in position of the spring
The first law for a closed system states that …
Q - W = ∆U + ∆KE + ∆PE
In the 1st law for a closed system were the boundary changes, we can say that work consists of …
other work + boundary work
An example of a closed system where the boundary changes is …
a piston cylinder device where the piston can move up or down
A piston cylinder device where the piston can move up or down is doing _____ work.
boundary work
In the 1st law for a closed system, with respect to the types of work, can be written as …
Q - W (other) = ∆H+ ∆KE + ∆PE
where ∆H (enthalpy) = ∆U + W (boundary)
Enthalpy takes into account …
boundary work of the closed system the and internal energy
In boundary work, we …
encounter a pressure to move and change the volume
In a fixed mass problem involving a piston, we can account for the boundary work embedded in the system by using …
the first law with respect to enthalpy instead of internal energy
Specific heat is …
the energy required to raise the temperature of 1 kg of material (that we trying to find the specific heat for) by 1 degree celsius
What you need to take into account when you deal with specific heat at constant pressure and specific heat at constant volume (for a gas or liquid) is …
at constant pressure, the boundary can move, so we need to take into account boundary work
you have expansion when heating that you need to take into account
The specific heat at constant pressure and specific heat at constant volume is _________ for a soild.
the same
The difference between specific heat at constant pressure and specific heat at constant volume (for a gas or liquid) is …
Cv = (du/dT)v Cp = (dh/dT)p and we need to account for boundary work of the moving boundary
Cv = (du/dT)v
we put it in brackets with a v because
that is how we denote or show that it is at constant volume
An example of a constant volume system is …
A box with heat being added and the temperature being monitored. The box does not change shape or expand, so we call it at constant volume.
Cp = (dh/dT)p
we put it in brackets with a v because
that is how we denote or show that it is at constant pressure
An example of a system at constant pressure is …
a piston with a load on top (to maintain the pressure) and a gas inside with heat being added and the temperature being monitored.
As heat is added to a piston with a load on top (to maintain the pressure) and a gas inside, what happens is …
the piston lifts off the mechanical stocks and boundary work occurs with this lifting. In addition, the system increases in size, which has an impact on the specific heat being measured.
Mechanical stalks in spanish
tallos mecánicos
Enthalpy and internal energy are functions of __________ alone.
temperature
We can use integration to find the change in enthalpy and internal energy for an ideal gas. Demonstrate how.
Cp = dh/dT -> Cp(T)*dT = dh -> ∫Cp(T)*dT = ∫dh = H Cv = du/dT -> Cv(T)*dT = du -> ∫Cv(T)*dT = ∫du = U
When using integration to find the change in enthalpy and internal energy for an ideal gas, if the temperature change we are dealing with is > 200˚C, what are the approximation that can be made?
we can approximate using the average specific heats:
u2-u1 = Cv(avg)*(T2-T1) h2-h1 = Cp(avg)*(T2-T1)
where Cv(avg) and Cp(avg) can be approximated from Tables A-2, A-17, A-18, 19,20, 21, 22, 23, and A-27, depending on the substance
When using integration to find the change in enthalpy and internal energy for an ideal gas, if the temperature change we are dealing with is < 200˚C, what are the approximation that can be made?
We need to use Tablas A-17, A-18, 19,20, 21, 22, and 23, which would allow us to calculate the change in internal energy or enthalpy of an ideal gas with respect to temperature
The relationship between the specific heat at constant pressure and the specific heating at constant volume with the gas constant, R, is ….
Cp = Cv + R
The ratio of specific heat, k, is given by …
k = Cp/Cv
The ratio of specific heat, k, is __________ for monatomic gases and _________ for diatomic gases.
- 667 for monatomic
1. 4 for diatomic gases and air
In an open system, mass crosses the boundaries.
True
The first law of thermodynamics with respect to an open system with steady flow states that …
Q̇ - Ẇ = Σṁf(he(vf)^21/2 * ghf) - Σṁi(hi(vi)^21/2 * ghi)
or
Q̇ - Ẇ = Σṁe(enthalpykinetic energypotential energy) leaving the system - Σṁe(enthalpykinetic energypotential energy) leaving the control volume
where Σṁe is the mass flux leaving the system
Q̇ = rate of heat transfer
Ẇ = rate of work
Σṁi = mass flux leaving the control volume
The rate of heat transfer, Q̇, is positive when …
heat is transferred into the control volume
The rate of work being done by the system, Ẇ, is positive when …
it is being done out of the control volume
Ẇ = power - shaft and electrical or other types of work.
True
Work being done by the system is expansion work and means that work is done _______ of the system.
out
Work done by the system is __________. (Positive/Negative)
Heat in is ___________. (Positive/Negative)
Positive
Positive
∆h of a stream =
h(out) - h(in)
Depending on the temperature of the ideal gas, we can approximate ∆h of a stream as
h(out) - h(in) = Cp(avg)*(Tout - Tin)
h(out) - h(in) = Cp(avg)*(Tout - Tin) works for any kind of gas.
FALSE!
h(out) - h(in) = Cp(avg)*(Tout - Tin) is only appropriate for ideal gases with ∆T< 200˚C
For ideal gases < 200˚C, we need _______ to approximate the value of Cp.
the steam tables
Common single input/ single output devices (steady flow devices) that make the first law for open system simpler are …
Nozzles Diffusers Turbines Compressors Throttling valves Mixing chambers Heat exchangers
Nozzles work by …
taking pressure and changing into kinetic energy to accelerate the fluid
Nozzle in spanish
la boquilla
la tobera
el inyector
Diffuser in spanish is …
el difusor
la difusora
A diffuser works by ….
taking kinetic energy and converting it to pressure of the fluid by decelerating the fluid
Turbine in spanish
turbina
Compressor in spanish
compresor
Turbines work by ….
getting work out of the fluid and the fluid changes property or state as a result of the work
Compressors work by …
doing work on the fluid, and the fluid changes state, usually through an increase in pressure and temperature
Throttling valve in spanish
Válvula de estrangulación
A throttling valve is …
a valve where you have a pressure drop, and in the process, you lose energy by having some of the energy converted to thermal energy and no mechanical work being done.
Throttling valves are found in refrigeration cycles.
True
Mixing chamber in spanish is …
Cámara de mezclado
A mixing chamber works by …
taking two fluid streams and mixing them to create a single output
Heat exchanger in spanish
intercambiador de calor
el termocambiador
For steady flow devices (single input/ single output devices), the first law can be modified as follows:
Q̇ - Ẇ = ṁ∫∆h + ∆KE + ∆PE
where ∆KE = 1/2 * V2^2-V1^2
∆PE = g * height 2 - height 1
1 = inlet state
2 = exit state
