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 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)


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