![m1=30 [kg/s]; m2= 5[kg/s]; m3=25 [kg/s]
p1=3000; p2=500; p3=15; x3=.9; p0=100
t1= 350; t2=200; t0=25; t0k=298
h1=enthalpy(steam,p=p1,t=t1)
s1=entropy(steam,p=p1,t=t1)
h0=enthalpy(water,p=p0,t=t0)
s0=entropy(water,p=p0,t=t0)
h2=enthalpy(steam,p=p2,t=t2)
s2=entropy(steam,p=p2,t=t2)
h3=enthalpy(steam,p=p3,x=x3)
s3=entropy(steam,p=p3,x=x3)
a1=h1-t0k*s1-(h0-t0k*s0) { availability at entry}
a2=h2-t0k*s2-(h0-t0k*s0)
a3=h3-t0k*s3-(h0-t0k*s0)
m1*h1=m2*h2+m3*h3+w3 { w3 = actual work output}
m1*s1+s_gen=m2*s2+m3*s3
w_rev=w3+t0k*s_gen
eta_2nd=w3/w_rev
h2s=enthalpy(steam,p=p2,s=s1)
h3s=enthalpy(steam,p=p3,s=s1)
m1*h1=m2*h2s+m3*h3s+w3_s {w3s = isentropic work of turbine}
eta_isen=w3/w3_s
30kg/s
3MPa, 350C
5kg/s
.5MPa, 200C
25kg/s
15kPa, 90%quality
1
2
3
𝑤cv](https://image.slidesharecdn.com/lect7-150420145113-conversion-gate01/85/Lect-7-thermo-6-320.jpg)
Lect 7 thermo
- 1. Maximum work or available energy from a system Initially a body of mass m, is at temp T, what is the maximum work that can be obtained out of it if the environment is at 𝑇0 which is lower than the body temperature T. 𝑇0 𝑄ℎ 𝑄𝑙 w 𝑚𝑐 𝑣 𝑇 𝑄ℎ = 𝑚𝑐 𝑣 𝑇 − 𝑇0 ; ∆𝑠 𝑏 = 𝑚𝑐 𝑣 ln 𝑇0 𝑇 ∆𝑠𝑠𝑢𝑟 = 𝑄𝑙/𝑇0 For max work output, ∆𝑠𝑠𝑢𝑟 + ∆𝑠 𝑏 = 0 𝑄𝑙 = −𝑇0 𝑚 𝑐 𝑣 ln 𝑇0 𝑇 ; 𝑤 = 𝑄ℎ − 𝑄𝑙 = 𝑚𝑐 𝑣 𝑇 − 𝑇0 − 𝑇0 𝑚𝑐 𝑣 ln 𝑇 𝑇0 𝑤 = 𝑚𝑐 𝑣 𝑇 − 𝑇0 − 𝑇0 ln 𝑇 𝑇0 = 𝑈1 − 𝑈0 − 𝑇0(𝑆1 − 𝑆0) = 𝑈1 − 𝑇0 𝑆1 − 𝑈0 − 𝑇0 𝑆0 = 𝐹1 − 𝐹0; 𝑤ℎ𝑒𝑟𝑒 𝐹 = 𝑈 − 𝑇0 𝑆, 𝐹 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑡ℎ𝑒 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑟 𝐻𝑒𝑙𝑚𝑜𝑙𝑡𝑧 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 If the system boundary expands against the atmospheric pressure p0 then F becomes: 𝜑 = 𝑈 − 𝑇0 𝑆 + 𝑝0 𝑉, 𝜑1 − 𝜑2 = 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑤𝑜𝑟𝑘 = 𝑤 𝑚𝑎𝑥
