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 celcius 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 $$\Delta S = \Delta Q/T$$

AM

 

[ogb p]

delta(S)=(40.3 m^3/s)*(10^6 g/m^3)*(4.186 J/g/K)*ln((288.5K)/(286K)) = 1.71 J/K

I would like to first clarify that the calculations done in the attempt at a solution are correct. The Carnot efficiency formula (Q_H/T_H = Q_L/T_L) and the heat transfer equation (Q = mc∆T) were appropriately used to determine the temperature increase of the river water downstream from the power plant and the entropy increase per kilogram of the downstream river water.

However, it is important to note that the assumptions made in this problem, such as the ideal Carnot engine and the constant river flow rate, may not accurately reflect real-world situations. In reality, power plants may not operate at ideal efficiency and river flow rates can vary.

Furthermore, the increase in temperature and entropy of the river water can have potential impacts on the surrounding ecosystem. As a scientist, it is important to consider the potential consequences of energy production on the environment and work towards finding sustainable and efficient solutions.

In conclusion, while the calculations done in this problem provide a theoretical answer, it is crucial to consider the practical implications and limitations of such a scenario in the real world.