Question

Water is pumped from a storage tank through a tube of 3.0 cm inside diameter at the rate of 0.001 m^3 /s. What is the kinetic energy per kg water in the tube?

(a) 2.00 J/kg

(b) 1.00 J/kg

(c) 5.00 J/kg

(d) 0.05 J/kg

This question was posed to me during an internship interview.

The above asked question is from Energy Balance Numericals Without Reactions topic in chapter Energy Balance of Bioprocess Engineering

Answer 1

The correct option is (b) 1.00 J/kg

To explain I would say: Kinetic energy Ek =  mv^2, tube dia. D = 3.0 cm, m = 1 kg.

Cross-section area of tube A = \(\frac{1}{4}\) πD^2 = \(\frac{1}{4}\) π(3.0/100 m)^2 = 7.0686×10^-4m^2

Average velocity of water v = Q/A = 0.001 m^3 /(7.0686×10^-4m^2) = 1.415 m/s

KE per kg = Ek/m = 1/2 v^2 = 1/2 (1.415 m/s)^2 = 1.00 m^2/s^2 = 1.00 J/kg.