● Dry air at 15 oC and 85 kPa flows over a 2m long wet surface with a free stream velocity of 3 m/s. Determine the mass transfer coefficient. • Because of low mass flux conditions, we can use dry air properties for the mixture at the specified temperature of 15°C and 1 atm, for which v= 1.47 *105 m²/s. The mass diffusivity of water vapor in air at 288 K is determined from Eq. below to be: T2.072 DAB 1.87 x 10-5 × P Answer:0.0115 m/s ● ●

● Dry air at 15 oC and 85 kPa flows over a 2m long wet surface with a free stream velocity of 3 m/s. Determine the mass transfer coefficient. • Because of low mass flux conditions, we can use dry air properties for the mixture at the specified temperature of 15°C and 1 atm, for which v= 1.47 *105 m²/s. The mass diffusivity of water vapor in air at 288 K is determined from Eq. below to be: T2.072 DAB 1.87 x 10-5 × P Answer:0.0115 m/s ● ●

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

See similar textbooks

Similar questions

Humid air at constant 101 kPa pressure, 30°C temperature and 40% relative humidity undergoes an isobaric cooling process unitl its temperature reaches 22°C. Calculate how much energy by heat is transferred from the air stream in kJ per kg of dry air. Give you answer to two significant figures and enter the numerical value only (eg. 0.0123, 0.123., 123, 1230, but not 0.0123 kJ/KGDA). 3
An air flow of 15,000 m3/h at 16 ºC and 40% relative humidity, you want to humidify up to conditions of 24 ºC and 80% relative humidity. The make-up water is at 18ºC. Calculate the necessary heat to be applied to the spray water. The specific volume of air at the inlet is 0.852 m3/kg
Moist air, which contains 2.30 mol% Hz0, at 25°Cand 100 kPa is to be dehumidified so thatduring its passage through a large cold room, ice formation can be avoided on the chilling coils in the room. Two suggestions have been offered: (1) Cooling the moist air atconstant pressure (2) Compressing the moist air atconstant temperature 1. Whatpressure should the moist air in process (2) reach for the process to obtain 60% percentremovalof water?
Moist air contains 0.025 kg water vapour per cubic metre of themixture at 313 K and 101.3 kPa. Calculate the following: The absolute humidity of the air
I need the answer as soon as possible
can u explane fast please
Answer are given all i need is the correct solution thank u
8. The foilowing observations were made during the trial of a boiler plant consisting of a battery of 6 lancashire boilers and an economiser: Calorific value of coal per kg= 30MJ/kg; Mass of feed water per kg of dry coal 9.1 kg; Equivalent evaporation from and at 100" C per kg of dry coal = 9.6 kg; Temperature of feed water to economiser 12" C: Temperature of feed water to boiler 105° C; Air temperature 13° C ; Temperature of flue gases entering economiser 370° C; Mass of flue gases entering economiser 18.2 kgkg of coal; Mean specific heat of flue gases = 1.005 kJ/kg K. %3D Find : 1. the cfficiency of the boiler alone; 2. the efficiency of the economiser alone ; and 3. the [Ans. 72.22% ; 54.43%; 84.1%l efficiency of the whole boiler plant.
None of above A mixture of water (density=1000 kg/m3) and vapor flowing with a mass rate= 7 kg/sec in a pipe of 1000 m long and 50 mm diameter, the gas fraction is 20%,then the volumetric rate of the water is: O 0.0056 m3/sec O 0.056 m3/sec O 0.56 m3/sec 5.6 m3/sec
Thermodynamics
Air at 35 m dry and 50 relative humidity enters a cooling coil at a rate of 500 (l/sec) and exits saturated at 23 m dry, calculate the cooling capacity of the coil (cooling belly)
Using a psychrometric diagram, describe the heating and humidification process from the following data. Initially the air was at a dry bulb temperature (DBT) of 30 ° C and RH60%. The air is heated to DBT 80 ° C. The heated air is then flowed through a humidifier to increase the RH to 25%. For this process, calculate,a. Change in absolute humidity of air from initial to final conditions = (kg water / kg air).b. The change in air enthalpy from initial to final conditions. = (kJ / kg air)