424519 Refrigeration Kylteknik 2018             Thermal and Flow Engineering      

         

Exercises 1 + 2      13.11 + 20.11 2018 

Questions 1‐11. 1. Ö96 – 2.1 A flow of R-12 at 1000 kPa, 100°C is sent through a throttling valve, lowering the pressure to 300 kPa. Will the temperature change? Heat exchange with the surroundings may be neglected.

2. D03 – example 3.2

3. Ö96 – 2.4

In a so-called bubble-cooler at 340 kPa pressure a flow of 0.024 kg/s R-12 from a throttling valve after a condenser that collects saturated R-12 condensate at 34°C is mixed with a flow of 0.018 kg/s R-12 vapour at 75°C, 340 kPa. Calculate the mass flows of liquid and vapour R-12 from the cooler, assuming steady state.

4. KJ05 – P8.3

A vapor-compression refrigeration cycle uses R-134a. Liquid at 1200 kPa exits the condenser at 40 ºC. The evaporator operates at a pressure of 240 kPa. The compressor isentropic efficiency is 75%. Determine the cycle coefficient of performance (COP) if the refrigerant leaves the evaporator as superheated vapor at 0, 5, 10, 15, and 20 ºC above the saturation temperature.

5. CB98 – question 10-100

6. CB98 – question 10-62

Sources: CB98: Y.A. Çengel, M.A. Boles “Thermodynamics. An Engineering Approach”, McGraw-Hill (1998) D03: İ. Dinçer “Refrigeration systems and applications” Wiley (2003) KJ05: D. Kaminski, M. Jensen ”Introduction to Thermal and Fluids Engineering”, Wiley (2005) Ö96: G. Öhman ”Kylteknik”, Åbo Akademi University (1996)

7. Xm 201. 8pts An office is heated using a heat pump system based on a vapour-compression cycle with methylchloride (CH3Cl, R-40) as refrigerant.

The evaporator operates at Te = -20 °C and exchanges heat with the surroundings; the radiators of the office heating act as condensor, operating at Tc = 30°C.

The cooling cycle is shown in the T,s figure: saturated vapour is compressed (1-2), from 1.23 bar to 6.79 bar; the vapour is condensed to a saturated liquid (2-3-4); the liquid is throttled (4-5) and vaporised in the evaporator (5-1), closing the cycle.

The following data is given for these states:

T °C p bar h kJ/kg s kJ/(kg∙K)

1 -20 1.23 812 1.575

2 97 6.79 909 1.600

3 30 6.79 837 1.385

4 30 6.79 466

5 -20 1.23

State 5 lies in the two-phase region; for the saturated liquid in equilibrium with vapour

1 at -20°C the specific enthalpy is h6 = 389 kJ/kg.

a. Calculate the heat of vaporisation of the R-40 at 30°C and -20°C, and calculate the mixture quality x for state 5 in the T,s diagram. (2 p.)

b. How much heat qin (kJ/kg refrigerant) is taken up from the surroundings; how much heat qout (kJ/kg refrigerant) is given off to the office; how much is the compressor work w (kJ/kg refrigerant) (2 p.)

c. What is the coefficient of performance COP of this heat pump? Compare it with the COP of a Carnot process heat pump operating between -20°C and 30°C. (2 p.)

d. Calculate the entropy difference s5-s4 and check if the entropy increases during the throttling step. Note: s5-s4 = (s5-s1) + (s1-s2) + (s2-s3) + (s3-s4) (3 p.)

e. Can you explain the name R-40 for this refrigerant (1 p.)?

