Chapter 8: Humidity and Humidity Chart (Solution)

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Problem 8.1: The vapor pressure of acetone at 285 K is 16.82 kPa. For a mixture of nitrogen gas saturated with the vapors of acetone at 285 K and 101.3 kPa, calculate (a) The mole percent of acetone in the mixture (b) The percent composition by weight of acetone (c) The mole ratio of acetone in the mixture (d) The weight ratio of acetone in the mixture (e) The amount of vapor in kilograms per m3 of the mixture.

Ans: (a) 16.6 % (b) 29.2 % (c) 0.1991 (d) 0.4127 (e) 0.4120 kg/m3

Explanation:

Problem 8.2: Carbon dioxide gas at 300 K and 100 kPa is saturated with water vapor. The vapor pressure of water at 300 K is 3.56 kPa. Calculate (a) The mole percent of water vapor in the mixture (b) The percent composition by weight (c) The mole ratio of water vapor in the mixture (d) The weight ratio of water vapor in the mixture (e) The amount of vapor in kilograms per m3 of the mixture.

Ans: (a) 3.56 % (b) H2O = 1.49 %, CO2 = 98.51 % (c) 0.0369 (d) 0.0151 (e) 0.0257 kg/m3

Explanation:

Problem 8.3: A gas holder contains dry nitrogen at 300 K and 101.3 kPa. Water is injected into the vessel till the gas is saturated. If the temperature after saturation is 300 K and the vapor pressure of water at 300 K is 3.56 kPa, calculate (a) The pressure inside the vessel and (b) The mole ratio of water to nitrogen in the saturated gas.

Ans: (a) 104.86 kPa (b) 0.0351

Explanation:

Problem 8.4: A gas holder contains 1000 cubic meters of gas saturated with water vapor at 300 K and 105 kPa. The vapor pressure of water at 300 K is 3.56 kPa. How many kilograms of water are present in the holder?

Ans: 25.71 kg

Explanation:

Problem 8.5: A certain exhaust gas consists of benzene vapor mixed with air. The partial pressure of benzene vapor is 300 Pa at a total pressure of 107 kPa and 298 K. Does this concentration exceed the ambient air quality stipulation of a maximum hydrocarbon concentration of 1.60×10-7 kg/m3?

Ans: Contains 0.946×10-2 kg benzene per m3

Explanation:

Problem 8.6: What is the minimum volume (cubic meters) of dry air at 293 K and 100 kPa necessary to evaporate 6 kg of ethyl alcohol if the total pressure remains constant at 100 kPa? Assume that the air is blown through the alcohol to remove it in such a way that the exit pressure of the air-alcohol mixture is 100 kPa. The vapor pressure of alcohol (kPa) is given as

\ln P^s=16.5092-\frac{3578.91}{T-50.50}
Ans: 51.93 m3

Explanation:

Problem 8.7: Dry air at 295 K and 100 kPa is bubbled through benzene. If the saturated air leaves at 300 K and 100 kPa, how many kilograms of benzene are evaporated per 100 m3 of entering air? The vapor pressure of benzene (kPa) is given by

\ln P^s=13.8858-\frac{2788.51}{T-52.36}
Ans: 50.99 kg

Explanation:

Problem 8.8: 10 kg KClO3 is decomposed and the oxygen evolved is collected over water at 300 K and 100 kPa. Determine the weight of saturated oxygen obtained if the vapor pressure of water at 300 K is 3.5 kPa.

Ans: 4 kg

Explanation:

Problem 8.9: Dry air is saturated with hexane vapors at 253 K and 101.3 kPa. The vapor pressure of hexane at 253 K is 1.85 kPa. What is the percent excess air for combustion?

Ans: 18.8 %

Explanation:

Problem 8.10: It is desired to prepare a saturated methanol vapour-air mixture containing 2 % methanol by bringing dry air in contact with methanol kept in a container at 100 kPa. What temperature should be used to achieve the desired composition? The vapor pressure of methanol is given by the Antoine equation

\ln P^s=16.5725-\frac{3626.25}{T-34.29}
Ans: 262.7 K

Explanation:

Problem 8.11:  Air at 303 K and 101.3 kPa is 60 percent saturated with water vapor. If it is desired that 90 percent of water is to be removed by compression at a constant temperature, what should be the pressure to which it is compressed? The vapor pressure of water at 303 K is 4.24 kPa.

Ans: 1623 kPa

Explanation:

Problem 8.12: A mixture containing benzene vapor and nitrogen in the weight ratio 0.1790:1 is found to have a percent humidity of 40 % at 101.325 kPa. What is its relative humidity?

Ans: 43.6 %

Explanation:

Problem 8.13: A mixture of benzene vapor and nitrogen gas at 105 kPa and 330 K contains benzene vapor to the extent that it exerts a partial pressure of 15 kPa. The vapor pressure of benzene is given by the Antoine equation

\ln P^s=13.8858-\frac{2788.51}{T-52.36}

Determine the following: (a) The mole fraction of benzene in the mixture (b) The weight fraction of benzene in the mixture (c) The molal humidity (d) The absolute humidity (e) The molal saturation humidity (f) The absolute saturation humidity (g) The mass of benzene in 100 m3 of the mixture (h) The percent humidity (i) The percent relative humidity

Ans: (a) 0.1429 (b) 0.3172 (c) 0.1667 (d) 0.4643 (e) 0.7988 (f) 2.2254 (g) 42.67 kg (h) 20.87 % (I) 32.17 %

Explanation:

Problem 8.14: A mixture of benzene vapor and air contains 10% benzene by volume at 300 K and 101.3 kPa. The vapor pressure of benzene at 300 K is 13.8 kPa. Calculate the following: (a) The partial pressure of benzene in the mixture (b) The weight fraction of benzene in the mixture (c) The molal humidity (d) The absolute humidity (e) The percent saturation (f) The percent relative saturation (g) The mass of air in 100 m3 of the mixture.

