OND N P (9.14) N-D N-D (9.15) If you use the same fluid for the model and the prototype Pppm and Hp- Hm. Canceling out the same physical properties and substituting Eq (8. 12 ) to Eq (8.13)yields mo P ,=10(P m0) (9.16 The equality of the reynolds number requires Np=0.0INm (9,17)P NDI ( ) 2   = (9.14) = (9.15) If you use the same fluid for the model and the prototype, ρp =ρm and μp=μm. Canceling out the same physical properties and substituting Eq.(8.12) to Eq. (8.13) yields (Pmo )P = 105 ( Pmo )m (9.16) The equality of the Reynolds number requires NP = 0.01Nm (9.17) m NDI ( ) 2   P I g N D ( ) 2 m I g N D ( ) 2 3 ( ) m P N N