For the pasteurization process, the cold milk with a 4°C temperature value enters the first heat exchanger system. The milk exits the first eat exchanger at 27 °C and enter the 2nd heat exchanger system and exit 54°C. Finally, the milk exit from the 3rd heat exchanger at 72°C (pasteurization temperature). The heat transfer is conducted by steam. The modeling procedures for system controlling is performed based on the initial cold milk temperature. For this system, a. Define process control elements and process dynamic. b. Feedforward or feedback system why? c. How could you express transfer function for this system

For the pasteurization process, the cold milk with a 4°C temperature value enters the first heat exchanger system. The milk exits the first eat exchanger at 27 °C and enter the 2nd heat exchanger system and exit 54°C. Finally, the milk exit from the 3rd heat exchanger at 72°C (pasteurization temperature). The heat transfer is conducted by steam. The modeling procedures for system controlling is performed based on the initial cold milk temperature. For this system, a. Define process control elements and process dynamic. b. Feedforward or feedback system why? c. How could you express transfer function for this system

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Consider the steady-state counterflow heat exchanger shown below. There are separate streams of air and water, and each stream experiences no noticeable change in pressure. Stray heat transfer with the surroundings and changes in kinetic and potential energy can be ignored. For the air, the ideal gas model can be applied and Rair = 0.287- For the operating conditions provided on kg-K the figure, determine: a. The temperature of the air at the outlet of the heat exchanger, T4, in [K] b. The rate of heat transfer between the air and the water, in [kW; , c. The rate of entropy production for the heat exchanger, in [kW/K] kg msteam = 12 P1 = 3 bar X1 = 1 P2 = P1 T2 = 200 P3 = 1 bar T3 = 1100 K P4 = P3 T, =? mair kg = 3.29
. A simple air cooled system is used for an aeroplane to take a load of 10 tons. Atmospheric temperature and pressure is 25°C and 0.9 atm respectively. Due to ramming the pressure of air is increased from 0.9 atm, to 1 atm. The pressure of air leaving the main compressor is 3.5 atm and its 50% heat is removed in the air-cooled heat exchanger and then it is passed through a evaporator for future cooling. The temperature of air is reduced by 10°C in the evaporator. Lastly the air is passed through cooling turbine and is supplied to the cooling cabin where the pressure is 1.03 atm. Assuming isentropic efficiency of the compressor and turbine are 75% and 70%, finda) Power required to take the load in the cooling cabinb) COP of the system.The temperature of air leaving the cabin should not exceed 25°C.
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DS Community Downl... Haritalar EOkuma list iş İÇİN Giriş yap 口口山凸9 İNGİLİZCE An air-cooled condenser is used to condense isobutane in a dual geothermal power plant. Isobutane, air entering at 22 oC with a flow of 18 kg/s (cp = 1.0 kJ / kg.K)is condensed at 85 oC. The total heat transfer coefficient for this heat exchanger is 2400 W/m2.K and surface area is 2.6 m2. What is the output temperature of the air? Çeviri yap: Google Bing
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Water melts as temp increases because : - chemical potential of the liquid is lower than that of gas - liquid and solid form eutectic mixture -chemical potential of the liquid and solid become equal
Heat-Transfer Coefficient in Single-Effect Evaporator. A feed of 4535 kg/h of a 2.0 wt % salt solution at 311 K enters continuously a single-effect evaporator and is being concentrated to 3.0%. The evaporation is at atmospheric pressure and the area of the evaporator is 69.7 m². Saturated steam at 383.2 K is supplied for heating. Since the solution is dilute, it can be assumed to have the same boiling point as water. The heat capacity of the feed can be taken as cp = 4.10 kJ/kg · K. Calculate the amounts of vapor and liquid product and the overall heat-transfer coefficient U. 8.4-1.
In an absorption refrigerator, the energy driving the process is sllPplied not as work, but as heat from a gas flame. (Such refrigerators commonly use propane as fuel, and are used in locations where electricity is unavailable. *) Let us define the following symbols, all taken to be positive by definition: Qf = heat input from flame Qe = heat extracted from inside refrigerator Qr = waste heat expelled to room Tf = temperature of flame Te temperature inside refrigerator Tr = room temperature Use the second law of thermodynamics to derive an upper limit on the COP, in terms of the temperatures Tf, Te, and Tr alone.