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Thermodinamica > Lecture 4 - 1st Law Open System > Flashcards

Lecture 4 - 1st Law Open System Flashcards

1
Q

Forms of heat transfer

A

conduction
convection
radiation

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2
Q

Conduction is where we have …

A

heat transfer going through a solid, and the heat transfer itself is driven by a temperature gradient within the solid.

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3
Q

We express one dimensional conduction using Fourier’s Law, which states …

A

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

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4
Q

In ______________ we have heat transfer between a solid and a fluid.

A

convective heat transfer

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5
Q

In convection, we have …

A

a flowing fluid carrying thermal energy within the fluid itself from the boundary of the system

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6
Q

The rate at which heat is transferred in convection is equal to …

A

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

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7
Q

In radiation heat transfer,

A

heat transfers without physical contact

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8
Q

Radiation is expressed as

A

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)
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9
Q

Forms of work

A 
Boundary work
Electrical work
Gravitational work
Shaft work
Spring work
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10
Q

Work crossing the boundary itself by heat transfer is called …

A

Boundary work

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11
Q

In electrical work, heat is transferred through wires and there is a voltage difference. All of this is doing _____.

A

work

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12
Q

Electrical work is equal to ..

A 
W = V*I*∆t
∆t = change inn time
I = current flow
V = voltage
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13
Q

Boundary work is expressed as

A 
W = 1∫2 Pdv
∫P = change in pressure at which the boundary is moving into with respect to the volume
dv = change in volume
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14
Q

In gravitational work, work is done …

A

through a potential field (gravitational) and a differential in position

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15
Q

Gravitational work is expressed as …

A

W = mg(h2-h1) = ∆Potential Energy

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16
Q

Acceleration work is when we …

A

change the velocity of the system

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17
Q

Acceleration work is equal to …

A

W = 1/2m(v2^2 - v1^2) = ∆kinetic energy of system

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18
Q

Shaft work is when …

A

you have a shaft going across the control boundary and the shaft is rotating at a given torque, thus doing work on the system

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19
Q

Propellers create _________ work and ____.

A

shaft work and heat

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20
Q

Shaft work

A

trabajo de eje

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21
Q

Propeller in spanish

A

hélice

el propulsor

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22
Q

Shaft work is expressed as …

A 
W = pi*2*n*τ
n = number of revolutions
τ = torque exerted on shaft
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23
Q

In spring work, we do work by …

A

extending a spring

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24
Q

Spring work is expressed as …

A

W = 1/2 * k * (x2^2 - x1^2)

k = spring constant
∆x = change in position of the spring
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25
The first law for a closed system states that ...
Q - W = ∆U + ∆KE + ∆PE
26
In the 1st law for a closed system were the boundary changes, we can say that work consists of ...
other work + boundary work
27
An example of a closed system where the boundary changes is ...
a piston cylinder device where the piston can move up or down
28
A piston cylinder device where the piston can move up or down is doing _____ work.
boundary work
29
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)
30
Enthalpy takes into account ...
boundary work of the closed system the and internal energy
31
In boundary work, we ...
encounter a pressure to move and change the volume
32
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
33
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
34
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
35
The specific heat at constant pressure and specific heat at constant volume is _________ for a soild.
the same
36
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 ```
37
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
38
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.
39
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
40
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.
41
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.
42
Mechanical stalks in spanish
tallos mecánicos
43
Enthalpy and internal energy are functions of __________ alone.
temperature
44
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 ```
45
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
46
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
47
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
48
The ratio of specific heat, k, is given by ...
k = Cp/Cv
49
The ratio of specific heat, k, is __________ for monatomic gases and _________ for diatomic gases.
1. 667 for monatomic | 1. 4 for diatomic gases and air
50
In an open system, mass crosses the boundaries.
True
51
The first law of thermodynamics with respect to an open system with steady flow states that ...
Q̇ - Ẇ = Σṁf(he*(vf)^2*1/2 * ghf) - Σṁi(hi*(vi)^2*1/2 * ghi) or Q̇ - Ẇ = Σṁe(enthalpy*kinetic energy*potential energy) leaving the system - Σṁe(enthalpy*kinetic energy*potential 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
52
The rate of heat transfer, Q̇, is positive when ...
heat is transferred into the control volume
53
The rate of work being done by the system, Ẇ, is positive when ...
it is being done out of the control volume
54
Ẇ = power - shaft and electrical or other types of work.
True
55
Work being done by the system is expansion work and means that work is done _______ of the system.
out
56
Work done by the system is __________. (Positive/Negative) | Heat in is ___________. (Positive/Negative)
Positive | Positive
57
∆h of a stream =
h(out) - h(in)
58
Depending on the temperature of the ideal gas, we can approximate ∆h of a stream as
h(out) - h(in) = Cp(avg)*(Tout - Tin)
59
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
60
For ideal gases < 200˚C, we need _______ to approximate the value of Cp.
the steam tables
61
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 ```
62
Nozzles work by ...
taking pressure and changing into kinetic energy to accelerate the fluid
63
Nozzle in spanish
la boquilla la tobera el inyector
64
Diffuser in spanish is ...
el difusor | la difusora
65
A diffuser works by ....
taking kinetic energy and converting it to pressure of the fluid by decelerating the fluid
66
Turbine in spanish
turbina
67
Compressor in spanish
compresor
68
Turbines work by ....
getting work out of the fluid and the fluid changes property or state as a result of the work
69
Compressors work by ...
doing work on the fluid, and the fluid changes state, usually through an increase in pressure and temperature
70
Throttling valve in spanish
Válvula de estrangulación
71
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.
72
Throttling valves are found in refrigeration cycles.
True
73
Mixing chamber in spanish is ...
Cámara de mezclado
74
A mixing chamber works by ...
taking two fluid streams and mixing them to create a single output
75
Heat exchanger in spanish
intercambiador de calor | el termocambiador
76
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

Thermodinamica (10 decks)

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