Ans: (a) 10.13 kPa (b) 0.2302 (c) 0.111 (d) 0.2990 (e) 70.44 % (f) 73.41 % (g) 106.2 kg

Explanation:

Problem 8.15: The percent saturation of a mixture of acetone vapor and nitrogen at 105 kPa and 300 K is found to be 80 %. The vapor pressure (kPa) of acetone is given by the Antoine equation with constants, A = 14.39155, B = 2795.817, and C = 43.198. Calculate the following: (a) The molal humidity (b) The absolute humidity (c) The partial pressure of acetone (d) The relative saturation (e) The dew point.

Ans: (a) 0.3710 (b) 0.7690 (c) 28.4 kPa (d) 85.4 % (e) 296.3 K

Explanation:

Problem 8.16: Moist air contains 0.025 kg water vapor per cubic metre of the mixture at 313 K and 101.3 kPa. Calculate the following: (a) The partial pressure of water vapor (b) The relative saturation (c) The absolute humidity of the air (d) The percent saturation (e) The temperature to which the mixture is heated so that its percent saturation becomes 10 %.
The vapor pressure of water (in kPa) is approximated by the Antoine equation as

\ln P^s=16.26205-\frac{3799.887}{T-46.854}
Ans: (a) 3.6122 kPa (b) 49.65 % (c) 0.023 (d) 47.79 % (e) 340.2 K

Explanation:

Problem 8.17: Air at 288 K and 100 kPa contains 1 kg of water vapor. Water is completely removed by adsorbing on silica gel. The dry air after adsorption measures 1000 m3 at 108 kPa and 293 K. What was the relative humidity of air before adsorption? The vapor pressure of water at 288 K is 1.7 kPa.

Ans: 7.35 %

Explanation:

Problem 8.18: The absolute humidity of a carbon dioxide-water vapor mixture at 310 K and 100 kPa is measured to be 0.022 kg water per kg dry carbon dioxide. Calculate the following: (a) The molal humidity (b) The percent relative saturation (c) The percent saturation (d) The temperature to which the gas is to be heated at constant pressure to reduce its percent saturation to 30 percent.
The vapor pressure of water in kPa is approximated by the Antoine equation as

\ln P^s=16.26205-\frac{3799.887}{T-46.854}
Ans: (a) 0.0537 (b) 82.42 % (c) 81.48 % (d) 327.45 K

Explanation:

Problem 8.19: A fuel gas saturated with water vapor at 288 K and 101.3 kPa has a heating value of 20000 kJ/m3 of the total gas. What will be the heating value of the gas in kJ/m3 of the gas at 295 K and 100 kPa if its relative saturation is only 50 %? The vapor pressure of water at 288 K and 295 K are respectively, 1.7 kPa and 2.6 kPa.

Ans: 19349 kJ/m3

Explanation:

Problem 8.20: A mixture of acetone vapor and air contains 8.3 % acetone by volume. Determine the dew point of the mixture if (a) The temperature and pressure are 300 K and 100 kPa (b) The temperature and pressure are 310 K and 90 kPa.
The vapor pressure of acetone (kPa) is given by the Antoine equation with constants, A = 14.39155, B = 2795,817, and C = 43.198.

Ans: (a) 271 K (b) 269 K

Explanation:

Problem 8.21: The dew point of air at an atmospheric pressure of 101.3 kPa is 300 K. The vapor pressure of water at 300 K is 3.56 kPa. Determine the mole fraction of water vapor in the air.

Ans: 0.035

Explanation:

Problem 8.22: If the dew point of air at 1.013 bar is 278 K, what will be the dew point at 10 bar? The Antoine constants of water are A = 16.26205, B = 3799.887, C = 46.854.

Ans: 315.4 K

Explanation:

Problem 8.23: Air at 298 K, 55% saturated with water vapor, is initially at 100 kPa. It is then compressed to 1000 kPa and cooled to a temperature so that 90 % of the water vapor is condensed. Calculate the following: (a) The final temperature of the air (b) The volume of final air per cubic meter of original air.
The vapor pressure of water (kPa) is given by the Antoine equation as

\ln P^s=16.26205-\frac{3799.887}{T-46.854}
Ans: (a) 288.9 K (b) 0.0954 m3

Explanation:

Problem 8.24: The dew point of a mixture of benzene vapor and dry air at 303 K and 100 kPa is found to be 288 K. The vapor pressure of benzene at 288 K is 7.8 kPa and at 303 K is 15.8 kPa. Determine the following: (a) The absolute humidity and molal absolute humidity of the mixture (b) The composition of the mixture in volume percent and weight percent (c) The relative saturation and the percent saturation.

Ans: (a) 0.2281, 0.0846 (b) Benzene 7.8 % (V), 18.6 % (W) (c) 49.37 %, 45.08 %

Explanation:

Problem 8.25: Methane is completely burned with air. After a portion of the water in the outlet gases from the burner which contains no oxygen is removed by absorption, the gas is found to contain 85% nitrogen at a temperature of 328 K and a pressure of 140 kPa. If more water is to be removed by condensation, determine the following: (a) The temperature to which this gas must be cooled at constant pressure (b) The pressure to which the gas must be compressed at constant temperature The vapor pressure of water in kPa is given by

\ln P^s=16.26205-\frac{3799.887}{T-46.854}
Ans: (a) 306.8 K (b) 420.9 kPa

Explanation:

Problem 8.26: A hydrocarbon fuel is burned with dry air. Determine the dew point of the flue gases at 533 K and 100 kPa if the dry flue gases analyzed 13.0 % CO2, 0.05 % CO, 2.5 % O2, and 84.45 % N2. The vapor pressure of water is given by the Antoine equation with A = 16.26205, B = 3799.887, and C = 46.854 when pressure is in kPa and temperature in K.

Ans: 322.9 K

Explanation:

Problem 8.27: Determine the mass of 10 m3 of air at 300 K and 115 kPa if the relative humidity is 65 % and the vapor pressure of water at 300 K is 3.56 kPa.

Ans: 13.25 kg

Explanation:

Problem 8.28: In a process in which benzene is used as a solvent, it is evaporated into dry nitrogen. The resulting mixture at a temperature of 300 K and a pressure of 101.3 kPa has a relative humidity of 60 %. It is required to recover 80 % of the benzene present by cooling to 283 K and compressing to a suitable pressure. The vapor pressure of benzene at 283 K is 6.02 kPa and that at 300 K is 13.81 kPa. To what pressure should the gas be compressed?

Ans: 344.22 kPa

Explanation:

Problem 8.29: Hydrogen saturated with the vapors of water at 101.3 kPa and 305 K is passed through a cooler in which the temperature of the gas is reduced to 280 K in order to condense a part of the vapor. The gas leaving the cooler is heated to 293 K. Calculate (a) The weight of water condensed per kg of moisture-free hydrogen and (b) The percent humidity of the gas in the final state.
The vapor pressure of water (in kPa) may be estimated using the equation

\ln P^s=16.26205-\frac{3799.887}{T-46.854}
Ans: (a) 0.3467 kg (b) 41.73 %

Explanation:

Problem 8.30: Air at 303 K, relative humidity 40 % is mixed adiabatically with air at 313 K, relative humidity 40 % in the ratio of 1 kg of the former to 2 kg of the latter on a dry basis. Find the final temperature and humidity of the air. The vapor pressure of water (in kPa) may be estimated using the equation

\ln P^s=16.26205-\frac{3799.887}{T-46.854}
Ans: 309.7 K, 0.0157 kg/kg of dry air

Explanation:

Problem 8.31: One cubic meter of a gas at STP with a dew point of 293 K is to be dried to a dew point of 278 K. How much water must be removed and what will be the volume of the gas at STP after drying? The vapor pressure of water is 0.872 kPa at 278 K and 2.337 kPa at 293 K.

Ans: 0.0117 kg, 0.9857 m3

Explanation:

Problem 8.32: Water is cooled by direct contact with ambient air in cooling towers. In a certain cooling operation, air enters the tower at 300 K with a partial pressure of water vapor of 670 Pa and leaves at 294 K with a partial pressure of water 2.2 kPa. The total pressure is 100 kPa. Calculate the following: (a) The relative humidity of air-water vapor at the inlet and exit of the tower (b) The percent composition by volume at the inlet and exit of the tower (c) The percent humidity of air at the inlet and exit of the tower (d) The mass of water vapor in kilograms at the inlet and exit per 100 m3 of dry air entering the tower (e) The mass of water vapor in kilograms at the inlet and exit per 100 m3 of wet air entering the tower.
The vapor pressure (in kPa) of water is given by

\ln P^S=16.26205-\frac{3799.887}{T-46.854}
Ans: (a) 19.2 %, 90.6 % (b) 0.67 % water, 2.2 % water (c) 18.63 %, 90.3 % (d) 0.487 kg, 1.625 kg (e) 0.484 kg, 1.614 kg

Explanation:

Problem 8.33: A mixture of nitrogen and methanol has a dew point of 313 K. The temperature and pressure of the mixture are 353 K and 200 kPa respectively. The vapor pressure of methanol (kPa) is

\ln P^S=16.5725-\frac{3626.55}{T-34.29}

Calculate the following: (a) The percent saturation of the mixture (b) Kilograms of alcohol per kilogram of nitrogen (c) Kilograms of alcohol per m3 of the mixture (d) Dew point of the mixture if the total pressure were 100 kPa (e) Pressure required to condense 60 % of the alcohol from the mixture at 300 K.

Ans: (a) 2.36 % (b) 0.244 (c) 0.384 (d) 298.9 K (e) 236.6 kPa

Explanation:

Problem 8.34: A mixture of acetone vapor and nitrogen gas at 300 K has a dew point of 283 K. It enters an evaporator where it picks up acetone vapor. The mixture leaving the evaporator is at 298 K and has a dew point of 293 K. The pressure is kept constant at 100 kPa. Determine the following: (a) The absolute humidity of the mixture entering and leaving the evaporator (b) The mass of acetone in kilograms evaporated per kilogram of nitrogen entering the evaporator (c) The mass of acetone in kilograms evaporated per 100 m3 of gas entering the evaporator (d) The volume of gas leaving the evaporator per 100 m3 of gas entering the unit.
The vapor pressure of acetone at 283 K is 15.4 kPa and 24.5 kPa at 293 K.

Ans: (a) 0.377 kg of acetone/kg dry of nitrogen, 0.6727 kg of acetone/kg dry nitrogen (b) 0.2953 kg (c) 28.05 kg (d) 111.3 m3

Explanation:

Problem 8.35: A mixture of air and acetone vapor at 303 K and 100 kPa has a dew point of 298 K. 90 percent of acetone in the mixture is to be condensed and removed by cooling and compression. If the mixture is cooled to 278 K, what should be the final pressure? The vapor pressure of acetone (kPa) may be estimated using the Antoine equation

\ln P^S=14.39155-\frac{2795.817}{T-43.198}
Ans: 285 kPa

Explanation:

Problem 8.36: Air supplied to a drier is at 340 K and 101.3 kPa and has a dew point of 278 K. The air leaves the drier at 310 K and 100 kPa with a dew point of 298 K. At what rate (m3 per hour measured at the drier inlet conditions) the air should be admitted to the drier if it is desired that 100 kg of water is removed per hour? The vapor pressure of water at 278 K is 0.87 kPa and at 298 K is 3.17 kPa.

Ans: 6498.35 m3/h

Explanation:

Problem 8.37: Air at a rate of 25 m3 per minute is supplied to a drier in which it removes moisture at a rate of 0.45 kg per minute. The air enters the drier at 333 K and 101.3 kPa with a percentage humidity of 10 and leaves at 310 K and 100 kPa. The vapor pressure of water at 310 K = 6.3 kPa and at 333 K = 19.9 kPa. (a) What is the percent humidity of air leaving the drier? (b) At what rate does air leave the drier in m3 per minute?

Ans: (a) 78.8 % (b) 24.23 m3/min

Explanation:

Problem 8.38: The atmosphere in the afternoon during a humid period is at 313 K and is 70 % saturated. During the night the temperature falls to 303 K. The pressure is 101.3 kPa. The vapor pressure of water at 303 K is 4.24 kPa and 7.38 kPa at 313 K. What percent of water in the afternoon air is deposited as dew during the night?

Ans: 20.5 %

Explanation:

Problem 8.39: A room of volume 100 m3 is to be maintained at 300 K and 60 % relative humidity. For this fresh air from the outside saturated at 260 K is to be admitted to the room at such a rate that the air is completely renewed each hour. The pressure is 100 kPa. The vapor pressure of water is 3.56 kPa at 300 K and 0.199 kPa at 260 K. Calculate the weight of water evaporated in one hour.

Ans: 1.65 kg/h

Explanation:

Problem 8.40: A mixture of air and water vapor has a relative humidity of 80 % at 293 K and 100 kPa. It is then cooled and compressed to 283 K and 250 kPa. The vapor pressure of water is 1.23 kPa at 283 K and 2.34 kPa at 293 K. Determine the following: (a) Molal humidity of air at initial and final conditions (b) Kilograms of water condensed per 100 m3 of initial wet air (c) Final volume of wet air per 100 m3 of initial air.

Ans: (a) 0.0191 kmol/kmol of dry air, 0.00494 kmol/kmol of dry air (b) 1.0276 kg (c) 38.10 m3

Explanation:

Problem 8.41: A gas mixture at 300 K and 100 kPa contains acetone vapor such that the percentage saturation is 70 %. The vapor pressure of acetone in kPa is given as

\ln P^S=14.39155-\frac{2795.817}{T-43.198}

If it is desired that 40 % of the acetone is to be removed by condensation, to what temperature the gas must be cooled at constant pressure?

Ans: 283.03 K

Explanation:

Problem 8.42: A mixture of nitrogen gas and benzene vapor at 300 K and 100 kPa has a percent saturation of 70 %. It is desired that 80 % of the benzene is to be removed by cooling and compression. If the gas is cooled to 273 K, what should be the pressure to which it is compressed? The vapor pressure of benzene at 273 K is 3.5 kPa and 13.8 kPa at 300 K.

Ans: 159.75 kPa

Explanation:

Problem 8.43: An acetone vapor-nitrogen gas mixture at 310 K and 101.3 kPa with a partial pressure of acetone of 1.5 kPa is passed over a solution containing acetone as part of a solvent-recovery operation. Acetone is to be recovered from the solution at a rate of 25 kg per hour by evaporating into the gas stream. If the gas stream leaves the unit at 287 K and 100 kPa with a percent saturation of 85 %, calculate the volumetric rate at which the gas is admitted in m3 per hour. The vapor pressure of acetone at 287 K is 8.86 kPa.

Ans: 164.53 m3/h

Explanation:

Problem 8.44: For an air-water vapor mixture with a dry-bulb temperature of 335 K and a wet-bulb temperature of 308 K, determine the humidity of air at 101.3 kPa The latent heat of vaporization of water at 308 K is 2419 kJ/kg and the psychrometric ratio is 0.950 kJ/kg K. The vapor pressure of water at 308 K is 5.62 kPa.

