Respuesta :
Answer:
The velocity of water at the tube inlet [tex]V_{1}[/tex] = 32 [tex]\frac{m}{s}[/tex]
The value of flow rate at inlet Q = 0.4415 [tex]\frac{m^{3} }{sec}[/tex]
Explanation:
Given data
[tex]P_{1}[/tex] = 7 M pa
[tex]P_{2}[/tex] = 6 M pa
[tex]T_{1}[/tex] = 65 °c = 337 K
[tex]T_{2}[/tex] = 450°c = 723 K
[tex]V_{2}[/tex] = 80 [tex]\frac{m}{s}[/tex]
Where 1 & 2 represents inlet & outlet conditions.
We know that mass flow rate through the boiler is constant so
Mass flow rate at inlet = mass flow rate at outlet
[tex]\rho_{1} A_{1} V_{1} = \rho_{2} A_{2} V_{2}[/tex]
Since area of the tube if the boiler is constant so
[tex]A_{1} = A_{2}[/tex]
⇒ [tex]\rho_{1} V_{1} = \rho_{2} V_{2}[/tex]
⇒ [tex]\frac{P_{1}V_{1} }{RT_{1} } = \frac{P_{2}V_{2} }{RT_{2} }[/tex]
⇒ [tex]\frac{P_{1}V_{1} }{T_{1} } = \frac{P_{2}V_{2} }{T_{2} }[/tex]
Put all the values in above equation
⇒ [tex]V_{1} = \frac{162240}{5061}[/tex]
⇒ [tex]V_{1}[/tex] = 32 [tex]\frac{m}{s}[/tex]
This is the velocity of water at the tube inlet.
Now volume flow rate at inlet [tex]Q = A_{1} V_{1}[/tex]
Area of the boiler tube is given by
⇒ [tex]A_{1} = A_{2} = 2\pi r^{3}t[/tex]
⇒ [tex]A_{1} =[/tex] 2 × 3.14 × [tex]0.13^{3}[/tex] × 1
⇒ [tex]A_{1} =[/tex] 0.0138 [tex]m^{3}[/tex]
Now flow rate at inlet Q = 0.0138 × 32
⇒ Q = 0.4415 [tex]\frac{m^{3} }{sec}[/tex]
This is the value of flow rate at inlet.
