Answer:
Throat diameter
=28.60 mm
Step-by-step explanation:
Bore diameter
⇒
![A_1=3.09* 10^(-3) m^2](https://img.qammunity.org/2020/formulas/engineering/college/cj7v6bkrohl737ig4qybzxt3qpp8ovrjtb.png)
Manometric deflection x=235 mm
Flow rate Q=240 Lt/min⇒ Q=.004
![(m^3)/(s)](https://img.qammunity.org/2020/formulas/engineering/college/neenpr9okmtwbywncn73ngnyhj7pdtkgq4.png)
Coefficient of discharge
=0.8
We know that discharge through venturi meter
![Q=C_d(A_1A_2√(2gh))/(√(A_1^2-A_2^2))](https://img.qammunity.org/2020/formulas/engineering/college/mnq502kifjie06usd72julwbf8f167z61a.png)
![h=x((S_m)/(S_w)-1)](https://img.qammunity.org/2020/formulas/engineering/college/9rsvnoqaeixgwuetyrgjviu43naltg42xk.png)
=13.6 for Hg and
=1 for water.
![h=0.235((13.6)/(1)-1)](https://img.qammunity.org/2020/formulas/engineering/college/cks1olnv7j0dshhak3ar04m871bjsv2vs8.png)
h=2.961 m
Now by putting the all value in
![Q=C_d(A_1A_2√(2gh))/(√(A_1^2-A_2^2))](https://img.qammunity.org/2020/formulas/engineering/college/mnq502kifjie06usd72julwbf8f167z61a.png)
![0.004=0.8* \frac{3.09* 10^(-3) A_2√(2* 9.81* 2.961)}{\sqrt{(3.09* 10^(-3))^2-A_2^2}}](https://img.qammunity.org/2020/formulas/engineering/college/jkvsym9avnz1fazfij550lhu2iqbo152a7.png)
![A_2=6.42* 10^(-4) m^2](https://img.qammunity.org/2020/formulas/engineering/college/9ak2q21ub1si5koea8t8fa7aqocyutvt4e.png)
⇒
=28.60 mm
So throat diameter
=28.60 mm