Ans: 0.0258 kg of water per kg of dry air

Explanation:

Problem 8.45: An air-water vapor sample has a dry-bulb temperature of 333 K and a wet-bulb temperature of 313 K. Compute the humidity of the sample given that the latent heat of vaporization of water at 313 K is 2407 kJ/kg and the psychrometric ratio is 0.950 kJ/kg K. Compare the result with the value read directly from the psychrometric chart. The vapor pressure of water at 313 K is 7.37 kPa.

Ans: 0.0408 kg of water per kg of dry air

Explanation:

Problem 8.46: A mixture of air and benzene vapor containing 0.7 kg benzene per kg dry air is to be cooled from 343 K to 288 K in a tubular cooler condenser. The specific heat of dry air is 1.005 kJ/kg K and the specific heat of benzene vapor is 1.1 kJ/kg K. The vapor pressure of benzene in kPa is given as

\ln P^S=13.8858-\frac{2788.51}{T-52.36}

The psychrometric ratio and the latent heat of vaporization of benzene in the range of interest may be taken to be constant at 1.7 kJ/kg K and 250 kJ/kg, respectively. The pressure is 101.3 kPa. Determine the following: (a) The wet-bulb temperature of the inlet stream (b) The humidity of the exit stream (c) The total amount of heat to be transferred per kg dry air.

Ans: (a) 314.1 K (b) 0.224 kg of benzene per kg of dry air (c) 216.7 kJ

Explanation:

Problem 8.47: The vapor pressure of benzene varies with temperature as

\ln P^S=13.8858-\frac{2788.51}{T-52.36}

The latent heat of vaporization of benzene at 293 K is 435.4 kJ/kg and the psychrometric ratio is 1.717 kJ/kg K. Calculate and plot the following: (a) The saturation curve for the benzene-air system at 101.3 kPa for the temperature range 283 K to 313 K with temperature increments of 10 K (b) The psychrometric line for the wet-bulb temperature of 293 K

Ans:

Explanation:

Problem 8.48: A toluene-air mixture at 330 K and 101.3 kPa contains 0.05 kg toluene vapour per kg dry air. The Antoine equation for vapour pressure of toluene is

\ln P^S=13.9987-\frac{3096.52}{T-52.67}

The specific heats of toluene vapor and air are, respectively, 1.256 kJ/kg K and 1.005 kJ/ kg K. Assume a constant value of 404 kJ/kg for the latent heat of vaporization of toluene. (a) Estimate the adiabatic saturation temperature (b) If the psychrometric ratio for the toluene-air system is 1.842 kJ/kg K, estimate the wet-bulb temperature.

Ans: (a) 299.3 K (b) 304 K

Explanation:

Problem 8.49: The vapor pressure and the latent heat of vaporization of carbon tetrachloride as a function of temperature are given below:

T (K)278300322333344350
PS (kPa)5.616.439.357.384.7101.3
λ (kJ/kg)215.7206.3197.5193.1188.6186.4

The specific heat of dry air and that of CCl4 vapor may be assumed constant. They are respectively, 1.005 kJ/kg K and 0.942 kJ/kg K. For a pressure of 101.3 kPa, calculate the following: (a) The adiabatic saturation temperature for dry air at 350 K (b) The adiabatic saturation temperature for air at 350 K and having a humidity of 1.5 kg vapor per kg dry air.

Ans: (a) 279 K (b) 312.3 K

Explanation:

Problem 8.50: The vapor pressure of chloroform (kPa) is given as

\ln P^S=13.9582-\frac{2696.79}{T-46.16}

Plot the 100 %, 50 %, and 25 % saturation curves for chloroform in air at 101.3 kPa.

Ans:

Explanation:

Problem 8.51: An air-water vapor sample at 101.3 kPa has a dry-bulb temperature of 333 K and an absolute humidity of 0.01 kg water vapor per kg dry air. Using the psychrometric chart, determine the following: (a) The percent saturation (b) The partial pressure of water vapor (c) The absolute saturation humidity at 333 K (d) The vapor pressure of water at 333 K (e) The percent relative saturation (f) The dew point of the system (g) The humid heat (h) The wet-bulb temperature.

Ans: (a) 6.25 % (b) 1.602 kPa (c) 0.16 kg of water/kg of dry air (d) 20.7 kPa (e) 7.73 % (f) 287.7 K (g) 1.0238 kJ/kg (h) 300.7 K

Explanation:

Problem 8.52: The wet-bulb and dry-bulb temperatures of air are 313 K and 333 K respectively. Determine the following using the psychrometric chart: (a) Absolute humidity (b) Molal humidity (c) The percent humidity (d) The humid volume (e) The dew point (f) The enthalpy of wet air.

Ans: (a) 0.04 (b) 0.0644 (c) 26.5 % (d) 1.0036 m3 (e) 310.2 K (f) 158.05 kJ/kg

Explanation:

Problem 8.53: The dry-bulb temperature and the dew point of air are 343 K and 308 K respectively. Estimate the following using the psychrometric chart: (a) Absolute humidity (b) Molal humidity (c) The percent humidity (d) The humid volume (e) The wet-bulb temperature (f) The enthalpy of wet air.

