Respuesta :
Answer:
a) The speed is 61.42 m/s
b) The drag force is 10.32 N
Explanation:
a) The Reynold´s number for the model and prototype is:
[tex]Re_{m} =\frac{p_{m}V_{m}L_{m} }{u_{m} }[/tex]
[tex]Re_{p} =\frac{p_{p}V_{p}L_{p} }{u_{p} }[/tex]
Equaling both Reynold's number:
[tex]\frac{p_{p}V_{p}L_{p} }{u_{p} }=\frac{p_{m}V_{m}L_{m} }{u_{m} }[/tex]
Clearing Vm:
[tex]V_{m} =\frac{p_{p}V_{p}L_{p} u_{m} }{u_{p} p_{m} L_{m} }=\frac{999.1*0.56*8*1.849x10^{-5} }{1.138x10^{-3}*1.184*1 } =61.42m/s[/tex]
b) The drag force is:
[tex]\frac{F_{Dm} }{p_{m}V_{m}^{2}L_{m}^{2} } =\frac{F_{Dp} }{p_{p}V_{p}^{2}L_{p}^{2} } \\F_{Dp} =\frac{F_{Dp}p_{p}V_{p}^{2}L_{p}^{2} }{p_{m}V_{m}^{2}L_{m}^{2} } \\F_{Dp}=\frac{2.3*999.1*0.56^{2} *8^{2} }{1.184*61.42^{2}*1^{2} } =10.32N[/tex]
Answer:
1. The air speed at which wind tunnel should be ran in order to achieve similarity = 61.423 m/s
2. The drag force on the prototype submarine at the conditions given above = 10.325 N.
Explanation:
1. Here we have
Density of water at 15 °C = 999.1 kg/m³
Density of air at 25 °C = 1.184 kg/m³
[tex]\mu_{Water}[/tex] at 15 °C = 1.138 × 10⁻³ kg/(m·s)
[tex]\mu_{Air}[/tex] at 25 °C = 1.849 × 10⁻⁵ kg/(m·s)
The formula is
[tex]\frac{V_m\times \rho_m \times L_m}{\mu_m} = \frac{V_p\times \rho_p \times L_p}{\mu_p}[/tex]
Where:
[tex]V_m[/tex] = Velocity of the model =
[tex]\rho_m[/tex] = Density of the model medium at the medium temperature = 1.184 kg/m³
[tex]L_m[/tex] = Length of the model = 1/8 × [tex]L_p[/tex] = 0.28 m
[tex]\mu_m[/tex] = Dynamic viscosity of model medium at the model medium temperature = 1.849 × 10⁻⁵ kg/(m·s)
[tex]V_p[/tex] = Velocity of the prototype = 0.560 m/s
[tex]\rho_p[/tex] = Density of the prototype medium at the medium temperature = 999.1 kg/m³
[tex]L_p[/tex] = Length of the prototype = 2.24 m
[tex]\mu_p[/tex] = Dynamic viscosity of prototype medium at the prototype medium temperature 1.138 × 10⁻³ kg/(m·s)
Therefore
[tex]{V_m}{} = \frac{V_p\times \rho_p \times L_p \times \mu_m}{\mu_p\times \rho_m \times L_m}[/tex]
[tex]{V_m}{} = \frac{0.560\times 999.1 \times 2.24 \times 1.849 \times 10^{-5}}{1.138 \times 10^{-3}\times 1.184 \times 0.28}[/tex] = 61.423 m/s
The air speed at which wind tunnel should be ran in order to achieve similarity = 61.423 m/s
2. The drag force on the prototype is given by
[tex]F_{D.p} = F_{D.m}(\frac{\rho_p}{\rho_m} )( \frac{V_p}{V_m})^2 ( \frac{L_p}{L_m})^2\\[/tex]
Where:
[tex]F_{D.p}[/tex] = Drag force of the prototype
[tex]F_{D.m}[/tex] = Drag force of the model
[tex]F_{D.p} =2.3(\frac{999.1 }{1.184} )( \frac{0.560 }{61.423 })^2 ( \frac{2.24 }{0.28})^2\\[/tex] = 10.325 N
The drag force on the prototype submarine at the conditions given above = 10.325 N.
