Thermodynamics question 1 (question about mass)

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SUMMARY

The discussion focuses on calculating the power required to overcome drag and rolling resistance forces using specific parameters in imperial units. Key formulas include the rolling resistance force (F_r = f * W) and drag force (F_d = C_d * A * (1/2) * ρ * v²). The user seeks clarification on converting weight from lbf to lbm and understanding the implications of these units in their calculations. The final power equation is established as P = (ΣF) * V, incorporating both resistance forces and velocity.

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  • Understanding of basic thermodynamics principles
  • Familiarity with imperial units (lbf, lbm, ft, etc.)
  • Knowledge of fluid dynamics concepts, specifically drag and rolling resistance
  • Ability to perform unit conversions and dimensional analysis
NEXT STEPS
  • Research "Unit conversion between lbf and lbm" for clarity on weight and mass in imperial units
  • Study "Fluid dynamics principles related to drag force" to deepen understanding of drag calculations
  • Explore "Rolling resistance in vehicle dynamics" to grasp its impact on power requirements
  • Learn "Power calculations in thermodynamics" for a comprehensive approach to energy assessments
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Students studying thermodynamics, engineers working on vehicle performance, and anyone involved in fluid dynamics calculations.

Saladsamurai
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Homework Statement



Alrighty then :smile:

I still cannot get used to texts that use lbf (lb force) and lbm (lb mass) and just plain old lb.
Metric is bomb.

I am trying to find the power needed too overcome drag and rolling resistance forces from this given info.

F_r=f\cdot W=Rolling resistance force
F_d=C_dA\frac{1}{2}\rho v^2=Drag force
Power=P=F\cdot V
Area=A=23.3 sq ft
C_d=.34
f=.02
\rho=.08\frac{lb}{ft^3}
v=55\frac{mi}{h}
weight=w=3550lbf <------how do I convert this term properly? I think I NEED TO

So, if P=F*V Total power is the sum of this product.

P=(\sum F)\cdot V

\Rightarrow P=(f*w+C_dA\frac{1}{2}\rho v^2)v

\Rightarrow P=(.02*3550lbf+.34*23.3ft^2*\frac{1}{2}*.08\frac{lb}{ft^3}*55^2\frac{mi^2}{h^2}*5280^2\frac{ft^2}{mi^2}*\frac{h^2}{3600^2s^s})*(55\frac{mi}{h}*\frac{5280ft}{mi}*\frac{h}{3600s})I cannot tell what to do with that "lbf" ? I know that tyhis is really simple, but I am not used to these units :redface:

Little help? Thanks!EDIT: Is a lbf just a lbm*ft/s^2 ? I am assuming that density is given in lbm even though it does not specify...
 
Last edited:
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Don't worry guys. I think I got it :smile:
 

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