Ans: (a) 0.035 kg/kg (b) 0.0564 kmol/kmol (c) 14 % (d) 1.0255 m3/kg (e) 313 K (f) 162.39 kJ/kg

Explanation:

Problem 8.54: Moist air at 313 K and 50 % saturation is heated to 362 K at a constant pressure of 101.3 kPa. Using the psychrometric chart, determine the heat added per cubic meter of initial air.

Ans: 55.81 kJ

Explanation:

Problem 8.55: An air stream at 101.3 kPa and 333 K is 25 % saturated with water vapor. Determine the humid volume and enthalpy of the air. If the air is adiabatically humidified to 70 % saturation, what will be its enthalpy in the final state?

Ans: 0.99875 m3/kg, 153 kJ/kg, 155 kJ/kg

Explanation:

Problem 8.56: Air with a dry bulb temperature of 363 K and a wet-bulb temperature of 308 K is passed over a wet material in a drier. The rate of evaporation is found to be 0.015 kg water per kg dry air. Assuming the process to be adiabatic, calculate the following using the psychrometric chart: (a) The absolute humidity, molal humidity, and percent saturation of air entering the drier. (b) The dry-bulb temperature, the wet-bulb temperature, the absolute humidity, and the percent saturation of air leaving the drier.

Ans: (a) 0.013 kg/kg of dry air, 0.0209 kmol/kmol of dry air, 1 % (b) 329 K, 0.028 kg/kg dry air, 22.5 %

Explanation:

Problem 8.57: The dry-bulb temperature and dew point of an air sample at 101.3 kPa are 328 K and 308 K respectively. Calculate the following using a psychrometric chart: (a) The absolute humidity, molal humidity, and percent saturation (b) The partial pressure of water vapor and relative saturation (c) The wet bulb temperature (d) The humid heat, humid volume, and total enthalpy.

Ans: (a) 0.035 kg/kg of dry air, 0.0564 kmol/kmol of dry air, 30 % (b) 5.407 kPa, 34.4 % (c) 310.7 K (d) 1.071 kJ/kg, 0.979 m3/kg, 150.3 kJ/kg

Explanation:

Problem 8.58: Construct a psychrometric chart for an acetone-air system in the temperature range of 260 K to 330 K which includes the following details. (a) 25 %, 50 %, 75 %, and 100 % saturation curves (b) Specific volume of dry air and humid volume of saturated gas (c) Enthalpy of dry and saturated gas in kJ/kg dry air referred to liquid acetone and dry air at 260 K (d) Adiabatic cooling line for an adiabatic saturation temperature of 300 K (e) Psychrometric line for a wet-bulb temperature of 300 K.
The vapor pressure (kPa) of acetone is given by the Antoine equation with constants, A = 14.5463, B = 2940.46, and C = 49.19. The specific heat of acetone vapor and air are 1.47 kJ/kg K and 1.005 kJ/kg K respectively. The latent heat of vaporization of acetone at 300 K is 544 kJ/kg and the psychrometric ratio for the air-acetone system is 1.7 kJ/kg K.

Ans:

Explanation:

Problem 8.59: Air at 313 K dry-bulb temperature and 295 K wet-bulb temperature is admitted to a spray chamber where water is sprayed into it adiabatically. If the exit air is at 300 K, determine, using the psychrometric chart, the quantity of water added per kg dry air.

Ans: 0.007 kg

Explanation:

Problem 8.60: It is desired to maintain the temperature of air entering a building at 308 K and humidity at 60 %. This is achieved by humidifying air available at 305 K with 65 % percent humidity to saturation adiabatically in a spray chamber and by heating the gas leaving the chamber to the desired final condition. Assume the pressure to be atmospheric. Using the psychrometric chart, determine the following: (a) The temperature of air leaving the spray chamber and (b) The amount of water picked up by one kg dry air passing through the chamber.

Ans: (a) 299.5 K (b) 0.004 kg

Explanation:

Problem 8.61: Air at 303 K and with 75 % saturation is to be prepared from a supply of warm summer air at 310.7 K and 35% humidity. This is achieved by spraying water into the inlet air and by heating the saturated air leaving the spray chamber to 303 K. Assume all the operations are at atmospheric pressure. Using the psychrometric chart, determine the following: (a) The temperature of the air leaving the chamber and (b) The amount of water evaporated per 1000 m3 of air entering the chamber.

Ans: (a) 298 K (b) 5.55 kg

Explanation:

Problem 8.62: An air stream at 101.3 kPa and 333 K is 26.5 % saturated with water vapor. It is sent to a spray chamber where it is adiabatically humidified. The temperature of the air leaving the chamber is 5 K above the adiabatic saturation temperature. Determine the following: (a) The absolute humidity of the initial air (b) The wet-bulb temperature of the air (c) The dew point of the initial air (d) The percent saturation of the air leaving the spray chamber (e) The humidity and temperature of the air leaving the chamber (f) The quantity of water picked up by 100 m3 of wet air admitted to the chamber.

Ans: (a) 0.04 kg/kg of dry air (b) 313 K (c) 310 K (d) 70 % (e) 0.046 kg/kg of dry air, 318 K (f) 0.598 kg

Explanation:

Problem 8.63: Air at 101.3 kPa has a dry-bulb temperature of 303 K and a wet-bulb temperature of 293 K. It is desired to heat the air in a preheater and then spray water into it adiabatically so that final air will be saturated at 308 K. Using the psychrometric chart, determine the following: (a) The humidity of the initial air (b) The percent humidity of the initial air (c) The temperature to which the air is to be heated in the preheater (d) The heat to be supplied per kg of dry air.

Ans: (a) 0.01 kg/kg dry air (b) 40 % (c) 370 K (d) 68.6 kJ

Explanation:

Problem 8.64: Air is dehumidified by spraying into it water whose temperature is below the dew point of the air. Air with a dew point of 294 K and a percent humidity of 40 % enters a dehumidifier and leaves at 287 K and with a dew point of 285 K. How much water is removed per kg dry air? How much heat is removed per kg dry air? Solve using a psychrometric chart.

Ans: 0.0085 kg, 41.3 kJ/kg

Explanation:

Problem 8.65: 10000 kg/h of air at 326 K and 20 % humidity is to be produced from the air at 301 K and 20 % humidity by first heating, then humidifying adiabatically to the desired humidity, and finally reheating the air to 326 K. The air leaves the adiabatic humidification chamber with 95 % saturation. Using the psychrometric chart, calculate the following: (a) The temperature to which air is preheated (b) The temperature of the air leaving the humidification chamber (c) The heat required in the preheater (d) The heat required in the preheater.

Ans: (a) 333 K (b) 298 K (c) 31.82 kJ (d) 27.6 kJ

Explanation:

Problem 8.66: Air with a dry-bulb temperature of 310 K and a wet-bulb temperature of 302.5 K is passed into an evaporator. The air leaves the evaporator at 325 K with a wet-bulb temperature of 320 K. The pressure is constant at 101.3 kPa. Calculate the following using psychrometric charts: (a) The molal humidity, percent saturation, and dew point of the initial air (b) The molal humidity, percent saturation, and the dew point of the final air (c) The amount of water evaporated per 100 m3 of wet air entering the evaporator.

Ans: (a) 0.0322, 50 %, 283.2 K (b) 0.1206, 85 %, 316.2 K (c) 6.07 kg

Explanation:

Problem 8.67:  It is desired that the air entering a building be at 308 K and 30 % saturated with water vapor. This is achieved by passing air at 298 K and 30 % saturated, through a series of water sprays where it is cooled and saturated. The air emerging from the spray chamber is then heated to 308 K. The water and air leave the chamber at the same temperature. Using a psychrometric chart, calculate the following: (a) The temperature of water (b) Humidity of air entering the building (c) The quantity of water condensed or evaporated in the spray chamber per kg moisture-free air.

Ans: (a) 289 K (b) 0.01 kg/kg of dry air (c) 0.004 kg

Explanation:

Problem 8.68: A wet material containing 70 % moisture is to be dried at a rate of 500 kg/h to 5 % moisture in a counter-current drier, both moisture content being on a wet basis. The air enters the drier at 333 K with a partial pressure of water of 1.0 kPa and leaves at 313 K and 70 % saturated. Calculate the air circulation rate in kg/h. The total pressure is 100 kPa and the vapor pressure of water at 313 K = 7.38 kPa.

Ans: 12122 kg/h wet air

Explanation:

Problem 8.69: A drier is to be designed for removing 100 kg/h of water from a wet material. The following conditions are expected for the air:
Inlet: temperature = 340 K, pressure = 101.3 kPa, dew point = 278 K
Exit: temperature = 310 K, pressure = 100.0 kPa, dew point = 298 K
Vapor pressure of water = 0.872 kPa (278 K), 3.166 kPa (298 K)
At what rate air must be admitted (m3/h) at the inlet conditions?

Ans: 6514.1 m3/h

Explanation:

Problem 8.70: Air at 333 K and pressure of 100 kPa, 10 % saturated with water vapor is sent to a drier at a rate of 20,000 m3/h where it picks up moisture from the solids at a rate of 25 kg/h. If the air leaves at 308 K and 98 kPa calculate the following: (a) The percent saturation of the air leaving the drier (b) The volume of air leaving the drier per hour.
The vapor pressure of water is 5.62 kPa (308 K), 19.92 kPa (333 K).

Ans: (a) 44.9 % (b) 18912.3 m3/h

Explanation:

Problem 8.71: A wet material is being dried by removing water at a rate of 100 kg in a drier. Water evaporates into an air stream entering the drier at 350 K and 101.3 kPa with a dew point of 278 K. The air leaves the drier at 308 K and 100 kPa with a relative humidity of 90 %. The vapor pressure of water at 278 K and 308 K are respectively 0.87 and 5.62 kPa. Determine the following: (a) The rate at which air is admitted, m3/h (b) The rate at which air leaves the drier, m3/h.

Ans: (a) 3604.6 m3/h (b) 3355.4 m3/h

Explanation:

Problem 8.72: Air at 365 K, 101.32 kPa, and a dew point of 283 K is supplied to a drier that operates adiabatically. Determine the following using psychrometric charts: (a) The minimum temperature to which the air gets cooled in the drier (b) On the basis of 10 m3 of air entering the drier, the maximum quantity of water evaporated in the drier (c) The volume of air leaving the drier for the conditions in part (b)

Ans: (a) 305 K (b) 0.2489 kg (c) 8.76 m3/h

Explanation:

Problem 8.73: Air enters a drier at 388 K, 101.3 kPa, and with a dew point of 286 K. The drier operates adiabatically. If the air leaves the drier saturated with water vapor, how many kilograms of water will be evaporated per 100 m3 of air entering? Solve using psychrometric charts.

