The air gas constant is 53.34 ft.lbf/Ibm°R

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a. 1. The air gas constant is 53.34 ft.lbf/Ibm°R. Convert this constant to Btu/Ibm.oR, joules/kg.K, and kcallkg.K.

  1. The universal gas constant is 1.986 Btu/lb mole. OR. Convert this constant to kcal/kg mole.K, ft.lbf/ lIb mole. oR, and joules/kg mole K.

b. 1. The thermal conductivity of aluminum is 200 W/m.oC. Convert this value to Btu/hr.ft.oP and kcaVsec.m.oC

  1. The thermal conductivity of copper is 50 Btu/h.ft. oF. Convert this value to W/m.oC and cal/sec.m.°c.

The Correct Answer and Explanation is :

Let’s tackle the conversions one at a time.

Part a.1: Convert the air gas constant (53.34 ft.lbf/Ibm.°R) to other units.

Given the air gas constant:
R = 53.34 ft.lbf/Ibm.°R

We need to convert this to:

  1. Btu/Ibm.°R
  2. Joules/kg.K
  3. kcal/kg.K

Conversion Factors:

  • 1 Btu = 1.35582 ft.lbf
  • 1 Joule = 0.73756 ft.lbf
  • 1 kcal = 4184 Joules
  • 1 lbm (pound mass) = 0.453592 kg
  • 1 °R = 5/9 °C
  • 1 °R = 1 °F (for practical purposes in this case)

Step-by-step conversion:

  1. Convert ft.lbf/Ibm.°R to Btu/Ibm.°R:
    [
    R = 53.34 \, \text{ft.lbf/Ibm.°R} \times \frac{1 \, \text{Btu}}{1.35582 \, \text{ft.lbf}} = 39.38 \, \text{Btu/Ibm.°R}
    ]
  2. Convert ft.lbf/Ibm.°R to Joules/kg.K:
    [
    R = 53.34 \, \text{ft.lbf/Ibm.°R} \times \frac{0.73756 \, \text{Joules}}{1 \, \text{ft.lbf}} \times \frac{1}{0.453592 \, \text{kg}} = 85.16 \, \text{Joules/kg.K}
    ]
  3. Convert ft.lbf/Ibm.°R to kcal/kg.K:
    [
    R = 53.34 \, \text{ft.lbf/Ibm.°R} \times \frac{0.73756 \, \text{Joules}}{1 \, \text{ft.lbf}} \times \frac{1}{0.453592 \, \text{kg}} \times \frac{1 \, \text{kcal}}{4184 \, \text{Joules}} = 0.0204 \, \text{kcal/kg.K}
    ]

Part a.2: Convert the universal gas constant (1.986 Btu/lb mole.°R) to other units.

Given the universal gas constant:
R = 1.986 Btu/lb mole.°R

We need to convert this to:

  1. kcal/kg mole.K
  2. ft.lbf/lb mole.°R
  3. Joules/kg mole.K

Conversion Factors:

  • 1 Btu = 1.35582 ft.lbf
  • 1 kcal = 4184 Joules
  • 1 lbm = 0.453592 kg

Step-by-step conversion:

  1. Convert Btu/lb mole.°R to kcal/kg mole.K:
    [
    R = 1.986 \, \text{Btu/lb mole.°R} \times \frac{1 \, \text{kcal}}{3.9683 \, \text{Btu}} \times \frac{1}{0.453592 \, \text{kg}} = 4.42 \, \text{kcal/kg mole.K}
    ]
  2. Convert Btu/lb mole.°R to ft.lbf/lb mole.°R:
    [
    R = 1.986 \, \text{Btu/lb mole.°R} \times 1.35582 \, \text{ft.lbf/Btu} = 2.69 \, \text{ft.lbf/lb mole.°R}
    ]
  3. Convert Btu/lb mole.°R to Joules/kg mole.K:
    [
    R = 1.986 \, \text{Btu/lb mole.°R} \times 1.35582 \, \text{ft.lbf/Btu} \times 0.73756 \, \text{Joules/ft.lbf} \times \frac{1}{0.453592 \, \text{kg}} = 830.3 \, \text{Joules/kg mole.K}
    ]

Part b.1: Convert the thermal conductivity of aluminum (200 W/m.°C) to other units.

