Heat machines

4. Heat machines#

4.1. Introduction#

Flow machines transfer energy to a flowing medium or extract it using mechanical energy. Heat machines also transfer and convert energy, but thermal energy plays a central role. This may involve adding heat (e.g. through combustion), removing heat (e.g. in a refrigerator), or transferring heat from one temperature level to another (e.g. with a heat pump). Our current energy supply is largely based on heat machines such as engines and gas turbines. Efficient use of energy is increasingly important. Existing fossil energy sources must be handled with the utmost care and, where possible, forms of renewable energy must be developed. In this chapter we deal with all processes in which thermal energy plays a role. Two main areas can be distinguished, namely energy storage and transport, and thermodynamics. Energy storage and transport play a particularly important role in the process industry, where substances must be cooled and heated. This is possible in large-scale installations, such as a waste incineration plant where the heat in flue gases is recovered via heat exchangers and is reused for heating houses or greenhouses.

Thermodynamics is used in the design of systems in which cyclic processes take place. Consider, for example, a refrigerator in which the cooling medium is pumped around and absorbs heat in the cooling space and releases heat in the rear radiator. Thermodynamics also plays a major role in drive train technology. Just think of the gasoline engine, the diesel engine, but also the aircraft engine and the rocket engine.

In this chapter you will become acquainted with some principles of thermal energy storage and heat transfer and with cyclic processes in thermodynamics.

4.2. Storage of heat#

When we add or remove heat from an object, the internal energy of that object changes.

If we assume that the object consists of a solid or a liquid, the following can happen:

  • A phase change occurs; the solid melts or the liquid evaporates. In either case, the temperature remains unchanged, or:

  • No phase transition occurs; the temperature of the object then increases or decreases.

In the latter case, the following applies to the temperature change:

(4.1)#\[ \Delta Q = m \cdot c \cdot \Delta T \]

In which:

  • \(\Delta Q\) is the added or removed heat in [J],

  • \(m\) is the mass of the object in [kg],

  • \(c\) is the specific heat capacity in [J kg\(^{-1}\) K\(^{-1}\)] and

  • \(\Delta T\) is the temperature change in [K] or [°C]

Instead of using the specific heat capacity \(c\), the heat capacity \(C\) of an object is also often used. Especially if the object consists of different materials. The heat capacity of an object determines how much energy must be supplied for a temperature rise of 1 K. In equation form:

(4.2)#\[ \Delta Q = C \cdot \Delta T \quad \textrm{[J]} = [\textrm{JK}^{-1}]\textrm{[K]}\]

For 1 kg of water, \(C = m c_{\textrm{water}} = 1 \cdot 4180 = 4180\,\textrm{J/K}\).

Heat storage tank

We use heat storage tanks to store excess heat that we can use later. We approach the storage tanks as cylindrical (see Fig. 4.1). The following applies to the dimensions of the storage tanks: \(h = 5.0\) m, \(R = 2.0\) m, and \(d = 3.0\) cm. The storage tanks are made of aluminum. Furthermore, the following is given: \(c_{\textrm{aluminium}} = 880 \, \textrm{J/(kg\,K)}\), \(c_\textrm{water} = 4180 \, \textrm{J/(kg\,K)}\).

For safety reasons, the temperature of the vessel must remain between 20 °C and 55 °C. What is the maximum amount of heat that can be stored in the vessel? Disregard heat losses to the outside world or assume that the storage tank is perfectly insulated.

_images/Ch6_Heat_Machines_6.1.png

Fig. 4.1 (a) Heat storage tanks (b) schematic representation of the storage tank with dimensions.#

The heat capacity of the tanks is equal to the heat capacity of the water plus the heat capacity of the tank wall:

