Question #298237

If the kinematic viscosity of water at S.TP is 1:0038 x 10 m²/a and the mass density is 998.23 kg/m³, calculate its absolute viscosity

1
Expert's answer
2022-02-16T14:54:02-0500

Absolute viscosity=(998.23kgm3\frac{kg}{m^3}) /(1:0038×\times 10m2/a)


Convert 998.23kgm3\frac{kg}{m^3} to g/mm3= 998.231000\frac{998.23}{1000}= 0.99823gmm3\frac{g}{mm^3}



1:0038×\times 10m2/a= 10m238a\frac{10m^2}{38a} = 5m219a\frac{5m^2}{19a}


Absolute viscosity= (0.99823gmm3\frac{g}{mm^3})/(5m219a\frac{5m^2}{19a} )


Absolute viscosity= 0.26269Pa.S


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS