Tải bản đầy đủ
2 Temperature, work and heat

2 Temperature, work and heat

Tải bản đầy đủ

2 Refrigeration and Air-Conditioning
Hot reservoir, T1
Q1

Q1
W
E

P

Q2

Q2
Cold reservoir, T0

Figure 1.1 Ideal heat engine, E, driving an ideal refrigerator (heat pump), P

temperature without any external work input, which is impossible. The relationship between Q1, Q2 and W depends only on the temperatures of the hot
and cold reservoirs. The French physicist Sadi Carnot (1796–1832) was the
first to predict that the relationship between work and heat is temperaturedependent, and the ideal refrigeration process is known as the Carnot cycle.
In order to find this relationship, temperature must be defined in a more fundamental way. The degrees on the thermometer are only an arbitrary scale.
Kelvin (1824–1907), together with other leading physicists of the period,
concluded that an absolute temperature scale can be defined in terms of the
efficiency of reversible engines.

Figure 1.2 William Thomson, appointed to the chair of natural philosophy at Glasgow
University, aged 22, published his paper on the absolute temperature scale two years later.
He became Lord Kelvin in 1892 (Glasgow University)

Ch001-H8519.indd 2

5/17/2008 2:39:23 PM

Fundamentals 3
The ideal ‘never-attainable-in-practice’ ratio of work output to heat input
(W/Q1) of the reversible engine E equals: Temperature Difference (T1 Ϫ T0)
divided by the Hot Reservoir Temperature (T1).
In Figure 1.1 the device P can be any refrigeration device we care to invent,
and the work of Kelvin tells us that the minimum work, W necessary to lift a
quantity of heat Q2 from temperature T0 to temperature T1 is given by:
W ϭ

Q2 (T1 Ϫ T0 )
T0

The temperatures must be measured on an absolute scale i.e. one that starts
at absolute zero. The Kelvin scale has the same degree intervals as the Celsius
scale, so that ice melts at ϩ273.16 K, and water at atmospheric pressure boils
at ϩ373.15 K. On the Celsius scale, absolute zero is –273.15°C. Refrigeration
‘efficiency’ is usually defined as the heat extracted divided by the work input.
This is called COP, coefficient of performance. The ideal or Carnot COP takes
its name from Sadi Carnot and is given by:
COP ϭ

T0
Q2
ϭ
(T1 Ϫ T0 )
W

E x a m p l e 1. 1
Heat is to be removed at a temperature of Ϫ5°C and rejected at a temperature of 35°C.
What is the Carnot or Ideal COP?
Convert the temperatures to absolute:
Ϫ5°C becomes 268 K and 35°C becomes 308 K (to the nearest K)
Carnot COP ϭ

268
ϭ 6.7
(308 Ϫ 268)

1.3 HEAT
Heat is one of the many forms of energy and is commonly generated from
chemical sources. The heat of a body is its thermal or internal energy, and a
change in this energy may show as a change of temperature or a change between
the solid, liquid and gaseous states.
Matter may also have other forms of energy, potential or kinetic, depending
on pressure, position and movement. Enthalpy is the sum of its internal energy
and flow work and is given by:
H ϭ u ϩ Pv
In the process where there is steady flow, the factor Pv will not change appreciably and the difference in enthalpy will be the quantity of heat gained or lost.

Ch001-H8519.indd 3

5/17/2008 2:39:23 PM

4 Refrigeration and Air-Conditioning
Enthalpy may be expressed as a total above absolute zero, or any other
base which is convenient. Tabulated enthalpies found in reference works are
often shown above a base temperature of Ϫ40°C, since this is also Ϫ40° on the
old Fahrenheit scale. In any calculation, this base condition should always be
checked to avoid the errors which will arise if two different bases are used.
If a change of enthalpy can be sensed as a change of temperature, it is called
sensible heat. This is expressed as specific heat capacity, i.e. the change in
enthalpy per degree of temperature change, in kJ/(kg K). If there is no change
of temperature but a change of state (solid to liquid, liquid to gas, or vice versa)
it is called latent heat. This is expressed as kJ/kg but it varies with the boiling
temperature, and so is usually qualified by this condition. The resulting total
changes can be shown on a temperature–enthalpy diagram (Figure 1.3).

