Question

A portion of an ammonia (NH) refrigeration cycle is shown in the figure below. Two streams enter a mixing chamber with one exit stream. One stream with mass flow rate my firsts enters a heat exchanger at Ti= 30 °C (state 1), is cooled to 15 °C (state 2) and then enters the mixer. The rate of heat transfer from the water side of the heat exchanger, Q7, is 100 kJ/s. The second stream with mass flow rate na -2 kg/s flows through a throttle before entering the mixing chamber to reduce the pressure to 300 kPa (state 4). The entrance temperature to the throttle is 20 °C (state 3). The ammonia enters the mixing chamber at state 4 through a 0.1 m diameter pipe. The pressure of the ammonia entenng and exiting the mixing chamber is 300 kPa, 1.e. P2 = P = Ps= 300 kPa. The kinetic energy at states 2 and 5 can be considered negligible but the kinetic energy at state 4 is not negligible. Treat ammonia as an ideal gas. The heat exchanger and mixing chamber are well insulated. a) Determine the mass flow rate, m. (3 Marks) b) Determine the temperature entering the mixing chamber at state 4. (2 Marks) c) Determine the entrance velocity, V4. (4 Marks) d) Determine the temperature at state 5. (5 Marks) M-2 kg/s T,-20 °c ® Throttling Valve D.-0.1 m © Ammonia T-30 °C Heat Exchanger Mixing Chamber T=15 °C P-P-P-300 kPa T. 6,-100 KW TO Water

Answer 1

for Ammonia Gas , specific heat capacity 9= 2.19 KIKOK (a) Using Energy Balance at heat exchanger - Heat transfer to watere = ma xC x (T1-T2) 2) 100 = my x2-19% (30-15) 3 iz = 3.044 Kg/s Ang 1. (b) at state 4, To - Saturation Tempereature at (Pu=300K Pal 1. because Ammonia would be in two-phase state sog Using saturation property table of Ammonia af 300 kPa (3bar) Satweation Temperature. Tsat = – 9.24 °C Tq = -9. 24°C CSScanned with cabasdannek

C first we have to find specific volume at state 4. for that we will go through following steps: STEP1 : at state 3, Enthalpy, hz = hf at 20°c = 274.26 K7kg During throttling process, Enthalpy remains constant, ha=h=274.26 KJ /kg Scanned with CamScanner

hfg at 3ookPa STEP2: Dryness fraction at state 4, Xo = .ha-hf at 300kPa = 274.26-137.42 1294.05 X4 = 0.1057 specific volume, Vq = Vs +Xq vfg m/ kg V4 = 0.00153 + 0.1037*0.4045 V4 = 0.0443 m kg Sony Velocity, Vi= ™ax V 2X0.0443 44 - 11.2.809 Wys) (d) Using Energy Balance at mixing chamber (maxha) + 1 mg Vg²+ inghe = mahs Note: ñg = math (mass flow rate ] balance = 2+3.044 his = 5.044 kg/s h2 = 14 90.56 KJ /kg at 300kPa $15°C) CSscanned with CamScanner L super heated table OO

Boy (2 x 274.26) + ( x2x11-283)+(3.044X1490.56) = 5.044 xh5 = 15 = 1042.6 KY/kg (satuweated 2-phese Region) Now- Temperature is = Tsat at 300 kPa 3-9.24°C CD Csscanned with CamScanner