Example 8: 900 Kg dry solid per hour isdried in a counter current continues dryerfrom 0.4 to 0.04 Kg H20/Kg wet solidmoisture content. The wet solid enters thedryer at 25 °C and leaves at 55 °C. Freshair at 25 °C and 0.01Kg vapor/Kg dry air ismixed with a part of the moist air leavingthe dryer and heated to a temperature of130 °C in a finned air heater and enters thedryer with 0.025 Kg/Kg alry air. Air leavingthe dryer at 85 °C and have a humidity0.055 Kg vaper/Kg dry air. At equilibriumthe wet solid weight is 908 Kg solid perhour.*=0.0088Calculate:- Heat loss from the dryer andthe rate of fresh air.Take the specific heat of the solidand moisture are 980 and 4.18J/Kg.Krespectively,A. =2500 KJ/Kg.Humid heat at 0.01 Kg vap/Kg dry=1.0238KJ/Kg. "C. Humid heat at 0.055 Kg/Kg1.1084 KJ/Kg. "C2.81:41 مАдO

Answered: Image /qna-images/answer/d0acb491-e78b-4a61-a577-f44c84eb20e0.jpg

Answered: Example 8: 900 Kg dry solid per hour is…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

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A furnace heating unit has an airflow of 2,500 CFM. The heating coil entering temperature is 35°F dry bulb temperature with a relative humidity of 30%. the furnace heating coil leaving temperature is 80°F dry bulb temperature. Plot the points and the process line on the psychrometric chart. Find the sensible, latent, and total loads of the coil. (hint: heating does not change the moisture content in the air) OUTSIDE AIR 2,500 CFM 35°F DRY BULB TEMP. 30% RELATIVE HUMIDITY FURNACE 80°F DRY BULB TEMP. 2 SUPPLY AIR
Evaporator single effect is used to upgrade the concentrate 10000 kg/h of tomato juice than 5% solids the total to 30% solids total. Juice enters the Evaporator at 15 °C. The Evaporator is operated with Steam (quality 80%) in 143.27 kPa. The condition of vacuum in the Evaporator allow the juice to a boil on the 75 °C. The hot kind of material dilute is 4.1 kJ/(kg °C) and the product the concentrate is a 3.1 kJ/(kg °C). Count a. The Rate of the needs of steam = ...kg/h. b. The economic steam if the temperature is condensate is released at 75 °C. = ...(kg water evaporated/kg of steam)
The juice of the fruit at 25 °C containing total solids 5% concentrated through evaporator, single effect. Evaporator operated in vacuum at the temperature of evaporation at 80 °C, and steam with a quality of 85% is supplied in 169.06 kPa. Concentration desired of the final product is 40% solids total. Rate product concentrated out of the evaporator is 1500 kg/h. The hot kind of juice the fruit is 4.05 kJ/(kg °C), and product concentrated 3.175 kJ/(kg °C). Count a. The rate of steam required is at = kg/h. b. The economic steam (Steam economy) if the temperature of the condensate released at 90 °C. = kg water evaporated/kg steam Help me, please :-(:'(
Moist air at 21 C° DBT and 15 C° WBT and standard atmospheric pressure is humid by water spray. If for each one kg of dry air passing through the duct 0.002 kg of wa 100 C° is injected and totally evaporated. Draw the psychometric and schematic figu process and calculate for the air leaving the a- The moisture content. spray chamber: b- The enthalpy c- Dry bulb temperature.
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4. GoBrownsite is an obscure element with a molar mass of 126.2 grams per mol. Below is the cooling curve for GoBrownsite. How much energy needs to be removed from a 250.0 g sample of GoBrownsite knowing the following. Heat of Condensation AH = -165 kJ/°C • Specific heat of liquid GoBrownsite = 197 kJ/g-°C Heat of Freezing AH = - 231 kJ/°C Cooling Curve 80 e) to (1) 60 40 20 -20 Temperature 40 -60 -80 -100 (1) to (s) -120 Time
A feed of 4535 kg/h of a 2% salt solution at 311 k enters continuously a single effect evaporator and is being concentrated to 3%. 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 over – all heat transfer coefficient.
1) A beaker with 10 cm in height is filled with liquid solvent n-hexane up to a level2 cm below the top (mouth) of the beaker at 25°C and 1 bar. With a fan, dry air at 25°C and 1 bar is blown gently across the mouth of the beaker, so that the hexane vapor transferred through the air and n-hexane gas-vapor mixture by diffusion. Hexane concentration at the mouth of the beaker is negligible. Assume that air does not dissolve in liquid hexane, and the liquid height in the beaker is constant within observation period, hence the system is in steady-state. Molecular diffusivity of n-hexane in air is DAB -2x10 m²/s at 25 °C and 1 bar, and vapor pressure of n-hexane is pa°-152 mmHg at 25 °C. a) Draw a schematic picture of the mass transfer in the beaker. b) Discuss the type of mass transfer in the gas-vapor mixture. c) Using the equation dyA N=-cD 4 + y , (N , + N,) + y, (N, +N, d z Derive the molar flux equation for the condition given, showing every step of integration. d) Calculate the total…
Using a psychrometric diagram, describe the heating and humidification process and the following data. Initially, the air at dry bulb temperature (DBT) 25 ° C and RH55% The air is heated to DBT 75 C. The heated air is then flowed through a humidifier to increase the RH to 25%. For this process, calculate, a Change in the absolute humidity of the air from initial to final conditions air b. The change in air enthalpy from initial to final conditions = (kg of water / kg (kJ / kg air
ans: 0.01925kg, 0.7123kg, 1.886kPa Use gas laws
19.1 10 SLE oz 19.1-22. Diffusion and Heterogeneous Reaction on a Surface. In a tube of radius R m filled with a liquid, dilute component A is diffusing in the nonflowing liquid phase repre- sented by 16 be A828 XUh She lE SE dc NA=-DAB 35 00 A АВ dz where z is distance along the tube axis. The inside wall of the tube exerts a catalytic effect and decomposes A so that the heterogeneous rate of decomposition on the wall in kg mol A/s is equal to kc Aw, where k is a first-order constant and Ay is the wall area in m2. Neglect any radial gradients (this means a uniform radial concentration). Derive the differential equation for unsteady state for diffusion and reaction for this system. [Hint: First make a mass balance for A for a Az length of tube as fol- lows: rate of input (diffusion) +rate of generation (heterogeneous) rate of outpur (diffusion)+ rate of accumulation.] sds a'сА 2k дс А A = D Ans. A R az3 АВ OTOt 19.2-1.

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