Steam is leaving a pressure cooker whose operating pressure is 20 psia. It is observed that the amount of liquid in the cooker has decreased by 0.6 gal in 45 minutes after the steady operating conditions are established, and the cross-sectional area of the exit opening is 0.15 in2. Determine (a) the mass flow rate of the steam and the exit velocity, (b) the total and flow energies of the steam per unit mass, and (c) the rate at which energy is leaving the cooker by steam.

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

their crystal patterns have been disrupted by stress

probably not if net force is zero