Respuesta :
Answer:
The exit temperature is 392.6 K and the volume flow rate is 3 m³/s
Explanation:
Given:
V₁ = 1 m³/s
T₁ = 20°C = 293 K
V₂ = 2 m³/s
T₂ = 200°C = 473 K
P₁ = P₂ = 100 kPa
P₃ = 100 kPa
Applying ideal gas law:
[tex]v_{1} =\frac{RT_{1} }{P_{1} } =\frac{0.287*293}{100} =0.841m^{3} /kg[/tex]
[tex]v_{2} =\frac{RT_{2} }{P_{2} } =\frac{0.287*473}{100} =1.358m^{3} /kg[/tex]
The mass flow rates are:
[tex]m_{1} =\frac{V_{1} }{v_{1} } =\frac{1}{0.841} =1.189kg/s[/tex]
[tex]m_{2} =\frac{V_{2} }{v_{2} } =\frac{2}{1.358} =1.473kg/s[/tex]
[tex]m_{3} =m_{1} +m_{2} =1.189+1.473=2.662kg/s[/tex]
The exit temperature is obtained from the energy expression:
[tex]m_{1} CpT_{1}+m_{2} CpT_{2} =m_{3} CpT_{3}\\T_{3} =\frac{m_{1}T_{1} }{m_{3} } +\frac{m_{2}T_{2} }{m_{3} }[/tex]
[tex]T_{3} =\frac{1.189*293}{2.662} +\frac{1.473*473}{2.662} =392.6K[/tex]
The specific volume at exit is:
[tex]v_{3} =\frac{RT_{3} }{P_{3} } =\frac{0.287*392.6}{100} =1.127m^{3} /kg[/tex]
The volume rate is:
[tex]V_{3} =m_{3} v_{3} =2.662*1.127=3m^{3} /s[/tex]
Answer:
[tex]T_{3} = 393.20\,K[/tex], [tex]\dot V_{3} = 3.002\,\frac{m^{3}}{s}[/tex]
Explanation:
This is a case of a mix chamber, which is modelled after the First Law of the Thermodynamics:
[tex]\dot m_{1} \cdot h_{1} + \dot m_{2} \cdot h_{2} - \dot m_{3} \cdot h_{3} = 0[/tex]
According to the Principle of Mass Conservation:
[tex]\dot m_{1} + \dot m_{2} - \dot m_{3} = 0[/tex]
Let assume that air behaves as an ideal gas. The density has the following expression:
[tex]P\cdot V = n\cdot R_{u}\cdot T[/tex]
[tex]P \cdot V = \frac{m}{M}\cdot R_{u}\cdot T[/tex]
[tex]\rho = \frac{P\cdot M}{R_{u}\cdot T}[/tex]
Densities at inlets are, respectively:
[tex]\rho_{1} = \frac{(100\,kPa)\cdot (28.02\,\frac{kg}{kmol} )}{(8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} )\cdot (293.15\,K)}[/tex]
[tex]\rho_{1} = 1.149\,\frac{kg}{m^{3}}[/tex]
[tex]\rho_{2} = \frac{(100\,kPa)\cdot (28.02\,\frac{kg}{kmol} )}{(8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} )\cdot (473.15\,K)}[/tex]
[tex]\rho_{2} = 0.712\,\frac{kg}{m^{3}}[/tex]
The mass flow at outlet is:
[tex]\dot m_{3} = (1.149\,\frac{kg}{m^{3}} )\cdot (1\,\frac{m^{3}}{s} ) + (0.712\,\frac{kg}{m^{3}} )\cdot (2\,\frac{m^{3}}{s} )[/tex]
[tex]\dot m_{3} = 2.573\,\frac{kg}{s}[/tex]
Specific enthalpies depends on temperature only. The required variable for inlet are obtained from property tables:
[tex]h_{1} = 293.31\,\frac{kJ}{kg}[/tex]
[tex]h_{2} = 475.46\,\frac{kJ}{kg}[/tex]
Specific enthalpy at outlet is:
[tex]h_{3} = \frac{\dot m_{1}\cdot h_{1} + \dot m_{2}\cdot h_{2}}{\dot m_{3}}[/tex]
[tex]h_{3} = \frac{(1.149\,\frac{kg}{s} )\cdot (293.31\,\frac{kJ}{kg} )+(1.424\,\frac{kg}{s} )\cdot (475.46\,\frac{kJ}{kg} )}{2.573\,\frac{kg}{s} }[/tex]
[tex]h_{3} = 394.11\,\frac{kJ}{kg}[/tex]
The exit temperature is:
[tex]T_{3} = 393.20\,K[/tex]
The density of air at outlet is:
[tex]\rho_{3} = \frac{(100\,kPa)\cdot (28.02\,\frac{kg}{kmol} )}{(8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} )\cdot (393.20\,K)}[/tex]
[tex]\rho_{3} = 0.857\,\frac{kg}{m^{3}}[/tex]
The volume flow rate at outlet is:
[tex]\dot V_{3} = \frac{\dot m_{3}}{\rho_{3}}[/tex]
[tex]\dot V_{3} = \frac{2.573\,\frac{kg}{s} }{0.857\,\frac{kg}{m^{3}} }[/tex]
[tex]\dot V_{3} = 3.002\,\frac{m^{3}}{s}[/tex]
