Set1s - ME 416 Computer Assisted Design of Thermal Systems...

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1 ME 416 Computer Assisted Design of Thermal Systems Solutions to Practice Problems #1 Design Methodology, Thermodynamic Derivatives, Curve Fitting 1. Some ideas (i) Build a new power plant (ii) Renovate current plants to make them more efficient (iii) Buy electricity from another power company (iv) Limit supply to users (v) Get growth limited by legislation All are functional, but (iv) is most probably unsatisfactory. 2. Some ideas (i) Add air ducts to blow air over brakes (ii) Apply a water spray to the brakes as needed (iii) Integrate the brakes into the engine cooling system 3. Here are my suggestions effectiveness (1.5) cost (1.0) weight (1.2) reliability (1.0) safety (1.0) 4. Assuming that s = s(T,P) we then have by the chain rule dP P s + dT T s = ds T P
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ME 416 CAD of Thermal Systems 2 By definition T c = T s P P and β v - = P s T From our problem statement we have c p = 0.67505 + 4.3156 x 10 -3 T - 1.498 x 10 -6 T 2 + 2.7403 x 10 -9 T 3 and ρ = 1037.3 - 0.38658 T - 5.7951 x 10 -4 T 2 + 5.384 x 10 -7 T 3 So we only need to evaluate P T v . Using the definition of specific volume, we can write T 1 - = T 1 = T v 1 = v P 2 P P ρ ρ ρ ρ Then taking the derivative and substituting ( 29 dP T 384x10 . 5 T 7951x10 . 5 T 38658 . 0 3 . 1037 T 384x10 . 5 T 7951x10 . 5 38658 . 0 - dT T 7403x10 . 2 T 498x10 . 1 3156x10 . 4 T 67505 . 0 = ds 2 3 7 2 4 2 7 4 2 9 6 3 + - - + - - + - + - - - - - - - We can now integrate this expression over temperature and pressure from the reference state to the state at which we want to evaluate the entropy. The coefficient of the pressure term is integrated only over pressure.
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ME 416 CAD of Thermal Systems 3 5. (a.) We must begin by eliminating h from the partial. Writing the defining differential
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Set1s - ME 416 Computer Assisted Design of Thermal Systems...

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