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) 0.0018lbm/s, 34.56ft/s

b) 74.4 btu/lbm, 1156.2 btu/lbm

c) 2.06 btu/s

Explanation:

The amount of liquid that gas evaporated

m = ΔV(liq)/V(f)

From table (A-5E) at 20psia,

V(f) = 0.01683 ft³/lbm

V(g) = 20.093 ft³/lbm

U(g) = 1081.8 btu/lbm

H(g) = 1156.2 btu/lbm

m = 0.6/0.01683 = 35.65 gal.lbm/ft³

Converting to lbm

35.65gal.lbm/ft³ * 0.13368 ft³/gal

m = 4.77lbm

Mass flow rate of the steam, m' = m/Δt

Converting time, t to seconds. 45mins = 45*60s = 2700s

m' = 4.77/2700 = 0.0018 lbm/s

Exit velocity of steam, is calculated as

V = m'V(g)/A(c)

V = 0.0018 * 20.093/0.15

V = 0.036/0.15

V = 0.24 * 144 = 34.56ft/s

Flow energy of the exit steam = E

E = PV = h - u

E(flow) = h(g) - u(g)

E(flow) = 1156.2 - 1081.8

E(flow) = 74.4 btu/lbm

Total energies of the exit steam =

E(total) = h + KE + PE

E(total) ~ h

E(total) = 1156.2 btu/lbm

Rate at which energy is leaving the cooker is

E' = m'E(total)

E'= 0.0018 * 1156.2

E' = 2.06 btu/s

unbalanced force

explanation: