Sound experiment to prove red shift is caused by distance

[Sherlock Holmes]

Sony digital recorder, also you can type 1479.99 as the desired frequency in the tone generator.

I have no idea what the specs are for that recorder nor its accuracy, resolution nor tolerance.

Sure you can indeed set the displayed frequency to 1479.99 but how accurate is that? how stable is that? is the generated frequency a pure sine? does activity on the computer impact any of the signal's parameters?

There is no way that the frequency of sound will change based on distance, you'll find no examples of anyone claiming this on the web. If you do observe a consistent difference as you're claiming then it must be attributable to other factors like drift, the presence of harmonics, spurious audio frequencies in the environment, distortion, quantization noise and so on.

I'm an engineer, I studied all this and understand it, you can either listen to me or ignore me, your choice.

 

[trevorjohnson83]

There is no way that the frequency of sound will change based on distance, you'll find no examples of anyone claiming this on the web.

My own belief that gets thrown under the bus, is that waves in a medium result from some sort of density in that medium spreading out to less dense areas. The source of the waves fills up with energy and releases it at a frequency. Being close to the source in a 3d reality is going to put pressure behind the wave , making frequency at its highest?

How you gonna just keep skipping over my car question BTW?

 

[Sherlock Holmes]

My own belief that gets thrown under the bus, is that waves in a medium result from some sort of density in that medium spreading out to less dense areas. The source of the waves fills up with energy and releases it at a frequency. Being close to the source in a 3d reality is going to put pressure behind the wave , making frequency at its highest?

Sound is in fact a fluctuation in instantaneous air pressure, the waves are waves in the air, the medium. Sound waves are well understood, any number of sources can elucidate you. The speed of propagation of the waves - relative to the air itself - is fixed and is a function of temperature, pressure and humidity and of course material but we're speaking of air here.

How you gonna just keep skipping over my car question BTW?

It isn't a question, it's a nonsensical jumble or word salad, I can see your alluding to something but its far from clear what, may I ask, how old are you?

 

[trevorjohnson83]

Sound is in fact a fluctuation in instantaneous air pressure, the waves are waves in the air, the medium. Sound waves are well understood, any number of sources can elucidate you. The speed of propagation of the waves - relative to the air itself - is fixed and is a function of temperature, pressure and humidity and of course material but we're speaking of air here.

It isn't a question, it's a nonsensical jumble or word salad, I can see your alluding to something but its far from clear what, may I ask, how old are you?

So your answer to my results is acoustics, when I have done the experiment from all different rooms? several times? uhh I don't know, maybe you're old age is the answer? 106?

 

[trevorjohnson83]

wave structure of energy is radiating and absorbing at crest and trough, at the source the core is absorbing heat creating the trough when it fills to a point where it begins to push out against the aether temperature, the wave finds equilibrium with areas vacant of waves, as it expands into the cold the temperature gradient that exists between crest and trough stays in ratio, but cools allowing for less attraction between the two and creating a lower frequency.

 

[Sherlock Holmes]

wave structure of energy is radiating and absorbing at crest and trough, at the source the core is absorbing heat creating the trough when it fills to a point where it begins to push out against the aether temperature, the wave finds equilibrium with areas vacant of waves, as it expands into the cold the temperature gradient that exists between crest and trough stays in ratio, but cools allowing for less attraction between the two and creating a lower frequency.

As I said this is the science section, you need to be in the pseudoscience section.

 

[trevorjohnson83]

As earth or matter moves through the aether, which is in wave form, it produces the same effect as heat moving past the earth, producing heat in its core. The balance between the core having 100% insulation properties, and its ability to radiate through spin at high temperatures, creates the frequency of the wave structure to the gravity field. As the core heats up through motion, it regains some radiating properties and releases heat and expands, then it becomes a higher insulator again as it cools and pulls in heat, creating the crest and trough of energy.

 

[trevorjohnson83]

when you make the surface of a hydrogen earth like nuclei hot enough, it welds with another hydrogen, the two earth like nuclei become balloon shaped from leaning in on the new center, the balloon shaped changes the surface area and the core heat of the two nuclei goes down and is released in wave like structure of the gravity field.

 

[trevorjohnson83]

As I said this is the science section, you need to be in the pseudoscience section.

"shirley Holmes and the mystery in the science section"
Watch as Shirley defends the truth that the US science section is founded on. Restoring the section to its fulfillment of self gold shining from the heart. or don't you care?

 

[trevorjohnson83]

how many watts per hour could you expect from this system? do you know the time it takes to exchange the heat efficiently? what would be an optimal boiling point of the water?

