Question

A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 of 0.355 - L cans per minute. At point 2 in the pipe, the gauge pressure is 152 kPa and the cross-sectional area is 8.00 cm^2. At point 1, 1.35m above point 2, the cross-sectional area is 2.00 cm^2. Find a-d, explaining with enough written words the understand the formulas and steps and why they are used.

(a) mass flow rate;

(b) volume flow rate;

(c) flow speeds at point 1 and 2;

(d) gauge pressure at point 1

Answer 1

given

Volume flow rate, dV/dt = 220*0.355 L/min

= 220*0.355/60 L/s

= 1.30 L/s

= 1.30*10^-3 m^3/s

P2_gauge = 152 kPa
A2 = 8.00 cm^2 = 8*10^-4 m^2
A1 = 2.00 cm^2 = 2*10^-4 m^2

a) mass flow rate, dm/dt = density*volume flowrate

= rho*dV/dt

= 1000*1.30*10^-3 kg/s

= 1.30 kg/s

b) Volume flow rate, dV/dt = 1.30*10^-3 m^3/s or 1.30 L/s

c) volume flow rate, dV/dt = A1*v1

v1 = (dV/dt)/A1

= 1.30*10^-3/(2.00*10^-4)

= 6.50 m/s

similalrly, v2 = (dV/dt)/A2

= 1.30*10^-3/(8.00*10^-4)

= 1.625 m/s

d) Now use, Bernoullis equation

P1_gauge + rho*g*h1 + (1/2)*rho*v1^2 = P2_gauge + rho*g*h2 + (1/2)*rho*v2^2

P1_gauge = P2_gauge + rho*g*(h2 - h1) + (1/2)*rho*(v2^2 - v1^2)

= 152*10^3 + 1000*9.8*(-1.35) + (1/2)*1000*(1.625^2 - 6.5^2)

= 1.19*10^5 pa

= 119 kPa