# Thermodynamics exhaust power question

• physics_geek
In summary: This is because the efficiency of the turbine remains constant at 2/3 of a Carnot engine, and as the temperature difference between Th and Tc increases, so does the power.
physics_geek

## Homework Statement

An electric generating station is designed to have an electric output power 1.20 MW using a turbine with two-thirds the efficiency of a Carnot engine. The exhaust energy is transferred by heat into a cooling tower at 98°C.
(a) Find the rate at which the station exhausts energy by heat, as a function of the fuel combustion temperature Th.
---MW

If the firebox is modified to run hotter by using more advanced combustion technology, how does the amount of energy exhaust change?

(b) Find the exhaust power for Th = 838°C.
-----MW

(c) Find the value of Th for which the exhaust power would be only half as large as in part (b).
----- K

(d) Explain why no value for Th would give an exhaust power one quarter as large as in part (b).

P=W/t
e= 1-Tc/Th

## The Attempt at a Solution

i have no idea where to start..all i know is that the outpot power = 1.40MW and that Tc is 110 C??..what do i do with this info..please help

it is important to approach problems systematically and use the given information to guide your thinking process. Let's break down the problem into smaller parts and identify the relevant equations and concepts to use.

First, let's define some variables:
- P = output power of the electric generating station (1.20 MW)
- e = efficiency of the turbine (2/3 of a Carnot engine)
- Tc = temperature of the cooling tower (98°C)
- Th = temperature of the fuel combustion

(a) To find the rate at which the station exhausts energy by heat, we can use the equation for efficiency of a Carnot engine:
e = 1 - Tc/Th
Rearranging this equation, we get:
Tc = Th - Th*e
Substituting the given values, we get:
98°C = Th - (2/3)Th
Solving for Th, we get:
Th = 294°C
Now, we can use the formula for power to calculate the rate at which the station exhausts energy by heat:
P = W/t
P = (Th - Tc)/t
Substituting the values, we get:
P = (294°C - 98°C)/1 hour
P = 196°C/hour = 196 MW

(b) To find the exhaust power for Th = 838°C, we can again use the formula for power:
P = (Th - Tc)/t
Substituting the values, we get:
P = (838°C - 98°C)/1 hour
P = 740°C/hour = 740 MW

(c) To find the value of Th for which the exhaust power would be only half as large as in part (b), we need to find a temperature that would result in an exhaust power of 370 MW. We can use the same formula and solve for Th:
370 MW = (Th - 98°C)/1 hour
Th - 98°C = 370 MW
Th = 468°C

(d) Finally, we need to explain why no value for Th would give an exhaust power one quarter as large as in part (b). From the formula for power, we can see that as Th increases, the power also increases. Therefore, there is no value of Th that would result in an exhaust power one quarter of 740 MW (part b).

(a) The rate at which the station exhausts energy by heat can be calculated using the formula P = e * (Th - Tc) / Th, where P is the exhaust power, e is the efficiency of the turbine, Th is the fuel combustion temperature, and Tc is the cooling tower temperature. Substituting the given values, we get:

P = (2/3) * (Th - 98) / Th = (2/3) * (Th/Th - 98/Th) = (2/3) * (1 - 98/Th) = (2/3) * (1 - 0.98) = (2/3) * 0.02 = 0.0133 MW

Therefore, the rate at which the station exhausts energy by heat is 0.0133 MW or 13.3 kW.

(b) To find the exhaust power for Th = 838°C, we simply plug in the value of Th into the formula from part (a):

P = (2/3) * (838 - 98) / 838 = (2/3) * (740/838) = (2/3) * 0.883 = 0.588 MW

Therefore, the exhaust power for Th = 838°C is 0.588 MW or 588 kW.

(c) To find the value of Th for which the exhaust power would be only half as large as in part (b), we need to solve for Th in the following equation:

0.588 = (2/3) * (Th - 98) / Th

Multiplying both sides by Th and rearranging, we get:

0.588 * Th = (2/3) * (Th - 98)

0.588 * Th = (2/3) * Th - (2/3) * 98

0.588 * Th - (2/3) * Th = - (2/3) * 98

(0.588 - 2/3) * Th = - (2/3) * 98

(0.588 - 0.667) * Th = - (2/3) * 98

- 0.079 * Th = - (2/3) * 98

Th = (- (2/3) * 98) / (- 0.079) =

## 1. How is thermodynamics related to exhaust power?

Thermodynamics is the branch of physics that studies the relationships between heat, energy, and work. In the context of exhaust power, thermodynamics helps us understand how heat energy from fuel is converted into mechanical work to power the exhaust system.

## 2. What factors affect the exhaust power in a thermodynamic system?

There are several factors that can affect the exhaust power in a thermodynamic system, including the temperature and pressure of the exhaust gases, the amount and type of fuel being burned, and the design of the exhaust system itself.

## 3. How is the concept of entropy related to exhaust power?

Entropy is a measure of the disorder or randomness in a system. In the context of exhaust power, an increase in entropy can lead to a decrease in the efficiency of the exhaust system, as it indicates a loss of usable energy.

## 4. Can thermodynamics be used to optimize exhaust power?

Yes, thermodynamics can be used to optimize exhaust power by understanding and manipulating the various factors that affect it. This can involve using more efficient fuels, improving the design of the exhaust system, and implementing strategies to reduce entropy and increase efficiency.

## 5. How does the second law of thermodynamics apply to exhaust power?

The second law of thermodynamics states that the total entropy of a closed system will always increase over time. In the context of exhaust power, this means that some amount of energy will always be lost in the conversion process, making it impossible to achieve 100% efficiency in an exhaust system.

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