(4.3)#\[\begin{split} C &= C_{\textrm{water}} + C_{\textrm{wall}} = m_{\textrm{water}} \cdot c_{\textrm{water}} + m_{\textrm{wall}} \cdot c_{\textrm{aluminium}} \\ m_{\textrm{water}} &= \rho_{\textrm{water}}V = \rho_{\textrm{water}} \cdot \pi(R-d)^2 \cdot h = 61.0 \cdot 10^3 \, \textrm{kg} \\ m_{\textrm{wall}} &= \rho_{\textrm{alu}}V = \rho_{\textrm{alu}} \cdot \left(\pi R^2 \cdot d + \left(\pi R^2 h - \pi (R-d)^2 h\right) + \pi R^2 \cdot d\right) \\ &= 2700 \left[2 \pi R^2 d + \pi \left(R^2-(R-d)^2\right)h\right] \approx 7.17 \cdot 10^3 \, \textrm{kg} \\ C &= m_{\textrm{water}} \cdot c_{\textrm{water}} + m_{\textrm{wall}} \cdot c_{\textrm{aluminium}} = 2.55 \cdot 10^8 + 6.31 \cdot 10^6 \approx 2.61 \cdot 10^8 \, \textrm{J/K} \end{split}\]

You can therefore store in a tank:

(4.4)#\[ \Delta Q = C \cdot \Delta T = 2.61 \cdot 10^8 (55-20) = 9.15 \cdot 10^9 \, \textrm{J} \]

Note

The contribution of the aluminum tank to the total heat capacity is low, namely 2.5%.

This is on the one hand because the mass of the aluminum casing is much less than that of the water.

Moreover, the specific heat of that aluminum is also much less than that of the water.

4.3. Heat transfer#

There are three basic mechanisms by which heat can be transferred, namely conduction, flow (convection) and radiation.

_images/Ch6_Heat_Machines_6.2.png

Fig. 4.2 (a) Conduction, (b) convection, (c) radiation.#

If you put a metal spoon in a pan filled with water, the whole spoon will become warm at some point. The heat is transported through the spoon through conduction. If you hold your hand next to the pan you will feel heat radiation. And if you keep your hand above the pan, you will feel heat transfer by convection from the rising warm air and vapor.

4.3.1. Heat flow by conduction#

If there is a temperature difference over a medium, the heat will flow from the area with the higher temperature to that with a lower temperature. In the context of this course, we only consider one-dimensional heat flow, so only in the \(x\) direction, as displayed in Fig. 4.3

_images/Ch6_Heat_Machines_6.3.png

Fig. 4.3 Heat flow by conduction.#

The heat-transfer rate through a medium is mathematically described by Fourier’s law:

(4.5)#\[ \dot{Q} = -k A \frac{dT}{dx} \quad [\textrm{W}] \]

In which:

  • \(\dot{Q}_{\textrm{rad}}\) is the emitted radiative power in [W],

  • \(\frac{dT}{dx}\) is the temperature gradient in the medium [K/m] or in [°C/m],

  • \(A\) is the area through which heat flows in [m\(^2\)] and

  • \(k\) is the thermal conductivity in [W/(m K)].

Joseph Fourier (1768-1830) [1]

Jean-Baptiste Joseph Fourier (21 March 1768 – 16 May 1830) was a French mathematician and physicist born in Auxerre and best known for initiating the investigation of Fourier series, which eventually developed into Fourier analysis and harmonic analysis, and their applications to problems of heat transfer and vibrations. The Fourier transform and Fourier’s law of conduction are also named in his honour. Fourier is also generally credited with the discovery of the greenhouse effect.

Fourier

Fig. 4.4 Portrait of Joseph Fourier.#

The minus sign indicates that the heat flow is directed from a high to a lower temperature. If the temperature decreases in the positive x direction, as is the case in Fig. 4.3, the temperature gradient dT/dx is negative and the heat flow (in the positive x direction) is positive.

The thermal conductivity \(k\) depends on the material. A good conductive material, such as silver or copper, has a high thermal conductivity. A poorly conductive (well-insulated) material, such as glass or air, has a low thermal conductivity. Below in Table 4.1 you can see several thermal conductivitys \(k\) of common materials.

