Heat Pump

By Prasad Patil

ENGD2005

P14183792

Submitted to: Dr. Ibrahim Abdalla

Table of Contents

1. Objectives 2

2. Introduction2

3. Background 2

3.1 Heat Pump Theory 3

3.2 Refrigerant 3

4. Types of heat pump 3

4.1 Air source 3

4.2 Water source 3

4.3 Ground source 3

5. Vapour compression cycle 3

6. Apparatus 3

7. Procedure 4

8. Data4

9. Calculation and Graph 4

10. Discussion5

11. Bibliography 6

1

By Prasad Patil

ENGD2005

P14183792

Submitted to: Dr. Ibrahim Abdalla

1. Objective

To scrutinize the actual and ideal C.O.P for a heat pump

2. Introduction

Heat pump is a device which is thermodynamically identical to refrigeration, which extracts heat energy from a source at a low temperature and transfers it to higher temperature. The difference between heat pump and refrigeration is where refrigeration provides useful cooling and heat pump provides useful heat. Heat pumps are available in many different sizes, shapes and types, where heat pump operating on vapour compression cycle are the most common, the other types includes thermoelectric devices and absorption cycle. Heat pump sizes may vary from some watts to megawatts output. The essential components of any heat pump are the heat exchangers through which heat is extracted and made available. There are different types of applications where heat pumps used are industrial application, domestic application and public and commercial buildings, the simple examples of heat pump application are pumping heat from cooling water provided to factories for space heating, whereas pumping air from outdoor to provide heat in small shops and houses.

3. Background

3.1 Heat pump theory

Heat pump consists of compressor, evaporator, condenser, expansion valve (throttle valve) which are connected by pipework. William Thomson (Lord Kelvin) proposed the first practical heat pump system, demonstrating that refrigeration machine can also be used for heating. Thomson’s proposed heat pump in which he used air as a working fluid, the ambient air is taken from outdoor and drawn in to the cylinder for expansion by reducing both its pressure and temperature, and then it is passed through air to air heat exchanger, where cooled air was able to pick up heat from the ambient air before entering the building for heating. The air was then compressed back to its atmospheric pressure, results in increase in temperature above the ambient level.

3.2 Refrigerant

It is a chemical used as a cooling mechanism in air conditioner and refrigeration and also the heat carrier which turns gas to liquid and vice versa. There are some common types of refrigerant used nowadays are halocarbons or Freon’s, Azeotropic refrigerants, Zeotropic refrigerants, Inorganic refrigerants like carbon dioxide, ammonia, water and air and Hydrocarbon refrigerants. In general refrigerants are known by their number e.g. refrigerant 12 or refrigerant 22 etc. this numbers are derived from chemical formula of the first digit one less than the carbon atoms.

2

4. Types of heat pump

4.1 Air source: in this type the outside ambient air is used as a heat source whereas the inside air extracts the heat from the condenser. In this system warm air is produced which is circulated by fans. The disadvantage of this system is when heat is required most, when the ambient temperature is cold, as a result coefficient of performance is low due to drop in source temperature.

4.2 Water source: Water is an excellent source of heat, where it is easily available in sufficient quantities which do not freeze even in winters. Where proper care must be taken for this heat pump and suitable heat exchangers material must be chosen to avoid corrosion.

4.3 Ground source: These heat pumps are also called geo thermal heat pumps. In this heat pump, pipes are buried under the ground level to extract heat from the ground, where heat is absorbed from the ground in to fluid and then it is passed through heat exchanger in to heat pump.

5. Vapour compression cycle

In this cycle heat pump is operated by constant change in the physical properties of the fluid through a cycle which is theoretically reverse of Carnot cycle, in which vapour are isothermally expanded at an absolute temperature which absorbs heat and then it is adiabatically compressed and further heated. Heat is then released thru isothermal compression and adiabatic expansion and then return to starting vapour condition.

6. Apparatus

Compressor, Evaporator, Condenser, Expansion valve (throttle valve),

Condenser (its where the heat is dissipated thereby turning the superheated vapour into saturated liquid)

Evaporator (absorbs heat which heats up the saturated liquid to vapour) Compressor ( compresses the vapour to high pressure which intern increases the

temperature of the vapour to become superheated vapour) Expansion valve(reduces the pressure of the liquid before entering the evaporator)

3

7. Procedure

The cooling water rates for both condenser (red cylinder) and evaporator (blue cylinder) is to be adjusted at 30kg/h

Pressure of 1000KN/m2 in condenser is provided by refrigerant expansion valve, when the compressor is operating

Approximately 30 min are allowed for equilibrium condition to reach before taking the readings

8. Data

a) compressor power: the time for one revolution was measuredTime taken for 1 revolution is 53 sec

b) Refrigerant properties Condenser inlet temperature = 70ºc Condenser outlet temperature = 27ºc Condenser pressure = 840 KPa Evaporator pressure = 100 KPa Evaporator inlet temperature = -14ºc Evaporator outlet temperature = 12ºc

c) cooling water properties

Cooling mass flow rate (M) = 50 KgKJ

Cooling water inlet temperature = 23ºc Cooling water outlet temperature = 35ºc C.P water = 4.19

9. Calculation & Graph

COPHp=Qcondensorenergy

Qcondensor=Qc .w Qc .w=[M∗C p∗∆T ]

[ 503600

KgKJ

]∗[4.19 KJKg . K ]∗[ (35−23 ) ]=0.46

Qcondensor=0.46 Kw

Energy consumed

4

E¿=[ 3600 Kws

166.67∗53 s ]=0.407Kw

C .O .Peff=0.460.47

=1

Since from the above data points to b plotted on the graph are

Point 1 (Pevap, Tevapout)

Point 2 (Pcond, Tcond in)

Point 3 (Pcon, Tcondout)

Point 3= Point 4

C .O .PHp=QcondWHp

=h2−h3

h2−h1 = 395−235

395−365 = 5.3

10. Discussion

The ideal coefficient of this experiment and the Mechanical Heat Pump is 5.3, which is not efficient due to the following reasons:

1. Mechanical Heat Pump is 15 years old and so the components are outdated. Hence does not provide efficient heat for space heating.

2. The pipework of the components has been corroded. Hence it is not efficient.

5

11. Bibliography

1. (Reay & Macmichael, 1988)

2. (Heap, 1983)

3. Heat Pump Handouts

6