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Question Number 136866 by mohammad17 last updated on 27/Mar/21

Commented by mohammad17 last updated on 27/Mar/21

help me sir

$${help}\:{me}\:{sir} \\ $$

Commented by mohammad17 last updated on 27/Mar/21

help me sir please i want Q2 ajust

$${help}\:{me}\:{sir}\:{please}\:{i}\:{want}\:{Q}\mathrm{2}\:{ajust} \\ $$

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

$$\left.\mathrm{1}\left.\right)\:\mathrm{A}\right) \\ $$$$\mathrm{Re}\:=\:\frac{{VL}_{{c}} }{\nu}\:\left(\mathrm{Reynolds}\:\mathrm{number}\right) \\ $$$$\left[\mathrm{Re}\right]\:=\:\frac{\left[{m}.{s}^{−\mathrm{1}} \right]×\left[{m}\right]}{\left[{m}^{\mathrm{2}} .{s}^{−\mathrm{1}} \right]}\:=\:\left[\frac{{m}^{\mathrm{2}} .{s}^{−\mathrm{1}} }{{m}^{\mathrm{2}} .{s}^{−\mathrm{1}} }\right]\::\:\mathrm{dimensionless} \\ $$$$\mathrm{Sp}.\mathrm{g}\:=\:\frac{\rho}{\rho_{{pure}\:{water}} }\:\left(\mathrm{or}\:\mathrm{Sp}.\mathrm{gr}\:\mathrm{specific}\:\mathrm{gravity}\right) \\ $$$$\left[\mathrm{Sp}.\mathrm{g}\right]\:=\:\frac{\left[{kg}.{m}^{−\mathrm{3}} \right]}{\left[{kg}.{m}^{−\mathrm{3}} \right]}\:=\:\left[\frac{{kg}.{m}^{−\mathrm{3}} }{{kg}.{m}^{−\mathrm{3}} }\right]\::\:\mathrm{dimensionless} \\ $$

Answered by Olaf last updated on 27/Mar/21

2)  Q = 1,95×10^(−3)  ft^3 /s = 5,52×10^(−5)  m^3 /s  D = 1in = 0,0254 m  ν = 1 c.p = 10^(−3)  Pa.s  V = (Q/S) = (Q/(π(D^2 /4))) = ((5,52×10^(−3) )/((π/4)×(0,0254)^2 )) = 11,73 m/s  Re = ((VD)/ν) = ((11,73×0,0254)/(10^(−3) )) = 297,9  0 < Re < 2000 ⇒ laminar flow

$$\left.\mathrm{2}\right) \\ $$$$\mathrm{Q}\:=\:\mathrm{1},\mathrm{95}×\mathrm{10}^{−\mathrm{3}} \:\mathrm{ft}^{\mathrm{3}} /{s}\:=\:\mathrm{5},\mathrm{52}×\mathrm{10}^{−\mathrm{5}} \:{m}^{\mathrm{3}} /{s} \\ $$$$\mathrm{D}\:=\:\mathrm{1in}\:=\:\mathrm{0},\mathrm{0254}\:{m} \\ $$$$\nu\:=\:\mathrm{1}\:\mathrm{c}.\mathrm{p}\:=\:\mathrm{10}^{−\mathrm{3}} \:\mathrm{Pa}.{s} \\ $$$$\mathrm{V}\:=\:\frac{\mathrm{Q}}{\mathrm{S}}\:=\:\frac{\mathrm{Q}}{\pi\frac{\mathrm{D}^{\mathrm{2}} }{\mathrm{4}}}\:=\:\frac{\mathrm{5},\mathrm{52}×\mathrm{10}^{−\mathrm{3}} }{\frac{\pi}{\mathrm{4}}×\left(\mathrm{0},\mathrm{0254}\right)^{\mathrm{2}} }\:=\:\mathrm{11},\mathrm{73}\:{m}/{s} \\ $$$$\mathrm{Re}\:=\:\frac{\mathrm{VD}}{\nu}\:=\:\frac{\mathrm{11},\mathrm{73}×\mathrm{0},\mathrm{0254}}{\mathrm{10}^{−\mathrm{3}} }\:=\:\mathrm{297},\mathrm{9} \\ $$$$\mathrm{0}\:<\:\mathrm{Re}\:<\:\mathrm{2000}\:\Rightarrow\:\mathrm{laminar}\:\mathrm{flow} \\ $$

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