You Yesterday, 1:13 PM Problem.21 Law-teat Transfer ideal Gas A 25 m closed rigi
ID: 1867752 • Letter: Y
Question
You Yesterday, 1:13 PM Problem.21 Law-teat Transfer ideal Gas A 25 m closed rigid tank contains oxywen as an ideal gas initially at 2C Ap ressure r comnected to this tank reads initially 500 kPa. Air, with a temperature of S'C and s heat transfer cowtmciens ws around this tank as shown in the figure below The surrounding air pressure is surface area of the tank is 2 m2.Oxygen is assumed to have a constant specific het 97 kPa. The total ratio k 1.4 a) b) Determine the mass of oxygen in the tank. (9 pts) Determine the rate of heat transfer from the tank to the surrounding air assuming that the tank wall temperature is maintained at 25°C. (6 pts) Determine the temperature of oxygen in "C after a duration of 1 minute. (10 pts) c) Pange 500 kPa Airflow 02 V 2.5 m3 T- 28°CExplanation / Answer
a) Mass of oxygen in the tank:
The pressure in the tank =P= 500 Kpa
Volume of the tank =V= 2.5 m3
temperature = T = 28 C =301 K
From Ideal gas equation PV = mRT.
R value for oxygen = 0.259 KJ//Kg
Mass = m = PV/RT = 500*2.5 / 0.259*301 = 16 kg.
b)
Surface area of tank = 2 m2.
Heat transfer coefficient = h = 50 W /m2K
Surrounding air temp = Ts = 5 C
Wall temp = Tw = 25 C
heat transfer = Q = hA(Tw - Ts) = 50*2(25-5) = 2000 W.
Heat lost rate by oxygen cylinder = 2000 W.
c) temp f oxygen after 1 minute.
Heat lost rate to surroundings = 2000 W.
heat lost in 1 minute = 2000 * 60 = 120000 W.
Mass of oxygen = m = 16 kg.
Specific heat at constant pressure of oxygen = 0.918 KJ/kg.
Heat loas = Q = mCp ( T1 - T2 )
Initial temp f oxygen = T1 = 301 K = 28 C
120 KW = 16*0.918 KJ/kg (28 - T2 ).
Final temp after 1 min = T2 = 20 C.
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