Respuesta :
Answer:
a) [tex]\frac{P}{\rho v^{3}D^{2} } = f(\frac{\Omega D}{v}, n)[/tex]
b) P = 2184.57 kW
c) [tex]\Omega = 122.5 rpm[/tex]
Explanation:
Pressure = P
Density = [tex]\rho[/tex]
Diameter = D
Velocity = v
Rate of rotation = Ω
number of blades = n
a) The relationship for the model is given by:
[tex]\frac{P}{\rho v^{3}D^{2} } = f(\frac{\Omega D}{v}, n)[/tex]
b) The Power developed geometrically and dynamically similar prototype:
The data corresponding to the model:
[tex]P_{mod} = 3.8 kW\\v_{mod} = 40 m/s\\D_{mod} = 50 cm = 0.5 m\\\rho_{mod} = 1.2255 kg/m^{3}\\ \Omega = 4200 rpm[/tex]
The data corresponding to the prototype
[tex]P_{prot} = ? \\v_{prot} = 35 m/s\\D_{prot} = 15 m\\\rho_{prot} = 1.1685 kg/m^{3}[/tex]
[tex](\frac{P}{\rho v^{3}D^{2} } )_{mod} = (\frac{P}{\rho v^{3}D^{2} } )_{prot}[/tex]
[tex]\frac{3.8}{1.2255 * 40^{3} * 0.5^{2} }= \frac{P}{1.1685 * 35^{3} * 15^{2} }\\P = \frac{3.8 * 1.1685 * 35^{3} * 15^{2}}{1.2255 * 40^{3} * 0.5^{2} }[/tex]
P = 2184.57 kW
c) To calculate the appropriate rotation rate for the prototype
[tex](\frac{\Omega D}{v})_{model} = (\frac{\Omega D}{v})_{prot}[/tex]
[tex]\frac{4200*0.5}{40}= \frac{\Omega * 15}{35}\\\Omega = \frac{4200*0.5*35}{40*15}\\\Omega = 122.5 rpm[/tex]
