Thermodynamics Heat Engine Problem

[smithnh]

Thank you in advance...

1. Homework Statement

Suppose a power plant delivers energy at 918MW using steam turbines. The steam goes into the turbines superheated at 626K and deposits its unused heat in river water at 286K. Assume that the turbine operates as an ideal Carnot engine. If the river flow rate is 40.4m^3/s, calculate the average temperature increase (in Celsius) of the river water downstream from the power plant. What is the entropy increase per kilogram of the downstream river water?


2. Homework Equations

(Q_H)/(T_H)=(Q_L)/(T_L)

delta(Q)=(m)(c)(delta(T))

delta(S)=(m)(c)ln((T_F)/(T_I))


3. The Attempt at a Solution

(918 MW)/(626K)=(Q_L)/(286K) so Q_L= 419.4 MW

(419400000 J/s)= (40.3 m^3/s)*(10^6 g/m^3)*(4.186 J/g/K)*delta(T)

so delta(T)= 2.5 celsius degrees right?

 

[Andrew Mason]

Thank you in advance...

1. Homework Statement

Suppose a power plant delivers energy at 918MW using steam turbines. The steam goes into the turbines superheated at 626K and deposits its unused heat in river water at 286K. Assume that the turbine operates as an ideal Carnot engine. If the river flow rate is 40.4m^3/s, calculate the average temperature increase (in Celsius) of the river water downstream from the power plant. What is the entropy increase per kilogram of the downstream river water?2. Homework Equations

(Q_H)/(T_H)=(Q_L)/(T_L)

delta(Q)=(m)(c)(delta(T))

delta(S)=(m)(c)ln((T_F)/(T_I))

$$\eta = \frac{W}{\Delta Q_h}$$ so

$$\Delta Q_h = \frac{W}{\eta}$$

Work out the efficiency from the temperature difference. Then determine the work/unit time from the output power. That will tell you the heat per unit time ([itex]dQ_h/dt[/tex]) delivered from the hot reservoir. You then work out what the heat flow to the cold reservoir is per unit time. Then work out the increase in temperature of the river water and the entropy change in the cold reservoir / unit mass using $$\Delta S = \Delta Q/T$$<br /> <br /> AM[/itex]