8 Xm 261. 10 p. A two-stage vapour-compression process, using refrigerant R-22, operates between the

temperatures 36 °C and -58°C, with intermediate temperature 0°C. The refrigerant mass

streams through the high temperature and low temperature heat exchangers are ṁ1 and

ṁ2, respectively. See the process schematic below, and assume ideal, isentropic

compressors. See also the log p, h diagram for R22 on the last page: hand in that page with your answer!

a. Give the values for the three pressure levels at which this process operates. b. Give the seven enthalpy values hA, hB, hC, hD, hE, hF, hG as indicated in the process schematic, and draw the process in the log p, h diagram. c. Calculate the quality, x (-), for the wet

refrigerant after throttle A-B (point B) and calculate the two enthalpy values hH and hI as indicated in the process schematic.

d. Calculate the combined power inputs P1 + P2 for the compressors, expressed as kJ per

mass stream ṁ1.

e. Calculate the ideal COPR, carnot and real COPR values (-) and from that the Carnot efficiency, ηCarnot (%) of this process.

36°C

-58°C

0°C

mixing point

throttle

throttle

gas/liquidseparator

Q1

Q2

Pcompressor 2

Pcompressor 1

A

B

C

D E

F

G H

I36°C

-58°C

0°C

mixing point

throttle

throttle

gas/liquidseparator

Q1

Q2

Pcompressor 2

Pcompressor 1

A

B

C

D E

F

G H

I

9 Xm 244. 10 pts See the schematic for

a steam-fired single-stage LiBr–water absorption chiller given in the figure . The chiller generates a cooling power of 3068 kW, denoted as “Useful effect (Cooling)”. It operates between the pressure levels ~ 0.88 kPa and ~ 9.8 kPa where the boiling point of water is ~ 5ºC and ~ 45ºC, respectively. The refrigerant is water, it is absorbed/desorbed in/from water-LiBr: a “concentrated” solution implies more water/less LiBr (X = mass fraction LiBr = 0.603), while “dilute” implies less water/more LiBr (X = 0.652). The table below provides the thermodynamic properties of the LiBr–water solution at the state point along the cycle. The second figure sketches the principal chiller components, heat flows and flow rates.

Based on this information and using mass and energy balances, determine: a. the ratio of solution mass flow to the generator to the refrigerant mass flow, ṁ5/ṁ6 , (which is

commonly called the circulation flow ratio CR) (2 p.) b. the refrigerant (water) mass flow ṁ6 = ṁ7 = ṁ8 (kg/s) and the mass flows of the concentrated

and dilute solutions ṁ4 = ṁ5 (kg/s) and ṁ1 = ṁ2 (kg/s), respectively; (3p.) c. the heat flows Qevap, Qcond, Qgen and Qabs (kW) , respectively; (3p.) d. the chiller’s COP and Carnot COPcarnot, defined as

and

(2 p.)

10. Xm248. 8 pts

A vapour absorption refrigeration system based on ammonia-water (see Figure above) has refrigeration capacity of 100 TR= 100 tons of refrigeration = 351.7 kW. The various state properties of the system shown below are given in the Table on the next page. Taking the heat rejection rate in the reflux condenser (Qd) as 88 kW, calculate: a. the mass flow rates ṁ10 = ṁ11 = ṁ12 = ṁ13 = ṁ14 = ṁ1 of (almost pure ammonia) solution

through the evaporator, of strong (ammonia) solution ṁ2 = ṁ3 = ṁ4 and of weak (ammonia) solution ṁ6 = ṁ7 = ṁ8, in kg/s (2 p.)

b. enthalpy values h1, h7, h8 and h13 not specified in the Table given on the next page (2 p) c. heat transfer rates at condenser, Qc, absorber, Qa, and generator, Qg, and solution pump

work, Wp, in kW (2 p.) d. system COP (2 p.), and e. plot the process in the Dühring plot for saturated states (!) of NH3/water mixtures given on