Temperature

Sensible heat of gas

373.15 K

Latent
heat of
melting

Latent heat of boiling

Sensible heat of liquid

273.16 K

Sensible heat of solid
334 kJ 419 kJ

2257 kJ

Enthalpy

Figure 1.3 Change of temperature (K) and state of water with enthalpy

E x a m p l e 1. 2
The specific enthalpy of water at 80°C, taken from 0°C base, is 334.91 kJ/kg. What is
the average specific heat capacity through the range 0–80°C?
334.91
ϭ 4.186 kJ/(kg K)
(80 Ϫ 0)

E x a m p l e 1. 3
If the latent heat of boiling water at 1.013 bar is 2257 kJ/kg, the quantity of heat which
must be added to 1 kg of water at 30°C in order to boil it is:
4.19 (100 Ϫ 30) ϩ 2257 ϭ 2550.3 kJ

Ch001-H8519.indd 4

5/17/2008 2:39:24 PM

Fundamentals 5

1.4 BOILING POINT
The temperature at which a liquid boils is not constant, but varies with the
pressure. Thus, while the boiling point of water is commonly taken as 100°C,
this is only true at a pressure of one standard atmosphere (1.013 bar) and,
by varying the pressure, the boiling point can be changed (Table 1.1). This
pressure–temperature property can be shown graphically (see Figure 1.4).

Table 1.1
Pressure (bar)

Boiling point (°C)

0.006

0

0.04

29

0.08

41.5

0.2

60.1

0.5

81.4

1.013

100.0

Critical
temperature
e

rv

Solid

Pressure

Liquid
cu
nt

i

g
ilin

po

Bo

Gas

Triple
point
Temperature

Figure 1.4 Change of state with pressure and temperature

The boiling point of a substance is limited by the critical temperature at the
upper end, beyond which it cannot exist as a liquid, and by the triple point at
the lower end, which is at the freezing temperature. Between these two limits, if
the liquid is at a pressure higher than its boiling pressure, it will remain a liquid
and will be subcooled below the saturation condition, while if the temperature

Ch001-H8519.indd 5

5/17/2008 2:39:24 PM

6 Refrigeration and Air-Conditioning
is higher than saturation, it will be a gas and superheated. If both liquid and
vapour are at rest in the same enclosure, and no other volatile substance is
present, the condition must lie on the saturation line.
At a pressure below the triple point pressure, the solid can change directly
to a gas (sublimation) and the gas can change directly to a solid, as in the formation of carbon dioxide snow from the released gas.
The liquid zone to the left of the boiling point line is subcooled liquid. In
refrigeration the term saturation is used to describe the liquid/vapour boundary,
saturated vapour being represented by a condition on the line and superheated
vapour below the line. More information on saturated properties for commonly
used refrigerants is given in Chapter 3.

1.5 GENERAL GAS LAWS
Many gases at low pressure, i.e. atmospheric pressure and below for water
vapour and up to several bar for gases such as nitrogen, oxygen and argon,
obey simple relations between their pressure, volume and temperature, with
sufficient accuracy for engineering purposes. Such gases are called ‘ideal’.
Boyle’s Law states that, for an ideal gas, the product of pressure and volume
at constant temperature is a constant:
pV ϭ constant

E x a m p l e 1. 4
A volume of an ideal gas in a cylinder and at atmospheric pressure is compressed to
half the volume at constant temperature. What is the new pressure?
p1V1 ϭ constant
ϭ p 2V2
V1
ϭ2
V2
so

p 2 ϭ 2 ϫ p1
ϭ 2 ϫ 1.013 25 bar (101 325 Pa)
ϭ 2.0265 bar (abs.)

Charles’ Law states that, for an ideal gas, the volume at constant pressure is
proportional to the absolute temperature:
V
ϭ constant
T

Ch001-H8519.indd 6

5/17/2008 2:39:24 PM

Fundamentals 7

E x a m p l e 1. 5
A mass of an ideal gas occupies 0.75 m3 at 20°C and is heated at constant pressure to
90°C. What is the final volume?
V2 ϭ V1 ϫ

T2
T1

= 0.75 ϫ

273 ϩ 90
(temperatures to the nearesst K)
273 ϩ 20

ϭ 0.93 m3
Boyle’s and Charles’ laws can be combined into the ideal gas equation:
pV ϭ (a constant) ϫ T
The constant is mass ϫ R, where R is the specific gas constant, so:
pV ϭ mRT

E x a m p l e 1. 6
What is the volume of 5 kg of an ideal gas, having a specific gas constant of 287 J/(kg K),
at a pressure of one standard atmosphere and at 25°C?
pV ϭ mRT
mRT

p
5 ϫ 287(273 ϩ 25)
ϭ
101 325
ϭ 4.22 m3

1.6 DALTON’S LAW
Dalton’s Law of partial pressures considers a mixture of two or more gases,
and states that the total pressure of the mixture is equal to the sum of the individual pressures, if each gas separately occupied the space.