Table 4.1 Several thermal conductivitys.#

Material

\(k\) [W/(m K)]

Silver (pure)

410

Copper (pure)

385

Aluminum (pure)

202

Nickel (pure)

93

Iron (pure)

73

Steel 1% C

43

Lead (pure)

35

Oak

0.17

Window glass

0.78

Glass wool

0.038

Water

0.556

Air

0.024

Carbon dioxide

0.0146

Insulation heat storage tank

We look again at the heat storage tank of Example 1. In that example, we have neglected heat loss to the environment. Now assume that there is only heat loss due to conduction. The water in the storage tank has a temperature of 55 \(°C\). The outside temperature is \(20 °C\). We assume that the heat can flow through all wall parts of the storage tank and that the temperature of the water in the tank is homogeneous. How large is the heat loss? Which material from the table would be better suited for insulation?

Solution

Over the wall of the storage tank there is a temperature difference of \(20 °C - 55 °C = -35 °C\). (Where the positive x direction is outwards.) The temperature decreases linearly, the gradient is:

(4.6)#\[ \frac{dT}{dx} = \frac{-35}{0.03} = -1167 \, °\textrm{C/m} \]

Furthermore, the tank is made of aluminium, with \(k = 202\,\textrm{W/(m K)}\). The area through which heat flows is:

(4.7)#\[ A = 2 \pi R h + 2 \cdot \pi R^2 = 2 \pi R(h+R) = 2 \pi \cdot 2.0 (5.0 + 2.0) = 87.96 \, \textrm{m}^2\]

Therefore, the heat flow from the storage vessel to the outside (the heat loss) is equal to:

(4.8)#\[ \dot{Q} = -k A \frac{dT}{dx} = -202 \cdot 87.96 \cdot -1167 = 2.07 \cdot 10^7 \, \textrm{W} \]

In order to keep the contents of the vessel at \(55 °C\), a heating system would have to supply approximately \(20.7\) MW of heat in this unrealistic uninsulated-wall model!

4.3.2. Heat transfer by radiation#

Any object above absolute zero emits thermal radiation. This is done with electromagnetic waves. At ordinary temperatures, most of this radiation lies in the infrared part of the spectrum. With infrared cameras you can visualize that heat. Such cameras are used, among other things, in night vision equipment.

_images/image11.png

Fig. 4.5 Infrared image showing the temperature distribution of a human.#

If an object becomes warm enough, it will also emit visible light. Just think of glowing steel or the glowing spiral in a radiant stove or toaster.

The emitted thermal-radiation power can be estimated with the Stefan–Boltzmann law:

(4.9)#\[ \dot{Q}_{\textrm{rad}} = A\,\sigma\,\varepsilon\,T^4 \quad [\textrm{W}] \]

In which:

  • \(\dot{Q}\) is the heat-transfer rate in [W],

  • \(A\) is the surface of the radiating object in [\(m^2\)],

  • \(\sigma = 5.67\cdot 10^{-8}\,\textrm{W}/(\textrm{m}^2\textrm{K}^4)\) is the Stefan–Boltzmann constant,

  • \(T\) is the absolute temperature of the radiating object, so always in [K],

  • \(\varepsilon\) is the emissivity of the surface [-].

Table 4.2 Some emission factors.#

Material

Emission factor \(\varepsilon\)

Copper oxide

0.74

Aluminum oxide

0.80

Paper

0.90

Glass

0.92

Wet ice

0.95

Exactly the same relationship applies to the amount of radiation that a small object absorbs, with the difference that \(T\) is then the temperature of the environment. If the object (indicated by index 1) is small relative to the environment (index 2), the net heat transfer from the object to the environment can be described as:

(4.10)#\[ \dot{Q}_{1\to2} = A_1\,\sigma\,\varepsilon_1\,(T_1^4-T_2^4) \quad [\textrm{W}] \]

In which \(A_1 \sigma \varepsilon_1 T_1^4\) is the amount of radiation the object emits and \(A_1 \sigma \varepsilon_1 T_2^4\) the amount of radiation the object absorbs.

Solar collector

In 2001 there were 15,000 households in the Netherlands that used solar collectors for domestic hot water. The solar collectors collect solar radiation and give heat to water that flows under the collectors. Around noon on an average spring day, such a collector absorbs \(1000\textrm{W/m}^2\) of solar energy.

