A Proposed Design of Ice Plant :
The plant can produce 30 tons of ice per day. Capacity/day = tons of ice / mass of ice block = 30000kg / 150kg/block = 200 blocks/day 1. To determine the size of ice to be produced. 2. To be able to know the kind of refrigerant to be used. 3. To know the specifications of the important equipment to be used in the design. Significance of the Study Methodology The design will use the international standard size of ice can, 11” (28cm) Long, 22” (56cm) Wide, 48” (122cm) High. The size of the ice block is approximately 330lbs (150kg). ICE CUBE SPECIFICATION BRINE TEMPERATURE The brine temperature may be computed by using the equation T= Where: T = brine temperature, °F a = thickness of ice block, inches n = no. of ice can / ton of ice / day W = mass of ice block, (lbs); use 330lbs For the value of n n = no. of ice can / ton of ice produced # of ice can = 200 say 400 to maximize and avoid shortage n = 400 blocks /30 tons n = 13 ice block / ton of ice For the value of T T = [ 32 – 582.3(11)² ]/(330)(13) = 15.58 °F or -9.12 °C FREEZING TEMPERATURE From the book Refrigeration Engineering by Macintire and Hutchinson, p. 571 freezing time can be evaluated using the equation and from the book Refrigeration by Chris Langley constant varying varies from 5.5 to 7: t= t = Freezing temperature C = Constant varying from 5.5 – 7
T = Brine temperature X = Thickness of ice t= t = 39.65 hrs COOLING LOAD CALCULATIONS: Raw Water Temperature = 80.6°F (27°C) Freezing Temperature = 32°F (0°C) Brine Temperature = 15.58°F (-9.12°C) Ice Temperature = 25.58°F (-3.5667 °C)
Q3 = raising the temperature of water from its freezing temperature to the brine temperature = (mass) (specific heat of water) (temperature difference) = (30,000 kg/day) (2.093 KJ/kg-°k) (0-9.12) (1/24) (1/3600) = 6.628 KW QT = (39.253 KW + 123.264 KW + 6.628 KW) = 169.145 KW For typical ice plant there is 15% allowance, therefore Qt = Qt + Qt (.15) = 169.145 + 169.145(.15) = 194.52 KW COOLING LOAD CALCULATIONS: Raw Water Temperature = 80.6°F (27°C) Freezing Temperature = 32°F (0°C) Brine Temperature = 15.58°F (-9.12°C) Ice Temperature = 25.58°F (-3.5667 °C)
Total heat load can be completed using QT= Q1 + Q2 + Q3 Where: Q1 = sensible heat required to lower temperature of water from its raw temperature to its freezing temperature. = (mass) (specific heat of water) (temperature difference) = (30,000 kg/day) (4.187 KJ/kg-°k) (27-0) (1/24) (1/3600) = 39.253 KW Q2 = changing the latent heat of water at its freezing temperature from liquid to solid = (mass) (latent heat of water) (30,000 kg/day) (355 KJ/kg) (1/24)(1/3600) = 123.264 KW NUMBER OF TUBE REQUIRED:
Use a tube made of B.I. Pipe, 1”D. the area of one length of the B.I. Pipe 1”D is 6.89 ft2. In determining the surface area of the condenser, using the equation from the book Principles of Refrigeration by R. J. Dossat, 2nd Editon page 316, Q=AxUxD where: Q = condenser capacity, Btu / hr; use 837,130.401 Btu/hr A = surface area of the condenser, ft2 D = temperature difference between the condensing refrigerant and the condensing medium, °F; use 10 °F U = overall heat transfer coefficient in Btu/hr-ft2-°F from the book Design Values of U for Metallic Surfaces of Refrigeration Engineering by Macintire & Hutchinson, page 263, the overall conductance factor for shell-and-tube condenser is 150 to 300 Btu/hrft2-°F. Use 200 Btu/hr-ft2-°F
