VIPSolutionsGive step-by-step solution with explanation and final answer: Act like a helpful tutor and exlain me : How much would one cubic foot of moist air weigh at sea level that has the following conditions: Tdrybulb = 9 5 F , Twetbulb = 8 4 F: A . 0 . 0 7 9 9 lbs B . 0 . 0 6 9 9 lbs C . 0 . 0 6 8 9 lbs D . 0 . 0 6 7 9 lbs E . 0 . 0 6 5 9 lbs
Question:
Give step-by-step solution with explanation and final answer: Act like a helpful tutor and exlain me : How much would one cubic foot of moist air weigh at sea level that has the following conditions: Tdrybulb = 9 5 F , Twetbulb = 8 4 F: A . 0 . 0 7 9 9 lbs B . 0 . 0 6 9 9 lbs C . 0 . 0 6 8 9 lbs D . 0 . 0 6 7 9 lbs E . 0 . 0 6 5 9 lbs
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Asked by: manish kumar
Created at: 2025-09-23 11:33:20
Asked by: manish kumar
Created at: 2025-09-23 11:33:20
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# Step-by-Step Solution: Weight of 1 Cubic Foot of Moist Air
Given:
- Dry bulb temperature (\( T_{db} \)) = 95°F
- Wet bulb temperature (\( T_{wb} \)) = 84°F
- Pressure = Sea level (standard atmospheric pressure, \( P = 14.696 \) psi or \( 29.92 \) inHg or \( 101.325 \) kPa)
- Options (in lbs):
A. 0.0799
B. 0.0699
C. 0.0689
D. 0.0679
E. 0.0659
We'll calculate the **weight (density) of 1 ft³ of moist air** step by step.
---
## 1. **Convert Temperatures to °F to °C**
\[
T_{db, ^\circ C} = \frac{5}{9}(T_{db, ^\circ F} - 32)
\]
\[
T_{wb, ^\circ C} = \frac{5}{9}(T_{wb, ^\circ F} - 32)
\]
- \( T_{db, ^\circ C} = \frac{5}{9}(95 - 32) = \frac{5}{9}(63) \approx 35.0^\circ C \)
- \( T_{wb, ^\circ C} = \frac{5}{9}(84 - 32) = \frac{5}{9}(52) \approx 28.9^\circ C \)
---
## 2. **Find Saturation Vapor Pressure at Wet Bulb (\( P_{ws} \))**
Use the simplified formula:
\[
P_{ws} = 0.6108 \times \exp\left(\frac{17.27 \times T_{wb}}{T_{wb} + 237.3}\right)
\]
where \( T_{wb} \) is in °C, \( P_{ws} \) in kPa.
\[
P_{ws} = 0.6108 \times \exp\left(\frac{17.27 \times 28.9}{28.9 + 237.3}\right)
\]
\[
= 0.6108 \times \exp\left(\frac{498.5}{266.2}\right)
\]
\[
= 0.6108 \times \exp(1.873)
\]
\[
= 0.6108 \times 6.509 \approx 3.98 \text{ kPa}
\]
---
## 3. **Calculate Actual Vapor Pressure (\( P_w \))**
\[
P_w = P_{ws} - A \cdot (P_{atm} - P_{ws}) \cdot (T_{db} - T_{wb})
\]
Where:
- \( A \approx 0.00066 \) per °C for normal barometric pressure
- \( P_{atm} \) in kPa = 101.325
- \( T_{db} - T_{wb} = 35.0 - 28.9 = 6.1^\circ C \)
\[
P_w = 3.98 - 0.00066 \cdot (101.325 - 3.98) \cdot 6.1
\]
\[
= 3.98 - 0.00066 \cdot 97.345 \cdot 6.1
\]
\[
= 3.98 - 0.00066 \cdot 593.8045
\]
\[
= 3.98 - 0.392
\]
\[
= 3.588 \text{ kPa}
\]
---
## 4. **Calculate Humidity Ratio (\( w \))**
\[
w = 0.622 \cdot \frac{P_w}{P_{atm} - P_w}
\]
\[
w = 0.622 \cdot \frac{3.588}{101.325 - 3.588}
\]
\[
= 0.622 \cdot \frac{3.588}{97.737}
\]
\[
= 0.622 \cdot 0.0367 \approx 0.0228 \text{ kg/kg dry air}
\]
---
## 5. **Calculate Moist Air Density (\( \rho \))**
\[
\rho = \frac{P_{da}}{R_{da} T} + \frac{P_w}{R_w T}
\]
But it's easier to use the following formula for US units:
\[
\rho = \frac{p_{atm} \times 29}{0.3704 \times (T_{db, ^\circ F} + 459.67) \times (1 + 1.6078w)}
\]
Where:
- \( p_{atm} = 14.696 \) psi (sea level)
- \( T_{db, ^\circ F} = 95 \)
- \( w = 0.0228 \)
\[
T_{db, ^\circ R} = 95 + 459.67 = 554.67^\circ R
\]
\[
\rho = \frac{14.696 \times 29}{0.3704 \times 554.67 \times (1 + 1.6078 \times 0.0228)}
\]
\[
= \frac{426.184}{0.3704 \times 554.67 \times (1.0367)}
\]
\[
= \frac{426.184}{205.393 \times 1.0367}
\]
\[
= \frac{426.184}{213.206}
\]
\[
= 2.0 \text{ lb/ft}^3
\]
This value is way too high; let's use the SI formula and convert.
### Standard air at 95°F (35°C) has a density of:
\[
\rho_{dry} = \frac{P}{R_{da} T}
\]
Where:
- \( P = 101325 \) Pa
- \( R_{da} = 287.05 \) J/kg·K
- \( T = 35 + 273.15 = 308.15 \) K
\[
\rho_{dry} = \frac{101325}{287.05 \times 308.15} = \frac{101325}{88455.9} \approx 1.146 \text{ kg/m}^3
\]
Now, correct for humidity:
\[
\rho = \frac{P_{da}}{R_{da} T} + \frac{P_w}{R_w T}
\]
Where:
- \( P_{da} = 101325 - 3588 = 97737 \) Pa
- \( R_w = 461.5 \) J/kg·K
\[
\rho = \frac{97737}{287.05 \times 308.15} + \frac{3588}{461.5 \times 308.15}
\]
\[
= \frac{97737}{88455.9} + \frac{3588}{142252.2}
\]
\[
= 1.105 + 0.0252 = 1.130 \text{ kg/m}^3
\]
Convert \( 1.130 \) kg/m³ to lb/ft³:
- \( 1 \) kg = \( 2.20462 \) lb
- \( 1 \) m³ = \( 35.3147 \) ft³
\[
1.130 \text{ kg/m}^3 \times \frac{2.20462}{35.3147} = 1.130 \times 0.06243 = 0.0705 \text{ lb/ft}^3
\]
---
## 6. **Final Answer**
**Closest answer choice:**
**B. 0.0699 lbs**
---
## **Summary Table**
| Step | Calculation | Result |
|------|-------------|--------|
| Convert T | \( 95^\circ F \to 35^\circ C \), \( 84^\circ F \to 28.9^\circ C \) | |
| Saturation vapor pressure | \( P_{ws} \) at 28.9°C | 3.98 kPa |
| Vapor pressure | \( P_w \) | 3.588 kPa |
| Humidity ratio | \( w \) | 0.0228 |
| Moist air density | \( \rho \) | 0.0705 lb/ft³ |
---
## **Final Answer**
\[
\boxed{0.0699 \text{ lbs (per cubic foot, choice B)}}
\]
---
**Explanation:**
The weight per cubic foot of moist air at the given conditions is about 0.070 lbs. This is less than dry air at lower temperatures because warm, moist air is less dense. The calculations involve psychrometric relationships between temperature, humidity, and air density.
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