We will try to get familiar with a heat pump, and try to determine its performance coefficient under different circumstances.

Download (0)

Full text

(1)

C4. Heat Pump C4. Heat Pump

I. OBJECTIVE OF THE EXPERIMENT

We will try to get familiar with a heat pump, and try to determine its performance coefficient under different circumstances.

II. INTRODUCTION

II.1. Thermodynamic Background

Thermodynamics study the connection between thermal and mechanical phenomena. In most thermodynamic problems, we are interested in systems, i.e. a set of bodies that undergo different transformations. These transformations are generally the result of work or heat transfer between systems, or between parts of a same system. As a result, we observe a temperature change, a variation in pressure or even a phase change. More generally, this results in a change in thermodynamic variables.

First principle: ΔU = ΔW + ΔQ

where ΔU represents the total energy variation between two states of the system (independently of the path used to get there) and ΔW and ΔQ correspond to the work and heat exchange between both states, while following a specific “path”. Even though this conservation law cannot be directly proven, numerous experiments and scientific conclusions based on it prove its validity. The first principle doesn’t exclude that within a given system, a certain amount of heat can be entirely transformed into work (ΔU = O so ΔW=-ΔQ). This, however, is not verified experimentally.

Second principle

The second principle limits the conversion of heat into work. Most notably, it states the impossibility for a body to transfer heat to a body at higher temperature without external work being applied. In other words, and according to experience, a system will have a natural tendency to reach a thermal equilibrium by transferring heat from the hot source to the cold source (fig. 1). To achieve the opposite transfer, we need an external source of energy (fig.2). This is how a heat pump works.

(2)

Fig. 1 : If both sources are put in contact, there is a natural heat transfer from the hot source to the cold source Tc > Tf.

Fig. 2 : If we would like a heat transfer from the cold source to the hot source (Tc > Tf), then a machine is required, that supplies a work W

II.2. Principle of the heat pump.

The basic idea behind a heat pump is the following: a condensable cooling fluid goes through a closed system. It will perform several heat exchanges with the surrounding environment, and will also undergo successive liquid/gas phase transformations back and forth. When it evaporates, it absorbs calories, and it loses some when condensing. For those transformations to happen in both heat exchange modules, that we will call respectively condenser (hot source, Tc) and evaporator (cold source Tf), we need to compress the gas between the heat exchange modules in one direction, and release it in the other.

Picking the fluid depends on the temperatures to reach, the characteristics of the compressor, the size of the heat exchange modules, the heat transfer coefficient of the cooling fluid, its latent heat, and finally its specific volume. The cold source (the evaporator, i.e. where we extract the heat) can be, for example, a water stream, the ground, or even air.

III. DESCRIPTION OF THE SETUP

The cooling fluid used in the heat pump is "Freon 12" whose boiling point is 243,37 K and latent heat 39,94 cal/g in one atmosphere of pressure.

Let’s now look into the different stages of the cycle, by referring ourselves to the heat pump diagram (Fig. 3):

The freon gases are compressed in a compressor (1), which strongly increases its temperature. In the condenser, the freon liquefies, releasing its latent heat to the liquid in the dewar (2). The liquefied freon the goes through a purificator, to eliminate all vapor bubbles still contained in it (3). The liquid then goes through an expansion valve (4) that releases a certain amount of liquid in the evaporator (5). The freon evaporates, and extracts the necessary heat from the liquid of the dewar (5). The low pressure found in the evaporator is created by the suction of the compressor. The correct setting of the expansion valve is regulated by a thermocouple (6) that measures the temperature difference between the cooling fluid entering and exiting the evaporator. If this difference isn’t large enough, the expansion valve reduces its throughput. The vaporized freon then re-enters the compressor, and the cycle starts over.

T

T

c

f

T

T

M W

f

c Q

Q

c

f

(3)

The amount of heat transferred from the evaporator to the condenser equates to approximately three times as much energy, than the electric energy consumed by the compressor. Despite this fact, this conclusion isn’t contrary to energy conservation laws. The remaining two thirds of energy come from the evaporator.

Fig. 3: Diagram of the heat pump

To quantify the performance of a heat pump, we define a performance coefficient or efficiency coefficient

ε

equal to the ratio of transferred heat to the amount of work provided to the system.

ε = Q

c

W

ε

> 1

W is the work supplied to the system, Qc the amount of heat transferred over one cycle and Tc andTf are the temperatures of the hot and cold source respectively.

The experimental performance coefficient

ε

is noticeably smaller than the theoretical value

ε

th due to some irreversible phenomena encountered in the real thermodynamic cycle.

ε

th

= T

c

T

c

− T

f

(4)

IV. THERMODYNAMIC DIAGRAMS

At low temperatures, gases must be treated like real fluids, whose behavior can be analyzed by a Clapeyron diagram (Fig..4).