- 2. cm 𝑇0 𝑄12 𝑄0 w12 𝑇ℎ Control mass process when the system receives heat from a source Maximum work or available energy 𝑈1 − 𝑈2 𝑆1 − 𝑆2 The system goes from 1-2, does work w12 and rejects heat 𝑄0 Lets us consider two situations: the body rejects heat either reversibly or irreversibly, but it receives heat reversibly. 𝑄12 − 𝑄0 𝑖𝑟𝑟 = 𝑈2 − 𝑈1 + 𝑤12 𝑎𝑐𝑡 𝑠 𝑔𝑒𝑛 = ∆𝑠𝑐𝑚 + ∆𝑠𝑠𝑢𝑟 + ∆𝑠𝑟𝑒𝑠 > 0, for irrev process 𝑠𝑔𝑒𝑛 𝑟𝑒𝑣 = 𝑆2 − 𝑆1 + 𝑄0 𝑟𝑒𝑣 𝑇0 − 𝑄12 𝑇ℎ = 0 for rev process 𝑄0 𝑟𝑒𝑣 = −𝑇0 𝑆2 − 𝑆1 + 𝑇0 𝑇ℎ Q12, From 1st law for the CM we get: 𝑄12 − 𝑄0 𝑟𝑒𝑣 = 𝑤12 𝑟𝑒𝑣 + 𝑈2 − 𝑈1 𝑤12 𝑟𝑒𝑣 = 𝑄12 − 𝑄0 𝑟𝑒𝑣 − 𝑈2 − 𝑈1 𝑤12 𝑟𝑒𝑣 = 𝑄12 + 𝑇0 𝑆2 − 𝑆1 − 𝑄12 𝑇0 𝑇ℎ − 𝑈2 − 𝑈1 𝑤12 𝑟𝑒𝑣 = 𝑄12 1 − 𝑇0 𝑇ℎ + 𝑈1 − 𝑇0 𝑆1 − 𝑈2 − 𝑇0 𝑆2 𝑤12 𝑟𝑒𝑣 = 𝑤 𝑚𝑎𝑥 = 𝑄12 1 − 𝑇0 𝑇ℎ + 𝐹1 − 𝐹2 sur res 𝑤12 𝑎𝑐𝑡 = 𝑄12 − 𝑄0 𝑖𝑟𝑟 − 𝑈2 − 𝑈1 𝐼 = 𝐼𝑟𝑟 = 𝑤12 𝑟𝑒𝑣 − 𝑤12 𝑎𝑐𝑡 = 𝑇0 𝑆2 − 𝑆1 − 𝑄12 𝑇0 𝑇ℎ + 𝑄0 𝑖𝑟𝑟 = 𝑇0 𝑆2 − 𝑆1 − 𝑄12 𝑇ℎ + 𝑄0 𝑖𝑟𝑟 𝑇0 = 𝑇0(∆𝑠𝑐𝑚 + ∆𝑠𝑟𝑒𝑠 + ∆𝑠𝑠𝑢𝑟) = 𝑇0∆𝑠 𝑢𝑛𝑖𝑣 = 𝑇0 𝑠𝑔𝑒𝑛 Measure of irreversibility
- 3. NH3 100C Example problem: 1kg of NH3 is contained in a spring loaded piston/cylinder as saturated liquid at -20C. Heat is added from a reservoir at 100C until a final state of 800kpa, 70C is reached. Find the work, heat transfer, and entropy generation assuming the process to be internally reversible. p1=pressure(r717,t=-20,x=0) v1=volume(r717,t=-20,x=0) u1=intenergy(r717,t=-20,x=0) s1=entropy(r717,t=-20,x=0) v2=volume(r717,p=800,t=70) u2=intenergy(r717,t=70,p=800) s2=entropy(r717,t=70,p=800) w12=(800+p1)*.5*(v2-v1) q12=w12+u2-u1 s_gen+q12/373=s2-s1
- 4. Maximum work in a flow process 1 2 w 𝑚 𝑚 q 𝑇ℎ 𝑇0𝑞0, Objective: to find the maximum work for the Case shown in picture 1st Law for the CV: 𝑚 ℎ1 + 𝑣1 2 2 + 𝑔𝑧1 + 𝑞 − 𝑞0 = 𝑚 ℎ2 + 𝑣2 2 2 + gz2 + w 𝑚𝑠1 + 𝑞 𝑇ℎ − 𝑞0 𝑇0 + 𝑠𝑔𝑒𝑛 = 𝑚𝑠2, 𝑠𝑔𝑒𝑛 = 0, 𝑓𝑜𝑟 𝑤 𝑡𝑜 𝑏𝑒 𝑚𝑎𝑥2nd Law for the CV 𝑚 𝑠1 − 𝑠2 + 𝑞 