Consider a rigid pressurized stainlesssteel SS304 propane ta

Consider a rigid, pressurized stainless-steel (SS-304) propane tank that weighs 18 lbs. (empty) and is initially at thermal equilibrium (Eq.) with its surroundings (latm, 22 degree C). The inner dimensions of the propane tank are roughly a cylinder (height: 13 1/4\" diameter: 10 1/4\"). The tank is 99% filled with propane and x=0.05. The welded handle and stand on the tank weighs 1 lbs. (combined) and valve on the tank weighs 1.5 lbs. and has an exit area of 1 cm^2. List below your references for the properties of propane and SS-304? What is the wall thickness of the propane tank in mm? What is inner volume of the propane tank in mL and gallons? What is the pressure inside the tank in psi and kPa? What is the mass of liquid propane in the tank in kg and lbs.? What is the volume of gaseous propane in the tank in mL? At t_0 = 0 seconds, the valve on the tank is instantly opened until equilibrium (Eq.) is reached @ t = t_Eq. What is the mass of air (in kg) in the tank @ Eq.?

Solution

solution

the weight of Propane Tank = 18 lbs

Pressure = P1 = 1atm = 760 torr

Temparature = T1 = 22 C

The Inner Dimensions of the Propane Tank Height = h = 13.25 inches

= 13.25 x 2.54

= 33.65 c.m

Diameter = d = 10.25 inches

= 10.25 x 2.54 =26.035 c.m

Radius = r = D/2 = 10.25 / 2 = 5.125 inches = 5.125 x 2.54 = 13.0175 c.m

C) Inner Volume of the propane tank = (3.14 ) r ^2 h = 1092.78132 cubic inches

F) volume of gaseous propane in the tank = 3.14 x 13.0175 x 13.0175 x 33.65

= 17904.8171137 cublic centimeters

Area of the cylindrical tank = 2x ( 3.14 x r^2 ) + h x ( 2 x 3.14 x r )

= (2 x 3.14 x 13.0175 x 13.0175 ) + 33.65 x 2x3.14 x13.0175

= (2 x 3.14 x 13.0175 ) x ( 13.175 + 33.65)

=( 81.7499) x (46.825)

= 3827 c.m²