Solution : Test 2 Thermodynamics CODE : C
Saturated liquid water contained in a closed rigid tank is cooled to a final state where the temperature is 55oC
and the masses of saturated vapor and liquid present are 0.05 and 1999.95 kg, respectively. Determine the heat
transfer for the process, in kJ
SOLUTION :
System : A Rigid tank contain water
Process : Constant Volume
Table Properties : Steam
Assumption : ∆EK = ∆EP = 0
Diagram :
T
Sat. liq
1
M ix. Phase
T2
T2
2
v
Final state : T2 = 55oC, mg2 = 0.05 kg and mf2 = 1999.95 kg
mtotal = 2000 kg, x2 = mg2/mtotal = 0.05 kg/ 2000kg = 0.000025
From Steam Table for Tsat = T2 = 55oC, vf2 = 0.001015 m3/kg, vg2 = 9.56 m3/kg, uf2 = 230.24 kJ/kg, ufg2 =
2219.10 kJ/kg,
so that :
v2 = vf2 + x2(vg2-vf2) = 0.001015 m3/kg + 0.000025(9.56-0.001015) m3/kg = 0.001254 m3/kg
u2 = uf2 + x2*ufg2 = 230.24 kJ/kg + 0.000025*2219.10 kJ/kg = 230.30 kJ/kg
Process took place at constant volume, so that v1 = v2 = 0.001254 m3/kg
Initial state is saturated liquid.
from interpolation of steam table for vf = v1 = 0.001254 m3/kg, we got u1 = uf = 1085.22 kJ/kg,
From energy conservation law :
∑Ein - ∑Eout =∆Esys or Qnet - Wnet = ∆Esys
for closed system Qnet - Wnet = ∆Usys + ∆EKsys + ∆EPsys
Wnet = 0, and ∆EK = ∆EP = 0 so that Qnet = ∆Usys = mtot(u2 -u1)
Qnet = 2000kg(230.30-1085.22)kJ/kg = -1709851.05 kJ
1709851.05 kJ heat transferred by the system to surrounding.
A two-phase liquid–vapor mixture of H2O with an initial quality of 30% is contained in a piston–cylinder assembly
as shown. The mass of the piston is 35 kg, and its diameter is 10 cm. The atmospheric pressure of the
surroundings is 100 kPa. The initial positions of the piston h 1 is 1 cm and the final positions of the piston h2 is
5.0 cm. As the water is heated, the pressure inside the cylinder remains constant until the piston hits the stops.
Heat transfer to the water continues until its pressure is 400 kPa. Friction between the piston and the cylinder
wall is negligible. Determine the total amount of heat transfer, in kJ. (use g = 10 m/s2)
SOLUTION:
System : Piston cylinder with upper stopper contain water
Process : combination of constant pressure and constant volume
Table properties : steam
Assumption : ∆EK = ∆EP = 0
Diagram :
P
3
1
2
v
x1 = 0.30
h1 = 1cm and h2 = 5.0cm
piston : mass = 35kg and dimeter = 10cm, area A = 0.008m2
Pressure from piston = P1 = Wpiston/A + Patm = (35kg*10m/s2)/0.008m2 (1kPa/1000Pa) + 100kPa = 144.56kPa
V1 = h1*A = 1/100 m*0.008m2 = 0.000079m3
V2 = h2*A = 5.0/100 m*0.008m2 = 0.000393m3
From steam table with using interpolation, we got
v1 = 0.3611m3/kg, and u1 = 1079.00kJ/kg
m = V1/v1 = 0.000079m3/0.3611m3/kg = 0.000218kg
Process 1-2 is a constant pressure process with P2 = P1 = 144.56kPa
Process 2-3 is a constant volume process, so that v3 = v2 = 1.8055 m3/kg
P3 = 400kPa, from steam table for Psat = 400kPa we got vg = 0.46m3/kg
because v3 > vg, final condition of water is in superheated phase.
from steam table of superheated for Psat = 400kPa and v3 = 1.8055m3/kg, we got u3 = 4668.00kJ/kg.
From energy conservation law :
∑Ein - ∑Eout =∆Esys or Qnet - Wnet = ∆Esys
for closed system Qnet - Wnet = ∆Usys + ∆EKsys + ∆EPsys
∆EK = ∆EP = 0 so that Qnet - Wnet = ∆Usys
3
2
3
W net =∫1 PdV =∫1 Pdv∫1 pdV =P V 2−V 10
Wnet = 400kPa(0.000393-0.000079)m3/kg = 0.045kJ
and Qnet = ∆Usy + Wnet = mtot(u3 -u1) + Wnet = kg(4668.00-1079.00)kJ/kg + 0.045kJ = 0.826kJ
Total amount of heat transfer to the system is 0.826 kJ
A rigid tank initially contains 4 kg of air at 400kPa, 290 K. The tank is connected by a valve to a piston–cylinder
assembly oriented vertically and containing 0.02m3 of air initially at 200 kPa, 290 K. now the valve is opened so
that a slow leak allows air to flow into the cylinder until the tank pressure falls to 200 kPa. The weight of the
piston and the pressure of the atmosphere maintain a constant pressure of 200 kPa in the cylinder; and owing
to heat transfer, the temperature stays constant at 290 K. For the air, determine the total amount of energy
transfer by work and by heat, each in kJ
SOLUTION:
System : An Air is contained in rigid tank air connected by a valve to a vertical piston–cylinder which containing
air too.
Process : constant temperature
Table properties : for Air Pcr = 3.77MPa, Ptank 400 kPa < Pcr so that air can be treated as an ideal gas.
Diagram :
Initial State :
Tank : mass = 4 kg, PT1 = 400 kPa, TT1 = 290 K
Cylinder-Piston : VP1 = 0.02 m3, PP1 = 200 kPa, TP1 = 290 K
from ideal gas relation m = PV/RT, for air R = 0.287 kJ/kg.K
mP1 = 200kPa*0.02m3/(0.287kJ/kg.K*K290) = 0.05kg
Total mass of air = 4kg + 0.05kg = 4.05kg
Initial condition of tank and cylinder-piston is different, but because the mass of air in cylinder-piston << the
mass of air in tank, we can assume the condition of air in cylinder-piston will be the same as the condition of air
in tank just after the valve is opened. So the initial condition become : mass of air = 4.05kg, T = 290K, P =
400kPa
Final state : mass of air = 4.05kg, T = 290K, P= 200kPa
From energy conservation law :
∑Ein - ∑Eout =∆Esys or Qnet - Wnet = ∆Esys
for closed system Qnet - Wnet = ∆Usys + ∆EKsys + ∆EPsys
∆EK = ∆EP = 0 so that Qnet - Wnet = ∆Usys
Because air can be treated as an ideal gas and the process take place at constant temperature, so that ∆Usys = 0
2
Q net =W net =∫1 PdV
from an ideal gas relation :
P=
2
2 mRT
mRT
P2
--> Q net =W net =∫ PdV =∫
dV
=mRTln
1
1
V
T
P1
Qnet = Wnet = 4.05kg*0.287kJ/kg.K*290K ln (400kPa/200kPa) = 108.35kJ
the total amount of energy transfer by work and by heat is the same, that is 108.35 kJ