A pipe 300 m long has a slope of 1 in 100 and tapers from 1.2 m diameter at the high end to 0.6 m diameter at the low end. Quantity of water flowing is 5400 liters per minute. If the pressure at the high end is 68.67 kPa, find the pressure at the low end. Neglect head loss.

P = 98052.64 Pa or 98.05 kPa

Explanation:

Using Bernoulli's equation;

P+rho*v^2/2+ rho*g*z = constant

at high end;

dia= 1.2 m

flow, Q = 5400 L/min = 0.09 m^3/s

therefore, velocity at the high end, v1 = Q/A =0.09/(pi()*(1.2/2)^2) = 0.08 m/s

pressure, P = 68.67 kPa

Solving for elevation, z

assume lower end is reference line. then higher end will be 'x' m high wrt to lower end.

x= 300*sin(tan^-1(1/100)) = 3 m

that means higher end will be 3 m above with respect to lower end

Similarly for lower end;

dia= 0.6 m

flow, Q = 5400 L/min = 0.09 m^3/s

therefore, velocity at the high end, v2 = Q/A =0.09/(pi()*(0.6/2)^2) = 0.318 m/s

assume pressure, P

z=0

put all values in the formula, we get;

68.67*10^3+1000*0.08^2/2+ 1000*9.81*3 =P+1000*0.318^2/2+ 0

solving this, we get;

P = 98052.64 Pa or 98.05 kPa

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