Superheated steam at 600 C and 4000 kPa is fed into a turbo expander at a rate o

Question

Superheated steam at 600 C and 4000 kPa is fed into a turbo expander at a rate of 10 kg/s. The steam expands and leaves the expander at 500 kPa pressure and 300 C. It is known that there is a heat loss of 50 kJ/sec. This heat loss can be assumed to take place at a surface temp of 500 C.

a) write down the appropriate forms of the energy balance and entropy balances for this case.

b) calculate the work produced per second.

c) Is the expansion reversible? if not calculate the entropy generation per second.

d) If the expansion were reversible, and assuming the same rate of heat loss, what would be the exit temperature if the exit pressure is kept at 500 kPa

Explanation / Answer

a)

Energy conservation:

h1 - h2 - q = w

where h1 and h2 are enthalpy before and after the turbine, q is heat transfer and w is turbine work.

Entropy conservation:

s1 - s2 + s_surr > 0

where s1 and s2 are entropies before and after the turbine and s_surr is the entropy increase of surroundings.

b)

From steam properties at 600 C and 4000 kPa, we get h1 = 3670 kJ/kg, s1 = 7.37 kJ/kg-K

From steam properties at 300 C and 500 kPa, we get h2 = 3060 kJ/kg, s2 = 7.46 kJ/kg-K

w = h1 - h2 - q

or W = m*(h1 - h2) - Q

= 10*(3670 - 3060) - 50

= 6050 kJ/s

c)

Entropy change of steam = m*(s2 - s1)

= 10*(7.46 - 7.37)

= 0.9 kJ/K/s

Entropy change of surrounding = Q_surr / T_surr

= 50 / (273 + 500)

= 0.0647 kJ/K/s

Net entropy generation = 0.0647 + 0.9 = 0.9647 kJ/K/s

d)

s2 - s1 = Q/T_surr

s2 - 7.37 = 0.0647

s2 = 7.4347 kJ/kg-K

From steam properties at P2 = 500 kPa and s2 = 7.4347 kJ/kg-K we get T2 = 293 deg C

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