Answer:
a) [tex]\dot m_{3} = 135\,\frac{lbm}{s}[/tex], b) [tex]h_{3}=168.965\,\frac{BTU}{lbm}[/tex], c) [tex]T = 200.829\,^{\textdegree}F[/tex]
Explanation:
a) The tank can be modelled by the Principle of Mass Conservation:
[tex]\dot m_{1} + \dot m_{2} - \dot m_{3} = 0[/tex]
The mass flow rate exiting the tank is:
[tex]\dot m_{3} = \dot m_{1} + \dot m_{2}[/tex]
[tex]\dot m_{3} = 125\,\frac{lbm}{s} + 10\,\frac{lbm}{s}[/tex]
[tex]\dot m_{3} = 135\,\frac{lbm}{s}[/tex]
b) An expression for the specific enthalpy at outlet is derived from the First Law of Thermodynamics:
[tex]\dot m_{1}\cdot h_{1} + \dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0[/tex]
[tex]h_{3} = \frac{\dot m_{1}\cdot h_{1}+\dot m_{2}\cdot h_{2}}{\dot m_{3}}[/tex]
Properties of water are obtained from tables:
[tex]h_{1}=180.16\,\frac{BTU}{lbm}[/tex]
[tex]h_{2}=28.08\,\frac{BTU}{lbm} + \left(0.01604\,\frac{ft^{3}}{lbm}\right)\cdot (14.7\,psia-0.25638\,psia)[/tex]
[tex]h_{2}=29.032\,\frac{BTU}{lbm}[/tex]
The specific enthalpy at outlet is:
[tex]h_{3}=\frac{(125\,\frac{lbm}{s} )\cdot (180.16\,\frac{BTU}{lbm} )+(10\,\frac{lbm}{s} )\cdot (29.032\,\frac{BTU}{lbm} )}{135\,\frac{lbm}{s} }[/tex]
[tex]h_{3}=168.965\,\frac{BTU}{lbm}[/tex]
c) After a quick interpolation from data availables on water tables, the final temperature is:
[tex]T = 200.829\,^{\textdegree}F[/tex]
Answer:
A
Explanation:I might not might be correct hmu if it is wrong im so sorry
