Ahorizontal pipe of diameter 0.985 m has a smooth constriction to a section of diameter 0.591 m . the density of oil flowing in the pipe is 821 kg/m3 . if the pressure in the pipe is 8100 n/m2 and in the constricted section is 6075 n/m2 , what is the rate at which oil is flowing

Ρgh is the gravitational component of Bernoulli's equation 
the pipe is horizontal 
so no change in h 
ρgh is equal on both sides 
cancels out 
now 
v (velocity) is inversely proportional to square of radius 
so 
v_2 / v_1 = (r_1 )^2 / (r_2)^2 = 1.1^2 / 0.66^2 = 2.78 
v_2 = 2.78 v_2 
and 
8130 N/m² + ((821 kg/m³) * (v_1)²)/2 = 6097.5 N/m² + ((821 kg/m³) * (2.78v_1)²)/2 
solve for v_1 
then 
R = 2 pi (1.1)^2 * v_1

A.adasnsnsnnskakakakwks