TEP 4255 Heat pumps: process and systems

sample solution 1

N TN U N orges teknisk-naturvitenskapelige universitet Department for Energy and Processes

TEP 4255

2011 Sample solution Exercise 1

Thermal analysis of a cooling plant in the food industry a)

Minimum power is considered as the Carnot process between room temperature and cooling water. Q 0 = 200 [kW]

Refrigeration room: .

Wmin  Q0 

at

tR1 = 0 [ºC]

tsur = 14 [ºC]

Tsur  TR1 287  273  200   10.25  kW  TR1 273

b)

Cooling room 1

Cooling room 2

Figur 1 Flow chart of a simple installation. Flow-chart of a simple one-step installation is shown in Figure 1. Based on the assumption that the use of a expansion valve provides dry saturated refrigerant at the outlet of the evaporator. (hard to do in practice) . The thermodynamic data plotted in the log-p-h diagram below are calculated using the Refrigeration calculator in Cool Pack. Isentropic efficiency = 0,68.

TEP 4255 Varmepumpende prosesser og systemer

Side 1

TEP 4255 Varmepumpende prosesser og systemer

Løsningsforslag 1

Diagram 1. H-log P with the process drawn(1-2-3-4).  Circulated refrigerant massflow: Q o 200 m R134a = = = 1,210kg/s h1 - h4 392.51 - 227.23 Determination of the compressor strokevolume: Theoretical:

V theo = m R134 a   g 

3



m  g ( t 1 ; p o ) =  g (-8 C  ; 2.17 bar  ) = 0.09186   kg 

m V theo = 1, 210  0.09186 = 0.1112   s

3



 m   = 400.1  h     3

Real:

  V theo V real 3 V theo = 400.1 = 460.0  m  = = V real V stroke  h   0.87   Determination of powerdemand: Theoretical: W theo = mR134a  (h2' - h1 ) h2'  hg (s2' ; pk )  hg (s1 ; pk )

Sample solution 1

page 2

TEP 4255 Varmepumpende prosesser og systemer

Løsningsforslag 1

h2’ is the enthalpy after the compressor with adiabatic reversible, ie isentropic compression: W theo = 1.210  ( 412.37 - 392.51 ) = 24.0  kW  Real :

 is  W theo

W real W theo = 24.0 = 35.3 kW   W real = 0.68  is

Actual performance of the condenser, provided all compressor losses are added to the refrigerant: Qc = m R134a  ( h 2 - h3 )

with

 is 

 h is h -h = 2' 1  h comp h 2 - h1

 kJ  h 2 = 421.72    kg  and

Qc = 1.210  ( 421.72 - 227.23 ) = 235.33  kW 

The power factor (COP) of the facility: Q 200  o = = 5.67 W real 35.3 c)

The reason why actual powerconsumption is higher than W min is due to 3 losses: 1) 2) 3)

Heat exchanger losses Process losses (choking and overheating) Compressor loss

In order to find the heat exchanger loss the power consumption, W C , is calculated for the Carnot process between t0= -8 [ºC] og tk= 20 [ºC] for the facility. T  T0 293  265 W C  Q 0 k  200   21.13 kW  T0 265 The difference between this power consumption and W for the process between 0 oC and 14 oC is:

Sample solution 1

min

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TEP 4255 Varmepumpende prosesser og systemer

Heatexchanger loss: Process loss: Compressor loss:

Løsningsforslag 1

 kW  Wprs  Wtheo  WC  24.0  21.13  2.87  kW  Wcomp.  Wreal  Wtheo  35.3  24.0  11.3  kW  Whx  WC  Wmin  21.13  10.25  10.88

Percentage distribution of losses:

Whx = W prs =

10.88 [kW]

43.4 %

2.87 [kW]

11.5 %

Wcompr =

11.3 [kW]

45.1 %

25.05 [kW] The process loss consist of expansion loss and overheating loss. Calculate s4 on the basis of x-value in Figure 1.

Choking or Expansion loss : Wchok  mR134 a  Tk  ( s4  s3 )

s3  1.0955kJ / kgK  , s4  1.1035kJ / kgK 

Wchok  1.21  293  (1.1035  1.0955) kW   2.85 kW 

Overheatningloss : Wovrh.  Wprs  Wchok  2.87  2.85  0.02  kW 

Diagram 2. T,s-diagram of a simple one-step process. The losses are represented by the shaded areas in the T,s-diagram in Diagram 2.

Sample solution 1

page 4

TEP 4255 Varmepumpende prosesser og systemer

d)

Løsningsforslag 1

Evaluation of the influence of different evaporation- and condensingtemperatures. Calculated for R134a and 3 different cases: t0/tk = -4/17, -8/20 and –12/23. Make use of Excel-sheets and refrigerant routines. Solution is presented below (study Excel-sheet for formulas).

Figur2 Bar chart showing the loss distribution. With ammonia instead of R134a the overheating loss will increase and the expansion loss is reduced. Sample solution 1

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TEP 4255 Varmepumpende prosesser og systemer

e)

Løsningsforslag 1

An installation with a compressor stroke volume oversized by 15%. The actual stroke volume of the compressor by the original process dimension b): 460.0 [m3/h]; 15 % bigger gives 529 [m3/h]. The print-out below is the results of the Excel-sheet and refrigerant routines. Capacity of compressor 529 Temperature condenser: 20 Entalphy inlet evaporator: 227,24 Condensation pressure: 5,716 Room temperature 0 Q 0 at -8 °C: 200 Evaporator temp Q0 t0 kW -12 300 -10 250 -8 200 -6 150 -4 100 -8,83

220,7

p0 h1 Bar kJ/kg 1,854 390,12 2,007 391,32 2,170 392,51 2,344 393,69 2,527 394,87

m3/h °C kJ/kg Bar °C kW

spec.vol m3/kg 0,10678 0,09899 0,09186 0,08535 0,07939

press.rel. lambda Q0 pk/p0 [-] kW 3,083 0,840 188,3 2,848 0,855 208,3 2,634 0,869 229,7 2,439 0,882 252,7 2,262 0,894 277,3

2,102 392,02 0,09473 2,720 0,863 Characteristics of evaporator and compressor 0,00 Dif

220,7

350

W k ,0 Q

300 250 200 150 Balancepoint

100

Evaporator

50

Compressor

0 -12

-10

-8

-6

-4

Fordampertemperatur, °C

Balancepoint at evaporation temperature = 8.83 °C og yteles = 220.7 kW

Sample solution 1

page 6