The solar collector has a temperature of \(60\) °C. Furthermore, it is stated that the emission factor \(\varepsilon\) of the solar collector is 0.90 and that the area \(A\) of the solar collector is \(3.0 \textrm{ m}^2\). The outside temperature is \(30 °C\).

_images/Ch6_Heat_Machines_6.5.png

Fig. 4.6 (a) Solar collector (source: Navem), (b) Schematic representation of the solar water heater.#

  1. Which forms of heat transfer play a role in the solar collector?

Solution

  1. The solar collector is warmer than the environment. The solar collector absorbs heat by radiation from the sun. The collector loses heat to the environment through radiation and via conduction and convection to the colder (flowing) ambient air. The collector also releases heat via conduction and convection to the colder water that flows through the collector.

  1. Calculate the amount of heat that the solar water heater loses via radiation to its environment? If we neglect other losses, how much heat is left to heat the water?

Solution

  1. The collector (\(60 °C\)) emits heat through radiation, with a power of:

(4.11)#\[ \varphi_{w,1} = A \cdot \sigma \cdot \varepsilon \cdot T^4_1 = 3.0 \cdot 5.67 \cdot 10^{-8} \cdot 0.90 \cdot 333^4 = 1882 \, \textrm{W}\]

The solar collector also absorbs heat from the environment (\(30\) °C) via radiation, with a power of:

(4.12)#\[ \varphi_{w,2} = A \cdot \sigma \cdot \varepsilon \cdot T^4_2 = 3.0 \cdot 5.67 \cdot 10^{-8} \cdot 0.9 \cdot 303^4 = 1290 \quad \textrm{ W} \]

The collector absorbs \(\dot{Q}_{\textrm{solar}} = AI = 3.0\,\textrm{m}^2 \cdot 1000\,\textrm{W/m}^2 = 3000\,\textrm{W}\) from the rays of the sun.

The net radiative loss is \(1882-1290=592\,\textrm{W}\). Neglecting all other losses, the remaining power available to heat the water is \(3000-592=2408\,\textrm{W}\).

4.3.3. Heat flow by convection#

If a fluid layer is stationary and a temperature difference exists across it, then there is conduction. Due to conduction, heat passes from the high to the low temperature. For steady one-dimensional conduction with constant properties, the temperature profile is linear, the temperature gradient is constant. This is shown in Fig. 4.3. If the medium flows at a certain speed, the temperature trend is different, see Fig. 4.7 When bulk fluid motion contributes to heat transfer, we speak of convection.

_images/Ch6_Heat_Machines_6.6.png

Fig. 4.7 Heat flow by convection in flowing air.#

With gaseous and liquid media (fluids), there is usually convection. If the flow is not imposed (by a pump or the wind, for example) spontaneous flow often arises due to the temperature difference. An example of this spontaneous flow is shown in Fig. 4.2(b). Initially, the water in the kettle is at rest. Due to the temperature difference, the warmer water flows up and the colder water flows down.

Calculating convection is a lot more complicated than calculating conduction. The exact course of the temperature gradient is not considered. For the situation in Fig. 4.7, where a hot plate gives off heat to a flowing liquid or gaseous medium, the heat flow can be calculated with:

(4.13)#\[ \dot{Q}_{\textrm{conv}} = h A (T_w-T_\infty) \quad [\textrm{W}] \]

In which:

  • \(\dot{Q}\) is the heat-transfer rate in [W],

  • \(A\) is the area through which heat flows in [m\(^2\)],

  • \(h\) is the convective heat-transfer coefficient [W/(m\(^2\) K)] and

  • \(T_w-T_\infty\) is the temperature difference between the wall and the bulk fluid [K].

The heat transfer coefficient \(h\) depends on the medium, but also on the flow velocity of the medium, the direction of flow and the type of flow (turbulent or laminar). The heat-transfer coefficient \(h\) is usually obtained from experiments or empirical correlations. The value of \(h\) will be given per situation within the framework of this course.

Flowing air can dissipate more heat than stationary air.

