MEC 2700: AEROSPACE ENGINEERING LAB 1

EXPERIMENT NO: 2

TITLE OF THE EXPERIMENT: HEAT PUMP

Date/ Day of Experiment: 11st April 2014 Due Date: 18th April 2014

Author’s Name:

Nur Adlina Binti Mat Nizam 1224256

Group Members:

Nur Aini Binti Zulkipli 1229182

Madihah Binti Mazlizam 1220020

Nur’Ain Binti Sapie 1229102

Name of Lecturer:

Dr Mirghani

OBJECTIVE

Water heat pump: To measure pressure and temperature in the circuit and in the

water reservoirs on the condenser side and the vaporizer side alternately. To calculate

energy taken up and released, also the volume concentration in the circuit and the

volumetric efficiency of the compressor.

Air-water heat pump: To measure vaporizer temperature and water bath temperature

on the condenser side under different operating conditions on the vaporizer side, ie.

natural air, cold blower and hot blower.

To determine the electric power consumed by the compressor and calculate the

coefficient of performance.

INTRODUCTION

There is no difference in principle between a heat pump (Figure 1) and a

refrigeration system (Figure 2). In a heat pump the heat which is rejected by the condenser

or heat exchanger is used for heating purposes. The condenser is therefore located within

the space to be heated, such as a room within a building. On the other hand, the evaporator

is located externally and draws its supply of heat from a source at a lower temperature than

that in the condenser. In practical, the heat source of heat pump is often the atmosphere,

but sometimes a river or soil is used instead. The only difference with air

conditioning/refrigeration system is that the heat pump system intended to cool a separate

source of heat and disposes the heat into the occupied area. Rating of heat pump is done by

the ratio of heat output to electrical input, which is called the Coefficient of Performance

(COP). Both of those systems can be summarized in figures below.

Figure 1- Heat Pump System Figure 2 - Air Conditioning/Refrigeration

System

CALCULATIONS

PART A: Water-water Heat Pump

1) Mass of water:

a) Condenser = 4.5L x 0.001m3 x 1000kg/m3

= 4.5kg

b) Vaporizer = 4.5L x 0.001m3 x 1000kg/m3

= 4.5kg

2) Graph of temperature vs time for all inlet and outlet.

Condenser

0 5 10 15 20 250

10

20

30

40

50

60

Temperature vs Time (Condenser)

θciθco

Time (min)

Tem

pera

ture

(o C

)

Vaporizer

0 5 10 15 20 250

5

10

15

20

25

30

Temperature vs Time (Vaporiser)

θviθvo

Time (min)

Tem

pera

ture

(o C

)

3) Calculations at t= 10mins

a) Vaporizer heat flow

Q̇o=c ∙mw ∙∆T 2

∆ t

¿4.187 kJkg ∙ K

×4.5kg× ¿¿

¿−0.345 kJs

b) Condenser heat flow

Q̇=c ∙mw ∙∆T2

∆ t

¿4.187 kJkg ∙ K

×4.5kg× (48+273)−(33+273)K

10m∈¿× 1min60 s

¿

¿0.471 kJs

c) Average compressor power, Pavg

Pavg=∑ P ∙∆ t

∑ ∆ t

¿ 2858.4W ∙min19min

¿150.44W

d) Performance at the condenser side

ε= Q̇Pavg

¿0.471 kJ

s0.150 kW

¿3.14

e) Volume flow at the vaporizer side

V̇=v ∙Q̇oh1−h3

*v=¿specific volume of the water at vaporizer

¿0.04348 m3

kg

¿0.04348 m3

kg×

0.345 kJs

406.61 kJkg

−219.03 kJkg

¿0.0800×10−3m3

s

f) Geometrical volume flow

Given

Vg = 5.08 cm3

f = 1450 min-1

V̇ g=V g ∙ f

¿5.08cm3×1450min−1× 1m3

1×106cm3×1min60 s

¿0.123×10−3m3

s

g) Volumetric efficiency of the compressor

λ= V̇V̇ g

¿0.0800×10−3 m3

s

0.123×10−3 m3

s

¿0. 650

PART B: Air-water Heat Pump

1. Graph of temperature versus time for all the results.

2. The average vaporizer temperature:

Natural air

T avg=∑T ∙∆ t

∑ ∆ t

¿ 101℃ ∙min20min

¿5.05℃ Hot blower

0 5 10 15 20 250

10

20

30

40

50

60

Temperature vs Time

natural air θ1hot blower θ1cold blower θ1

Time (min)

Tem

pera

ture

(o C

)

T avg=∑T ∙∆ t

∑ ∆ t

¿ 286℃∙min18min

¿15.89℃

Cold blower

T avg=∑T ∙∆ t

∑ ∆ t

¿ 217℃∙min20min

¿10.85℃

3. Condenser heat flow:

Natural Air

Q̇=c ∙mw ∙∆T∆ t

¿4.187 kJkg ∙ K ×4.5kg×

(32+273 )−(20+273)K20min × 1min

60 s

¿0.188 kJs

Hot blower

Q̇=c ∙mw ∙∆T∆ t

¿4.187 kJkg ∙ K ×4.5kg×

( 48+273 )−(20+273)K18min × 1min

60 s

¿0.488 kJs

Cold blower

Q̇=c ∙mw ∙∆T∆ t

¿4.187 kJkg ∙ K ×4.5kg×

( 43+273 )−(20+273)K20min × 1min

60 s

¿0.361 kJs

4. The performance:

Natural Air

ε= Q̇Pavg

¿0.188 kJ

s0.0907 kW

¿2.073

Hot blower

ε= Q̇Pavg

¿0.488 kJ

s0.133 kW

¿3.669

Cold blower

ε= Q̇Pavg

¿0.361 kJ

s0.121kW

¿2.983

DISCUSSION

Heat pump cycle is based on thermodynamic theory, which states that heat moves from

low pressure cold air to high pressure hot air. The benefit of using heat pump is its

effectiveness in heating cold air in short interval of time. The effectiveness in heating

depends on how well the heat pump transfer the hot air to cold air and the power required

to do so. The refrigeration cycle uses a fluid known as refrigerant. The purpose of using this

fluid is to transfer heat from one place to another. The refrigerant boils at much lower

temperature than water at the same pressure. The heat pump and refrigerant cycles are

through compression, condensation, evaporation and expansion. Both system can run

effectively if the temperature difference is higher and the heat lost to surrounding

considerably low.

Based on the experiment we were conducted, this experiment is divided into two parts,

part A for water-water heat pump and part B for air-water heat pump.

In part A, we used two water reservoirs, one at the condenser side while the other one at

the vaporizer side. Then, we have to record the power reading, pressure and temperature

for both sides. Based on the plotted graph, for condenser side, it shows increasing trend as

time increases while decreasing trend as time increases for vaporizer side. Next, we

calculated the vaporizer heat flow from equation Q̇o=c ∙mw ∙∆T 2

∆ t and condenser heat flow

from equation Q̇=c ∙mw ∙∆T2

∆ t. Then, the average compressor power (1), performance at the

condenser side (2), volume flow at the vaporizer side (3), geometrical volume flow (4) and

volumetric efficiency of the compressor (5) from these equations:

Pavg=∑ P ∙∆ t

∑ ∆t ---- (1) ε=

Q̇Pavg

----(2) V̇=v ∙Q̇oh1−h3

-----(3) V̇ g=V g ∙ f ----(4)λ=V̇V̇ g

----(5)

In part B, we only used one water reservoir but in three conditions, by natural air, hot

blower and cold blower. From this experiment, we obtained power reading and

temperature at the vaporizer outlet and the condenser water temperature. Then, from the

plotted graph, we can conclude that the temperature is directly proportional to time for all

conditions. As time increase, the temperature also increase. Next, we calculate the average

vaporizer temperature by using this equation T avg=∑T ∙∆ t

∑ ∆ t for all conditions. The condenser

heat flow is then calculated using the equation Q̇=c ∙mw ∙∆T∆ t and ε=

Q̇Pavg

is used to find the

performance. From the calculation, hot blower shows the highest temperature at vaporizer

outlet, condenser and performance compares to the two, which are natural air and cold

blower. This shows that when air is blown, the effect on heat pump process is less compared

to static air regardless of their hotness and coolness.

There are some errors encountered during the experiment, the first one is parallax error

during the recording of temperature from thermometer since the eyes are not in line with

the thermometer. Next, the power reader machine produce unconstant result, the value

keep on changing.

CONCLUSION

As the conclusion, the objectives of this experiment are achieved. We are able to

measure pressure and temperature in the circuit and in the water reservoirs on the

condenser side and the vaporizer side alternately and energy taken up and released, also

the volume concentration in the circuit and the volumetric efficiency of the compressor.

Next, we are also able to measure vaporizer temperature and water bath temperature on

the condenser side under different operating conditions on the vaporizer side ( natural air,

cold blower and hot blower). And lastly, the electric power consumed by the compressor

and calculate the coefficient of performance are determined from this experiment.