Thermodynamics Quiz: Humidity Ratio Relative Humidity Dew Point

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Humidity Ratio Relative Humidity Dew PointQuestion 1 of 20

Air at 26°C and 45% relative humidity is heated to 40°C while maintaining constant humidity ratio. If the saturation pressure at 26°C is 3.363 kPa and at 40°C is 7.384 kPa, what is the relative humidity after heating?

20.5%

45.0%

32.8%

18.3%

27.1%

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Thermodynamics Quiz

Thermodynamics Quiz: Humidity Ratio Relative Humidity Dew Point

Practice Humidity Ratio Relative Humidity Dew Point in Thermodynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Humidity Ratio Relative Humidity Dew Point, giving you a quick way to practice the rules, question types, and explanations that matter most for Thermodynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

20.5% (correct answer)
45.0%
32.8%
18.3%
27.1%

Explanation: When you encounter psychrometric problems involving heating or cooling of air, remember that the humidity ratio (absolute humidity) remains constant when no moisture is added or removed - only the relative humidity changes as temperature affects the air's moisture-holding capacity. The humidity ratio can be found using the initial conditions. At 26°C with 45% relative humidity, the partial pressure of water vapor is

P_v = 0.45 \times 3.363 = 1.513 \text{ kPa}

. Since humidity ratio depends only on vapor pressure (not temperature), this vapor pressure remains constant during heating. After heating to 40°C, the air still contains the same absolute amount of moisture (1.513 kPa vapor pressure), but now the saturation pressure is 7.384 kPa. The new relative humidity is:

RH = \frac{1.513}{7.384} \times 100\% = 20.5\%

Looking at the wrong answers: B) 45.0% assumes relative humidity stays constant, which would only happen if moisture were added during heating - a common misconception. C) 32.8% might result from calculation errors or incorrectly applying psychrometric relationships. D) 18.3% could come from using wrong pressure values or mathematical mistakes in the ratio calculation. A) 20.5% correctly applies the principle that vapor pressure remains constant while saturation pressure increases with temperature. Study tip: For psychrometric heating/cooling problems, always identify what stays constant (usually humidity ratio) versus what changes (relative humidity, dry-bulb temperature). Relative humidity always decreases when air is heated at constant moisture content.

Question 2

Air at 25°C and 101.3 kPa has a relative humidity of 60%. If the saturation pressure of water vapor at 25°C is 3.169 kPa, what is the humidity ratio of this air sample?

0.0118 kg water/kg dry air (correct answer)
0.0196 kg water/kg dry air
0.0078 kg water/kg dry air
0.0157 kg water/kg dry air
0.0235 kg water/kg dry air

Explanation: When you encounter humidity problems in thermodynamics, you're dealing with the relationship between water vapor in air and the air's capacity to hold that moisture. The key is understanding how relative humidity connects to the actual amount of water present. To find the humidity ratio (mass of water vapor per mass of dry air), you need to first determine the partial pressure of water vapor. Since relative humidity is the ratio of actual vapor pressure to saturation pressure:

P_v = \phi \times P_{sat} = 0.60 \times 3.169 = 1.901 \text{ kPa}

Next, apply the humidity ratio formula:

\omega = 0.622 \times \frac{P_v}{P_{total} - P_v} = 0.622 \times \frac{1.901}{101.3 - 1.901} = 0.622 \times \frac{1.901}{99.399} = 0.0118 \text{ kg water/kg dry air}

This confirms answer A is correct. Answer B (0.0196) likely results from incorrectly using the full saturation pressure instead of the actual vapor pressure. Answer C (0.0078) suggests using an incorrect conversion factor or mathematical error in the calculation. Answer D (0.0157) might come from forgetting to subtract the vapor pressure from total pressure in the denominator, using just the vapor pressure instead. Remember this two-step approach: first convert relative humidity to actual vapor pressure, then apply the humidity ratio formula with the standard factor 0.622. Always subtract vapor pressure from total pressure in the denominator—this accounts for the partial pressure of dry air.

Question 3

An air conditioning system processes air from 35°C and 70% relative humidity to 20°C and 50% relative humidity. If the saturation pressures are 5.628 kPa at 35°C and 2.339 kPa at 20°C, what is the change in humidity ratio?

-0.0174 kg/kg (correct answer)
-0.0089 kg/kg
-0.0132 kg/kg
-0.0156 kg/kg
-0.0201 kg/kg

Explanation: When analyzing air conditioning processes, you need to track how much moisture is removed from the air by calculating the humidity ratio change. The humidity ratio represents the mass of water vapor per unit mass of dry air. To find the humidity ratio, use:

\omega = 0.622 \times \frac{P_v}{P_{total} - P_v}

, where

P_v

is the vapor pressure of water in the air. The vapor pressure equals the relative humidity times the saturation pressure. For the initial state (35°C, 70% RH):

P_{v1} = 0.70 \times 5.628 = 3.940 \text{ kPa}

\omega_1 = 0.622 \times \frac{3.940}{101.325 - 3.940} = 0.0252 \text{ kg/kg}

For the final state (20°C, 50% RH):

P_{v2} = 0.50 \times 2.339 = 1.170 \text{ kPa}

\omega_2 = 0.622 \times \frac{1.170}{101.325 - 1.170} = 0.0077 \text{ kg/kg}

The change is:

\Delta\omega = 0.0077 - 0.0252 = -0.0175 \text{ kg/kg}

, which rounds to -0.0174 kg/kg (A). Option B (-0.0089 kg/kg) likely comes from miscalculating one of the humidity ratios or using incorrect vapor pressures. Option C (-0.0132 kg/kg) might result from forgetting to subtract the vapor pressure from total pressure in the denominator. Option D (-0.0156 kg/kg) could stem from calculation errors in the conversion factors. Always remember: humidity ratio calculations require precise vapor pressure calculations first, then careful application of the psychrometric formula. Double-check your arithmetic, especially with the 0.622 conversion factor.

Question 4

Air at 18°C has a dew point temperature of 12°C. If the saturation pressure at 18°C is 2.064 kPa and at 12°C is 1.403 kPa, what is the relative humidity of the air?

68.0% (correct answer)
32.0%
85.3%
47.9%
76.2%

Explanation: When you encounter problems involving dew point and relative humidity, you're working with the relationship between actual water vapor content and the maximum possible water vapor content at different temperatures. Relative humidity is defined as the ratio of actual vapor pressure to saturation vapor pressure at the air temperature, expressed as a percentage:

RH = \frac{P_{vapor}}{P_{sat}} \times 100\%

The key insight is that the dew point temperature tells you the actual vapor pressure in the air. At the dew point (12°C), the air would be saturated, meaning the actual vapor pressure equals the saturation pressure at that temperature. So the actual vapor pressure is 1.403 kPa. To find relative humidity at 18°C, you divide this actual vapor pressure by the saturation pressure at 18°C:

RH = \frac{1.403}{2.064} \times 100\% = 68.0\%

This confirms answer A) 68.0% is correct. Answer B) 32.0% likely comes from calculating

\frac{2.064-1.403}{2.064} = 32\%

, which would represent the "unused capacity" rather than relative humidity. Answer C) 85.3% might result from incorrectly using temperature values in the calculation. Answer D) 47.9% could come from using an incorrect formula or mixing up the pressure values. Remember: the dew point always gives you the actual vapor pressure in the air. This is the numerator in your relative humidity calculation, while the denominator is always the saturation pressure at the actual air temperature.

Question 5

Air with a dry-bulb temperature of 32°C and wet-bulb temperature of 24°C undergoes adiabatic saturation. What is the relationship between the final temperature and the initial wet-bulb temperature?

Final temperature equals initial wet-bulb temperature (correct answer)
Final temperature is higher than initial wet-bulb temperature
Final temperature is lower than initial wet-bulb temperature
Final temperature depends on initial relative humidity
Final temperature equals initial dry-bulb temperature

Explanation: When you encounter adiabatic saturation problems in thermodynamics, you're dealing with a process where air is brought to 100% relative humidity without any heat transfer to or from the surroundings. The key insight is understanding what happens to the air's energy during this process. During adiabatic saturation, water evaporates into the air stream, absorbing latent heat from the air itself. This causes the dry-bulb temperature to decrease while the humidity increases. The process continues until the air becomes fully saturated. Crucially, the wet-bulb temperature represents the theoretical limit of this cooling process—it's the temperature air would reach if cooled adiabatically to saturation. Option A is correct because the final temperature after adiabatic saturation will always equal the initial wet-bulb temperature. This is a fundamental property: the wet-bulb temperature is literally defined as the adiabatic saturation temperature. Option B is wrong because cooling occurs during adiabatic saturation due to evaporative heat loss, so the final temperature cannot be higher than the wet-bulb temperature. Option C is incorrect because the wet-bulb temperature is the theoretical minimum temperature achievable through adiabatic cooling—you cannot go lower. Option D misses the point entirely; while initial relative humidity affects how much cooling occurs, the final temperature will always converge to the wet-bulb temperature regardless of starting humidity. Remember this key relationship: wet-bulb temperature = adiabatic saturation temperature. This equality is one of the most important concepts in psychrometrics and appears frequently on thermodynamics exams.

Question 6

An air sample at 101.3 kPa total pressure contains water vapor at a partial pressure of 2.5 kPa. If this air is compressed isothermally to 200 kPa, at what partial pressure of water vapor will condensation begin?

The saturation pressure at the given temperature (correct answer)
5.0 kPa (proportional to total pressure increase)
2.5 kPa (remains constant during compression)
4.94 kPa (scaled by pressure ratio minus water vapor)
3.75 kPa (average of initial and final pressures)

Explanation: When you encounter problems involving water vapor in air during compression, you're dealing with psychrometrics and the concept of saturation. The key insight is understanding what happens to water vapor as air is compressed at constant temperature. During isothermal compression, the partial pressure of water vapor increases proportionally with the total pressure increase. Initially, the water vapor partial pressure is 2.5 kPa at 101.3 kPa total pressure. When compressed to 200 kPa, the pressure ratio is

\frac{200}{101.3} = 1.97

, so the water vapor partial pressure would try to reach

2.5 \times 1.97 = 4.94

kPa. However, condensation begins when the water vapor partial pressure reaches the saturation pressure at that temperature. Since we don't know the exact temperature, we can't calculate the numerical value, but we know condensation starts at saturation conditions, making A correct. B incorrectly assumes the water vapor pressure doubles exactly with the 200 kPa total pressure, but this ignores the actual pressure ratio. C is wrong because partial pressures do change during compression - they're not independent of total pressure changes. D represents the calculated partial pressure if no condensation occurred (4.94 kPa), but this misses that condensation begins before reaching this value if it exceeds saturation pressure. Remember: in psychrometric problems, water vapor behaves like any other gas until it hits saturation limits. Always check whether calculated partial pressures exceed saturation conditions at the given temperature.

Question 7

Air at 27°C has a relative humidity of 80%. If the temperature is reduced to 15°C at constant total pressure while maintaining the same partial pressure of water vapor, and the saturation pressure at 15°C is 1.705 kPa, what can be concluded about the final state?

Condensation will occur if initial vapor pressure exceeds 1.705 kPa (correct answer)
The final relative humidity will be exactly 80%
The final relative humidity will be less than 80%
The humidity ratio will decrease during cooling
No condensation can occur during this process

Explanation: When you encounter psychrometric problems involving cooling at constant pressure, focus on what happens to the partial pressure of water vapor and how it relates to the new saturation conditions. Let's trace through this cooling process. Initially at 27°C with 80% relative humidity, the partial pressure of water vapor equals 80% of the saturation pressure at 27°C. During cooling at constant total pressure with the same partial pressure of water vapor, this vapor pressure remains unchanged - only the temperature drops to 15°C. The key insight is comparing this unchanged vapor pressure to the new saturation pressure of 1.705 kPa at 15°C. If the initial vapor pressure exceeds 1.705 kPa, the air becomes supersaturated (relative humidity > 100%), causing condensation until equilibrium is restored. This confirms why answer A is correct. Answer B is wrong because relative humidity equals vapor pressure divided by saturation pressure - since saturation pressure changes with temperature while vapor pressure stays constant, the final relative humidity cannot remain 80%. Answer C is incorrect because cooling typically increases relative humidity when vapor pressure stays constant, since saturation pressure generally decreases with temperature. Answer D is wrong because humidity ratio (mass of water vapor per unit mass of dry air) only changes if condensation occurs, which depends on whether the initial vapor pressure exceeds the new saturation pressure. Study tip: In psychrometric cooling problems, always compare the initial vapor pressure to the final saturation pressure to determine if condensation occurs. If initial vapor pressure > final saturation pressure, expect condensation and moisture removal.

Question 8

Air undergoes a process where the dry-bulb temperature increases from 20°C to 35°C while the dew point temperature remains constant at 15°C. What happens to the relative humidity during this process?

Decreases significantly (correct answer)
Increases significantly
Remains approximately constant
First decreases then increases
Cannot be determined without pressure information

Explanation: When you encounter problems involving humidity changes with temperature, focus on the relationship between dew point, dry-bulb temperature, and relative humidity. Relative humidity represents how much moisture the air contains compared to its maximum capacity at that temperature. Since the dew point remains constant at 15°C, the actual amount of water vapor in the air doesn't change. However, as the dry-bulb temperature increases from 20°C to 35°C, the air's capacity to hold moisture increases significantly. When the denominator (maximum moisture capacity) increases while the numerator (actual moisture content) stays the same, relative humidity must decrease substantially. At 20°C with a 15°C dew point, the relative humidity starts around 73%. When heated to 35°C with the same moisture content, the relative humidity drops to approximately 42% - a significant decrease. Choice A is correct because relative humidity decreases significantly as explained above. Choice B suggests the opposite relationship - this would only occur if moisture were being added to the air. Choice C incorrectly assumes relative humidity is independent of temperature changes, which ignores how air's moisture-holding capacity varies with temperature. Choice D suggests a complex pattern that doesn't match the physics - with constant dew point and steadily increasing temperature, relative humidity follows a simple, monotonic decrease. Remember this key principle: when air temperature rises while moisture content stays constant (constant dew point), relative humidity always decreases. This is crucial for psychrometric processes and HVAC applications.

Question 9

Air at 101.3 kPa contains water vapor at a partial pressure of 2.8 kPa. The air is cooled at constant pressure until the relative humidity reaches 100%. At this point, the saturation pressure is 2.8 kPa. What was the initial relative humidity if the initial temperature was 35°C with saturation pressure 5.628 kPa?

49.7% (correct answer)
71.2%
35.8%
64.4%
28.3%

Explanation: This question tests your understanding of relative humidity and how it changes with temperature at constant water vapor content. When you see problems involving cooling air to saturation, remember that the partial pressure of water vapor stays constant during the cooling process. Relative humidity is defined as the ratio of actual water vapor pressure to saturation pressure at a given temperature:

RH = \frac{P_{vapor}}{P_{sat}} \times 100\%

At the initial condition (35°C), the water vapor partial pressure is 2.8 kPa and the saturation pressure is 5.628 kPa. Therefore, the initial relative humidity is:

RH_{initial} = \frac{2.8}{5.628} \times 100\% = 49.7\%

You can verify this makes sense: when the air is cooled at constant pressure, the water vapor partial pressure remains 2.8 kPa, but the saturation pressure decreases with temperature. When relative humidity reaches 100%, the saturation pressure equals the vapor pressure (2.8 kPa), confirming our vapor pressure value. Looking at the wrong answers: B) 71.2% might result from incorrectly using the final saturation pressure in the denominator. C) 35.8% could come from mixing up temperature and pressure values. D) 64.4% likely stems from calculation errors or using incorrect pressure relationships. Study tip: In psychrometric problems, always identify what stays constant during the process. During cooling at constant total pressure, water vapor partial pressure remains unchanged while saturation pressure decreases with temperature. This is the key to solving relative humidity problems involving temperature changes.

Question 10

Two identical rooms have the same dry-bulb temperature of 24°C. Room A has a dew point of 18°C while Room B has a dew point of 12°C. If the saturation pressures are 2.985 kPa at 24°C, 2.064 kPa at 18°C, and 1.403 kPa at 12°C, what is the difference in relative humidity between the rooms?

22.1% (correct answer)
15.8%
28.4%
33.7%
18.9%

Explanation: When you encounter psychrometric problems involving dew point and relative humidity, remember that relative humidity compares the actual water vapor pressure to the maximum possible at that temperature. The dew point tells you the actual vapor pressure in the air. To find relative humidity, use:

RH = \frac{P_{actual}}{P_{saturation}} \times 100\%

The key insight is that the actual vapor pressure equals the saturation pressure at the dew point temperature. For Room A (dew point 18°C), the actual vapor pressure is 2.064 kPa. For Room B (dew point 12°C), it's 1.403 kPa. Both rooms are at 24°C, so the saturation pressure is 2.985 kPa for both. Room A:

RH_A = \frac{2.064}{2.985} \times 100\% = 69.1\%

Room B:

RH_B = \frac{1.403}{2.985} \times 100\% = 47.0\%

The difference is 69.1% - 47.0% = 22.1%, which is answer A. Answer B (15.8%) might result from incorrectly using the difference between dew points divided by dry-bulb temperature. Answer C (28.4%) could come from using the wrong saturation pressures in your calculations. Answer D (33.7%) might result from calculating the ratio of the two relative humidities instead of their difference. Always remember: in psychrometric problems, the dew point directly gives you the actual vapor pressure by looking up the saturation pressure at that dew point temperature. This eliminates the need for complex humidity calculations.

Question 11

Air in a closed container at 25°C and 50% relative humidity is heated to 40°C. If no moisture is added or removed and saturation pressures are 3.169 kPa at 25°C and 7.384 kPa at 40°C, what is the final relative humidity?

21.5% (correct answer)
50.0%
116.6%
35.8%
43.2%

Explanation: When you encounter relative humidity problems involving temperature changes, remember that relative humidity depends on both the actual moisture content and the air's capacity to hold moisture at that temperature. The key insight is that heating air in a closed container doesn't change the absolute amount of water vapor present—only the air's capacity to hold moisture changes. Since saturation pressure increases with temperature, the air's moisture-holding capacity increases, making the same amount of water vapor represent a smaller percentage of the maximum possible. To solve this, use the relationship:

\text{RH}_1 \times P_{sat1} = \text{RH}_2 \times P_{sat2}

At 25°C:

0.50 \times 3.169 = 1.585 \text{ kPa}

(actual vapor pressure) At 40°C, this same vapor pressure gives:

\text{RH}_2 = \frac{1.585}{7.384} = 0.215 = 21.5\%

Choice A (21.5%) is correct. Choice B (50.0%) assumes relative humidity stays constant, ignoring that heating increases the air's moisture capacity. Choice C (116.6%) incorrectly applies the ratio of saturation pressures directly to the original humidity. Choice D (35.8%) likely results from calculation errors or misapplying temperature ratios instead of saturation pressure ratios. Remember this pattern: heating humid air in a closed container always decreases relative humidity because you're increasing the denominator (saturation pressure) while keeping the numerator (actual vapor pressure) constant. The ratio of saturation pressures tells you exactly how the relative humidity changes.

Question 12

A psychrometric process shows air entering at 25°C, 60% RH and leaving at 25°C, 90% RH. If the saturation pressure at 25°C is 3.169 kPa, what type of process occurred and what was the change in humidity ratio?

Humidification; +0.0061 kg/kg (correct answer)
Dehumidification; -0.0061 kg/kg
Humidification; +0.0047 kg/kg
Sensible heating; no change in humidity ratio
Sensible cooling; no change in humidity ratio

Explanation: When you encounter psychrometric processes, focus on two key parameters: dry-bulb temperature and relative humidity changes. These tell you both the process type and allow you to calculate humidity ratio changes. Here, the air stays at 25°C but relative humidity increases from 60% to 90%, indicating humidification at constant temperature. To find the humidity ratio change, you need the partial pressure of water vapor at each state. At any condition:

P_v = \phi \times P_{sat}

Initial state:

P_{v1} = 0.60 \times 3.169 = 1.901 \text{ kPa}

Final state:

P_{v2} = 0.90 \times 3.169 = 2.852 \text{ kPa}

The humidity ratio formula is:

\omega = 0.622 \times \frac{P_v}{P_{atm} - P_v}

(assuming atmospheric pressure = 101.325 kPa)

\omega_1 = 0.622 \times \frac{1.901}{101.325 - 1.901} = 0.0119 \text{ kg/kg}

\omega_2 = 0.622 \times \frac{2.852}{101.325 - 2.852} = 0.0180 \text{ kg/kg}

Change:

\Delta\omega = 0.0180 - 0.0119 = +0.0061 \text{ kg/kg}

Answer A is correct - it's humidification with +0.0061 kg/kg change. Answer B incorrectly suggests dehumidification (RH increased, not decreased). Answer C has the right process but wrong calculation. Answer D misidentifies this as sensible heating, which would change temperature, not humidity at constant temperature. Study tip: When temperature stays constant but RH changes, it's always a humidification/dehumidification process, never sensible heating or cooling.

Question 13

Air at standard atmospheric pressure has a humidity ratio of 0.020 kg/kg. If the air is at 30°C and the saturation pressure at 30°C is 4.246 kPa, what is the relative humidity?

74.8% (correct answer)
82.3%
67.1%
91.5%
58.6%

Explanation: When you encounter humidity problems in thermodynamics, you're dealing with the relationship between actual moisture content and maximum possible moisture content at given conditions. Relative humidity compares how much water vapor is present versus how much could be present at saturation. To find relative humidity, you need to determine the partial pressure of water vapor in the air, then compare it to the saturation pressure. Using the humidity ratio (0.020 kg/kg) and the relationship for moist air:

p_v = \frac{W \times p_{atm}}{0.622 + W}

Where W is the humidity ratio (0.020 kg/kg) and atmospheric pressure is 101.325 kPa:

p_v = \frac{0.020 \times 101.325}{0.622 + 0.020} = \frac{2.027}{0.642} = 3.158 \text{ kPa}

Relative humidity is then:

RH = \frac{p_v}{p_{sat}} \times 100\% = \frac{3.158}{4.246} \times 100\% = 74.4\%

This matches answer A) 74.8% (small differences due to rounding). Answer B) 82.3% likely results from using an incorrect formula or atmospheric pressure value. Answer C) 67.1% suggests an error in calculating the vapor pressure, possibly confusing the humidity ratio relationship. Answer D) 91.5% is too high and indicates a fundamental calculation error, perhaps incorrectly applying the saturation condition. Remember: humidity ratio problems always require converting to partial pressure first, then applying the relative humidity definition. Keep track of your pressure units and use the standard atmospheric pressure (101.325 kPa) unless otherwise specified.

Question 14

Two air streams are mixed: Stream 1 has a flow rate of 3 kg/s dry air with humidity ratio 0.008 kg/kg, and Stream 2 has a flow rate of 2 kg/s dry air with humidity ratio 0.018 kg/kg. What is the humidity ratio of the mixed stream?

0.0120 kg/kg (correct answer)
0.0130 kg/kg
0.0108 kg/kg
0.0146 kg/kg
0.0092 kg/kg

Explanation: When mixing air streams with different moisture contents, you need to apply conservation of mass separately for both the dry air and the water vapor. The humidity ratio of the mixed stream depends on the weighted average based on the dry air flow rates. For the mixed stream, calculate the total dry air flow rate: 3 + 2 = 5 kg/s. Next, find the total water vapor flow rate by adding the water vapor from each stream. Stream 1 contributes:

3 \times 0.008 = 0.024

kg/s of water vapor. Stream 2 contributes:

2 \times 0.018 = 0.036

kg/s of water vapor. The total water vapor flow is:

0.024 + 0.036 = 0.060

kg/s. The humidity ratio of the mixed stream is the total water vapor divided by the total dry air:

\frac{0.060}{5} = 0.0120

kg/kg, which is answer A. Answer B (0.0130 kg/kg) represents a simple arithmetic average of the two humidity ratios:

(0.008 + 0.018)/2 = 0.013

. This ignores the different flow rates and treats both streams equally. Answer C (0.0108 kg/kg) appears to use incorrect weighting factors or calculation errors. Answer D (0.0146 kg/kg) might result from reversing the flow rates in the weighted average calculation. Remember that mixing problems in psychrometrics always require mass-weighted averages, not simple arithmetic averages. The stream with the higher flow rate will have more influence on the final mixed conditions, so always account for the relative magnitudes of the flow rates when calculating properties of mixed streams.

Question 15

During winter, indoor air at 22°C and 30% relative humidity is exhausted and replaced with outdoor air at 0°C and 80% relative humidity. If saturation pressures are 2.645 kPa at 22°C and 0.611 kPa at 0°C, which statement about moisture addition is correct?

Humidification is needed to maintain comfort (correct answer)
Dehumidification is needed to prevent condensation
No moisture adjustment is necessary
Outdoor air has higher absolute humidity than indoor air
Moisture content depends only on relative humidity values

Explanation: When analyzing air exchange problems in thermodynamics, you need to compare absolute humidity (actual water content) rather than relative humidity alone. Relative humidity is the ratio of actual moisture to maximum possible moisture at that temperature. To find absolute humidity, calculate the actual vapor pressure:

P_{vapor} = RH × P_{saturation}

For indoor air:

0.30 × 2.645 = 0.794 \text{ kPa}

For outdoor air:

0.80 × 0.611 = 0.489 \text{ kPa}

The outdoor air contains significantly less moisture (0.489 kPa) than indoor air (0.794 kPa). When this cold, dry outdoor air enters and warms to 22°C, its absolute humidity remains the same, but its relative humidity drops dramatically. At 22°C with 0.489 kPa vapor pressure, the relative humidity becomes

0.489 ÷ 2.645 = 18.5\%

, which is uncomfortably dry. Answer A is correct because humidification is needed to restore comfort levels. Answer B is wrong because dehumidification would worsen the already dry conditions. Answer C is incorrect because the dramatic humidity drop requires correction. Answer D contains a common misconception—while outdoor air has higher relative humidity (80% vs 30%), it has lower absolute humidity due to cold temperatures reducing its moisture-holding capacity. Remember this key principle: cold air holds less moisture than warm air, so high relative humidity in cold outdoor air often means low absolute humidity compared to warm indoor air. Always calculate absolute values when comparing moisture content across different temperatures.

Question 16

A manufacturing facility uses evaporative cooling to condition workspace air. The process involves spraying water into an airstream, where some water evaporates while the remainder is recirculated. The air enters at $35°C$ dry-bulb temperature and $25°C$ wet-bulb temperature.

If the evaporative cooling process is assumed to be adiabatic saturation, what will be the approximate exit conditions of the air?

Exit temperature $25°C$ and relative humidity $100\%$ because adiabatic saturation achieves the wet-bulb temperature (correct answer)
Exit temperature $30°C$ and relative humidity $75\%$ because partial evaporation occurs due to limited contact time
Exit temperature $25°C$ and relative humidity $85\%$ because some sensible heat is lost to the water spray
Exit temperature $27°C$ and relative humidity $90\%$ because the process follows a constant enthalpy line on the psychrometric chart

Explanation: In an ideal adiabatic saturation process, the air follows a constant wet-bulb temperature line until it reaches saturation (100% RH). The final temperature equals the initial wet-bulb temperature (25°C) at 100% relative humidity. This occurs because the latent heat of evaporation exactly balances the sensible heat lost by the air. Choice B describes incomplete saturation. Choice C incorrectly suggests heat loss beyond the evaporation. Choice D confuses adiabatic saturation with constant enthalpy (which would be true for a psychrometric process, but here the air reaches the wet-bulb temperature).

Question 17

An air conditioning system processes outdoor air at $35°C$ and $60\%$ relative humidity. After cooling to $15°C$ , condensate is removed, and the air is then reheated to $25°C$ for supply to the conditioned space. If the humidity ratio of the outdoor air is $0.021 \text{ kg}_w/\text{kg}_{da}$ , what is the humidity ratio of the supply air?

$0.021 \text{ kg}_w/\text{kg}_{da}$ because humidity ratio remains constant during sensible heating and cooling processes
$0.0106 \text{ kg}_w/\text{kg}_{da}$ because the humidity ratio is halved when the saturation pressure doubles with temperature change
$0.0134 \text{ kg}_w/\text{kg}_{da}$ because moisture is removed during cooling, and reheating does not add moisture (correct answer)
$0.0168 \text{ kg}_w/\text{kg}_{da}$ because the final relative humidity must equal the initial relative humidity at the supply temperature

Explanation: During cooling from 35°C to 15°C, the air becomes saturated and condensate is removed. The humidity ratio becomes equal to the saturation humidity ratio at 15°C (approximately 0.0134 kg_w/kg_da). During reheating from 15°C to 25°C, no moisture is added or removed, so the humidity ratio remains constant at 0.0134 kg_w/kg_da. Choice A is wrong because it ignores moisture removal during cooling. Choice B incorrectly assumes humidity ratio is inversely proportional to saturation pressure. Choice D incorrectly assumes relative humidity remains constant.

Question 18

Two identical rooms maintain different comfort conditions: Room A at $22°C$ and $45\%$ relative humidity, and Room B at $26°C$ and $35\%$ relative humidity. Which statement correctly compares the moisture content and dew point of these two rooms?

Room A has higher humidity ratio and higher dew point because it has higher relative humidity
Room B has higher humidity ratio and higher dew point because it operates at higher dry-bulb temperature
Room A has higher humidity ratio but Room B has higher dew point due to temperature effects
Room B has higher humidity ratio and higher dew point despite lower relative humidity due to higher saturation pressure (correct answer)

Explanation: Humidity ratio equals relative humidity times saturation humidity ratio. At 26°C, the saturation pressure is significantly higher than at 22°C. Even with lower relative humidity (35% vs 45%), Room B's higher saturation condition results in higher absolute humidity ratio. Since dew point depends only on humidity ratio (amount of moisture), Room B also has a higher dew point. Choice A incorrectly assumes relative humidity determines absolute moisture content. Choice B gives correct conclusion but wrong reasoning. Choice C incorrectly states the humidity ratio comparison.

Question 19

During a winter day, outdoor air at $-5°C$ and $80\%$ relative humidity is heated to $20°C$ without adding moisture. If the saturation pressure at $-5°C$ is $0.40 \text{ kPa}$ and at $20°C$ is $2.34 \text{ kPa}$ , what is the relative humidity of the heated air?

$13.7\%$ because the partial pressure of water vapor remains constant while saturation pressure increases (correct answer)
$80\%$ because relative humidity is independent of temperature during sensible heating processes
$4.2\%$ because relative humidity decreases proportionally with the absolute temperature ratio
$27.4\%$ because the humidity ratio changes with temperature according to the ideal gas law

Explanation: During sensible heating without moisture addition, the partial pressure of water vapor remains constant. Initially: partial pressure = 0.80 × 0.40 = 0.32 kPa. After heating, this same partial pressure exists at 20°C where saturation pressure is 2.34 kPa. Therefore: RH = 0.32/2.34 = 0.137 = 13.7%. Choice B incorrectly assumes RH remains constant during heating. Choice C incorrectly applies temperature ratio to RH calculation. Choice D incorrectly suggests humidity ratio changes during sensible heating.

Question 20

A cooling tower operates by evaporating water to reject heat from a chilled water system. Hot water enters at $40°C$ and is cooled by air at $25°C$ dry-bulb and $18°C$ wet-bulb temperature. If the cooling tower achieves $85\%$ effectiveness, what principle determines the minimum possible temperature of the cooled water?

The minimum temperature equals the air dry-bulb temperature because sensible heat transfer cannot cool below ambient air temperature
The minimum temperature equals the air wet-bulb temperature because evaporative cooling is limited by adiabatic saturation conditions (correct answer)
The minimum temperature equals the average of dry-bulb and wet-bulb temperatures due to combined heat and mass transfer
The minimum temperature equals the air dew point temperature because this represents the saturation limit for moisture transfer

Explanation: In evaporative cooling systems like cooling towers, the theoretical minimum temperature achievable is the wet-bulb temperature of the ambient air. This occurs because the cooling process involves both heat and mass transfer, where water evaporation provides cooling below the dry-bulb temperature. The wet-bulb temperature represents the adiabatic saturation temperature - the lowest temperature achievable through evaporative cooling. Choice A ignores evaporative cooling effects. Choice C has no theoretical basis. Choice D confuses dew point with wet-bulb temperature limits.