Homework # 8

6.76
Remember that both the pump and turbine are adiabatic. The work of the pump
is v1(P2 - P1) = 0.001017 m3/kg*(10,000-20) = 10.15 kJ/kg. The work of the
turbine you get from the energy balance so w = h1 - h2. The h2 = hg @ 20
kPa. The h1 you get from the fact that s1 = s2 = sg at 20 kPa = 7.9085
kJ/kg-K. Then h1-h2 = 4712.8 - 2609.7 = 2103.1 kJ/kg. The ratio becomes 207.2.

6.111
a) First find Qin = mCp(Tout - Tin) = 0.25 kg/s*4.18 kJ/kg-C*(45-15) =
31.35 kW.
b) Then you need the final temperature of the hot water. Since Qin = -Qout,
then 31.35 kW = -mC(DT). That means the final T = 97.5 °C. So the Sgen =
mcoldCln(T2/T1) + mhotC ln(T2/T1) = 0.25 * 4.18 kJ/kg-K*ln(318/288) +
3*4.19*ln(97.5+273/373) = 0.0190 kW/K
6.122
This is exactly like an example problem in the book.  
a) First find Q = kA(DT/x) = 0.69 W/m°C)(5*6m2)15/0.3 = 1035 W.
b) Sgen = Qout/Tout - Qin/Tin = 1035 W/278 - 1035/293 = 0.191 W/K.

6.130
I did this problem in class. There is a typo, where the vfg was wrong to
find out x2.  Otherwise you can follow what I did.

6.133
a) The energy balance is 0 = h2 - h1 + (V22 - V12/2). Find h2 = 3230.9
kJ/kg - (3202- 702/2 * 1kJ/kg/1000 m2/s2 = 3182.15 kJ/kg. Watch the
conversion! This and the P = 2 MPa fixes the T at 370.4 °C°.
b) The entropy balance just becomes Sgen = m(s2 - s1). The mass flow rate
is found from 1/v1 * A1 V1 = 1/.09936 m3/kg * (7 X 10-4 m2)(70 m/s) = 0.493
kg/s. Then Sgen = 0.493 kg/s * (7.0260 - 6.9212) kJ/kg-K.