Given Information:
Inlet Temperature = T₁ = 700° F
Exit Temperature = T₂ = 645° F
Inlet Velocity = v₁ = 82 ft/s
Specific heat of air = cp = 0.253 Btu/lbm.R
Required Information:
Exit velocity = v₂ = ?
Answer:
Exit velocity = [tex]v_{2}= 838.75[/tex] [tex]ft/s[/tex]
Explanation:
As we know the energy balance equation is given by
[tex]E_{in} = E_{out}[/tex]
[tex]m(h_{1} + \frac{v_{1}^2 }{2}) = Q_{out} + m(h_{2} + \frac{v_{2}^2 }{2})[/tex]
[tex](h_{1} + \frac{v_{1}^2 }{2}) = (h_{2} + \frac{v_{2}^2 }{2})[/tex]
[tex]v_{2}^2= v_{1} ^2 +2(h_{1} - h_{2} )[/tex]
[tex]v_{2}= \sqrt{ v_{1} ^2 +2(h_{1} - h_{2} )}[/tex]
[tex]v_{2}= \sqrt{ v_{1} ^2 +2c_{p} (T_{1} - T_{2} )}[/tex]
1 Btu/lbm is equal to 25037 ft²/s²
[tex]v_{2}= \sqrt{ (82 ft/s)^2 +2*0.253\frac{Btu}{lbm.R} (700 - 645)R *25037\frac{\frac{ft^2}{s^2}}{1\frac{Btu}{lbm}} }[/tex]
[tex]v_{2}= 838.75[/tex] [tex]ft/s[/tex]
