An air conditioner cools humid air Temperature 1 =25 celcius Temperature 2 = 10 celcius Relative humidity 1 = 70% Determine the final relative humidity (2) and calculate: 1) the energy removed 2) the water mass extracted from the air
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Hi please show working? Thankyou !
An air conditioner cools humid air
Temperature 1 =25 celcius
Temperature 2 = 10 celcius
Relative humidity 1 = 70%
Determine the final relative humidity (2) and calculate:
1) the energy removed
2) the water mass extracted from the air
Step by step
Solved in 3 steps with 2 images
- Define absolute humidity and relative humidity. When an air conditioner is operated insummer, lots of water needs to be drained out from it. Briefly explain this situation inrelation to psychrometry.าา 3.15.) In air conditioning unit, 71,000 cfm at 80'F dry bulb 60% th and standard atmospheric pressure, enter the unit. The air leaves the wait at 57°F dry bulb and 90% relative humidity. Calculate the following a.) Cooling capacity of the air conditioning unit, in Btu/h 6.) rate of water removal from the unit. (.) sensible heat load on the conditioner, in Btu/h di) latent heat load on the conditioning in Blu/h e.) The dew point of the air leaving the conditioner.HUMIDIFIER: MASS BALANCE On a hot summer day, the air is at 30.6 ° C and 80% relative humidity. The air conditioning in a building supplies 30 m3 / min of air at 13 ° C to keep the air inside the building at an average temperature of 24 ° C and 40% relative humidity. If the ventilation switch of the air conditioner is “open”, the air from outside enters the unit as shown in the drawing (I): The air conditioner cools the air to a temperature low enough to condense the required amount of water and reheat it to 13 ° C (unsaturated current), at which point it has the same absolute humidity as the air in the building. Calculate: a. the speed at which condenses water (kg / min) b. the temperature at which the air must be cooled to condense water at this rate. Do the calculations without using the psychrometric chart (although 1 atm of current pressure can be assumed). If the ventilation switch is closed, as normal, the air from inside the building would be recirculated in the air…
- A summer air-conditioning shown below is required to maintain a certain room at specified conditions. The layout of the system is indicated on the next page and the specifications are listed below.System specifications:Required room temperature: 23⁰C, 50% Relative humidityAtmospheric pressure: 101 kPaOutside air conditions: 30⁰C, 80% Relative humidity Sensible heat gain: 5 kWLatent heat gain: 4 kWTemperature rise across fan: 1.5⁰CTemperature of air leaving cooler: 12⁰CTemperature of air entering room: 18⁰CApparatus dew point of cooler: 7⁰CHumidification is a process which involves . water vapor per Kilogram of dry air Select one: a. Cooling b. Increasing c. Reducing d. HeatingNo card B a m OO ON 94)| 1:50 mlms.hu.edu.jo/mod/qu 3 A 476 m rigid tank contains saturated Refrigerant-134a mixture at 700 kPa with 27% quality. A valve at the bottom of the tank is now opened, and only liquid is withdrawn from the tank. Heat is transferred to the refrigerant such that the pressure inside the tank remains constant. The valve is closed when no liquid is left in the tank and vapor starts to come out. What is the amount of R-134a that exited from the tank "kg"? R-134a Sat. vapor P = 700 kPa Sat. Liquid Select one: A. 42.72 B. 49.05 C. 45.49 D. 35.99 E. 39.55
- a) Determine the amount of work required to compressed 1.5 kg of air at a pressure of 7 barg(gauge pressure). The isentropic index is 1.2 and the air intake is at 1 bar, 20 degrees Celcius & Relative Humidity =65%.i) Find the temperature at the end of the compression.(ii) The discharged air is passed through an R134a refrigerant dryer, the temperature of air coming out of the dryer is 40-degree celsius. The air then settles inside a reservoir at a temperature of 25-degree Celcius. Find the humidity variation throughout the production of compressed air.How can I calculate the pressure 2 of a vapor compression refrigeration system, if in state one my pressure is 0.101. How can I calculate the increase in pressure in state 2? the system is located in a zone at 24C with a humidity of 90%. My refrigerant to use is 134aAir enters a humidifier at 20.25 C and 30% RH. The air is at a constant pressure of 101.325 kPa. If the air leaves the humidifier at 80% RH, what is: The exit dry-bulb temperature of the air in C The saturation effectiveness in % of the humidifier
- Fill the following blanks with a right word: k): During simple cooling process of air, specific humidity remains constant but its relative humidityA summer air-conditioning shown below is required to maintain a certain room at specified conditions. The layout of the system is indicated on the next page and the specifications are listed below.System specifications:Required room temperature: 23⁰C, 50% Relative humidityAtmospheric pressure: 101 kPaOutside air conditions: 30⁰C, 80% Relative humiditySensible heat gain: 5 kWLatent heat gain: 4 kWTemperature rise across fan: 1.5⁰CTemperature of air leaving cooler: 12⁰CTemperature of air entering room: 18⁰CApparatus dew point of cooler: 7⁰C Draw the processes on a psychrometric chart and determine the following: 1.The ratio of mass flow rate of re-circulated air to the mass flow rate of air supplied to the room. 2.The heat removed by the cooling coil.3.The heat supplied in the re-heating coil. 4.The cooling coil bypass factor.Frost accumulates on the evaporators of forced-draft refrigerators because A. they are generally located within a house. B. they are generally located in high-humidity areas. C. they are low-temperature appliances. D. there is a heat exchange with the suction line.