Respuesta :
Answer:
a) 251.31 m/s and 55.29 m/s
b) The mass flow rate is 0.0396 kg/s
c) The rate of entropy production is 0.0144 kW/K
Explanation:
a) The steady state is:
mi = mo
[tex]\frac{A_{i}V_{i} }{v_{i} } =\frac{A_{o}V_{o} }{v_{o} } \\V_{i} =V_{o}(\frac{v_{i}}{v_{o}} )=V_{o}(\frac{T_{i}P_{o} }{T_{o}P_{i} } )=V_{o}(\frac{330*120}{300*600}) =0.22V_{o}[/tex]
The energy balance is:
[tex]h_{i}+\frac{V_{i}^{2} }{2} =h_{2}+\frac{V_{o}^{2} }{2} \\h_{i}-h_{o}+(\frac{V_{i}^{2}-V_{o}^{2} }{2} )=0\\V_{o}=\sqrt{\frac{2(h_{i}-h_{o})}{0.9516} } =\sqrt{\frac{2*(330.24-300.19)x10^{3} }{0.9516} } =251.31m/s[/tex]
Vi = 0.22 * 251.31 = 55.29 m/s
b) The mass flow rate is:
[tex]m=\frac{A_{o}V_{o}}{v_{o}} =\frac{\pi d^{2}V_{o}P_{o} }{4RT_{o}} =\frac{\pi *(0.012^{2})*251.31*120x10^{3} }{4*287*300} =0.0396kg/s[/tex]
c) The entropy produced is equal to:
[tex]\frac{Q}{T} +S_{gen} =m(s_{2} -s_{1} )\\0+S_{gen} =m(s_{2} -s_{1} )\\S_{gen} =m(s_{2} -s_{1} )\\S_{gen}=0.0396*(c_{p} ln\frac{T_{o}}{T_{i}} -Rln\frac{P_{o}}{P_{i}} )=0.0396*(1.004ln\frac{300}{330} -0.287ln\frac{120}{600} )=0.0144kW/K[/tex]
The inlet velocity, exit velocity, mass flow rate, rate of entropy production are respectively;
a) V₁ = 55.38 m/s and V₂ = 251.73 m/s
b) m' = 0.04 kg/s
c) S_gen = 0.0144 kW/.k
What is Entropy and mass flow rate?
We are given;
Diameter of Pipe; D = 1.2 cm = 0.012 m
Inlet pressure; p₁ = 600 kPa
inlet temperature; T₁ = 330 K
exit pressure; p₂ = 120 kPa = 120000 Pa
exit temperature; T₂ = 300 K
A) Applying the condition at steady rate;
m'₁ = m'₂
Thus;
A₁V₁/v₁ = A₂V₂/v₂
Area of the pipe is the same and so it cancels out to give;
V₁/v₁ = V₂/v₂
We know that v = RT/Pv
Thus;
V₁ = (p₂T₁V₂)/p₁T₂
V₁ = (120 * 330 * V₂)/(600 * 300)
V₁ = 0.22V₂
From energy balance equation;
V₂ = √(2(h₁ - h₂)/V₁)
From ideal gas properties table, we know that;
at T₁ = 330 K, h₁ = 330.34 * 10³ J/kg
Similarly, at T₂ = 300 K, h₂ = 300.19 * 10³ J/kg
Plugging in the relevant values and solving will give us;
V₂ = 251.73 m/s and V₁ = 55.38 m/s
B) The mass flow rate is gotten from;
m' = p₂A₂V₂/RT₂
where A₂ = πd²/4 = π(0.012²)/4 = 0.00045 m²
Thus;
m' = (120000 * 0.00045 * 251.73)/(287 * 300)
m' = 0.04 kg/s
C) The entropy produced is gotten from the formula;
S_gen = m'((c_p * In T₂/T₁) - (R In P₂/P₁))
where;
c_p = 1.004 kj/kg.k
R = 0.287 kJ/kg.k
Thus;
S_gen = 0.04((1.004 * In 300/330) - (0.287 In 120/600))
S_gen = 0.0144 kW/.k
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