estimate the energy head lost along a short length of pipe suddenly enlarging a diameter of 350mm to 700mm which discharges 0.7(m^-3) of water per second .
the solution given is Q1= Q2 = 0.7(m^3)/s
0.7 = pi ((350x10^-3)^2 ) V1 / 4 , V1= 7.28m/s
0.7 = pi ((700x10^-3)^2 ) V2 / 4 , v2 = 1.82m/s
[(v1)^2 ] / 2g = [(v2)^2 ] / 2g +head loss
head loss = 2.53m
The Attempt at a Solution
the bernoulli's equation is P1/ ρg + (V1^2) /2g + z1 = P2/ ρg + (V2^2) /2g + z2, by writing [(v1)^2 ] / 2g = [(v2)^2 ] / 2g +head loss , the author assume P1 = P2 , how could it be possible ? since the inlet is smaller than outlet , the pressure at inlet should be lower , right ? the velocity at inlet is smaller