A) \[4\,km{{s}^{-\,1}}\]
B) \[4\,km{{h}^{-\,1}}\]
C) \[0.4\,km{{s}^{-\,1}}\]
D) \[0.4\,km\,{{h}^{-\,1}}\]
Correct Answer: C
Solution :
To keep situations similar, we must keep reynolds number of flow same in both cases. |
\[\Rightarrow \] \[{{N}_{R}}={{\left( \frac{\rho vD}{\eta } \right)}_{\text{model}}}={{\left( \frac{\rho vD}{\eta } \right)}_{\text{car}}}\] |
\[\Rightarrow \] \[{{v}_{m}}{{D}_{m}}={{v}_{c}}{{D}_{c}}\] |
\[\Rightarrow \] \[{{v}_{m}}={{v}_{c}}\cdot \frac{{{D}_{c}}}{{{D}_{m}}}=15\times \left( \frac{550}{20} \right)=0.4\frac{km}{s}\] |
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