A=Q/UxD = 837,130.401 Btu/hr-ft2-°F/225(10°F) = 372.057 ft2 A = 372.057 + 15% allowance = 372.057 + 372.057(.15) = 427.867 ft2 no. of tubes = A / surface area of the tube used m = Qr/ (h1 – h3) = 245.277KW / 1443.9 kJ/kg – 390.587 kJ/kg m = .233 kg/s Volume flow rate, f F = m/ρof ammonia = 681.92 kg/m3 f = (0.233kg/s) / (681.92 kg/m3) f = 3.415x10-4 m3/s
Pipe Size Q = AV (use velocity of 2 m/s from INDUSTRIAL REFRIGERATION HANDBOOK, CHAPTER 9 PIPE SIZING, page 349.) 3.415x10-4 m3/s = [πD2 (2 m/s)]/ 4 D = 0.0147 m = 0.557 in (use B.I pipe with nominal diameter of 1 in) use 1.315in diameter from (http://gotocfr.com/tools/pipe-schedule.pdf) = (1.315in) (1 ft/12 in) (20 ft) = 6.89 ft2 per length of B.I. Pipe = 427.867 ft2 / 6.89 ft2/ length = 62.14 length of B.I. pipe = 63 length of B.I. Pipe 1” D ENTHALPIES The enthalpies can be calculated by knowing this two temperatures from the compressor model # VZ6AVT(HASEGAWA), the
Evaporating temperature (-15°C) and the Condensing temperature (40°C). The refrigerant is ammonia; we can refer to the table A-3 page 420 RAC by Stoecker & Jones. By h1 h3 S1 h2 h4
interpolation: = 1443.9 kJ / kg = hf @ 40°C = 390.587 kJ / kg = S2 = Sg @ -15 °C = 5.827 kJ/kg K = h1 = 1,557 kpa = 1720 kJ/ kg = h3
CONDENSER Qr = mr ( h2 – h3 ) where: Qr = heat rejected Qa = mr ( h1 – h2 ) mr = Qa / ( h1-h3 ) = 194.52 / (1443.9 – 390.587) = .1845 kg/s Qr = .1845 kg/s (1720 -390.587 ) = 245.277 KW Mass of cooling water Assuming: Water outlet temperature = 30 - 32°C Water inlet temperature = 26 - 28°C mw = Qr / cp ( T ) = 245.277 / 4.187 kJ/kg – K) ( 31 – 27) = 14.65 kg/s FLOW RATE f = Mw/Pw = 14.65 kg/s / 1000 kg/m3 = .0146 m3/s
Based on the specification of the Model HVU-VZ6AMT(HASEGAWA) is appropriate in this design with a capacity of 217 kW and operates at a maximum speed of 1075 RPM. Other data from the manufacturer’s specification are listed below. Capacity = 216.98 kW No. of cylinder = 6 Cylinder: Bore mm) = 132 Stroke (mm) = 106 Speed = 1075 RPM Piston Displacement = 561.38 m3/hr Refrigerant Connection
Suction (mm) = 100 Discharge (mm) = 75 Standard Motor = 45 kW COMPRESSOR The compressor used in this design is a flanged motor compressor, single stage and the refrigerant used is ammonia. The selection of the compressor is based on the heat load. Q = 194.52 kW FOR TYPE OF BELT Power transmitted = 60.35 hp (45 kW) Speed of Motor = 1075 rpm Design Power = power transmitted x normal torque Normal torque = 1.2 from table 7.1 machine elements in mechanical engineering by Robert L. Mott Design Power = (60.35) (1.2) = 72.42 hp For sheave pulley Most commercial available sheaves are cast iron, which should be limited to 6500 ft/min Belt speed. In this design use belt = 4000 ft/min Vd = belt speed Dd = Driver sheave = 12 Vd/ n= 12(4000) / 3.1416(1075) = 14.21
For length of V belt From table 7-8 Machine Elements in Mechanical Design by Robert L. Mott the center distance is ranging D2 < C < 3 (D2 + D1) D2 < C < 3 (21.1 + 13.9) 21.1 < C > 105 For the interest of conserving speed use C = 30 From the book Machine Elements in Mechanical Design by Robert L. Mott, equation 7-3 L = 2C + /2 (D2 + D1) + ( D2 – D1 )/ 4 = 2(30) + ( /2) (21.1 + 13.9) + (21.1 – 13.9/4(105)) = 136.04 inches From table 7-2 Machine Elements in Mechanical Design by Robert L. Mott select standard belt = 140 inches
