Question

Flow rate of a fluid ( $\rho =1000kg/m^3$) in a small diameter tube is $800m{m}^3/s$. The length and diameter of the tube are 2 m and 0.5 mm, respectively. The pressure drop in 2 m length of is equal to 2 MPa. The viscosity of the fluid is:

• A. $0.025N-s/m^2$
• B. $0.012N-s/m^2$
• C. $0.0019N-s/m^2$
• D. $0.0010N-s/m^2$

Answer 1

The correct option is C $0.0019N-s/m^2$

$\text{Pressure drop},P_1-P_2=\frac{32\mu VL}{D^2}$

$V=\frac{Q}{A}=\frac{800\times 4}{\pi \times (0.5{)}^2}=4074.4mm/s=4.07m/s$

$2\times {10}^6=\frac{32\times \mu \times 4.07\times 2}{(0.5\times {10}^{-3}{)}^2}$

$\mu =0.0019N-s/m^2$