Question

Required information NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. 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 0.145 in2. The properties of saturated liquid water and water vapor at 20 psia are ve 0.01683 ft3/lbm, vg= 20.093 ft/bm, ug=1081.8 Btu/bm, and ng= 1156.2 Btu/lbm.

Determine the mass flow rate of the steam and the exit velocity. (You must provide an answer before moving on to the next part.)
The mass flow rate of the steam is _____x 10-3 lbm/s.
The exit velocity is_____ ft/s.

Answer 1

The mass flow rate of the steam is 398.4 lbm/min. The exit velocity is 4.57 ft/s.

Step-by-step explanation:

To determine the mass flow rate of the steam, we can use the equation:

Mass flow rate = (Change in liquid volume / Change in time) x Density of saturated liquid water

Mass flow rate = (0.6 gal / 45 min) x 231 ft3/gal x 62.4 lbm/ft3

Mass flow rate = 398.4 lbm/min

The exit velocity can be calculated using the equation:

Exit velocity = Mass flow rate / (Density of saturated vapor water x Cross-sectional area of the exit opening)

Exit velocity = 398.4 lbm/min / (20.093 ft3/lbm x 0.145 in2 x (1 ft/12 in)2)

Exit velocity = 274.1 ft/min

Converting the exit velocity to ft/s:

Exit velocity = 274.1 ft/min x (1 min/60 s)

Exit velocity = 4.57 ft/s

Answer 2

The mass flow rate of the steam is 0.001547 lb/s. The exit velocity is 444.95 ft/s.

Step-by-step explanation:

To determine the mass flow rate of the steam, we can use the equation:

Mass flow rate = change in liquid volume / time = (0.6 gal) / (45 min) = 0.0133 gal/min

Converting the mass flow rate from gallons per minute to pounds per second:

Mass flow rate = (0.0133 gal/min) * (8.34 lb/gal) * (1 min/60 s) = 0.001547 lb/s

To find the exit velocity, we can use the equation:

Exit velocity = mass flow rate / (density * cross-sectional area)

Given that the density of water vapor at 20 psia is 0.01683 ft3/lbm and the cross-sectional area of the exit opening is 0.145 in2:

Exit velocity = (0.001547 lb/s) / (0.01683 ft3/lbm * (0.145 in2 * (1 ft/12 in)2)) = 444.95 ft/s