1. Consider a rigid tank of volume 15 m3. This tank is connected to a steam supp
ID: 1939737 • Letter: 1
Question
1. Consider a rigid tank of volume 15 m3. This tank is connected to a steam supply line. Thetank is initially evacuated. Steam at 15 bar and 280 0C is now allowed to flow into tank until the
pressure inside the tank reaches 15 bar. The tank is perfectly insulated. Determine
a) the amount of mass of the steam flowed into tank in kg,
b) the temperature of the steam in the tank in 0C.
Answers: 71.1 kg; 425 0C.
2. Refrigerant-134a is contained in a 4 ft3 tank at 160 lpf/in2. At the beginning, the saturated
vapour occupies a volume of 2 ft3 and the remaining portion is filled with the saturated liquid. A
valve is opened to allow escaping some saturated vapour. When the valve is closed, the volume
of the liquid is 1 ft3. During this process, the pressure inside the tank remains constant.
Determine
a) the mass of the refrigerant that has escaped in lb,
b) the heat transferred during this process in Btu.
Answers: 67.4 lb; 4835.1 Btu
Explanation / Answer
1. Conservation of energy for control volume
State 1 : P1=15 bar , T1=280C , h1=2991.85 kJ/kg
m1h1 = m2u2 (m1=m2)
u2 = h1 = 2991.85 kJ/kg
State 1 : P2=15 bar , u2=2991.85 kJ/kg -> T2 = 425C v2=0.211m3/kg
mass = V/v = 15/0.211 = 71.09 kg
2. Before opening the valve,
mass of sat vapor = Volume of sat vapor / specific volume of sat vapor at 160psi = 2/0.2896 = 6.91lb
mass of sat liq = Volume of sat liq/ specific volume of sat liq at 160psi = 2/0.01414 = 141.44 lb
Total mass = 148.35 lb
quality x1=6.91/(141.44+6.91) = 0.047
u1= x1(ug-uf)+uf
u1= 0.047(107.33-47.39)+47.39 = 50.21Btu/lb
hg1= 115.96 Btu/lb
After opening the valve,
mass of sat vapor = Volume of sat vapor / specific volume of sat vapor at 160psi = 3/0.2896 = 10.36lb
mass of sat liq = Volume of sat liq/ specific volume of sat liq at 160psi = 1/0.01414 = 70.72 lb
Total mass = 81.08 lb
Mass escaped = 148.35 - 81.08 = 67.3 lb
quality x2=10.36/(70.72+10.36) = 0.128
u2= 0.128(107.33-47.39)+47.39 = 55.06Btu/lb
Conservation of energy: m2u2-m1u1 = Q - mouthg1
Q=m2u2-m1u1+mouthg1= 81.08(55.06) - 148.35(50.21) + 67.3(115.96) = 4788 btu (heat trasferred into the tank)
(Small error from conversion since I did not have the table in english unit)
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