Answer:
a) 34.05ft/s
b).1156.2BTU/lbm
c) 2.04BTU/s
Explanation:
Amount of liquid that has evaporated, m = ◇Vliq/ Vf
We replace the values to make conversion
m = (0.6gal/ 0.01683ft^3/lbm) × (0.13368ft^3/1gal)
m = 4.755lb
The mass flow rate of exit steam is given by:
m' = m/◇t
We replace values to make conversion
m' =( 4.766lb/45min) = 0.1059lb/min × 1min/60s
m' = 0.001765lb/s
The exit velocity V = m'/pA = m'Vg/A
We replace values to make conversion
V =[ (0.001765lbm/s)(20.093ft^3/lbm) /(0.15 in^2)]× (144in^2/1ft^2)
V = 34.05ft/s
b) The total and flow energies per unit mass is given by:
Eflow= Pv = h - u
We replace the values to make conversion
Eflow = 1156.2 - 1081.8
Eflow = 74.4BTU/lbm
Therefore theta= h + ke + pe
Theta approximately =h = 1156.2BTU/lbs
c) The rate at which energy is leaving the cooler by steam is given by:
Emass = m'theta
Emass = (0.001765)×(1156.2)
Emass = 2.04BTU/s