What can you say about the physical meaning and units of \(h_{\text{air}}\) compared with \(k_{\text{air}}\)?

4.4. Energy conversion, heat and efficiency#

Many processes are conceivable that add or remove heat to a system. In the previous section we looked at the basic mechanisms on which heat can be transported. In this section we consider the process and its return. We define the efficiency of a heating or cooling process as follows:

(4.14)#\[ \eta = \frac{E_{\textrm{useful}}}{E_{\textrm{input}}} \]

Hot plate

A pan with 2.0 L of water is standing on a hotplate. At t = 0 the water has a temperature of \(20 °C\). The hotplate has an electrical power of 1000 W. The efficiency of the hotplate is 80%. The water must be heated from \(20 °C\) to \(70 °C\). How long is the heating time? Neglect the heat capacity of the pan. (Why is that allowed?)

_images/Ch6_Heat_Machines_6.7.png

Fig. 4.8 Hot plate heats pan with water.#

The amount of heat supplied to the water is per second:

(4.15)#\[ P_{\textrm{added}} = \eta \cdot P_{\textrm{used}} = 0.80 \cdot 1000 = 800 \, \textrm{W} \]

The increase in temperature per second is equal to:

(4.16)#\[ \frac{dT}{dt} = \frac{P_{\textrm{added}}}{mc} = \frac{800}{2.0 \cdot 4.18\cdot10^3} \approx 0.096\,\textrm{K/s} \]

The heating time is then equal to: \(\Delta t_{\textrm{warm-up}} = \frac{70-20}{0.096} = 5.2 \cdot 10^2 \, \textrm{s}\)

Note

Note the difference between the symbols: \(t\) for time and \(T\) for temperature.

Heat exchanger

In a gas-fired water heater, water is heated by flames. The cold water flows through the tubes and is heated at the exit by the heat supplied from the flames. It is stated that the combustion yields 45 kW. The heat-transfer efficiency to the flowing water is 75%.

_images/Ch6_Heat_Machines_6.8.png

Fig. 4.9 Heat exchanger with heater flames.#

  1. Which heat transfer mechanisms play a role here?

Solution

  1. There is radiation: the tubes absorb heat by radiation from the flames. There is conduction: the pipes conduct the heat to the water. And there is convection: the hot combustion gases flow past the hot pipes and the hot pipes release heat to the flowing water.

  1. How much heat is supplied to the flowing water per second?

Solution

  1. \(0.75 \cdot 45\, \textrm{kJ} = 33.75 \,\textrm{kJ}\) is supplied to the water every second.

  1. 0.1 kg of cold water is supplied per second and 0.1 kg of heated water leaves the system. In other words: the mass flow rate is 0.1 kg/s. The total internal water volume of the heat exchanger is 5 L. Initially the water is 10 °C. What is the final temperature? What assumptions do you make to solve this easily?

Solution

  1. In steady operation, use an energy balance on the flowing water:

\[ \dot{Q} = \dot{m}\,c_p\,(T_{\textrm{out}}-T_{\textrm{in}}). \]

Thus,

\[ T_{\textrm{out}} = 10 + \frac{33.75\cdot10^3}{0.1\cdot4.18\cdot10^3} \approx 90.7\,^\circ\textrm{C}. \]

This assumes steady state, negligible heat loss to the surroundings, and approximately constant water properties.

Note

If you want more water per second (you increase the mass flow) and you do not increase the capacity of the burner, the temperature of the outgoing water drops! As the heat exchanger becomes hotter, it will radiate heat itself to its (albeit also warm) environment!

4.5. Thermodynamics#

4.5.1. Ideal gas law#

For an ideal gas, the gas law can be deduced by conducting three experiments.

  • In the first experiment, a fixed amount of gas in a cylinder is compressed while, while the temperature remains constant: isothermal compression.

  • In the second experiment, a fixed amount of gas is heated in a cylinder while the pressure is kept constant: isobaric expansion.

  • In the third experiment, a fixed amount of gas is heated in a cylinder, while the volume is kept constant: isochoric heating.

The three experiments are shown schematically in Fig. 4.10.

