HEAT TRANSFER – EVAPORATION & HEAT EXCHANGE SYSTEM
just answer the questions for the lab manual.
AEN340. Principles of Food Engineering Fall 2019 Note: Specific heat of water ( C p v ) at 45°C and water at atmospheric conditions, ( C pw ) are 4.176 kJ/kg.K and 4.18 kJ/kg.K, respectively; Latent heat of vaporization, Lv at 45°C is 2394.75 kJ/kg. LAB 4: HEAT TRANSFER – EVAPORATION & HEAT EXCHANGE SYSTEM (50 points) Due on/before 12 PM, November 12, 2019 The objectives of this Lab session are first, to demonstrate evaporation process; and secondly to estimate heat transfer coefficient and other stream parameters in tubular heat exchangers. Procedure: You are provided a single effect evaporator. In the set up that comprises of two tubular heat exchangers, there is heat transfer occurring between the product stream and steam in the first (See schematic below) tubular heat exchanger (on the right), and also an exchange of heat from the vapor released from heated product under vacuum to cold water stream flowing through the section of the system. Materials: Caliper, stop-watch, ruler/tape, containers, refractometer, large scale and apple juice. Required: 1. Draw schematic and graph diagrams of the two heat exchangers showing the direction of mass flow, temperatures at the inlet and exit points, all depicted with simple acronyms/symbols. The acronyms/ symbols should be well defined on a separate page/ line away from the diagrams. Note parameters that are NOT available for measurements. 2. Determine the quantity of apple juice sucked into the evaporator by gravimetric method, and note the time for the evaporation process. 3. Determine the final mass of the concentrated apple juice after operation for about 60 - 70 min. 4. Obtain the brix of the juice before and after evaporation (concentration). 5. At the operating condition (after at least 15 min of operation), collect data for temperatures and mass flow rates where possible. Note also the operating pressures (All in SI unit). 6. Measure the diameter and length of the annular pipe in the heat exchanger (on the right) that conveys the product into the evaporator. Questions: Vapor-Cold Water Heat Exchanger – Left Side Hex: 1. Determine rate of heat transfer, q using the cold stream. 2. Measure mass flow rate of the VAPOR-CONDENSATE stream. 3. Using q obtained from the cold water stream, determine mass flow rate of the VAPORCONDENSATE stream and compare to the measured value. 4. Determine the log mean temperature difference, ∆T lm and thermal resistance of the system. Product (Apple Juice)-Steam Heat Exchanger – Right Side Hex: 1. Calculate the log mean temperature difference, ∆T l m . 2. Determine the rate of evaporation by dividing the mass difference of apple juice before and after evaporation with time. 3. Compare the evaporator temperature to saturation temperature at the operating pressure (vacuum) of the evaporator. 4. Estimate the outside area available for heat transfer in the heat exchanger. General Questions: 1. Determine the brix of the apple juice before and after evaporation (concentration). 2. Explain the importance of evaporation under vacuum. Answer can be itemized. Give at least 2 points. 3. What will be the impact of increasing the flow rate of product stream on heat transfer rate while every other condition remains constant? Show this using the enthalpy/sensible heat equation.
: Specific heat of water ( C p v ) at 45°C and water at atmospheric conditions, ( C pw ) are 4.176 kJ/kg.K and 4.18 kJ/kg.K, respectively; Latent heat of vaporization, Lv at 45°C is 2394.75 kJ/kg.
- Determine rate of heat transfer (q) using the cold stream.
- q = MC∆T
- Where=
- q = heat transfer
- M = Mass of apple
- C = Specific heat of Apple
- ∆T = Change in Temperature
- Q = (1802.9 grams)(4.186 joule/gram °C)(42°C - 36° C)
- q = (702.2grams)( 3.64 J/g°C)(6°C)
- q = 15,336.04 J
- Measure mass flow rate of the VAPOR-CONDENSATE stream.
- m = ρvA
- Where =
- m = mass flow rate
- ρ = density of the Fluid
- mass/volume
- 1802.9g/500cm3
- 3.6058 g/cm3
- v = Velocity of the fluid
- 0.0018 m/s
- A = area or cross section
- A=2πrh+2πr2
- A = 2π(22.8mm)(698.5mm)+2π(22.8mm)2
- A = 1.03×105
- m = (1802.9g)( 0.0018 m/s)(1.03×105)
- m = 0.066851532 g/s
- Using q obtained from the cold water stream, determine mass flow rate of the VAPORCONDENSATE stream and compare to the measured value.
- a. q = 15,336.04 J
- Determine the log mean temperature difference (∆Tlm)
- a. LMTD
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