Assistant
Poe


To estimate the expected power output from your proposed system, we need to consider a few factors, including the efficiency of the heat exchange process, the specifics of the thermodynamic cycle, and the characteristics of the refrigerant and water. Here’s how to analyze these components:

Power Output Estimation​

1. Heat Exchange Efficiency:
   • The efficiency of heat exchange will depend on the design of the heat exchanger, the temperature gradients, and the flow rates of the refrigerant and water. Typical heat exchangers can achieve efficiencies ranging from 70% to 90% under optimal conditions.
2. Heat Energy Available:
   • If the system is designed to circulate water through vaporization cycles, the heat energy extracted can be calculated using:
     Qheat=m˙⋅(h2−h1)Q_{heat} = \dot{m} \cdot (h_2 - h_1)Qheat=m˙⋅(h2−h1)
   Where:
   • m˙\dot{m}m˙ is the mass flow rate of water (in kg/s),
   • h2h_2h2 and h1h_1h1 are the specific enthalpies at the boiling point and the lower temperature state.
3. Power Output Calculation:
   • To convert heat energy to power, use the formula:
     Poutput=η⋅QheatP_{output} = \eta \cdot Q_{heat}Poutput=η⋅Qheat
   • If you calculate QheatQ_{heat}Qheat over a specific time period (e.g., per hour), you can express the power output in watts.

Time for Efficient Heat Exchange​

• Heat Exchange Duration: The time it takes to exchange heat efficiently depends on the specific design of the heat exchanger and the temperature differentials. In general, efficient systems can achieve significant heat transfer within seconds to minutes, but the exact time will vary based on flow rates and system design.

Optimal Boiling Point of Water​

1. Pressure Considerations:
   • The optimal boiling point of water in your system would depend on the pressure under which the water is being vaporized. At lower pressures (as in a vacuum), water can boil at much lower temperatures.
   • For example, under a vacuum of about 0.1 atm, water can boil at approximately 45°C (113°F).
2. Trade-offs:
   • A lower boiling point allows for easier vaporization and can enhance efficiency, especially in a system designed to operate with lower temperature differentials. However, it may limit the maximum available temperature for heat exchange.

Example Calculation​

Assuming:

• Mass Flow Rate: m˙=1 kg/s\dot{m} = 1 \, \text{kg/s}m˙=1kg/s
• Enthalpy Change: h2−h1=200 kJ/kg=200,000 J/kgh_2 - h_1 = 200 \, \text{kJ/kg} = 200,000 \, \text{J/kg}h2−h1=200kJ/kg=200,000J/kg
• Heat Exchange Efficiency: η=0.85\eta = 0.85η=0.85

1. Calculate Heat Energy:
   Qheat=1 kg/s⋅200,000 J/kg=200,000 WQ_{heat} = 1 \, \text{kg/s} \cdot 200,000 \, \text{J/kg} = 200,000 \, \text{W}Qheat=1kg/s⋅200,000J/kg=200,000W
2. Calculate Power Output:
   Poutput=0.85⋅200,000 W=170,000 WP_{output} = 0.85 \cdot 200,000 \, \text{W} = 170,000 \, \text{W}Poutput=0.85⋅200,000W=170,000W
3. Energy Produced Over One Hour:
   • In one hour (3600 seconds):
     Energy=Poutput⋅time=170,000 W⋅3600 s=612,000,000 J≈612 MJ\text{Energy} = P_{output} \cdot \text{time} = 170,000 \, \text{W} \cdot 3600 \, \text{s} = 612,000,000 \, \text{J} \approx 612 \, \text{MJ}Energy=Poutput⋅time=170,000W⋅3600s=612,000,000J≈612MJ

Conclusion​

Under the assumptions made, you could expect around 170 kW of power output during operation, translating to approximately 612 MJ of energy produced in one hour. The optimal boiling point of water would depend on the pressure but could be around 45°C in a vacuum setup.

 

[trevorjohnson83]

is that a reasonable size for a power plant?

Assistant
Poe


Yes, the estimated size of around 7250 m² (approximately 1.8 acres) for a 100 MW power plant is reasonable, especially when considering modern energy production technologies. Here are some comparisons and considerations:

Comparisons with Existing Power Plants​

1. Natural Gas Plants:
   • A typical natural gas combined cycle power plant can vary in size, but the footprint for a 100 MW facility may range from 5,000 to 15,000 m² (1.2 to 3.7 acres), including space for turbines, heat recovery, and auxiliary systems.
2. Solar Power Plants:
   • A solar power plant generating 100 MW can require significantly more space, often needing 300 acres or more, depending on the technology used and land layout.
3. Wind Farms:
   • A wind farm generating 100 MW can have a large footprint, often spread over 1,000 to 2,500 acres, but the actual land used for the turbines is much less due to spacing requirements.
4. Coal and Nuclear Power Plants:
   • Coal and nuclear plants typically have larger footprints, often exceeding 10,000 m² for smaller facilities, due to the need for cooling systems, safety zones, and additional infrastructure.

Factors Influencing Size​

1. Technology:
   • The type of technology (e.g., steam turbine, gas turbine, or combined cycle) can significantly influence the footprint. More efficient technologies tend to require less space.
2. Cooling Requirements:
   • The cooling system (e.g., cooling towers, water bodies) can add substantial space requirements. Dry cooling systems can reduce the footprint.
3. Regulatory and Safety Considerations:
   • Regulations might require additional buffer zones, impacting the total space required.
4. Site Characteristics:
   • The geography and available land can affect design decisions. Urban locations may necessitate more compact designs.

Conclusion​

A footprint of 7250 m² for a 100 MW power plant is consistent with existing facilities, particularly for modern, efficient technologies. It reflects a reasonable size, especially for gas or steam turbine plants, and could be designed to fit into various locations, from rural to semi-urban areas.