Question

Consider a steady flow Carnot cycle with water as the working fluid. The maximum and minimum temperatures in the cycle are 350 and 50 C. The quality of water is 0.891 at the beginning of the heat rejection process and 0.1 at the end. Determine:

(a) the thermal efficiency (how much percent).
(b) the pressure at the turbine inlet, and
(c) the network output

a) Etath,C= %
b) P2=MPa
c) wnet = kJ/kg

Answer 1

Answer:

a) Etath = 48.2 %

b) P2 = 1.195 MPa

c) wnet = 1749.14 KJ/kg

Explanation:

Given:

T1 = T2 = 50 , TL = 50 + 273 = 323 K

T3 = T4 = 350 , TH = 350 + 273 = 623 K

x3 =0.891

x4 = 0.1

Part a

Thermal efficiency (Etath) of carnot cycle is:

Etath = 1 - (TL / TH )

Etath = 1 - (323) / (623) = 48.2 %

Part b

Note:

From A - 4 Table

T1 = 50 C @sat

sf = 0.7038 KJ/kg.K

sfg = 7.3710 KJ/Kg.K

s2 = s3 = sf + x3 * (sfg)

s2 = s3 = 0.7038 + 0.891*7.3710 = 7.271361 KJ/kg.K

Thus,

@ T2 = 350 C  ; sg = 5.2114 KJ/kg.K

s2 = 7.271361 KJ/kg.K

s2 > sg@T = 350 C, hence super-heated region.

Use Table A-6

T2 = 350 C

s2 = 7.271361 KJ/kg.K

P2 = 1.195 MPa (using interpolation)

part c

wnet can be calculated by finding the enclosed area on a T-s diagram:

s4= sf + x4*sfg =  0.7038 + 0.1*7.3710 = 1.4409 KJ/kg.K

wnet = (TH - TL)*(s3 - s4)

wnet = (623 - 223) * (7.271361 - 1.4409) = 1749.14 KJ/kg