Please show work for this problem ! (if writing on paper please take a clear pic
ID: 1544071 • Letter: P
Question
Please show work for this problem ! (if writing on paper please take a clear picture)
Water flows in the sloped pipe shown in the figure. At point 1, the cross sectional area is 3.0 cm^2 and at point 2, the cross sectional area is 2.0 cm^2. The fluid in the manometer is mercury, which has a density 13.6 times the density of water. If the distance between points 1 and 2 is 26.2 cm and the height h is 1.0 cm, how fast is the water moving at point 1? At point 2? (Treat the water as an ideal fluid in laminar flow. Use 9.8 m/s^2 for gravitation acceleration.) v_1 = ___ v_2 = ___Explanation / Answer
from equation of continuity
volume flow rate is constant
A1*v1 = A2*v2
v2/v1 = A1/A2 = 3/2
v2 = 1.5*v1
from bernoullis principle
P1 + (1/2)*rho_water*v1^2 + rho_water*g*x1 = P2 + (1/2)*rho_water*v2^2 + rho_water*g*x2
P1 - P2 = (1/2)*rho_water*(v2^2-v1^2) + rho_water*g((x2-x1)
rho_mercury*g*h = (1/2)*rho_water*(v2^2-v1^2) + rho_water*g((x2-x1)
rho_mercury = 13.6*rho_water
x2 -x1 = d*sin30 = 26.2*sin30 = 13.1 cm = 0.131 m
13.6*9.8*0.01 = (1/2)*(1.5^2*v1^2 - v1^2) + (9.8*0.131)
v1 = 0.28 m/s <<<<-----answer
v2 = 0.42 m/s <<<<-----answer