Answer:
Head loss=0.00366 ft
Step-by-step explanation:
Given :Water flow rate Q=0.15
![(ft^(3))/(sec)](https://img.qammunity.org/2020/formulas/engineering/college/he3lvglprugycy743qm4tlphcf9upxj99a.png)
= 6 inch=0.5 ft
=2 inch=0.1667 ft
As we know that Q=AV
![A_(1)* V_(1)=A_(2)* V_(2)](https://img.qammunity.org/2020/formulas/engineering/college/8mx9wecg7oxw2q17lg78v0t4toejj6k5yb.png)
So
![V_(2)=(Q)/(A_2)](https://img.qammunity.org/2020/formulas/engineering/college/6a1ypvdc4dy8fc4aen7ydfet0zs2y6bkkc.png)
![V_(2)=(.015)/((3.14)/(4)* 0.1667^(2))](https://img.qammunity.org/2020/formulas/engineering/college/p04ycn7hz8oi62ykg7hheialzfp7hxzpkx.png)
=0.687 ft/sec
We know that Head loss due to sudden contraction
![h_(l)=K(V_(2)^2)/(2g)](https://img.qammunity.org/2020/formulas/engineering/college/fpnpi969xvb5tl79b63bqrz60ruf5gb0ht.png)
If nothing is given then take K=0.5
So head loss
![h_(l)=(0.5)\frac{{0.687}^2}{2* 32.18}](https://img.qammunity.org/2020/formulas/engineering/college/r9uifufsxjjurop4gagb8ap5gja8w47eaa.png)
=0.00366 ft
So head loss=0.00366 ft