1. A 25 mm hose is used to siphon water from a tank. a. How long does it take to
ID: 2994352 • Letter: 1
Question
1. A 25 mm hose is used to siphon water from a tank.
a. How long does it take to fill a 10 liter bucket?
b. As water rises in a siphon tube its pressure drops. It is important that the pressure remains above saturation pressure so that the water does not vaporize and vapor lock the siphon. If the water is 70
A 25 mm hose is used to siphon water from a tank. How long does it take to fill a 10 liter bucket? As water rises in a siphon tube its pressure drops. It is important that the pressure remains above saturation pressure so that the water does not vaporize and vapor lock the siphon. If the water is 70 degree C determine the maximum value of h so that the pressure at the top of the siphon does not exceed saturation pressure (for those who have not had thermo, saturation pressure is listed in Table A3 as an absolute pressure).Explanation / Answer
Applying Bernoulis equation to the free surface (1) and exit of the tube (2)
P1 + rho*g*z1 + 1/2*rho*v1^2 = P2 + rho*g*z2 + 1/2*rho*v2^2
v1 = 0
P1 = P2 = Patm
Let
z1 = 0
z2 = -3
Now
1/2*rho*v2^2 + rho*g*z2 = 0
v2 = Sqrt(2*9.81*3) = 7.672 m/s
a)
Volume of the bucket to be filled (V) = 10 lit = 10^-2 m^3
Let t be the time required
Let A be the area of the pipe
Volumetric flow rate,Q = A*v2 = pi/4*25^2*10^-6*7.672
Q = 3.764 *10^-3 m^3/s
Therefore
Q*t = V
t = 2.66 s
b)
Psat (at T=70 deg) = 31.2 kPa
Therefore at Top of Siphon,Pressure should be always greater than Psat(P>Psat)
For maximum h
P = Psat
Applying Bernoulis equation at free surface (1) and at the top surface of siphon (3)
P1 + rho*g*z1 + 1/2*rho*v1^2 = P3 + rho*g*z3 + 1/2*rho*v3^2
P1 = Patm = 101 kPa
P3 = Psat = 31.2 kPa
v1 = 0
v3 = v2 = 7.672 m/s
Let
z1 = 0
z2 = h
101*10^3 = 31.2*10^3 + 1000*9.81*h + 1/2*1000*7.672^2
h = 4.12 m
Therfore maximum value of h = 4.12 m
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