424519 Refrigeration Kylteknik 2017             Thermal and Flow Engineering      

         

Exercises 3 + 4      23.2+ 2.3 2017 11. CB98 – question 10-29

12. Ö96 – question 1.2 In the so-called Linde process, in which air is cooled from atmospheric pressure, 290 K to a liquid form at 80 K, the energy input is given to be 2.10 kWh electricity per kg air. The difference in specific enthalpy for air as gas at 290 K and liquid at 80 K is 419 kJ/kg, while the difference in specific entropy is 3.87 kJ/(kg·K). Calculate (if possible) the total efficiency ηt based on this information.

13. Ö96 – question 6.5 Calculate the mechanical power needed per kg liquid air at 100 kPa for a) a reversible process (entropy generation = 0), and b) a simple Linde process with reversible isothermal compression to 20 MPa. Air inlet temperature (= surroundings temperature) is 290 K. Use the enclosed (next page) pressure – enthalpy diagram for air (Ö96-Figure 6.2)

For / till 13. Pressure – enthalpy diagram for air (Ö96-Figure 6.2) (reference value: h=0 for saturated liquid at 1 bar)

Sources: CB98: Y.A. Çengel, M.A. Boles “Thermodynamics. An Engineering Approach”, McGraw-Hill (1998) Ö96: G. Öhman ”Kylteknik”, Åbo Akademi University (1996)

14. Xm225. In a vapour-compression refrigeration process, CO2 is used as the refrigerant. See the process scheme: compressed superheated gas at T1 = 10°C, p1 = 25 bar is compressed with isentropic efficiency ηc isentrop = 0.7 to p2 = 90 bar. The compressed gas is then cooled with ambient air (Tair = 20°C) to T3 = 30°C, followed by heat transfer (3→4), at constant pressure, with the low pressure stream (1→6). The cooled pressurised gas is throttled to p5 = p6 = p1. At state 6, the CO2 is a gas/liquid mixture with gas mass fraction (“quality”) x6 = 0.9, at T6 = T5 = -12°C, p6 = 25 bar. Use the table below for CO2 data.

a. Calculate the enthalpies at state 6, h6 , and at state 4, h4 . (2p.)

b. Calculate the compressor work needed per kg refrigerant (kJ/kg). (3 p.)

c. Calculate the mass fraction gas in the mixture at state 5, x5 (kg/kg). (2p.)

d. Calculate the COP of this refrigerator for cooling effect Q56 . (1½ p.)

e. Calculate the refrigerant mass flow ṁ (kg/s) for a 10 kW cooling effect (1½p.)

CO2

´ = saturated liquid ´´= saturated gas Critical point data : Tc = 31,0 C pc = 7.38 MPa ρc = 476.6 kg/m3 hc = 332.25 kJ/kg sc = 1443.6 J/(kg·K)

Q23

Q56

15. Xm203. Air enters an evaporative cooler at 30C and 20% relative humidity.

How much liquid water (as g/kg of dry air) at 15C must be evaporated to cool the air

to 20C and a relative humidity of 70%? (5 p.)

Neglect the enthalpy brought in with the liquid water. Use the psychometric chart for (humid) air given below.

16 Xm226.

A 50 kg box of cherries at 10ºC having a water content of 77 %-wt is to be frozen to a temperature of -20ºC. Assume respiration only until the temperature has dropped to 0ºC (use a value for heat of respiration for the average temperature 5ºC). a. Neglecting respiration, calculate the amount of heat that must be removed (kJ). b. If the cooling from 10ºC to 0ºC is to be accomplished within 1 hour, calculate

the minimum cooling/freezing capacity of the refrigerator (W) and the time needed to reach -20ºC (hours).

17. Xm247. A stream of high pressure nitrogen,

N2, is used to produce liquid nitrogen (stream 4). See the figure. High pressure N2 (stream 1) is cooled (stream 2) and throttled to a two-phase mixture at atmospheric pressure. Saturated liquid (stream 4) is extracted after separation from gas (stream 3) which is passed through the heat exchanger to cool the incoming stream. The heated gas (stream 5) is released to the surroundings. Neglect pressure drop in the heat exchanger for both streams.

(continues on next page) Using data for N2 as given in the h, log p diagram enclosed:

a. What is the mass flow ratio γ = ṁ4/ṁ1 (kg/s)/(kg/s) of liquid product per incoming N2? b. What is the temperature T2 of the nitrogen in state 2? c. What is the entropy generation, Ṡgen, for this process, per kg input high pressure N2? Use the enthalpy diagram for nitrogen (R-728) given below.

p1 = 10 MPaT1 = 27ºC

18. Xm228. Ammonia (NH3) is stored in liquid form, in phase equilibrium with a small gas volume, in a storage vessel at atmospheric pressure (plow = 1 bar). The heat insulation is good but not perfect and a heat flow Qin = 843 W enters the storage from the surroundings. As a result, a small mass amount ṁvap (kg/s) is continuously vaporised. A vapour-compression process is used to liquefy the ammonia that has vaporised. A stream ṁ (kg/s) of saturated ammonia vapour (state 1) is continuously withdrawn from the vessel and is compressed with isentropic efficiency ηc isentrop = 0.85 to phigh = 12 bar (state 2). After that the stream is cooled at constant pressure to a saturated liquid (state 3) in a heat exchanger, extracting a hear flow Qout from the stream. Finally the pressure of the stream is reduced to plow = 1 bar, giving a partially liquefied stream (state 4) which is returned to the storage vessel. a. See the log p, h diagram at the end of the exam: draw the process cycle in the diagram and

hand it in with your answers. b. Calculate the enthalpy difference (h1 – h4) and calculate the mass fraction γ (kg/kg) of

stream ṁ that is liquefied. c. Assuming steady-state operation, calculate the stream ṁ (kg/s) that is extracted from the

vessel, and the stream ṁvap (kg/s) of ammoniak that evaporates in the storage vessel as a result of heat Qin.

d. Calculate the enthalpy difference (h2-h1) (W) and calculate the electric power P (W) needed for this compression if the total efficiency ηt of the compressor is 80 %.

19 Xm227. (Solar cooling)

a. Calculate how much cooling (in W) can be achieved with a reversible absorption refrigeration system driven by the sun. The heat from the sun is collected with a flat plate solar collector (A = 5m2) on the 16 of February (day = 47), at the equator, at noon so that the collector surface is perpendicular to the sun. It is assumed that the output temperature from the collector is 90 °C. This heat is then used in a single-effect absorption cooler. The temperature of the refrigerated space is Tref = 10 °C and the temperature of the ambient is Tamb = 34 °C.

b. Compare the result from a.) with an irreversible refrigeration case, where 50% of the incident energy is collected by the solar panel and the COP of the single-effect absorption cooler is 0.7.

20 Xm242. A cooling tower is used for cooling the condenser water of a refrigeration system that has a heat rejection rate of 100 kW. In the cooling tower air enters at 35ºC (dry bulb temperature) and 24ºC (wet bulb temperature) and leaves the cooling tower at 26ºC (dry bulb temperature) with a relative humidity of 95%. See the psychrometric chart given with Question 15.

a. Give the absolute humidity ω (kg H2O/kg dry air) and the enthalpy h (kJ/kg dry air) for the inlet and outlet air streams.

b. What is the required flow rate of wet and dry air at the inlet to the cooling tower in kg/s and in m3/s.

c. What is the amount of make-up water to be supplied, in kg/s.

The temperature of make-up water is at 30ºC; its enthalpy (hw) may be taken constant as 125.4 kJ/kg. Assume the barometric pressure to be 1 atm.