Replace the steam in Problem 2.11 with liquid water. If the specific internal energy of the liquid qater is 45 Btu/lbm, determine the total internal energy, the kinetic energy, and the potential energy of 2.50 lbm of water in the pipe
Given: m = 2.50 lbm; z = 15ft; V = 200 ft/s; u = 45 Btu/lbm
Assume: g = 32.174 ft/s2
What will be an ideal response?
U = mu = (2.50 lbm) (45 Btu/lbm) = 113 Btu
KE = ½ mV2/gc = ½ (2.50 lbm)(200 ft/s)2/(32.174 lbm-ft/lbf-s2) = 1550 ft-lbf = 2.00 Btu
PE = mgz/gc = (2.50 lbm)(32.174 ft/s2)(15 ft)/(32.174 lbm-ft/lbf-s2) = 37.5 ft-lbf = 0.0482 Btu
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