For the flow shown in Fig. P$3\mathrm{.}68,$ estimate the pressure $P_1$ and velocity $V_1$ if $V_2=20\frac{m}{s}$ and $H=10cm.(Hg=$ The

Symbol of Mercury$).$ DO NOT ANSWER, BUT EXPALIN THE MANOMETER EQUATION USING THE

IMAGE BELOW AND ANSWER THE QUESTION

$H=10cm=0\mathrm{.}10m$

${}_{Hg}=13\mathrm{.}6{}_w$

${}_w=1$

$V_2=20\frac{m}{s}$

$V_1=$

Bernoulli's Equation:

$\frac{P_1}{}+\frac{V_1^2}{2g}+z_1=\frac{P_2}{}+\frac{V_2^2}{2g}+z_2$

$\frac{P_1}{}+\frac{V_1^2}{2g}=\frac{V_2^2}{2g}$

Manometer Equation:

$P_1+{}_w(h-H)+{}_{Hg}H=P_2+\frac{V_2^2}{2g}*{}_w+{}_{w}h$

$P_1+{}_{w}h-{}_{w}H+{}_{Hg}H=P_2+\frac{V_2^2}{2g}*{}_w+{}_{w}h$

$P_1-{}_{w}H+{}_{Hg}H=\frac{V_2^2}{2g}*{}_w$

ANSWER

THESE

QUESTION

Define a manometer. Why is pitot

tube added in this problem? Why

did they add mercury into the Pitot

tube? In context of this question,

how does the manometer function?

Why do we consider the velocity

head for the manometer equation?

Why is the velocity head multiplied

by the specific weight of water?

How does this have to do anything

with the picture? Try to organize

your thoughts and provide a detail

response. Please relate to the

picture.

IMPORTANT: I NEED TO

KNOW WHY VELOCITY HEAD

IS THERE? WHY IS THE

SPECIFC WEIGHT OF WATER

MULTIPLED WITH VELOCITY

HEAD? WHERE IN THE

PICTURE WOULD WE KNOW

TO ADD VELOCITY HEAD?

Substitution of Manometer Equation into Bernoulli's Equation: