Answer:
V=33.66 m/s
![Re=448.8* 10^6](https://img.qammunity.org/2020/formulas/engineering/college/b8qmtylho4ba9tj62iaazgkj5ka4esq1lu.png)
Re>4000, The flow is turbulent flow.
Step-by-step explanation:
Given that
Pressure difference = 50 mm of Hg
We know that density of Hg=136000
![Kg/m^3](https://img.qammunity.org/2020/formulas/engineering/college/gasx9m1vowicna8z9nee738dc1p8wvxqp4.png)
ΔP= 13.6 x 1000 x 0.05 Pa
ΔP=680 Pa
Diameter of tunnel = 200 mm
Property of air at 25°C
ρ=1.2
![Kg/m^3](https://img.qammunity.org/2020/formulas/engineering/college/gasx9m1vowicna8z9nee738dc1p8wvxqp4.png)
Dynamic viscosity
![\mu =1.8* 10^(-8)\ Pa.s](https://img.qammunity.org/2020/formulas/engineering/college/5sfwebacc9d1239yqq8ykr8qzebew8wdmq.png)
Velocity of fluid given as
![V=\sqrt{(2\Delta P)/(\rho_(air))}](https://img.qammunity.org/2020/formulas/engineering/college/2mzynv3ezfvphegfjp0rw52ie1lz8wz2cr.png)
![V=\sqrt{(2* 680)/(1.2)}](https://img.qammunity.org/2020/formulas/engineering/college/rshoxxjffxvs27pblzsydhkfbpc8edzxbc.png)
V=33.66 m/s
Reynolds number
![Re=(\rho _(air)Vd)/(\mu )](https://img.qammunity.org/2020/formulas/engineering/college/ikcfjjtdxkuywl8l1mdzk72hw11i6qwwrz.png)
![Re=(1.2* 33.66* 0.2)/(1.8* 10^(-8))](https://img.qammunity.org/2020/formulas/engineering/college/64fshytccspejis9v8z94n3c10dnkk86rr.png)
![Re=448.8* 10^6](https://img.qammunity.org/2020/formulas/engineering/college/b8qmtylho4ba9tj62iaazgkj5ka4esq1lu.png)
Re>4000,So the flow is turbulent flow.