Lecture 4 - 1st Law Open System Flashcards

1
Q

Forms of heat transfer

A

conduction
convection
radiation

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

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

A

convective heat transfer

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

In convection, we have …

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

In radiation heat transfer,

A

heat transfers without physical contact

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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)
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

Forms of work

A
Boundary work
Electrical work
Gravitational work
Shaft work
Spring work
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

Work crossing the boundary itself by heat transfer is called …

A

Boundary work

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

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

A

work

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

Electrical work is equal to ..

A
W = V*I*∆t
∆t = change inn time
I = current flow
V = voltage
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

In gravitational work, work is done …

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

Gravitational work is expressed as …

A

W = mg(h2-h1) = ∆Potential Energy

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

Acceleration work is when we …

A

change the velocity of the system

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

Acceleration work is equal to …

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

Propellers create _________ work and ____.

A

shaft work and heat

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

Shaft work

A

trabajo de eje

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q

Propeller in spanish

A

hélice

el propulsor

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

Shaft work is expressed as …

A
W = pi*2*n*τ
n = number of revolutions
τ = torque exerted on shaft
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

In spring work, we do work by …

A

extending a spring

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
25
Q

The first law for a closed system states that …

A

Q - W = ∆U + ∆KE + ∆PE

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
26
Q

In the 1st law for a closed system were the boundary changes, we can say that work consists of …

A

other work + boundary work

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
27
Q

An example of a closed system where the boundary changes is …

A

a piston cylinder device where the piston can move up or down

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
28
Q

A piston cylinder device where the piston can move up or down is doing _____ work.

A

boundary work

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
29
Q

In the 1st law for a closed system, with respect to the types of work, can be written as …

A

Q - W (other) = ∆H+ ∆KE + ∆PE

where ∆H (enthalpy) = ∆U + W (boundary)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
30
Q

Enthalpy takes into account …

A

boundary work of the closed system the and internal energy

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
31
Q

In boundary work, we …

A

encounter a pressure to move and change the volume

32
Q

In a fixed mass problem involving a piston, we can account for the boundary work embedded in the system by using …

A

the first law with respect to enthalpy instead of internal energy

33
Q

Specific heat is …

A

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
Q

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 …

A

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
Q

The specific heat at constant pressure and specific heat at constant volume is _________ for a soild.

A

the same

36
Q

The difference between specific heat at constant pressure and specific heat at constant volume (for a gas or liquid) is …

A
Cv = (du/dT)v
Cp = (dh/dT)p and we need to account for boundary work of the moving boundary
37
Q

Cv = (du/dT)v

we put it in brackets with a v because

A

that is how we denote or show that it is at constant volume

38
Q

An example of a constant volume system is …

A

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
Q

Cp = (dh/dT)p

we put it in brackets with a v because

A

that is how we denote or show that it is at constant pressure

40
Q

An example of a system at constant pressure is …

A

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
Q

As heat is added to a piston with a load on top (to maintain the pressure) and a gas inside, what happens is …

A

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
Q

Mechanical stalks in spanish

A

tallos mecánicos

43
Q

Enthalpy and internal energy are functions of __________ alone.

A

temperature

44
Q

We can use integration to find the change in enthalpy and internal energy for an ideal gas. Demonstrate how.

A
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
Q

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?

A

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
Q

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?

A

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
Q

The relationship between the specific heat at constant pressure and the specific heating at constant volume with the gas constant, R, is ….

A

Cp = Cv + R

48
Q

The ratio of specific heat, k, is given by …

A

k = Cp/Cv

49
Q

The ratio of specific heat, k, is __________ for monatomic gases and _________ for diatomic gases.

A
  1. 667 for monatomic

1. 4 for diatomic gases and air

50
Q

In an open system, mass crosses the boundaries.

A

True

51
Q

The first law of thermodynamics with respect to an open system with steady flow states that …

A

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

52
Q

The rate of heat transfer, Q̇, is positive when …

A

heat is transferred into the control volume

53
Q

The rate of work being done by the system, Ẇ, is positive when …

A

it is being done out of the control volume

54
Q

Ẇ = power - shaft and electrical or other types of work.

A

True

55
Q

Work being done by the system is expansion work and means that work is done _______ of the system.

A

out

56
Q

Work done by the system is __________. (Positive/Negative)

Heat in is ___________. (Positive/Negative)

A

Positive

Positive

57
Q

∆h of a stream =

A

h(out) - h(in)

58
Q

Depending on the temperature of the ideal gas, we can approximate ∆h of a stream as

A

h(out) - h(in) = Cp(avg)*(Tout - Tin)

59
Q

h(out) - h(in) = Cp(avg)*(Tout - Tin) works for any kind of gas.

A

FALSE!

h(out) - h(in) = Cp(avg)*(Tout - Tin) is only appropriate for ideal gases with ∆T< 200˚C

60
Q

For ideal gases < 200˚C, we need _______ to approximate the value of Cp.

A

the steam tables

61
Q

Common single input/ single output devices (steady flow devices) that make the first law for open system simpler are …

A
Nozzles
Diffusers
Turbines
Compressors
Throttling valves
Mixing chambers
Heat exchangers
62
Q

Nozzles work by …

A

taking pressure and changing into kinetic energy to accelerate the fluid

63
Q

Nozzle in spanish

A

la boquilla
la tobera
el inyector

64
Q

Diffuser in spanish is …

A

el difusor

la difusora

65
Q

A diffuser works by ….

A

taking kinetic energy and converting it to pressure of the fluid by decelerating the fluid

66
Q

Turbine in spanish

A

turbina

67
Q

Compressor in spanish

A

compresor

68
Q

Turbines work by ….

A

getting work out of the fluid and the fluid changes property or state as a result of the work

69
Q

Compressors work by …

A

doing work on the fluid, and the fluid changes state, usually through an increase in pressure and temperature

70
Q

Throttling valve in spanish

A

Válvula de estrangulación

71
Q

A throttling valve is …

A

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
Q

Throttling valves are found in refrigeration cycles.

A

True

73
Q

Mixing chamber in spanish is …

A

Cámara de mezclado

74
Q

A mixing chamber works by …

A

taking two fluid streams and mixing them to create a single output

75
Q

Heat exchanger in spanish

A

intercambiador de calor

el termocambiador

76
Q

For steady flow devices (single input/ single output devices), the first law can be modified as follows:

A

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