_images/Ch6_Heat_Machines_6.9.png

Fig. 4.10 Isothermal compression, isobaric heating, isochoric heating.#

For a fixed amount of an ideal gas, the following relation applies:

(4.17)#\[ pV=nRT \quad \Rightarrow \quad \frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2} \quad \text{for fixed }n \]

4.5.2. \(p\)-\(V\) diagram#

In thermodynamics, it is common to display the state of the gas in a p-V diagram. In Fig. 4.11 the experiments of Fig. 4.10 are plotted in \(p\)-\(V\) diagrams. You see an isothermal (constant temperature), an isobaric (constant pressure) and an isochoric (constant volume) process, respectively.

_images/Ch6_Heat_Machines_6.10.png

Fig. 4.11 (a) Isothermal, (b) isobaric, (c) isochoric.#

Note

By which relationships, which can be derived from (4.17), are these three graphs described?

4.6. Cyclic process#

A cyclic process is a process that consists of several process steps. For those process steps, it applies that after performing the final step, the state of the gas is again identical to the initial state. This applies, for example, to the processes of a cylinder of a (car) engine in which one has to deal with a compression step, a combustion step, an expansion step and an exhaust/intake step. But a refrigerator also works that way; or a heat pump or a gas turbine. With a gas, the state is described by variables such as pressure, volume, and temperature. In a \(p-V\) diagram, therefore, a cyclic process always forms a closed loop.

Recyclic process

A simple cyclic process is shown in Fig. 4.12

_images/Ch6_Heat_Machines_6.11.png

Fig. 4.12 Cyclic process.#

At the end of process 6, the state is identical to start state 1: this is called a cyclic process. The following happens during the cyclic process:

Process/state

Description

1

The piston rests on the lower locking pins.

1 \(\to\) 2

A mass is placed on the piston. The piston rests on the locking pins.

2 \(\to\) 3

Heat is supplied. The pressure increases until the gas pressure overcomes the load. The volume remains the same.

3 \(\to\) 4

Heat is supplied. The pressure no longer increases, but the gas expands until the piston just presses against the upper locking pins.

4 \(\to\) 5

The added mass is removed. The piston rests against the upper locking pins.

5 \(\to\) 6

Heat is extracted from the gas. The pressure decreases until the pressure is just equal to the outside air pressure plus the pressure of the weight of the piston.

6 \(\to\) 1

Heat is further extracted from the gas. The pressure no longer decreases, but the volume of the gas decreases. If the piston rests correctly on the locking pins, the heat dissipation will be stopped.

If we plot this cyclic process in a p-V diagram we get:

_images/Ch6_Heat_Machines_6.12.png

Fig. 4.13 \(p-V\)-diagram of the cyclic process.#

4.6.1. Work of a gas#

In the cyclic process of Example 6 the gas does work during \(3\to4\). After all, a (constant) force is exerted over a distance traveled. This work is performed by the gas. The gas performs the following work:

(4.18)#\[\begin{split} \begin{aligned} W & = F \cdot \Delta x &\textrm{[J]} & =\textrm{[N]}\cdot\textrm{[m]}\\ & = pA \cdot \Delta x && = [\textrm{Nm}^{-2}][\textrm{m}^2]\cdot\textrm{[m]}\\ & = p \Delta V && = [\textrm{Nm}^{-2}][\textrm{m}^3] \end{aligned} \end{split}\]

If the force is not constant during the displacement, or if the pressure is not constant during the expansion of the gas, then the following applies to the work of the gas:

(4.19)#\[ W = \int_{V_1}^{V_2} p\, d V \]

In a \(p-V\) diagram, this is equal to the area under the graph. If the volume of the gas increases, the gas performs positive work, if the volume decreases (then the gas is compressed), the gas performs negative work.

Work in a cyclic process

On which routes is work carried out by the gas in the cyclic process of Example 6?

Is that negative or positive work?

Solution

The gas only performs work if it changes its volume. From \(3 \to 4\) the gas performs positive work (expansion), from \(5 \to 6\) the gas performs negative work (compression).