𝑇ℎ = 𝑞0 𝑇0 𝑞0 = 𝑚 𝑠1 − 𝑠2 𝑇0 + 𝑞𝑇0 𝑇ℎ (1) (2) Now use eqn (3) in (1) to obtain max work as: (3) 𝑤 𝑚𝑎𝑥 = 𝑚{ ℎ1 − 𝑇0 𝑠1 − ℎ2 − 𝑇0 𝑠2 + 𝑣1 2 − 𝑣2 2 2 + 𝑔 𝑧1 − 𝑧2 } + 𝑞 1 − 𝑇0 𝑇ℎ 1 = h − T0s1 + v1 2 2 + gz1; Keenan function 𝑤 𝑚𝑎𝑥 = 𝑚 1 − 2 + 𝑞 1 − 𝑇0 𝑇ℎ Second law efficiency: 𝑤 𝑎𝑐𝑡/𝑤 𝑚𝑎𝑥 = 𝑤 𝑎𝑐𝑡/(𝑖 − 𝑒)
- 5. T S 1 2 2s P=p1 P=p2 (ℎ1−ℎ2)/(ℎ1 − ℎ2𝑠) = _𝑖𝑠𝑒𝑛, 𝑡𝑢𝑟 For the case of a turbine expansion P=p2 1 2s 2 T S P=p1 For the case of a compressor (ℎ1−ℎ2𝑠)/(ℎ1 − ℎ2) = _𝑖𝑠𝑒𝑛, 𝑐𝑜𝑚𝑝 An insulated steam turbine, receives 30kg/s of steam at 3 Mpa, 350C . In the turbine where the pressure is 500kpa, steam is beld off at the rate of 5kg/s, for processing equipment. The temperature of this steam is 200C. The balance of the steam leaves the turbine at 15kpa, 90% quality. Find the availability per kg of steam at the entry, the isentropic efficiency and the second law efficiency of the turbine. What is the actual work output of the turbine Ans: 1110 kj/kg, eta_isen=.7975, eta_2nd=.8176, w_act=20144 kw Example problem
- 6. m1=30 [kg/s]; m2= 5[kg/s]; m3=25 [kg/s] p1=3000; p2=500; p3=15; x3=.9; p0=100 t1= 350; t2=200; t0=25; t0k=298 h1=enthalpy(steam,p=p1,t=t1) s1=entropy(steam,p=p1,t=t1) h0=enthalpy(water,p=p0,t=t0) s0=entropy(water,p=p0,t=t0) h2=enthalpy(steam,p=p2,t=t2) s2=entropy(steam,p=p2,t=t2) h3=enthalpy(steam,p=p3,x=x3) s3=entropy(steam,p=p3,x=x3) a1=h1-t0k*s1-(h0-t0k*s0) { availability at entry} a2=h2-t0k*s2-(h0-t0k*s0) a3=h3-t0k*s3-(h0-t0k*s0) m1*h1=m2*h2+m3*h3+w3 { w3 = actual work output} m1*s1+s_gen=m2*s2+m3*s3 w_rev=w3+t0k*s_gen eta_2nd=w3/w_rev h2s=enthalpy(steam,p=p2,s=s1) h3s=enthalpy(steam,p=p3,s=s1) m1*h1=m2*h2s+m3*h3s+w3_s {w3s = isentropic work of turbine} eta_isen=w3/w3_s 30kg/s 3MPa, 350C 5kg/s .5MPa, 200C 25kg/s 15kPa, 90%quality 1 2 3 𝑤cv