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Convert: a. 10 ft lbf/lbm to J/kg b. 500 °R to K c. 1.987 Btu/lb mol °R to cal/g mol K d. 100 BTU/h ft² °F to kW/m² °C e. 0.002 Kg/(m s) to lbm/(ft s)

The Correct Answer and Explanation is:

Here’s how we can convert each of these units step by step:

a) Convert 10 ft·lbf/lbm to J/kg

The conversion factor between foot-pounds and joules is:

• 1 ft\cdotplbf=1.35582 J1 \, \text{ft·lbf} = 1.35582 \, \text{J}1ft\cdotplbf=1.35582J

And for pounds mass to kilograms:

• 1 lbm=0.453592 kg1 \, \text{lbm} = 0.453592 \, \text{kg}1lbm=0.453592kg

Now, the conversion: 10 ft\cdotplbf/lbm=10 ft\cdotplbf1 lbm×1.35582 J1 ft\cdotplbf×1 lbm0.453592 kg10 \, \text{ft·lbf/lbm} = \frac{10 \, \text{ft·lbf}}{1 \, \text{lbm}} \times \frac{1.35582 \, \text{J}}{1 \, \text{ft·lbf}} \times \frac{1 \, \text{lbm}}{0.453592 \, \text{kg}}10ft\cdotplbf/lbm=1lbm10ft\cdotplbf​×1ft\cdotplbf1.35582J​×0.453592kg1lbm​ =10×1.355820.453592=29.9 J/kg= \frac{10 \times 1.35582}{0.453592} = 29.9 \, \text{J/kg}=0.45359210×1.35582​=29.9J/kg

b) Convert 500 °R to K

The conversion formula between Rankine (°R) and Kelvin (K) is: K=59×(°R−459.67)K = \frac{5}{9} \times \left(°R – 459.67 \right)K=95​×(°R−459.67)

Now for the conversion: 500 °R=59×(500−459.67)=59×40.33=22.4 K500 \, \text{°R} = \frac{5}{9} \times (500 – 459.67) = \frac{5}{9} \times 40.33 = 22.4 \, \text{K}500°R=95​×(500−459.67)=95​×40.33=22.4K

c) Convert 1.987 Btu/lb·mol·°R to cal/g·mol·K

We need to know the conversion factors:

• 1 Btu=252.164 cal1 \, \text{Btu} = 252.164 \, \text{cal}1Btu=252.164cal
• 1 lb\cdotpmol=453.592 g\cdotpmol1 \, \text{lb·mol} = 453.592 \, \text{g·mol}1lb\cdotpmol=453.592g\cdotpmol
• 1 °R=1 °F=5/9 K1 \, \text{°R} = 1 \, \text{°F} = 5/9 \, \text{K}1°R=1°F=5/9K

So: 1.987 Btu/lb\cdotpmol\cdotp°R=1.987×252.164453.592 cal/g\cdotpmol\cdotpK1.987 \, \text{Btu/lb·mol·°R} = 1.987 \times \frac{252.164}{453.592} \, \text{cal/g·mol·K}1.987Btu/lb\cdotpmol\cdotp°R=1.987×453.592252.164​cal/g\cdotpmol\cdotpK =1.987×0.556=1.1 cal/g\cdotpmol\cdotpK= 1.987 \times 0.556 = 1.1 \, \text{cal/g·mol·K}=1.987×0.556=1.1cal/g\cdotpmol\cdotpK

d) Convert 100 BTU/h·ft²·°F to kW/m²·°C

We use the following conversion factors:

• 1 BTU/h=0.293071 W1 \, \text{BTU/h} = 0.293071 \, \text{W}1BTU/h=0.293071W
• 1 ft²=0.092903 m²1 \, \text{ft²} = 0.092903 \, \text{m²}1ft²=0.092903m²
• 1 °F=5/9 °C1 \, \text{°F} = 5/9 \, \text{°C}1°F=5/9°C

So: 100 BTU/h\cdotpft²\cdotp°F=100×0.293071×10.092903×95 kW/m²\cdotp°C100 \, \text{BTU/h·ft²·°F} = 100 \times 0.293071 \times \frac{1}{0.092903} \times \frac{9}{5} \, \text{kW/m²·°C}100BTU/h\cdotpft²\cdotp°F=100×0.293071×0.0929031​×59​kW/m²\cdotp°C =100×0.293071×10.764×1.8=565.5 kW/m²\cdotp°C= 100 \times 0.293071 \times 10.764 \times 1.8 = 565.5 \, \text{kW/m²·°C}=100×0.293071×10.764×1.8=565.5kW/m²\cdotp°C

e) Convert 0.002 kg/(m·s) to lbm/(ft·s)

We need to use the following conversion factors:

• 1 kg=2.20462 lbm1 \, \text{kg} = 2.20462 \, \text{lbm}1kg=2.20462lbm
• 1 m=3.28084 ft1 \, \text{m} = 3.28084 \, \text{ft}1m=3.28084ft

Now for the conversion: 0.002 kg/(m\cdotps)=0.002×2.204623.28084 lbm/(ft\cdotps)0.002 \, \text{kg/(m·s)} = 0.002 \times \frac{2.20462}{3.28084} \, \text{lbm/(ft·s)}0.002kg/(m\cdotps)=0.002×3.280842.20462​lbm/(ft\cdotps) =0.002×0.671=0.00134 lbm/(ft\cdotps)= 0.002 \times 0.671 = 0.00134 \, \text{lbm/(ft·s)}=0.002×0.671=0.00134lbm/(ft\cdotps)

Summary:

• a) 10 ft·lbf/lbm = 29.9 J/kg
• b) 500 °R = 22.4 K
• c) 1.987 Btu/lb·mol·°R = 1.1 cal/g·mol·K
• d) 100 BTU/h·ft²·°F = 565.5 kW/m²·°C
• e) 0.002 kg/(m·s) = 0.00134 lbm/(ft·s)