Given the thermal conductivity of aluminum:
K = 200 W/m.°C

We need to convert this to:

  1. Btu/hr.ft.°F
  2. kcal/sec.m.°C

Conversion Factors:

  • 1 W = 3.41214 Btu/hr
  • 1 W = 0.8598 cal/sec
  • 1 meter = 3.28084 feet
  • 1 °C = 1.8 °F (temperature difference is equivalent)

Step-by-step conversion:

  1. Convert W/m.°C to Btu/hr.ft.°F:
    [
    K = 200 \, \text{W/m.°C} \times \frac{3.41214 \, \text{Btu/hr}}{1 \, \text{W}} \times \frac{1}{3.28084 \, \text{ft}} \times \frac{1}{1.8 \, \text{°C/°F}} = 1.28 \, \text{Btu/hr.ft.°F}
    ]
  2. Convert W/m.°C to kcal/sec.m.°C:
    [
    K = 200 \, \text{W/m.°C} \times 0.8598 \, \text{cal/sec} \times \frac{1}{1000} = 0.17196 \, \text{kcal/sec.m.°C}
    ]

Part b.2: Convert the thermal conductivity of copper (50 Btu/h.ft.°F) to other units.

Given the thermal conductivity of copper:
K = 50 Btu/h.ft.°F

We need to convert this to:

  1. W/m.°C
  2. cal/sec.m.°C

Conversion Factors:

  • 1 Btu = 1055.06 J
  • 1 hour = 3600 seconds
  • 1 foot = 0.3048 meters
  • 1 °F = 5/9 °C (temperature difference)

Step-by-step conversion:

  1. Convert Btu/h.ft.°F to W/m.°C:
    [
    K = 50 \, \text{Btu/h.ft.°F} \times \frac{1055.06 \, \text{J/Btu}}{1 \, \text{Btu}} \times \frac{1 \, \text{h}}{3600 \, \text{s}} \times \frac{1}{0.3048^2 \, \text{m}^2} \times \frac{1}{5/9 \, \text{°F/°C}} = 235.87 \, \text{W/m.°C}
    ]
  2. Convert Btu/h.ft.°F to cal/sec.m.°C:
    [
    K = 50 \, \text{Btu/h.ft.°F} \times 0.8598 \, \text{cal/sec} \times \frac{1}{0.3048^2 \, \text{m}^2} \times \frac{1}{5/9 \, \text{°F/°C}} = 0.02847 \, \text{cal/sec.m.°C}
    ]

Final Results:

  • Air gas constant:
  • 39.38 Btu/Ibm.°R
  • 85.16 Joules/kg.K
  • 0.0204 kcal/kg.K
  • Universal gas constant:
  • 4.42 kcal/kg mole.K
  • 2.69 ft.lbf/lb mole.°R
  • 830.3 Joules/kg mole.K
  • Thermal conductivity of aluminum:
  • 1.28 Btu/hr.ft.°F
  • 0.17196 kcal/sec.m.°C
  • Thermal conductivity of copper:
  • 235.87 W/m.°C
  • 0.02847 cal/sec.m.°C

Explanation:

These conversions are necessary when dealing with various units in thermodynamics, particularly when working with different measurement systems (e.g., Imperial vs. SI units). It’s important to use proper conversion factors to ensure consistency in calculations, as thermodynamic properties like gas constants and thermal conductivity can differ in their expression depending on the unit system. Understanding these conversions is fundamental to accurately analyzing and applying thermodynamic principles in real-world engineering applications.

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