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ここに一つあって、

それをすり減らして毎日生きてる。多分。


いろいろな、

色や形や音や馨や、


飛んできては過ぎ去って、

砂煙を上げて、消えて。



歩き方が分からなくなった日は立ち止まって、


何を探して、




いまは、まぁるくてやらかいもの。


昔は硬くて四角いもの。



ご存知ありませんか？


追記） みつけたかもしれない。

3.3)

 3.3 Answer)

a) Let kinetic energy increased be equal to k

k = 1/2m(v₂²-v₁¹) = 216000 = 216 kJ/kg


b) Let the change in enthalpy be equal to Δh

Δh = {(u₂-u₁)+(p₂v₂-p₁v₁)}×10³ =  {(125-292)+(40×1/0.8-700×1/6.0)}10³
= -234.666...×10³ = -234 kJ/kg


c) Change in flow energy (i.e, pV-work) f is,

f = p₂v₂-p₁v₁ = -66.66666 = -66.7 kJ/kg


d) Let the changes in potential energy per unit of mass be equal to u. u is given by,

u = g (h₂-h₁) = 98 J/kg


e) The mass flow rate per 1 sec. is,

R = (0.15/2)²×π×60×6.0 = 6.3617.... =6.4kg/s

(cf. (0.124/2)²×π×660×0.8 = 6.3762....)



f) The changes in total energy per 1 sec. between point (1) and (2) is,

ΔE = R(k+u+Δh) = -144572 = -115 kW


e) From the answer of question (f), the flow gas loses 115kJ/s between (1) and (2), 
in the other words, the gas works 115kW on the surroundings except heat gained.
Thus, the power transferred to the surroundings between (1) and (2) is given by 
following equation.

W = 50×6.4 + 115 = 435 kW

h) From the relation between specific heats and gas constant of ideal gas, we get,

Cp-Cv = R  
      Cp = R + Cv
    ∴Cp = 280 + 700 = 980 J/(kg・K)

A homework....

3.1 Answer)

Total energy E_t=1/2mv²+mgz+U+pV

Potential energy charges are negligible, so we get,
E_t = 1/2mv²+H (H - Enthalpy)

Entering energy per 1 second is given by following equation
E₁=1/2mv₁²+H₁

The energy theta exiting flow has is given by following equation
E₂=1/2mv₂²+H₂


We can neglect the heat loss, we get the power generated per 1 sec. by following equation

P_[J/s]   = E₁ - E₂ = 1.8［1/2(20²-38²)+(3140-2500)×10³］=1151060.4 = 1151 kW