For actual center distance The center distance assumed earlier is just for the purpose of solving L,
now that we have L, we can calculate the actual center distance by the equation 7-4 and 7-5 from the book Machine Elements in Mechanical Design by Robert L. Mott C= B = 4L – 6.28 ( D2 + D1 ) = 4(136.04) – 6.28 (21.1 + 13.9) = 324.36 Solving for C C= = 40.38 Number of Belts Ө1 = 180°– 2 sin-1 [(D2 – D1) / 2C] = 180°– 2 sin-1 [(21.1 – 10.8) / 40.38] = 154° Determine the correction factors from Figure 7-14 and Figure 7-15 Elements in Mechanical Design by Robert L. Mott For Ө1 = 154° CӨ = 0.93; For L = 140 in; CL = 1.0 Corrected Power = CӨ x L x CL = (0.93) (27.69) (1.0) = 25.75 hp Number of Belts = 439.94 / 19.32 = 2.8 (use 3 belts)
LIQUID RECIEVER The size of the liquid receiver is such that it can store the total volume of the refrigerant when pumping down the system. From the compressor capacity Compressor Capacity = m (h2 – h3) Compressor capacity = 194.52 kW m = mass flow rate m = 194.52 / (1720 – 390.5870) m = .14632 kg/s total required mass of ammonia refrigerant system is = .14632 kg/s (15 min) (60sec/min) = 131.688 kg = 290.322 lb V = m / ρ brine ρ brine = 39.96 lbs/ ft3 from table RAC by Stoecker & Jones V = 290.322 lb / 39.96 lbs/ft3 ACCUMULATOR
Normally accumulator is should be less than 30% to 40% of the total evaporator capacity. (use 35% allowance). D = .062 ft R = .031 ft V = Volume of evaporator coil V = π (.031)2 (20) (203) = 12.257 ft3 x (.35) = 4.29 ft3 BRINE AGIGATOR The brine agitator is installed in the end portion of the brine tank because it is more convenient when removal is required for repair. The brine agitator should have the capacity of 60 to 70 gallons/ton/min. (use 65gallons/ton/min). P = ϒ Qb H Qb = 65 gallons/ton/min (30 tons) = 1950 gallons/min (1/60) = 32.5 gal/s or .123 m3/s P = (1.755)(9.81)(.123)(1) = 2.117 kW or 2.84 hp FREEZING TANK Freezing tank or brine tank is provided with a race way, which the evaporator coil is installed. In this design the coil race way has a width of 39.37 inches. Base = base of ice x 20 blocks/row + 39.37 + .5 inch gap between blocks x 20 blocks = 269.27 inches or 7 meters Length = length of ice x 20 blocks + 1 inch gap x 20 gaps = 470 inches or 12 meters PUSH TROLLEY AND ELECTRIC HOIST Ice blocks are too heavy for human labor to remove it from brine tank so an electric hoist is needed. Weight of ice block = 150 kg Number of ice can per row = 20 Total weight per row = 150 (20) = 3000 kg Add 15% for the additional load = 3000 kg x 1.15 = 3450 kg Therefore, Use a 3.5 ton hoist COOLING TOWER The selection of cooling tower is based on the computed quantity of the condensing medium, which is water.
Quantity of condensing medium = flow rate of water in condenser = .014 m3/ s = 50.4 m3/ hr From AOSUA table of specification model AB-60 is selected with the following specifications: Nominal flow = 60 m3 / hr Motor = 1.5kW (2hp) Fan diameter = 990 mm Water inlet diameter = 100 mm Water outlet diameter = 100 mm Net weight = 368 kg
