Question Number 136866 by mohammad17 last updated on 27/Mar/21
Commented by mohammad17 last updated on 27/Mar/21
$${help}\:{me}\:{sir} \\ $$
Commented by mohammad17 last updated on 27/Mar/21
$${help}\:{me}\:{sir}\:{please}\:{i}\:{want}\:{Q}\mathrm{2}\:{ajust} \\ $$
Answered by Olaf last updated on 27/Mar/21
![1) A) Re = ((VL_c )/ν) (Reynolds number) [Re] = (([m.s^(−1) ]×[m])/([m^2 .s^(−1) ])) = [((m^2 .s^(−1) )/(m^2 .s^(−1) ))] : dimensionless Sp.g = (ρ/ρ_(pure water) ) (or Sp.gr specific gravity) [Sp.g] = (([kg.m^(−3) ])/([kg.m^(−3) ])) = [((kg.m^(−3) )/(kg.m^(−3) ))] : dimensionless](https://www.tinkutara.com/question/Q136875.png)
$$\left.\mathrm{1}\left.\right)\:\mathrm{A}\right) \\ $$$$\mathrm{Re}\:=\:\frac{{VL}_{{c}} }{\nu}\:\left(\mathrm{Reynolds}\:\mathrm{number}\right) \\ $$$$\left[\mathrm{Re}\right]\:=\:\frac{\left[{m}.{s}^{−\mathrm{1}} \right]×\left[{m}\right]}{\left[{m}^{\mathrm{2}} .{s}^{−\mathrm{1}} \right]}\:=\:\left[\frac{{m}^{\mathrm{2}} .{s}^{−\mathrm{1}} }{{m}^{\mathrm{2}} .{s}^{−\mathrm{1}} }\right]\::\:\mathrm{dimensionless} \\ $$$$\mathrm{Sp}.\mathrm{g}\:=\:\frac{\rho}{\rho_{{pure}\:{water}} }\:\left(\mathrm{or}\:\mathrm{Sp}.\mathrm{gr}\:\mathrm{specific}\:\mathrm{gravity}\right) \\ $$$$\left[\mathrm{Sp}.\mathrm{g}\right]\:=\:\frac{\left[{kg}.{m}^{−\mathrm{3}} \right]}{\left[{kg}.{m}^{−\mathrm{3}} \right]}\:=\:\left[\frac{{kg}.{m}^{−\mathrm{3}} }{{kg}.{m}^{−\mathrm{3}} }\right]\::\:\mathrm{dimensionless} \\ $$
Answered by Olaf last updated on 27/Mar/21
$$\left.\mathrm{2}\right) \\ $$$$\mathrm{Q}\:=\:\mathrm{1},\mathrm{95}×\mathrm{10}^{−\mathrm{3}} \:\mathrm{ft}^{\mathrm{3}} /{s}\:=\:\mathrm{5},\mathrm{52}×\mathrm{10}^{−\mathrm{5}} \:{m}^{\mathrm{3}} /{s} \\ $$$$\mathrm{D}\:=\:\mathrm{1in}\:=\:\mathrm{0},\mathrm{0254}\:{m} \\ $$$$\nu\:=\:\mathrm{1}\:\mathrm{c}.\mathrm{p}\:=\:\mathrm{10}^{−\mathrm{3}} \:\mathrm{Pa}.{s} \\ $$$$\mathrm{V}\:=\:\frac{\mathrm{Q}}{\mathrm{S}}\:=\:\frac{\mathrm{Q}}{\pi\frac{\mathrm{D}^{\mathrm{2}} }{\mathrm{4}}}\:=\:\frac{\mathrm{5},\mathrm{52}×\mathrm{10}^{−\mathrm{3}} }{\frac{\pi}{\mathrm{4}}×\left(\mathrm{0},\mathrm{0254}\right)^{\mathrm{2}} }\:=\:\mathrm{11},\mathrm{73}\:{m}/{s} \\ $$$$\mathrm{Re}\:=\:\frac{\mathrm{VD}}{\nu}\:=\:\frac{\mathrm{11},\mathrm{73}×\mathrm{0},\mathrm{0254}}{\mathrm{10}^{−\mathrm{3}} }\:=\:\mathrm{297},\mathrm{9} \\ $$$$\mathrm{0}\:<\:\mathrm{Re}\:<\:\mathrm{2000}\:\Rightarrow\:\mathrm{laminar}\:\mathrm{flow} \\ $$