Consider a horizontal pipe of diameter 20 cm. Steam enters the pipe as a saturat
ID: 1856689 • Letter: C
Question
Consider a horizontal pipe of diameter 20 cm. Steam enters the pipe as a saturated vapor at 0.5 MPa with a velocity of 12 m/s and exits at 0.45 MPa with a quality of 95%. Heat transfer from the pipe to the surroundings at 300 K takes place at an average outer surface temperature of 400 K. For operation at steady state, determine: Steam enters (a) the velocity at the exit, in m/s (b) the rate of heat transfer from the pipe, in kW. (c) the rate of entropy generation, in kW/K, for a control volume comprising only the pipe and its contents. (d) The rate of entropy generation, in kW/K, for an enlarged control volume that includes the pipe and enough of its immediate surroundings so that heat transfer from the control volume occurs at 300 K. (e) Why do the answers of part (c) and (d) differ? Please show all work, ThanksExplanation / Answer
Cross-section area A = pi/4*0.2^2 = 0.0314 m^2
From steam tables, corresponding to P1 = 0.5 MPa and quality x1 = 1 (sat. vapor), we get h1 = 2750 kJ/kg and v1 = 0.375 m3/kg, s1 = 6820 J/kg-K
And, for exit : At P2 = 0.45 MPa and quality = 0.95, we get h2 = 2640 kJ/kg and v2 = 0.393 m3/kg, s2 = 6600 J/kg-K
Mass flow rate m = A*V1 / v1
m = 0.0314*12 / 0.375 = 1.005 kg/s
a)
Heat transfer q = T(s2 - s1)
q = 400*(6600 - 6820)
q = -88000 J/kg = -88 kJ/kg
Steady flow energy eqn: h1 + V1^2 / 2 + q = h2 + V2^2 / 2
2750*10^3 + 12^2 / 2 -88000 = 2640*10^3 + V2^2 / 2
Solving, V2 = 210.1 m/s
b)
Rate of heat transfer Q = mq = -1.005*88 = -88.44 kW
c)
Rate of entropy generation for pipe and its contents = -88.44 / 400 = -0.2211 kW/K
d)
Rate of entropy generation for pipe and its surroundings = -88.44 / 300 = -0.2948 kW/K
e)
Part c and d differ because the heat transfer in c is being considered to be at 400 K while in d it is at 300 K.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.