Saturated liquid water is compressed from 100 kPa (State 1) to 2.5 MPa (State 2)
ID: 1862842 • Letter: S
Question
Saturated liquid water is compressed from 100 kPa (State 1) to 2.5 MPa (State 2) in an
adiabatic pump having 85% isentropic efficiency.
(a) Calculate the specific work (kJ/kg) required for the compression of liquid water.
(b) Determine the entropy generation (kJ/kg-K) for the pump.
Saturated water vapor is compressed from 100 kPa (State 3) to 2.5 MPa (State 4) in an
adiabatic compressor having 85% isentropic efficiency.
(c) Calculate the specific work (kJ/kg) required for the compression of water vapor.
(d) Determine the entropy generation (kJ/kg-K) for the compressor.
(e) Comment on the relative magnitudes of the specific work and entropy generation for the pump and the compressor.
Please answer (a)-(e) and show all relevant calculations, assumptions and diagrams. Thank you!
Answers: (a) Not attached; (b) +0.0012 kJ/kg-K; (c) -950.18 kJ/kg; (d) Not attached; (e) Not attached
Explanation / Answer
At P = 100 kPa
Specific vol = v1 = vf = 0.001043 m^3/kg
T = 99.62 deg = 372.62 K
For PUMP, the Specific volume is constant from inlet to exit
Specific work,w1 = -v1(P2-P1) = -0.001043*(2500-100) = -2.5 KJ/kg
For 85% efficiency,we require more work for compression
w = w1/0.85 = -2.95 KJ/kg
Extra work = 0.45 KJ/kg
Entropy generation = 0.45/372.62 = 1.207 J/kg-K or 0.001207 KJ/kg-K
b)Here Specific volume varies from inlet to exit
At P1 =100 kPa
v1 = vg = 1.694 m^3/kg
T1 =99.62 K
h1 = hg = 2675.46 KJ/kg
s1 = sg = 7.3593 KJ/kg-K
At P2 =2500 kPa
Let's first slove for Isoentropic condition
s2 = s1 = 7.3593 KJ/kg-K
From steam tables
s = 7.3233 ---- h = 3462.04 KJ/kg-------T = 500 deg
s = 7.5960----- h = 3686.8 KJ/kg --------T = 600 deg
s = 7.3593------h2s = ?---------------------T2s = ?
LInear interpolating
h2 = 3686.8 +(3686.8-3462.04)*(7.3593-7.596)/(7.596-7.3233)
h2s = 3491.71 KJ/kg
T2s = 600 +(600-500)*(7.3593-7.596)/(7.596-7.3233)
T2s = 513.2
Now
Efficiency = (h1-h2s)/(h1-h2a) = (2675.46-3491.71)/(2675.46-h2a) = 0.85
h2a = 3635.75 KJ/kg
w = -(h2a-h1) = -(3635.75-2675.46) = -960.3 KJ/kg
Actual exit temp is greater than T2s becuase of irreversabilities
h2a = 3635.75 and P2 = 2500 kPa
From steam tables
h = 3468.8 ----- T2a = 600 deg------s2 = 7.596
h = 3686.25 ------T2a = 700 deg------s2 = 7.8435
h2a = 3635.75 ------T2a =?
Linear Interpolating
T2a = 700 +(700-600)*(3635.75-3686.25)/(3686.25-3468.8)
T2a = 676.78 deg
s2a = 7.8435 +(7.8435-7.596)*(3635.75-3686.25)/(3686.25-3468.8)
s2a = 7.786
Entropy generation(s) = s2a-s1 = 7.786-7.3593
s = 0.427 KJ/kg-K
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