Question

A 27 kg iron block initially at 375 C is quenched in an insulated tank that contains 130kg of water at 26 C. Assume the water that vaporizes during the process condenses back in the tank and the surroundings are at 12 C and 125 kPa. The specific heats of iron and water are 0.45 kJ/kgK and 4.18 kJ/kgK, respectively. a: Determine the final equilibrium temperature. b: Determine the entropy change of the combined system at the initial state. c: Determine the exergy of the combined system at the initial state. d: Determine the waste work potential during this process.

Answer 1

a). Applying the energy balance,

`$\Delta E_(sys)=E_(in)-E_(out)$`

`$0=\Delta U$`

`$0=(\Delta U)_(iron) + (\Delta U)_(water)$`

`$0=[mc(T_f-T_i)_(iron)] + [mc(T_f-T_i)_(water)]$`

`$0 = 27 * 0.45 * (T_f - 375) + 130 * 4.18 * (T_f-26)$`

`$t_f=33.63^\circ C$`

b). The entropy change of iron.

`$\Delta s_(iron) = mc \ln\left((T_f)/(T_i) \right)$`

`$ = 27 * 0.45\ \ln\left((33.63 + 273)/(375 + 273) \right)$`

= -9.09 kJ-K

Entropy change of water :

`$\Delta s_(water) = mc \ \ln\left((T_f)/(T_i) \right)$`

`$ = 130 * 4.18\ \ln\left((33.63 + 273)/(26 + 273) \right)$`

= 10.76 kJ-K

So, the total entropy change during the process is :

`$\Delta s_(tot) = \Delta s_(iron) + \Delta s_(water) $`

= -9.09 + 10.76

= 1.67 kJ-K

c). Exergy of the combined system at initial state,

`$X=(U-U_(0)) - T_0(S-S_0)+P_0(V-V_0)$`

`$X=mc (T-T_0) - T_0 \ mc \ \ln \left((T)/(T_0) \right)+0$`

`$X=mc\left((T-T_0)-T_0 \ ln \left((T)/(T_0) \right)\right)$`

`$X_(iron, i) = 27 * 0.45\left(((375+273)-(12+273))-(12+273) \ln (375+273)/(12+273)\right)$`

`$X_(iron, i) =63.94 \ kJ$`

`$X_(water, i) = 130 * 4.18\left(((26+273)-(12+273))-(12+273) \ln (26+273)/(12+273)\right)$`

`$X_(water, i) =-13.22 \ kJ$`

Therefore, energy of the combined system at the initial state is

`$X_(initial)=X_(iron,i) +X_(water, i)$`

= 63.94 -13.22

= 50.72 kJ

Similarly, Exergy of the combined system at initial state,

`$X=(U_f-U_(0)) - T_0(S_f-S_0)+P_0(V_f-V_0)$`

`$X=mc\left((T_f-T_0)-T_0 \ ln \left((T_f)/(T_0) \right)\right)$`

`$X_(iron, f) = 27 * 0.45\left(((33.63+273)-(12+273))-(12+273) \ln (33.63+273)/(12+273)\right)$`

`$X_(iron, f) = 216.39 \ kJ$`

`$X_(water, f) = 130 * 4.18\left(((33.63+273)-(12+273))-(12+273) \ln (33.63+273)/(12+273)\right)$`

`$X_(water, f) =-9677.95\ kJ$`

Thus, energy or the combined system at the final state is :

`$X_(final)=X_(iron,f) +X_{water, f$`

= 216.39 - 9677.95

= -9461.56 kJ

d). The wasted work

`$X_(in) - X_(out)-X_(destroyed) = \Delta X_(sys)$`

`$0-X_(destroyed) = $`

`$X_(destroyed) = X_(initial) - X_(final)$`

= 50.72 + 9461.56

= 9512.22 kJ