2004-2005 Academic Year
THERMODYNAMICS I (ME 206) MIDTERM-1
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28.03.2005
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Duration: 120 Min. (Open Book Exam)
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Q-1) Refrigerant-12 is contained in a piston/cylinder arrangement at 2 MPa, 150°C with
a massless piston against the stops, at which point V = 0,5 m3. The side above the piston
is connected by an open valve to an air line at 10°C, 450 kPa, shown in Fig.1. The whole
setup now cools to the surrounding temperature of 10°C. Find the heat transfer and show
the process in a P–v diagram.
Q-2) An insulated cylinder is divided into two parts of 1 m3 each by an initially locked
piston, as shown in Fig.2. Side A has air at 200 kPa, 300 K, and side B has air at 1,0MPa,
1000 K. The piston is now unlocked so it is free to move, and it conducts heat so the air
comes to a uniform temperature TA = TB. Find the mass in both A and B, and the final T
and P.
Q-3) A vessel having a volume of 5 m3 contains 0,05 m3 of saturated liquid water and
4,95 m3 of saturated water vapor at 0,1 MPa (See Fig.3). Heat is transferred until the
vessel is filled with saturated vapor. Determine the heat transfer for this process.
Fig.1
Fig.2
Fig.3
SOLUTIONS
Q-1) C.V.: R-12. Control mass. Continuity: m = constant,
Energy Eq. m(u2 - u1) = Q - W
Process:
F↓ = F↑ = P A = PairA + Fstop
if V < Vstop Fstop = 0
This is illustrated in the P-v diagram shown below.
State 1: v1 = 0,01265 m3/kg, u1 = 252,1 kJ/kg
m = V/v = 39,523 kg
State 2: T2 and on line compressed liquid, see figure below.
v2 ≅ vf = 0,000733 m3/kg V2 = 0,02897 m3; u2 = uf = 45,06 kJ/kg
W12 = ∫PdV = Plift(V2 - V1) = 450 (0,02897 – 0,5) = -212,0 kJ ;
Energy Eq.
Q = 39,526 (45,06 – 252,1) - 212 = -8395 kJ
Q-2) C.V. A + B Force balance on piston: PAA = PBA
So the final state in A and B is the same.
State 1A: Table A.17 uA1 = 214,07 kJ/kg,
mA = PA1VA1/RTA1 = 200 × 1/(0,287 × 300) = 2,323 kg
State 1B: Table A.17 uB1 = 758,94 kJ/kg,
mB = PB1VB1/RTB1 = 1000 × 1/(0,287 × 1000) = 3,484 kg
For chosen C.V. Q12 = 0 , W12 = 0 so the energy equation becomes
mA(u2 - u1)A + mB(u2 - u1)B = 0
(mA + mB)u2 = mAuA1 + mBuB1
= 2,323 × 214,07 + 3,484 × 758,94 = 3141 kJ
u2 = 3141/(3,484 + 2,323) = 540,973 kJ/kg
From interpolation in Table A.17: T2 = 736 K
P = (mA + mB)RT2/Vtot = 5,807 kg × 0,287 kJ/kg K × 736 K/ 2 m3 = 613 kPa
Q-3)
Q12=U2-U1
m1 liq =Vliq/vf=0,05/0,001043=47,94 kg
m1 vap= Vvap/vg=4,95/1,6940=2,92 kg
So
U1= m1 liq u1 liq+ m1 vap u1 vap
=47,94(417,36)+2,92(2506,1)=27 326 kJ
State 2 is saturated vapor, x=100%
m= m1 liq+ m1 vap=47,94+2,92=50,86 kg
v2=V/m= 5/50,86 = 0,09831 m3/kg
in Table B1.2 , we find by interpolation, P2=2,03 MPa vg=0,09831 m3/kg Then
u2=2600,5 kJ/kg
U2=mu2=50,86(2600,5)=132 261 kJ
Q12=132261-27326=104 935 kJ