Question

Saturated steam at 100 C is heated to 400 C. Use the steam table to determine the required heat input if continuous stream flowing at 150 kg/min undergoes this process at constant pressure.

Answer 1

ΔH + ΔE_k + ΔE_p = Q -W_s    (eq. 7.4-15) all flowrates

ΔE_k , ΔE_p , W =0      ΔU = Q

H(400 degrees C, 1 atm) = 3278 kJ / kg   (table B.7)

H(100 degrees C, sat'd 1 atm) = 2676 kJ / kg   (table B.5)

stream in: 100 kg H2O (v)/s

                100degrees C, sat'd

stream 1 out:   100 kg H2O (v)/s

                      400degrees C, 1 atm

also (into system): Q(kW)

Q = (100kg/s)[(3278-2676)kJ/kg](1000 J/kJ) = 6.02 x10⁷ J/s

b) ΔU + ΔE_k + ΔE_p = Q -W   (eq. 7.3-4)

ΔE_k , ΔE_p , W =0     ΔU = Q

Table B.5 → U(100 degrees C, 1 atm) = 2507 kJ/kg

V(100 degrees C, 1 atm) = 1.673 m³ / kg = V(400degrees C, P_final)

Interpolate in table B.7 to find P at which V = 1.673 at 400degrees C, and then interpolate again to find U at 400 degrees Cand that pressure:

V = 1.673 m³ / g → P_final =1.0 + 4.0[(3.11-1.673) / (3.11-0.617)] = 3.3 bar

U(400 degrees C, 3.3 bar) = 2966 kJ/kg

Q = ΔU = mΔU = 100 kg [(2966 - 2507)kJ/kg](1000J/kJ) = 4.59 x 10⁷ J

The difference is the net energy needed to move the fluidthrough the system (flow work).

(The energy change associated with the pressure change in Part(b) is insignificant.)