Work in a cyclic process (2)

Calculate the work done by the gas in the cycle of Fig. 4.3 (a). What is the net work per cycle that the gas delivers?

_images/Ch6_Heat_Machines_6.13.png

Fig. 4.14 (a) Cyclic process, (b) positive work, (c) negative work.#

Solution

From 2 to 3, the gas performs positive work during expansion:

\(W_{2 \to 3} = p \cdot \Delta V = 200\, \textrm{kPa} \cdot (2-1)\, \textrm{m}^3= 200\, \textrm{kJ}\)

From 4 to 1 the gas performs negative work during compression:

\(W_{4 \to 1} = 100\, \textrm{kPa} \cdot (1-2) [\textrm{m}^3]= -100 \, \textrm{kJ}\)

From 1 to 2 and from 3 to 4 are isochoric steps, the volume does not change, and the gas does not perform any work.

Overall, the gas performs in this cycle \(W_{\textrm{cycle}} = 200\, \textrm{kJ} - 100\, \textrm{kJ} = 100\, \textrm{kJ}\) net positive work.

For a closed system, heat supplied to the gas can increase its internal energy and/or be converted into work done by the gas. With the sign convention that heat added to the gas and work done by the gas are positive, energy conservation gives:

\[ \mathrm{d}Q = \mathrm{d}U + \mathrm{d}W. \]

This equation is known as the first law of thermodynamics. Over a complete thermodynamic cycle, only part of the supplied heat can generally be converted into net work.

4.7. What lies ahead#

In this chapter you have become acquainted with thermal energy storage and heat transfer. You can use the theory that you have learned here to make an initial estimate of heat storage and heat transfer processes. Heat transfer via convection has not yet been discussed in detail but will be discussed extensively in the remainder of the curriculum. Convection plays an important role in the development of heat exchangers, for example. All kinds of design choices (material, pipe diameter, pipe roughness) all influence the heat transfer. These design choices also influence the choice of pumps that the media must pump through the heat exchanger. The field of energy technology is closely linked to the field of process engineering constructions (liquids and flow) and combustion technology.

4.8. Problems#

For the following exercises use symbols as long as possible and only in the end fill in the values given.

Context

Heat is present everywhere in our environment, such as the heat in the outside air, the surface water or the soil. However, this heat cannot be used immediately because the temperature is too low. With a heat pump, it is possible to efficiently upgrade this heat to a suitable temperature level. Just as a water pump can pump water from a low to a high level, a heat pump “pumps” heat from a low to a high-temperature level. An example of such a heat pump is present in the Geohuis of the TU/e. The Geohuis is a building in which installations for climate control can be installed. With the aid of the available measuring instruments, these installations can be analyzed for their performance. Current testing facilities include a heat pump in which energy from the environment is ‘pumped up’ to higher temperatures for heating purposes.

_images/Thermal_machines_ex10.jpg

Fig. 4.19 The Geohuis in the DUBO-park at the TU/e (left), the heat pump at the Geohuis (right).#

The heat pump

There are many different types of heat pumps, but the method of operation is always the same. The process of a heat pump can be divided into four steps:

  • \(1\rightarrow 2\) A liquid with a boiling point lower than the temperature of the environment temperature is used for the transportation of heat. This liquid evaporates under the influence of the heat from the environment. This way, heat is extracted from the environment by the evaporating liquid. (This phenomenon can also be observed with a spray can for, for example, deodorant or hair spray, which gets cold if used).

  • \(2\rightarrow 3\) The vapor is then compressed by a compressor. This increases the temperature and pressure of the vapor. (This phenomenon can also be observed when inflating a bicycle tire, where the bottom of the pump can get very hot.) Because the pressure has increased, the temperature of the boiling point has also increased (think of a pressure cooker).

  • \(3\rightarrow 4\) . The vapor then condenses into a liquid, releasing heat. The heat released is usable because the temperature is higher than in the first step. As an example, this heat can be used for central heating.

  • \(4\rightarrow 1\) Finally, the liquid flows via a thermal expansion valve in which the pressure and temperature are lowered back to the evaporator. At the evaporator, the process starts all over again.

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