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Cyrus
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2-48) The 60-W fan of a central heating system is to circulate air through the ducts. The analysis of the flow shows that the fan needs to raise the pressure of air by 50Pa to maintain flow. The fan is located in a horizontal flow section where the diameter is 30cm at both inlet and outlet. Determine the highest possible average flow velocity in the duct.
Hmmm, I have this equation for flow energy on time rate basis:
\Delta \dot{E_{mec}} = \dot{m} \Delta e_{mec} = \dot{m}( \frac{P_2 - P_1}{ \rho} + \frac{V_2^2 - V_1 ^2} {2} + g(z_2 - z_1 ))
I know delta P, but I don't know delta V, and delta z is zero. Hmmmm... I guess I can assume no heat loss, and that \Delta \dot{E} = 60W, so that knocks off another unknown. The density of air is not given, and it has to be assumed incompressable. I guess I'll have to look that up in the index. The problem is I don't know \dot{m}, nor do I know the inlet or outlet flow velocity.....hmmmm more thinking to myself, perhaps I do know \dot{m} I can substitute into it \rho A_c V_{avg} Because mass flow in = mass flow out, via conservation of mass. Oh, and I was given delta P across the inlet and outlet to be 5KPa....OOOOOOOOOOOOOOO...that picture is NOT for this problem. That 5kPa and the picture is for the previous problem. Strike that. AHHHHHHHHH! KEY WORD, *MAINTAIN FLOW*, which means V-in = V -out. Hence, the second fractional term goes to zero also! Yippie! Now my \rho cancel out too!
So the anwser is:
\dot{E} = \frac { \pi d^2} {4} V_{avg} \Delta P
Plug in and I get: 16.976 m/s
Seem good to you?
The devil was in the details in this problem.
Hmmm, I have this equation for flow energy on time rate basis:
\Delta \dot{E_{mec}} = \dot{m} \Delta e_{mec} = \dot{m}( \frac{P_2 - P_1}{ \rho} + \frac{V_2^2 - V_1 ^2} {2} + g(z_2 - z_1 ))
I know delta P, but I don't know delta V, and delta z is zero. Hmmmm... I guess I can assume no heat loss, and that \Delta \dot{E} = 60W, so that knocks off another unknown. The density of air is not given, and it has to be assumed incompressable. I guess I'll have to look that up in the index. The problem is I don't know \dot{m}, nor do I know the inlet or outlet flow velocity.....hmmmm more thinking to myself, perhaps I do know \dot{m} I can substitute into it \rho A_c V_{avg} Because mass flow in = mass flow out, via conservation of mass. Oh, and I was given delta P across the inlet and outlet to be 5KPa....OOOOOOOOOOOOOOO...that picture is NOT for this problem. That 5kPa and the picture is for the previous problem. Strike that. AHHHHHHHHH! KEY WORD, *MAINTAIN FLOW*, which means V-in = V -out. Hence, the second fractional term goes to zero also! Yippie! Now my \rho cancel out too!
So the anwser is:
\dot{E} = \frac { \pi d^2} {4} V_{avg} \Delta P
Plug in and I get: 16.976 m/s
Seem good to you?
The devil was in the details in this problem.
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