Question

A heat exchanger (boiler) produces dry steam at 100 from feed water at 30 at a rate of 1.5kg/s. The specific heat capacity of water is 4187 J/kgK and its specific latent heat of vaporization is 2,257,000 J/kg

Answer 1

Answer on Question #77748-Engineering- Material Science Engineering

A heat exchanger produces dry steam at 100 C from feed water at 30C at a rate of 1.5 kgs^-1. The heat exchanger receives heat energy at a rate of 600 kW from the fuel used. The specific heat capacity of water is 4187 Jkg^-1K^-1 and its specific latent heat of vaporisation is 2257000 Jkg^-1.

A) Determine the heat energy received per kilogram of steam produced.

B) Determine the output power of the heat exchanger and its thermal efficiency.

Solution

$$\frac{dm}{dt} = 1.5 \frac{kg}{s}$$ $$T_c = 30\,C$$ $$T_h = 100\,C$$ $$P = 600\,kW$$ $$C = 4187 \frac{J}{kg\,K}$$ $$L = 2257000 \frac{J}{kg}$$

A)

$$Q = mC(T_h - T_c) + mL$$ $$dQ = C(T_h - T_c)dm + Ldm$$ $$\frac{dQ}{dm} = C(T_h - T_c) + L = 4187 \frac{J}{kg\,K} (100\,C - 30\,C) + 2257000 \frac{J}{kg} = 2550000 \frac{J}{kg}$$

B)

$$P_0 = \frac{dQ}{dt} = (T_h - T_c) \frac{dm}{dt} + L \frac{dm}{dt} = \left(\frac{dm}{dt}\right) [C(T_h - T_c) + L] = \left(\frac{dm}{dt}\right) \left(\frac{dQ}{dm}\right) = 1.5 \frac{kg}{s} \cdot 2550000 \frac{J}{kg}$$ $$= 3.825 \cdot 10^6 \frac{J}{s} = 3825\,kW.$$ $$e = \frac{P}{P_0} = \frac{600}{3825} = 0.157 \text{ or } 15.7\%$$

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