Thermodynamics Heat Engine Problem

AI Thread Summary
The discussion revolves around calculating the average temperature increase of river water downstream from a power plant using steam turbines operating as an ideal Carnot engine. Given the power output of 918 MW and the temperatures of the steam and river water, the heat rejected to the river is determined to be 419.4 MW. The flow rate of the river is 40.4 m³/s, leading to a calculated temperature increase of approximately 2.5 degrees Celsius. Additionally, the entropy change per kilogram of the river water is to be evaluated using the relevant thermodynamic equations. The calculations emphasize the efficiency of the heat engine and the relationship between heat transfer and temperature change.
smithnh
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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?
 
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smithnh said:
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 (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
 
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