the next page. (2 p.) Hand in that page with the diagram with your answers. Process description, for more clarity, if needed: the figure, low temperature and low pressure vapour (almost pure ammonia) at state 14 leaves the evaporator, exchanges heat with the condensed liquid in Heat Exchanger-I and enters the absorber at state 1. This refrigerant is absorbed by the weak solution (weak in ammonia) coming from the solution expansion valve, state 8. The heat of absorption, Qa is rejected to an external heat sink. Next the strong solution that is now rich in ammonia leaves the absorber at state 2 and is pumped by the solution pump to generator pressure, state 3. This high pressure solution is then pre-heated in the solution heat exchanger (Heat Exchanger-II) to state 4. The preheated solution at state 4 enters the generator and exchanges heat and mass with the hot vapour flowing out of the generator in the rectification column. In the generator, heat is supplied to the solution (Qg). As a result vapour of ammonia and water are generated in the generator. As mentioned, this hot vapour with five to ten percent of water exchanges heat and mass with the rich solution descending from the top. During this process, the temperature of the vapour and its water content are reduced. This vapour at state 5 then enters the dephlegmator, where most of the water vapour in the mixture is removed by cooling and condensation. Since this process is exothermic, heat (Qd) is rejected to an external heat sink in the dephlegmator. The resulting vapour at state 10, which is almost pure ammonia (mass fraction greater than 99 percent) then enters the condenser and is condensed by rejecting heat of condensation, Qc to an external heat sink. The condensed liquid at state 11 is subcooled to state 12 in the subcooling heat exchanger by rejecting heat to the low temperature, low

Note: the “dephlegmator” condenses portions of the vapours rich in high boiling point constituents and returns the condensate to the column as reflux

pressure vapour coming from the evaporator. The subcooled, high pressure liquid is then throttled in the refrigerant expansion valve to state 13. The low temperature, low pressure and low quality refrigerant then enters the evaporator, extracts heat from the refrigerated space (Qe) and leaves the evaporator at state 14. From here it enters the subcooling heat exchanger to complete the refrigerant cycle. Now, the condensed water in the dephlegmator at state 9 flows down into the rectifying column along with rich solution and exchanges heat and mass with the vapour moving upwards. The hot solution that is now weak in refrigerant at state 6 flows into the solution heat exchanger where it is cooled to state 7 by preheating the rich solution. The weak, but high pressure solution at state 7 is then throttled in the solution expansion valve to state 8, from where it enters the absorber to complete its cycle.

ÅA ChemEng Refrigeration exam 15-5-2013 248. Name: Matriculate number:

11. Xm316. 10 pts.    

A heat pump system –  see  the Figure  ‐  is based on a modified  Joule  cycle with  two 

compressors. Heat q2 is taken in while heat q1 is given off by the system.  Ambient air 

(p2 = 1 bar) air is compressed in compressor C1 to an intermediate pressure level, p = 

pm.  After  C1,  heat  q1a  is  rejected  via  heat  exchanger  HEX  A.  The  air  is  then  further 

compressed to the pressure  level p1 = 4 bar by compressor C2,  followed by heat q1b 

rejected  via  heat  exchanger  HEX  B.  The  turbine,  T,  that  follows  (to  bring  back  the 

pressure to 1 bar) is driving compressor C1 while compressor C2 is driven by an electric 

motor.  

The  working  media  is  air  with  a  specific  heat,  cp,  of  1006  J/(kg∙K).  The  isentropic 

exponent  is  cp/cv  =  1,4,  and  the  isentropic  efficiencies  of  both  the  turbine  and  the 

compressors are 0.9. The temperatures Tc = Te = 30°C, while the ambient temperature 

Ta = ‐5 °C. Assume that the air can be treated as an ideal gas and calculate:  

 

a.   The temperature of the air after the turbine, Tf , with 1 decimal accuracy. 

b.  The intermediate pressure level, pm  (note: power generated by T = power used by 

C1). 

c.   The heating coefficient of performance COPHP. 

 

Advice: when calculating Tb ,Td or Tf, calculate first isentropic process outlet values Tb,is, 

Td,is or Tf,is and from that calculate Tb, Td or Tf.