Question

The power input to the compressor of an ammonia vapor compression plant is 8.2 kW. The mechanical efficiency is 85%. The ammonia is dry saturated vapor at -5^oC at inlet to the compressor. After compression, the vapor is at 1000 kPa. The compressor has an isentropic efficiency of 90%. The condenser produces saturated liquid. Calculate the coefficient of performance for the refrigerator and for the heat pump for this plant.

Answer 1

* Evaporator -50% and saturated vapous. expansion Valve w Compressor ad saturated liquid. P₂ = P2 = 1000 Kla Condensor P2 = 1000 kPa ste I 72 At point , m do kgk Ti= -5°c and saturated vopows. property table of Ammonia: h = 1599.8 kJ and S, = 6·1592 kJ kg foo isentropic compression, Si= S2= 6:1592 KS как. has-hi hza-hi as niso =0.90 and P2 =1000kPa At 52=6.1592 kJ kg k = 1741.2 kJ h2s kg = 0.90 17 41.2 - 1599.8 h2a - 1599.8 1741:2-1599.8 hza- 1599.8 0.90 i hra 157.44 + 1599.8 - 17 57024 KJ kg vul the ad **st these al uellhaus

.: Powere applied by compressor = m ( 12a-hu) m. (1757.24 - 1599.8) KJ Powere input to compressor = 8.2 kW Kg efficieny - 85% So Net powere given to compressor = 0.85 X 8'2 KW = 6.97 kW so m (1757.24 - 1599.8) = 6.97 kW m = 0.0443 kg S. Now, at B3 = 1000 kPa (as condensere operates at constant pressure) and saturated lequid. M₂ = 460.32 KJ kg Since enpansion is woenthalpic process, h₂ = hu = 460.32 kJ kg So, heat transfere in evaporator m (hi- hul -0.0443 0.004 3 kg (1599-8 -460-32) KJ : kg = 50.47 89 KW

Now, Desired effect cop of refrigerator Work input = 50.4789 kW 0.0443 X (haa-hi) 50.4789 kW 0.0443x(1757-24-1599.8) kW = 7.24 ... СЫР 7:24 op of refrigerator Meat pump = (COP) refingercator +1 COP = 7.24 + 1 = 8.24 .: COP for Pleat pump = 8.24 Any