Fig. 4: Clapeyron diagram

Mathematically, the real gas can be described by the Van der Waals equation:

for 1 mol of gas

where describes the attraction between molecules, and b represents the specific volume of individual molecules. We can also calculate Tc, i.e. the temperature of the isotherm that has a horizontal tangent at its inflexion point, where V = Vc.

et

thus

The gas’ liquefaction is only possible if T < Tc. From the diagram in figure 4, let’s suppose two isotherms T1 et T2 below Tc, and let’s make the Van der Waals gas follow the cyclic transformations ABCDA: A -> B adiabatic compression ; B -> B' isobaric cooling, followed by an isothermal liquefaction T1 (B' -> C') and isobaric cooling of the liquid (C' -> C); C -> D adiabatic relaxation; (D -> D') isothermal evaporation on T2, D' -> A followed by an isobaric heating.

Over a cycle, the gas extracts heat during the evaporation on the isotherm T2, and releases that heat during the liquefaction on the isotherm T1. Work is supplied to the gas during the adiabatic compression (A -> B).

P + a

V2 V - b = R T a

V2

∂P

∂VT = Tc

= 0 ⇒ - R Tc Vc - b2 + 2 a

Vc3 = 0 ∂2P

2VT = Tc

= 0 ⇒ 2 R Tc Vc - b3 - 6a

Vc4 = 0

Tc = 8 a

27 R b , Vc = 3 b et Pc = a 27 b2

(5)

Mollier Diagram: lnP vs. h (Fig. 5).

When working on a heat pump or cooling machine, the state changes of the gas are described using Mollier diagrams, on which we plot pressure P on a logarithmic y-axis, and enthalpy on the x-axis.

Fig. 5: Mollier diagram (lnP vs. h) of freon R12 gas.

Recall that the enthalpy H is defined by the internal energy U, the pressure P and the volume V, according to the equation: H = U + P V

In this graph, curves of constant pressure are horizontal lines, and curves of constant enthalpy are vertical lines. In the liquid-gas region, i.e. under the saturation line, isothermal lines are, like isobaric lines, horizontal. Curves of constant gas to liquid ratio intersect isothermal curves according to a linear graduation.

While the heat pump is running, we measure the condensation and evaporation pressures Pcond and Pevap, as well as the temperatures of the gas before compression, after compression, and before relaxation T1, T2 and T3. These five parameters allow us to plot the thermodynamic cycle (T1->T2-> T3 ) on the Mollier diagram. (Fig. 5). Isotherms in the part of Mollier’s diagram describing liquid are vertical lines!

From this, we can deduce three value of the specific enthalpy. h1 before compression, h2 after compression, and finally h3 before evaporation. These three values allow us to determine the maximal performance coefficient that can be achieved using the given gas.

The thermodynamic limit is:

(6)

ε

M

= h

2

− h

3

h

2

− h

1

Ex.: The cycle in figure 5 yields:

ε

M

= 394 − 206 394 − 354 = 4.7

V. SUGGESTED EXPERIMENTS

Measuring the performance coefficient under different circumstances

1) Let the heat pump run for 30 minutes, using a finite reservoir. Every minute, take note of the temperatures in both dewars, the temperatures of the gas (T1,T2,T3), the condensation and evaporation pressures, as well as the electrical power supplied to the compressor (for a total of, 5 temperature, 2 pressures, 1 voltage and 1 current).

a. Plot a graph of the temperature variations over time, and calculate the amount of heat transferred from the cold source to the hot source. Compare it with the work the compressor used.

b. Make a Mollier diagram of the cycle in the steady state (copies of the diagram are available at the lab). Determine the gas’ maximal performance coefficient.

c. Determine the heat pump’s performance coefficient, and compare it to the best result we could expect (Carnot cycle). Discuss.

d. If there is enough time, start over now that the compressor is hot (but do par 2) first). Did the heat pump perform better?

2) Do the same experiment, but this time with an infinite reservoir (after having changed the water in both containers)

3) Comment on the signification of the three different performance coefficients Using the setup (see fig. 6)

Fill the dewars with 5 liters of water, by hitting the filling switch. Then, start the mixing. To do so, put both “mix-empty” switches onto mix (brassage), and activate the pumping in both fillers (using the pumping switches). Verify that the temperature has stabilized.

Plug in the compressor in order to be able to read the amount of power that it uses. Launch the compressor and the stopwatch simultaneously.

DO NOT let the water in cold source freeze. Stop the compressor before reaching 0°C.

For the infinite reservoir, place the “mix-empty” switch of the cold source on mix, and let the water flow (filling switch) in order for the volume to remain constant.

(7)

Fig.6 : Image of the heat pump

VI. BIBLIOGRAPHY

1.FEYNMAN : "Cours de Physique - mécanique 2" (p. 264 et suivantes) InterEditions, Paris

2.A. HOUBER ECHTS : "La thermodynamique technique", tomes 1 et 2.

Vander Editeurs.

3.G. BRUHAT : "Cours de Physique Générale - Thermodynamique".

Masson & Cie, Editeurs.

Figure

Updating...

References

Related subjects :