E x a m p l e 1. 7
A cubic metre of air contains 0.906 kg of nitrogen of specific gas constant 297 J/(kg K),
0.278 kg of oxygen of specific gas constant 260 J/(kg K) and 0.015 kg of argon of specific
gas constant 208 J/(kg K). What will be the total pressure at 20°C?
pV ϭ mRT
V ϭ 1 m3
p ϭ mRT
so

Ch001-H8519.indd 7

5/17/2008 2:39:24 PM

8 Refrigeration and Air-Conditioning
For the nitrogen
For the oxygen
For the argon

pN ϭ 0.906 ϫ 297 ϫ 293.15 ϭ 78 881 Pa
pO ϭ 0.278 ϫ 260 ϫ 293.15 ϭ 21 189 Pa
915 Pa
pA ϭ 0.015 ϫ 208 ϫ 293.15 ϭ
Total pressure ϭ 100 985 Pa
(1.009 85 bar)

The properties of refrigerant fluids at the pressures and temperatures of interest to refrigeration engineers exhibit considerable deviation from the ideal gas
laws. It is therefore necessary to use tabulated or computer-based information
for thermodynamic calculations.

1.7 HEAT TRANSFER
Heat will move from a hot body to a colder one, and can do so by the following
methods:
1. Conduction. Direct from one body touching the other, or through a
continuous mass
2. Convection. By means of a heat-carrying fluid moving between one and
the other
3. Radiation. Mainly by infrared waves (but also in the visible band,
e.g. solar radiation), which are independent of contact or an
intermediate fluid.
Conduction through a homogeneous material is expressed directly by its
area, thickness and a conduction coefficient. For a large plane surface, ignoring
heat transfer near the edges:
area ϫ thermal conductivity
thickness
Aϫk
ϭ
L

Conductance ϭ

and the heat conducted is
Q f ϭ conductance ϫ (T1 Ϫ T2 )

E x a m p l e 1. 8
A brick wall, 225 mm thick and having a thermal conductivity of 0.60 W/(m K), measures
10 m long by 3 m high, and has a temperature difference between the inside and
outside faces of 25 K. What is the rate of heat conduction?
10 ϫ 3 ϫ 0.60 ϫ 25
0.225
ϭ 2000 W (or 2 kW)

Qf ϭ

Ch001-H8519.indd 8

5/17/2008 2:39:24 PM

Fundamentals 9
Thermal conductivities, in watts per metre Kelvin, for various common materials are as in Table 1.2. Conductivities for other materials can be found from
standard reference works.

Table 1.2
Material

Thermal conductivity
(W/(m K))

Copper

200

Mild steel

50

Concrete

1.5

Water

0.62

Cork

0.040

Expanded polystyrene

0.034

Polyurethane foam

0.026

Still air

0.026

Convection requires a fluid, either liquid or gaseous, which is free to move
between the hot and cold bodies. This mode of heat transfer is complex and
depends firstly on whether the flow of fluid is ‘natural’, i.e. caused by thermal
currents set up in the fluid as it expands, or ‘forced’ by fans or pumps. Other
parameters are the density, specific heat capacity and viscosity of the fluid and
the shape of the interacting surface.
With so many variables, expressions for convective heat flow cannot be as
simple as those for conduction. The interpretation of observed data has been
made possible by the use of a number of dimensionless groups which combine
the variables and which can then be used to estimate convective heat flow.
The main groups used in such estimates are as shown in Table 1.3. A typical combination of these numbers is that for turbulent flow in pipes expressing
the heat transfer rate in terms of the flow characteristic and fluid properties:
Nu ϭ 0.023 (Re)0.8 (Pr)0.4
The calculation of every heat transfer coefficient for a refrigeration or airconditioning system would be a very time-consuming process, even with modern
methods of calculation. Formulas based on these factors will be found in standard
reference works, expressed in terms of heat transfer coefficients under different
conditions of fluid flow.

Ch001-H8519.indd 9

5/17/2008 2:39:25 PM

10 Refrigeration and Air-Conditioning
Table 1.3
Group

Symbol

Reynolds

Re

ρvx
μ

Velocity of fluid, v
Density of fluid, ρ
Viscosity of fluid, μ
Dimension of surface, x

Forced flow in
pipes

Nusselt

Nu

hx
k

Thermal conductivity of fluid, k
Dimension of surface, x
Heat transfer coefficient, h

Convection heat
transfer rate

Prandtl

Pr

Specific heat capacity of fluid, Cp
Viscosity of fluid, μ
Thermal conductivity of fluid, k

Fluid properties

Coefficient of expansion of fluid, β
Density of fluid, ρ
Viscosity of fluid, μ
Force of gravity, g
Temperature difference, θ
Dimension of surface, x

