new doc20140430075946776.pdf (1 page) Q.1 Thermodynamic concepts: a) An inventor
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Question
new doc20140430075946776.pdf (1 page) Q.1 Thermodynamic concepts: a) An inventor reports having created a cyclic engine which receives 325 kJ from a 1000 K thermal reservoir and rejects 125 kJ to a 400 K reservoir, while producing 200 kJ of work or output. Is this possible? b) A system consists of 2 m3 of gaseous water, initially at , 10 kPa, contained in a closed rigid tank. Energy is transferred to the system from a reservoir at until the temperature of the water is . The temperature of the surroundings where heat transfer occurs is a constant . Determine the heat transfer, the change in entropy, and the amount of entropy produced. Q.2 Carnot cycle: An engine which processes 1 kg of Helium gas through a Carnot cycle suffers a mechanical break. The resulting performance is as shown. where the lower isotherm has become an iso- bar. Assume CPIG. a) For the original Carnot cycle, find the efficiency, the heat transfer in, and the world done. b) For the broken cycle, find the heat transfer out, the not work, and the new efficiency. Q.3 As shown, a steam turbine at steady state is controlled by throttling the steam to a lower pressure be- fore it enters the turbine. Before throttling, the pressure and temperature are, respectively. 1.4 MPa and . After throttling, the pressure is 1 MPa. At the turbine exit, the steam is at 10 kPa and a quality of 90%. Heat transfer with the surroundings and all kinetic and potential energy effects can be ignored.Explanation / Answer
First law:
Q_in - W = Q_out
We have Q_in = 325 kJ, W = 200 kJ, and Q_out = 125 kJ
Since 325 - 200 = 125, hence it satisfies 1st law.
Carnot efficiency = 1 - T_low / T_high
= 1 - 400 / 1000
= 0.6
Actual efficiency = W / Q_in
= 200 / 325
= 0.615
Since, actual eff > Carnot eff, this is not possible.
b)
From steam properties at T1 = 50 deg C and P1 = 10 kPa we get v1 = 14.9 m^3/kg, u1 = 2440 kJ/kg, h1 = 2590 kJ/kg, s1 = 8.17 kJ/kg-K
From steam properties at T2 = 200 deg C and v2 = v1 = 14.9 m^3/kg, we get P2 = 14.6 kPa, u2 = 2660 kJ/kg, h2 = 2880 kJ/kg, s2 = 8.73 kJ/kg-K
Mass m = V/v = 2 / 14.9 = 0.1342 kg
Entropy change = m*(s2 - s1)
= 0.1342*(8.73 - 8.17)
= 0.07517 kJ/K
Heat transfer Q = W + (U2 - U1)
= 0 + 0.1342*(2660 - 2440)
= 29.5 kJ
Entropy change of surroundings = -Q / T_surr
= -29.5 / (300 + 273)
= -0.0515 kJ/K
Entropy production = 0.07517 - 0.0515 = 0.0236 kJ/K
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