Ans: 3.15 kg

Explanation:

Problem 8.74: Air is supplied to a drier at a dry-bulb temperature of 305 K and a wet-bulb temperature of 298 K. It is heated to 363 K before admitting to the drier. The drier is operated adiabatically. Air leaves the drier fully saturated. Using the psychrometric chart, determine the following: (a) The humidity, percent humidity, and the dew point of the initial air (b) The heat to be supplied in the heater per 100 m3 of initial air (c) Water evaporated per 100 m3 of air entering the drier (d) The temperature of air leaving the drier.

Ans: (a) 0.017 kg/kg of dry air, 58 %, 295 K (b) 6782.5 kJ (c) 2.706 kg (d) 310 K

Explanation:

Problem 8.75: An adiabatic drier is used to dry paper. The drying air enters at 371 K and 101.3 kPa with a dew point of 298 K. Measurements show that 2 kg of water is evaporated per 100 m3 of wet inlet air. Calculate the following using a psychrometric chart: (a) The exit air wet-bulb and dry-bulb temperatures (b) The exit air percent humidity (c) The volume of exit air per 100 m3 of inlet wet air.

Ans: (a) 312 K, 323 K (b) 48 % (c) 90.07 m3

Explanation:

Problem 8.76: A rotary drier operating at 101.3 kPa dries 500 kg/h of wet grain at 295 K from 10 % moisture to 1 % moisture. Air-flowing counter current to the wet material enters at 380 K dry-bulb and 316 K wet-bulb temperature and leaves at 325 K dry-bulb temperature saturated with water vapor. The grains leave the drier at 316 K and its specific heat is 0.75 kJ/kg K. Using the psychrometric chart, determine the following: (a) The humidity of the air entering and leaving the drier (b) Water removed in kg/h (c) The heat input to the drier.

Ans: (a) 0.028 and 0.096 kg/kg of dry air (b) 45.45 kg/h (c) 82612 kJ/h

Explanation:

Problem 8.77: Air, with a dry-bulb temperature of 311 K and a wet-bulb temperature of 300 K, is scrubbed with water to remove dust. The water is maintained at 298 K. The air leaving the scrubber is in equilibrium with water. It is then heated to 366 K in an air pre-heater and admitted to an adiabatic rotary drier. The air leaves the drier at 322 K. The material to be dried enters and leaves the drier at 319 K. The material loses moisture at a rate of 0.05 kg of water per kg of product. The total product is 1000 kg/h. Using the psychrometric chart, determine the following: (a) The humidity and percent humidity of the initial air, air after scrubbing, after preheating, and air leaving the drier (b) The total weight of dry air used per hour (c) The total volume of air leaving the drier.

Ans: (a) 0.0175, 44 %; 0.02, 100 %; 0.02, 0.9 %; 0.038, 45 % (b) 2778 kg (c) 2686 m3

Explanation:

Problem 8.78: Wet material is spread over trays stacked one above the other. Air containing 0.005 kg water/kg dry air is heated to 323 K and passed over the lower tray. It leaves the tray at 60 % humidity and is then reheated to 323 K and passed over the upper tray after which the air leaves the drier. On the assumption that the material on each tray has reached the wet-bulb temperature and that heat losses from the drier can be neglected, determine (a) The temperature of the material on the trays (b) The amount of water removed in kg/h if 1000 m3/h of moist air leaves the drier (c) The temperature to which the air is to be preheated, if the final condition of air is to be achieved in the first tray itself.

Ans: (a) 295.7 K, 300.7 K (b) 16.69 kg/h (c) 342.7 K

Explanation:

Problem 8.79: A laboratory drier produces 50 kg of dry solid containing 6 weight percent water from a wet feed material containing 20.5 % water. The drier operates adiabatically, with the wet solid entering the drier at the wet-bulb temperature of the dryer air. A total of 500 m3/h of hot air at 365.7 K and 101.3 kPa with a dew point of 303.2 K is supplied to the drier. Calculate the following: (a) The temperature of the air leaving the drier (b) The percent humidity of the air leaving the drier.

Ans: (a) 318.65 K (b) 70 %

Explanation:

Problem 8.80: Air at 101.3 kPa enters an adiabatic drier at 372 K with a dew point of 287 K and leaves at 80 % humidity. Wet paper enters the drier with 25 % moisture and leaves with 5 % moisture. Determine the following: (a) The temperature of the air leaving the drier (b) Water evaporated in kilograms per 100 cubic meters of air entering (c) Kilograms of finished product per 100 cubic meters of air entering.

Ans: (a) 311 K (b) 2.43 kg (c) 9.11 kg

Explanation:

Problem 8.81: It is required to evaporate 100 kg/h of water from a wet solid. The drying is carried out in an adiabatic drier. Air is available at 101.3 kPa and 295 K at 70 % humidity. Its temperature is raised by passing through a preheater to such a value that the air leaving the drier is at 311.7 K and 90 % saturated. Calculate the following: (a) The adiabatic saturation temperature of the air leaving the drier (b) The temperature of the air leaving the preheater (c) The rate of circulation of moist air (kg/h) (d) The volume of moist air before preheating.

Ans: (a) 310 K (b) 382.2 K (c) 3393.3 kg/h (d) 2851.9 m3/h

Explanation: