Question:
Vapor pressure of water at various temperature is given in the table.Temperature (in K) 284 289 294 299 310 Vapor pressure of water (in kPa) 1.28 1.80 2.50 3.40 6.40
An air and water vapor mixture at 100 kPa with a relative humidity of 20% has a dry-bulb temperature of 310 K. Assume latent heat of vaporization for water is 44 kJ mol-1 and specific heat capacity of the air and water vapor mixture is 0.035 kJ mol-1K-1. Which one of the following is the closest to its wet-bulb temperature (in K)?
Vapor pressure of water at various temperature is given in the table.
| Temperature (in K) | 284 | 289 | 294 | 299 | 310 |
|---|---|---|---|---|---|
| Vapor pressure of water (in kPa) | 1.28 | 1.80 | 2.50 | 3.40 | 6.40 |
An air and water vapor mixture at 100 kPa with a relative humidity of 20% has a dry-bulb temperature of 310 K. Assume latent heat of vaporization for water is 44 kJ mol-1 and specific heat capacity of the air and water vapor mixture is 0.035 kJ mol-1K-1. Which one of the following is the closest to its wet-bulb temperature (in K)?
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Find the actual water vapor mole fraction from the relative humidity, then find which tabulated temperature makes the sensible heat given up by the gas equal to the latent heat needed to saturate it.
Updated On: Aug 10, 2026
- 284
- 294
- 310
- 321
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The Correct Option is B
Solution and Explanation
Step 1: Find the actual partial pressure of water vapor.
At \(T_{db}=310\ K\), \(p_{sat}=6.40\) kPa. \(p_v = 0.20 \times 6.40 = 1.28\) kPa
Step 2: Convert to mole fraction.
\(y = 1.28/100 = 0.0128\)
Step 3: Write the wet-bulb energy balance.
\[C_s(T_{db}-T_{wb}) = \lambda(y_{sat}(T_{wb}) - y)\]
Step 4: Test 294 K.
\(p_{sat}(294)=2.50\), \(y_{sat}=0.0250\). Left side: \(0.035\times16=0.560\). Right side: \(44\times0.0122=0.537\). Close agreement given the coarse table.
Step 5: Rule out other options.
284 K and 310 K give large mismatches; 321 K exceeds dry-bulb, impossible for unsaturated air.
Step 6: Conclusion.
\[ \boxed{T_{wb} \approx 294\ \text{K}} \]
At \(T_{db}=310\ K\), \(p_{sat}=6.40\) kPa. \(p_v = 0.20 \times 6.40 = 1.28\) kPa
Step 2: Convert to mole fraction.
\(y = 1.28/100 = 0.0128\)
Step 3: Write the wet-bulb energy balance.
\[C_s(T_{db}-T_{wb}) = \lambda(y_{sat}(T_{wb}) - y)\]
Step 4: Test 294 K.
\(p_{sat}(294)=2.50\), \(y_{sat}=0.0250\). Left side: \(0.035\times16=0.560\). Right side: \(44\times0.0122=0.537\). Close agreement given the coarse table.
Step 5: Rule out other options.
284 K and 310 K give large mismatches; 321 K exceeds dry-bulb, impossible for unsaturated air.
Step 6: Conclusion.
\[ \boxed{T_{wb} \approx 294\ \text{K}} \]
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