Natural convection

Cp μ
k

Grashof

Gr

β g ρ 2 x 3θ
μ2

Parameters

Typical
Relevance

Number

Where heat is conducted through a plane solid which is between two fluids,
there will be the convective resistances at the surfaces. The overall heat transfer must take all of these resistances into account, and the unit transmittance,
or ‘U’ value is given by:
Rt ϭ Ri ϩ Rc ϩ Ro
U ϭ 1/ Rt
where Rt ϭ total thermal resistance
Ri ϭ inside convective resistance
Rc ϭ conductive resistacne
RO ϭ outside convective resistance

E x a m p l e 1. 9
A brick wall, plastered on one face, has a thermal conductance of 2.8 W/(m2 K), an
inside surface resistance of 0.3 (m2 K)/W, and an outside surface resistance of 0.05
(m2 K)/W. What is the overall transmittance?

Ch001-H8519.indd 10

5/17/2008 2:39:25 PM

Fundamentals 11
Rt ϭ Ri ϩ Rc ϩ Ro
1
ϭ 0.3 ϩ
ϩ 0.05
2.8
ϭ 0.707
U ϭ 1.414 W/(m2 K)

Typical overall thermal transmittances are:
Insulated cavity brick wall, 260 mm thick, sheltered
exposure on outside
Chilled water inside copper tube, forced draught
air flow outside
Condensing ammonia gas inside steel tube, thin
film of water outside

0.69 W/(m2 K)
15–28 W/(m2 K)
450–470 W/(m2 K)

Special note should be taken of the influence of geometrical shape, where
other than plain surfaces are involved.
The overall thermal transmittance, U, is used to calculate the total heat flow.
For a plane surface of area A and a steady temperature difference ΔT, it is
Q f ϭ A ϫ U ϫ ΔT
If a non-volatile fluid is being heated or cooled, the sensible heat will change and
therefore the temperature, so that the ΔT across the heat exchanger wall will not
be constant. Since the rate of temperature change (heat flow) will be proportional
to the ΔT at any one point, the space–temperature curve will be exponential. In
a case where the cooling medium is an evaporating liquid, the temperature of
this liquid will remain substantially constant throughout the process, since it is
absorbing latent heat, and the cooling curve will be as shown in Figure 1.5.

TA

Co

ole

d

m

ed

ium

Ra
ch te of
an te
ge mp
era
tur
ΔT
e

ΔTmax

TB

In

Cooling medium

ΔTmin

Out

Figure 1.5 Changing temperature difference of a cooled fluid

Ch001-H8519.indd 11

5/17/2008 2:39:25 PM

12 Refrigeration and Air-Conditioning
Providing that the flow rates are steady, the heat transfer coefficients do not
vary and the specific heat capacities are constant throughout the working range,
the average temperature difference over the length of the curve is given by:
ΔT ϭ

ΔTmax Ϫ ΔTmin
ln (ΔTmax / ΔTmin )

This is applicable to any heat transfer where either or both the media change in
temperature (see Figure 1.6). This derived term is the logarithmic mean temperature difference (LMTD) and can be used as ΔT in the general equation, providing
U is constant throughout the cooling range, or an average figure is known, giving
Q f ϭ A ϫ U ϫ LMTD

TA in

TR
Ai

r

TR

TA out
ΔTmin

refrigerant
(a)

TA in
ΔTmin
Ai

r

Tw out

ΔTmax

ΔTmax
Evaporating

Condensing
refrigerant

r

te
Wa

ΔTmax
Tw out

TA out
Water

ΔTmin
Tw in

Tw in
(b)

(c)

Figure 1.6 Temperature change. (a) Refrigerant cooling fluid. (b) Fluid cooling refrigerant.
(c) Two fluids

E x a m p l e 1. 10
A fluid evaporates at 3°C and cools water from 11.5°C to 6.4°C. What is the logarithmic
mean temperature difference and what is the heat transfer if it has a surface area of
420 m2 and the thermal transmittance is 110 W/(m2 K)?
ΔTmax ϭ 11.5 Ϫ 3 ϭ 8.5 K
ΔTmin ϭ 6.4 Ϫ 3 ϭ 3.4 K
8.5 Ϫ 3.4
LMTD ϭ
ln(8.5 / 3.4)
ϭ 5.566 K
Qf ϭ 420 ϫ 110 ϫ 5.566
ϭ 257 000 W or 257 kW
In practice, many of these values will vary. A pressure drop along a pipe
carrying boiling or condensing fluid will cause a change in the saturation temperature. With some liquids, the heat transfer values will change with temperature. For these reasons, the LMTD formula does not apply accurately to all
heat transfer applications.

Ch001-H8519.indd 12

5/17/2008 2:39:25 PM