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Physics of Wind Turbines
Over a thousand years ago, windmills were in operation in Persia and China, see TelosNet and Wikipedia. Post mills appeared in Europe in the twelfth century, and by the end of the thirteenth century the tower mill, on which only the timber cap rotated rather than the whole body of the mill, had been introduced. In the United States, the development of the water-pumping windmill was a major factor in allowing farming and ranching across vast areas in the mid-nineteenth century. These windpumps (sometimes called Western mills) are still common in America and Australia. They have a rotor with about 30 vanes (or blades) and the ability to turn themselves slowly. Of the 200,000 windmills existing in Europe in the mid-nineteenth century, only one in ten remained a century later. The old windmills have since been replaced by steam and internal combustion engines. However, since the end of the last century the number of wind turbines is growing steadily, and they are beginning to take an important role in power generation in many countries.

We first show that for all wind turbines, wind power is proportional to wind speed cubed. Wind energy is the kinetic energy of the moving air. The kinetic energy of a mass m with the velocity v is

The air mass m can be determined from the air density ρ and the air volume V according to

Then,

Power is energy divided by time. We consider a small time, Δt, in which the air particles travel a distance s = v Δt to flow through. We multiply the distance with the rotor area of the wind turbine, A, resulting in a volume of

which drives the wind turbine for the small period of time. Then the wind power is given as

The wind power increases with the cube of the wind speed. In other words: doubling the wind speed gives eight times the wind power. Therefore, the selection of a "windy" location is very important for a wind turbine.

The effective usable wind power is less than indicated by the above equation. The wind speed behind the wind turbine can not be zero, since no air could follow. Therefore, only a part of the kinetic energy can be extracted. Consider the following picture:

The wind speed before the wind turbine is larger than after. Because the mass flow must be continuous, A v = constant, the area A2 after the wind turbine is bigger than the area A1 before. The effective power is the difference between the two wind powers:

If the difference of both speeds is zero, we have no net efficiency. If the difference is too big, the air flow through the rotor is hindered too much. The power coefficient cp characterizes the relative drawing power:

To derive the above equation, the following was assumed: A1v1 = A2v2 = A (v1+v2) / 2. We designate the ratio v2/v1 on the right side of the equation with x. To find the value of x that gives the maximum value of CP, we take the derivative with respect to x and set it to zero. This gives a maximum when x = 1/3. Maximum drawing power is then obtained for v2 = v1 / 3, and the ideal power coefficient is given by

Another wind turbine located too close behind would be driven only by slower air. Therefore, wind farms in the prevailing wind direction need a minimum distance of eight times the rotor diameter. The usual diameter of new wind turbines is 140 m with an installed capacity of 5 MW onshore. 200 m diameter and 10 MW will come soon offshore.

The installed capacity or rated power of a wind turbine corresponds to an electrical power output of a speed between 12 and 16 m/s, with optimal wind conditions. For safety reasons, the plant does not produce greater power at the high wind conditions than those for which it is designed. During storms, the plant is switched off. Throughout the year, a workload of 23% can be reached inland. This increases to 28% on the coast and 43% offshore.

More details can be found in the Internet pages wind-works.org and in the pages of the American Clean Power Association (ACP).

For 2023, IRENA reports an installed wind power capacity of 1017 GW worldwide and 148 GW in the United States. The Alta Wind Energy Center in California has been the largest wind farm in the United States since 2013, with a capacity of 1.6 GW. The electricity produced from wind power in the United States amounts about 10% of all generated electrical energy. Detailed information about the present state in the US can be found in Wikipedia and in a 2024 PNNL report.

A crucial point about wind power is that the times of peak electricity demand and the times of optimal wind conditions rarely coincide. Thus, other electric power producers with short lead times and a well developed electricity distribution system are necessary to supplement wind power generation.

Why have the wind turbines of today lost one blade in comparison to the old four-blade windmills?
The rotor power Pmech = 2π M n is proportional to the torque M acting on the shaft and the rotation frequency n. The latter is influenced by the tip speed ratio λ, which is calculated according to λ = vu / v1 from the ratio of peripheral speed (tip speed) vu of the rotor and the wind speed v1. The torque M increases with the number of blades. It is therefore largest for the many-vaned Western mills, smaller for wind mills with four blades, and smallest for today’s wind turbines with 3 blades. However, every blade, as it rotates, reduces the wind speed for the following blades. This "wind shadow" effect increases with the number of blades. The optimal tip speed ratio is about one for the Western mill, barely over 2 for the four-bladed type, and 7−8 for the three-bladed rotors. At their optimal tip speed ratio, three-bladed rotors achieve a cp value of 48% and come closer to the ideal value of 59% than wind turbines with 4 blades. For wind turbines with two blades or weight-balanced one-bladed rotor configurations, the yield is smaller in spite of a higher tip speed ratio, because of the smaller torque M. Therefore, wind turbines today have three blades.

AI Overview     Designing a thermal updraft, or solar chimney, for a home requires custom calculations rather than using standard chimney sizing for a fireplace. While a standard fireplace chimney is sized for hot, fast-moving exhaust, a solar chimney is designed for much slower, cooler air movement and is influenced by external factors. Residential versions are often used for passive cooling or ventilation, and achieving a useful updraft may require a significantly taller chimney than is typical for combustion. 

Key principles of residential thermal updraft Solar chimney function: The chimney creates an updraft by heating the air inside it to a higher temperature than the air outside. The rising, buoyant warm air then draws in cooler air from within the house.Optimal placement: To maximize heating, the chimney should be installed on an exterior wall with the most sun exposure, preferably a southern wall in the Northern Hemisphere. For best performance, it should also be integrated into the main structure, with as little exposure to outside temperatures as possible.Passive cooling and heating:

Passive cooling: When used for passive cooling, a solar chimney draws hot air out of the home, pulling in cooler air from a lower intake point. Dark-colored or ceramic materials at the top of the chimney can help increase heating to enhance this effect.Passive heating: When combined with a heat source like a Trombe wall, a solar chimney can pull heated air back into the house. Sizing and calculation considerations Standard chimney-sizing rules, such as the 3-2-10 rule for safety and drafting, are primarily for combustion appliances and do not apply to a passive thermal chimney. For a residential-scale thermal updraft, consider these factors: 

Chimney dimensions Height: A taller chimney creates a stronger draft because of the greater pressure differential between the top and bottom. Residential solar chimney concepts often recommend heights of 20 meters (about 65 feet) or more to function optimally.

Cross-sectional area: The area must be large enough to allow for good airflow, but if it is too large, the air will lose heat too quickly, weakening the draft. For passive ventilation, the air velocity will be much slower than in a combustion flue.Ratio of area to height: The ideal sizing depends on a complex interplay of the cross-sectional area and height.

As an example, one engineering calculation for natural draft uses the formula for airflow rate (\(Q\)), which depends on the chimney height (\(H\)), cross-sectional area (\(A\)), and the temperature difference between the inside (\(T_{i}\)) and outside (\(T_{o}\)) air.\(Q=A\times \sqrt{2\times g\times H\times \frac{T_{i}-T_{o}}{T_{o}}}\) 

Example calculation for natural draft ventilation To calculate the airflow rate (\(Q\)) for a simple, two-story house with an 8-meter high chimney (from ground floor to top duct) in a climate where the inside temperature (\(T_{i}\)) is 20°C and the outside temperature (\(T_{o}\)) is -10°C: 

Step 1: Convert temperatures to Kelvin \(T_{o}=-10\degree C+273.15=263.15K\)\(T_{i}=20\degree C+273.15=293.15K\)

Step 2: Define variables \(H\) (chimney height) = 8 m\(g\) (gravitational acceleration) = 9.81 m/s² 

Step 3: Calculate the density of the air Outside air density (\(\rho _{o}\)):\(\rho _{o}=\rho _{standard}\times \frac{T_{standard}}{T_{o}}=1.293\frac{kg}{m^{3}}\times \frac{273.15K}{263.15K}\approx 1.342\frac{kg}{m^{3}}\)Inside air density (\(\rho _{i}\)):\(\rho _{i}=\rho _{standard}\times \frac{T_{standard}}{T_{i}}=1.293\frac{kg}{m^{3}}\times \frac{273.15K}{293.15K}\approx 1.205\frac{kg}{m^{3}}\) 

Step 4: Calculate the velocity and flow rate (simplified) A simplified formula for velocity (\(v\)) shows that it is driven by the pressure difference created by the height and temperature differential. The airflow rate is then calculated based on this velocity and the cross-sectional area of the chimney.The final calculation for a 0.2m (approx. 8 inches) diameter duct in the example yields an airflow of 0.12 m³/s. Practical design considerations 

Insulation: Insulating the chimney is crucial to maintain high internal temperatures and a strong draft. Placing the chimney within the house's warm envelope is more effective than placing it on an exterior wall.Location: The optimal location depends on the purpose. For passive heating, a central location might be effective, while for ventilation, a location that draws air effectively from the living space is best.Professional consultation: Due to the complexity and reliance on external factors, consulting with an engineer or experienced chimney professional is recommended for a residential-scale updraft thermal system.

Buckley Rumford Fireplaces
Chimney Draft

11/5/23
Chimney draft can be complicated but the principles are well known. If we consider three factors that affect draft separately, we can build chimneys to vent fireplaces that will work. The three factors are:

1) Pressure differential - the basic mechanism of draft.
2) Flow capacity or rate - getting enough draft.
3) The chimney in relationship to the house.

1) Pressure differential - the basic mechanism of draft.
The combustion flue gases inside chimneys are hotter than the ambient outside air and therefore less dense and lighter than the ambient air. The column of warm air inside the chimney, being more buoyant, rises in the chimney, reducing the pressure at the bottom of the chimney so that ambient air flows into the combustion zone and moves the flue gases up and out of the chimney. That movement or flow of combustion air and flue gas is called "natural draught", "natural ventilation", "chimney effect", or "stack effect". The taller the stack, the more draught is created. The equation below provides an approximation of the pressure difference, ΔP, (between the bottom and the top of the flue gas stack) that is created by the draught:

where:

ΔP = available pressure difference, in Pa
C = 0.0342
a = atmospheric pressure, in Pa
h = height of the chimney, in m
To = absolute outside air temperature, in K
Ti = absolute average temperature of the flue gas, in K
The above equation is an approximation because it assumes that the molar mass of the flue gas and the outside air are equal and that the pressure drop through the flue gas stack is quite small. Both assumptions are fairly good but not exactly accurate.
The important thing for us is to note that draft is directly proportional to the height of the chimney, the temperature of the flue gasses and the atmospheric pressure. A twenty foot tall chimney creates a pressure differential twice as great as a ten foot chimney. A chimney at sea level where a = 14.7 pounds per sq.ft. has almost one and a half times the pressure differential as an equal chimney at 10,000 feet in elevation where a - 10 pounds per sq.ft.

The pressure differential is only part of the question. Notice that flue size doesn't matter at all.

2) Flow capacity or rate - getting enough draft.
How much air flows into the bottom of the chimney is estimated by the formula below.

where:

Q = flue gas flow rate, m³/s
A = cross-sectional area of chimney, m²
C = discharge coefficient (usually from 0.65 to 0.70)
g = gravitational acceleration at sea level, 9.807 m/s²
H = height of chimney, m
Ti = absolute average temperature of the flue gas in the stack, K
To = absolute outside air temperature, K
Again, without getting too concerned about the math or the units, the important thing for us is to note that the flow is directly proportional to the cross-sectional area of chimney and to the square root of the height of the chimney and the temperature differential multiplied by a big number. In other words the height of the chimney and the temperature of the flue gasses, while directly proportional to the pressure differential in the static case above are a little less important due to friction or drag and energy it takes to accelerate the gasses.
3) The chimney in relationship to the house.
That's all great if the chimney were standing out in the middle of a field. But most fireplaces are inside houses. Houses are filled with warm air relative to ambient air in the heating season and, like in a chimney, the warm air in a house is less dense than the ambient cooler air and tends to rise. In other words the house itself can act as a chimney. If you open an upstairs window the warm air in the house would rise and flow out of the window while creating negative pressure downstairs that would cause air to flow in an open downstairs window. Complicate this situation with various fans and other appliances that draw air out of the house.

In a house what matters is the pressure differential between the air in the bottom of the chimney compared with the pressure at the same level inside the house. If the house is taller than the chimney the house could draft better than the chimney and outside air will flow down the chimney into the house. If an appliance, like a kitchen fan, is turned on that could reduce the indoor air pressure and cause a downdraft even if the chimney is tall enough.

Flow matters too. In a tightly sealed up house with no fans on and no leaks or open windows there may not be an excessively negative indoor air pressure but there also needs to be enough flow rate (Q) for combustion and to carry away the smoke in the fireplace.

Effects of Altitude
Altitude affects the pressure differential probably because pressure vs altitude is logarithmic rather than lineal. A 30 foot tall chimney at sea level spans a greater pressure differential than it does at 10,000 feet.

Altitude has no effect on flow rate, but flue size does.

The rule-of-thumb advice is usually to increase the flue area at high altitudes but, if you want to compensate for a decreased pressure differential, the chimney height should be increased.

In our smoky fireplace recommendations we recommend under normal living conditions (furnace, bath and kitchen fans on or off windows open or closed, etc.) there should be an updraft in a cold chimney with no fire in the fireplace. This is the pressure differential test.

We also recommend that adequate outside air be ducted into the house to make up for all the air being leaked out or exhausted by all the appliances and fans likely to be on - and enough extra for the fireplace - at least one cubic foot per minute (CFM) for every square inch of flue area. This is to test for adequate flow.

As a practical matter we recommend that no altitude adjustment is needed because what really matters is chimney draft compared with indoor air pressure. Just like at sea level, if the chimney draws when cold while simulating the one CFM per sq.in. of flue area with a fan, the fireplace will work fine at any altitude. But at altitude if you have trouble getting a pressure differential (updraft) you can increase the height of the chimney and if you can't get enough flow you can use a larger flue size.

The Lunar Module's crew mass is approximately 144 kg (317 lb), representing the combined weight of two astronauts in their suits and equipment. This crew mass is carried by the Lunar Module's two-person ascent stage, which was designed for space operations on and around the Moon.

Key Specifications Related to Crew Mass
Crew Capacity: 2
Total Mass (Ascent Stage, Loaded): Approximately 4,780 kg (10,538 lb)
Crew Mass (Approximate): 144 kg (317 lb)
Pressurized Volume: 6.7 m³ (235 cu ft)
Habitable Volume: 4.5 m³ (160 cu ft)

The crew mass is an important component that influences the overall mass of the spacecraft and is factored into its performance and mission design.

During the Apollo missions, the two-person lunar module (LM) crew was provisioned with a food supply that weighed approximately 0.77 kg (1.7 lbs) per man per day, including packaging. The extremely small weight and volume envelope of the spacecraft required that most foods be dehydrated.

Food system specifications
Daily caloric intake: Each crew member received approximately 2,800 calories per day, divided into three meals.
Dehydrated food: About 80% of the food weight was removed by dehydrating it, with water from the LM's fuel cells used to rehydrate the meals during the mission. The freeze-drying process made the food extremely light and compact, with an almost indefinite shelf life.

Packaging: Meals were packed in four-ply, laminated film pouches to protect against moisture and oxygen invasion. Color-coded Velcro patches were attached to meal packages for easy identification.
Food preparation: Unlike the Command Module, the Lunar Module had no hot water dispenser, so food consumed on the lunar surface was rehydrated with cold water.

Bite-sized cubes: Astronauts also ate ready-to-eat, bite-sized food cubes coated with gelatin to prevent crumbs, which could interfere with sensitive equipment.
Snacks: Some missions included high-nutrient food bars, which astronauts could nibble on while wearing their helmets during moonwalks

The Apollo Lunar Module's crew oxygen supply varied by mission, with early missions using less than 100 pounds of gaseous oxygen for lunar descent and surface operations. The supply was separated into tanks in the descent stage for surface activities and tanks in the ascent stage for the return to the command module. Later, "J-missions" were equipped with additional capacity for longer stays.

Oxygen system mass specifications
Apollo 11 (Early mission)

Descent stage: The LM descent stage housed two oxygen tanks, each containing 48 pounds of gaseous oxygen at 2,690 psia. This supply covered descent, the surface stay, EVA backpack refills, and cabin repressurizations.

Ascent stage: For the rendezvous and docking after lunar liftoff, the ascent stage had two tanks, each containing 2.43 pounds of oxygen at 840 psia.
Extended "J-missions" (e.g., Apollo 15, 16, 17)

The final three Apollo missions carried greater oxygen capacity to support longer stays on the lunar surface. While the specific additional mass is not detailed in sources, the increased duration from a few hours (Apollo 11) to nearly three days (Apollo 17) necessitated a larger oxygen reserve.
Oxygen system context

The LM oxygen system was designed to support the astronauts for a limited time away from the Command and Service Module (CSM), which housed the primary liquid oxygen supply for the journey to and from the Moon.
Oxygen use

Cabin pressurization: The LM's atmosphere was pure oxygen at a pressure of 4.8 pounds per square inch.
Suit repressurization: Oxygen was used to repressurize the cabin and refill the Portable Life Support Systems (PLSS) worn by the astronauts during their moonwalks.

Metabolic consumption: The crew consumed oxygen for breathing. For instance, the descent tank on early missions was sized to allow for six PLSS refills and four cabin repressurizations.

Mass summary
Component     Mass (pounds)    Details    Reference
Descent Stage Tank (x2)    96    Each tank held 48 lbs of gaseous oxygen.   
Ascent Stage Tank (x2)    4.86    Each tank held 2.43 lbs of oxygen for the trip back to the CSM.

Lunar launch mass is highly variable, depending on the payload and mission, but a typical ratio can be seen with the Apollo Lunar Module, where its total mass of 15,103 kg included 2,376 kg of propellant in the ascent stage and more in the descent stage. Modern Starship designs aim to bring significantly larger payloads, with a lunar variant requiring approximately 1,500 tons of propellant.
Apollo Lunar Module (Example)
Total Mass: 15,103 kg (including fuel and oxidizer)
Ascent Stage Propellant: 2,376 kg
Descent Stage Propellant: 8,200 kg (DPS propellants)

Lunar Lander Engines: These are specialized engines designed for the precise maneuvers required for landing on and taking off from the Moon's surface. Examples include:
Apollo Lunar Module Descent Engine: Developed by Thompson Ramo Wooldridge Inc (TRW), this engine could be throttled between 1,000 and 10,000 pounds of thrust, making it the most powerful LM engine. It was crucial for controlled landings and could be throttled for hovering and precise site selection.

Apollo Lunar Module Ascent Engine (LMAE): Built by Bell Aerosystems and later redesigned by Rocketdyne, this engine lifted the ascent stage off the Moon's surface and rendezvoused with the orbiting Command Module. It produced 3,500 pounds of thrust.
Reaction Control System (RCS) Thrusters: These small thrusters are used for attitude control and minor orbital adjustments. The Apollo LM had sixteen RCS thrusters, each with 100 pounds-force (440 N).

Here's a breakdown of how the figure relates to the Apollo program:
LLO Departure Vehicle: In the context of the Apollo missions, the LM was the vehicle that departed from Low Lunar Orbit (LLO) to land on the moon. The term "LLO departure vehicle" refers to this stage of the mission.

Total propellant: The Apollo 11 mission's LM had a total propellant load of over 23,000 kg, or approximately 53,000 pounds.

Fuel vs. Propellant: The LM used hypergolic propellants, meaning the fuel (Aerozine 50) and oxidizer (nitrogen tetroxide) ignited on contact, without a separate ignition system.

Propellant breakdown for a typical Apollo LM
The actual mass of propellant was divided between the two stages of the LM. While the exact figures varied by mission, for Apollo 11 the propellant breakdown was:

Descent stage propellant: 18,184 lbs (approx. 8,250 kg). This was used to descend from LLO to the lunar surface.
Ascent stage propellant: 5,238 lbs (approx. 2,376 kg). This was used to launch off the lunar surface and rendezvous with the Command and Service Module (CSM) in lunar orbit

The propulsive landing of a rocket involves using its engines to decelerate and control its descent, ensuring a safe landing. This process is governed by physics principles, including the Tsiolkovsky rocket equation and Newton's laws of motion. Here's a simplified explanation of the key equation used in such scenarios:

Key Equation for Propulsive Landing
The velocity change (Δv) required for a rocket to decelerate and land safely is derived from the Tsiolkovsky rocket equation:
$$ \Delta v = v_e \ln\left(\frac{m_0}{m_f}\right) $$
Where:

Δv: Change in velocity needed for landing.
v_e: Effective exhaust velocity (specific impulse × gravitational acceleration).
m_0: Initial mass of the rocket (including remaining fuel).
m_f: Final mass of the rocket (after burning fuel).

Additional Considerations for Landing

Thrust-to-Weight Ratio: The rocket must generate enough thrust to counteract gravity and decelerate. The thrust-to-weight ratio is given by:
$$ T/W = \frac{T}{m \cdot g} $$
Where:

T: Thrust produced by the engines.
m: Current mass of the rocket.
g: Gravitational acceleration (≈9.81 m/s² on Earth).

Burn Time: The duration of the engine burn to achieve the required Δv:
$$ t = \frac{m_0 - m_f}{\dot{m}} $$
Where:

t: Burn time.
\dot{m}: Mass flow rate of the propellant.

Controlled Descent: The rocket must balance thrust and gravity to achieve a soft landing. This involves precise control of the throttle and orientation.

Practical Application
For a successful landing, engineers simulate and calculate:

The required Δv based on the rocket's altitude, velocity, and mass.
The optimal burn time and thrust profile to ensure a smooth deceleration.
Real-time adjustments using onboard sensors and guidance systems.

This combination of physics and engineering ensures rockets like SpaceX's Falcon 9 can land safely and be reused!

There is no single "LLO 20,000 kg moon landing equation"; instead, a complex set of equations and principles govern a moon landing, including the Tsiolkovsky rocket equation to determine propellant requirements and orbital mechanics to calculate trajectories for rendezvous and descent. The Apollo Lunar Module, weighing approximately 15,200 kg, used these principles for its lunar orbit and descent, which would not be a simple formula.
Key Concepts and Equations
Tsiolkovsky Rocket Equation:
This fundamental equation relates the change in velocity (delta-v) of a rocket to its exhaust velocity and the ratio of its initial to final mass.
Δv = Ve * ln(m0 / mf)
Δv is the change in velocity.
Ve is the exhaust velocity.
ln is the natural logarithm.
m0 is the initial mass of the spacecraft.
mf is the final mass (after fuel is expended).
Orbital Mechanics:
Complex equations are used to predict and control the spacecraft's path in orbit around the Earth and Moon, ensuring a successful rendezvous and landing.
How it Applies to a Moon Landing
1. Mission Planning:
Engineers use these equations to determine how much fuel is needed for a lunar module with a specific payload, such as a 20,000 kg lander, to reach the Moon, enter lunar orbit, and then land safely.
2. Descent & Landing:
The equations calculate the necessary adjustments in velocity and thrust to slow the lunar module from orbital speed to a soft landing on the Moon's surface.
3. Ascent & Return:
The same principles are applied for the ascent to rejoin the command module and the return journey to Earth.
Example of Mass
The Apollo Lunar Module (LEM) had a standard launch mass of approximately 15,200 kg (33,500 lb).

AI Overview     To determine the engine thrust required for a 20,000 kg lunar landing, you would use Newton's Second Law of Motion. The required thrust must be greater than the gravitational force of the Moon acting on the lander to allow for a controlled descent. A simplified calculation is provided below, followed by more complex considerations. Equation for required thrust The most basic equation for the thrust (\(F_{T}\)) required for a controlled lunar descent is:\(F_{T}>m\times g_{L}\) Where: \(F_{T}\) is the engine thrust (in Newtons)\(m\) is the mass of the lander (in kg)\(g_{L}\) is the acceleration due to gravity on the Moon, which is approximately 1.62 m/s². For a 20,000 kg lander, the minimum thrust would be:\(F_{T}>20,000\text{\ kg}\times 1.62\text{\ m/s}^{2}\)\(F_{T}>32,400\text{\ N}\) or 32.4 kN Key considerations and complexities Controlled descent: This equation calculates the minimum thrust needed to hover. For a controlled landing, the engine must be "throttleable"—meaning its thrust can be adjusted. To slow the descent and land softly, the engine's thrust must be greater than gravity. The Apollo Lunar Module, for example, had a descent engine with a thrust that could be varied, or "deep throttled," over a wide range.Mass is not constant: A rocket's total mass decreases as it burns propellant. This means less thrust is needed to counteract gravity as the spacecraft empties its fuel. Modern landers must be designed to accommodate this change in mass and the resulting change in thrust requirements.Total "delta-v" (ΔV): The total change in velocity required for the mission is a more comprehensive metric than just the instantaneous thrust. The Tsiolkovsky Rocket Equation calculates the total ΔV capability of a spacecraft based on its propellant and engine efficiency. A lunar landing requires a specific ΔV budget, typically around 2,050 m/s for the descent phase.Thrust-to-weight ratio (TWR): Engineers use the TWR to describe engine performance relative to the spacecraft's mass. The TWR is typically calculated relative to Earth's gravity but can also be expressed for a lunar landing. For a lunar descent, an optimal TWR at the start of the burn might be in the range of 0.35 to 0.5 (using Earth weight)

Reported by offtherock    Not So Free Chat » Oil, Peak Oil, etc. » Post #233655
Reason

Solar is currently only about 2% of world energy production.
Its the cheapest way to make energy, cleanest and the fastest growing.
But still, only a few percentages.

So therefore, something not being powered by solar, isn't really an argument for anything.
Most thing are powered by something that isn't solar.

We come from a dirty past.
But dirty past isn't an argument for a dirty future.

AI Overview

Creating a solar panel requires a significant amount of energy, often referred to as embodied energy, which accounts for the energy consumed throughout the entire manufacturing process, from raw material extraction to assembly and transportation.

Here's a breakdown of the key aspects related to the embedded energy in solar panels:

Silicon Processing:
The production of metallurgical-grade silicon from quartz sand through carbothermic reduction consumes about 20 kWh/kg. Further refinement into electronic-grade silicon adds another 100 kWh/kg.
Melting and crystallizing this silicon into single crystal ingots consumes 290 kWh/kg.
Overall, after accounting for material losses during these steps, approximately 460 kWh of energy are embodied into each kilogram of silicon single crystal.

Solar Cell Production:
Slicing the silicon ingots into wafers and subsequent processing steps, including high-temperature diffusion, oxidation, deposition, and annealing, add another 120 kWh/m² of embodied energy.

Module Assembly: Assembling the solar cells with a glass front panel, encapsulant, copper ribbon, foil back cover, and aluminum frame adds 190 kWh/m² of embodied energy.

Support Structure: The type of support structure for the solar panels also impacts the embodied energy. Rooftop installations generally require less additional energy (around 200 kWh/m²) compared to ground-mounted systems requiring concrete, cement, and steel (up to 500 kWh/m²).

Miscellaneous Components: Other components like inverters, wiring, and potentially batteries for energy storage add additional, variable amounts of embodied energy.

Energy payback time
The crucial question is how long it takes for a solar panel to generate the same amount of energy that was used to produce it, which is known as the energy payback time (EPBT).
For silicon solar cells, the energy payback time typically ranges from 1.65 to 4.12 years.
Newer technologies, such as organic and perovskite solar cells, can have even shorter payback times, potentially less than half a year.

Utility-scale PV systems in the United States have estimated energy payback times ranging from 0.5 to 1.2 years, according to a recent NREL study.

After the energy payback period, the solar panels generate clean energy for the remainder of their lifespan, which is typically 25 to 30 years or more.

In essence, while manufacturing solar panels requires energy, particularly in the silicon processing stage, the energy they generate over their lifespan significantly outweighs the initial energy investment.

AI Overview
Solar Panel Sizes, Dimensions & Wattage for Businesses | GSE

The average mass of a solar panel typically falls within the range of 40 to 60 pounds (18 to 27 kilograms).
Residential solar panels, often containing 60 cells, generally weigh around 40 pounds.
Larger, commercial-sized panels with 72 cells can weigh 50 pounds or more. The exact weight can vary based on factors like size, materials used, and specific panel design.

Here's a more detailed breakdown:
Residential solar panels (60-cell): Approximately 40 pounds (18 kg).
Commercial solar panels (72-cell): Around 50 pounds (23 kg) or more.

Factors influencing weight:
Size: Larger panels generally weigh more.
Materials: Different materials like the frame, glass, and back sheet can affect the overall weight.

Panel type: Monocrystalline panels may be heavier than polycrystalline or thin-film panels.
Additional weight considerations: The weight estimate doesn't include the weight of mounting hardware or racking systems, which can add to the total system weight.

AI Overview
Northrop Grumman's Cygnus Spacecraft: A Decade of Delivering ...
Cygnus spacecrafts have evolved from a standard configuration with ~2,750 kg capacity and 18 m³ volume to an enhanced version with ~3,750 kg and 27 m³, and now an upgraded "Mission B" with a capacity of 5,000 kg and a 36 m³ pressurized volume. These figures vary depending on the specific mission and version, with the enhanced and Mission B versions featuring longer pressurized modules to accommodate the increased cargo.

Evolution of Cygnus Cargo Specifications
Standard Version:
Payload Capacity: 2,750 kg
Pressurized Volume: 18 m³
Enhanced Configuration:
Payload Capacity: 3,750 kg
Pressurized Volume: 27 m³
Mission B (Upgraded Version):
Payload Capacity: 5,000 kg
Pressurized Volume: 36 m³
Key Details
What they carry:
Cygnus spacecrafts deliver food, water, spare parts, repairs, and scientific investigations to the International Space Station (ISS).
Design Evolution:
The spacecraft's design has been updated by Northrop Grumman to meet evolving customer needs, including increased cargo capacity.
Late Load Capability:
Enhanced and Mission B Cygnus versions allow for "late load" capability, meaning special equipment and science needs can be added shortly before launch.
Return Cargo:
Cygnus spacecrafts dispose of ISS waste by burning up upon atmospheric reentry.

he Northrop Grumman Cygnus cargo spacecraft typically carries around 800 kg of hypergolic propellant.

The fuel is stored in the Service Module and powers the main engine and smaller thrusters for navigation and orbit adjustments. Cygnus propellant specifications Propellant type:

The Cygnus uses hypergolic propellant, a mixture of hydrazine (\(N_{2}H_{4}\)) fuel and nitrogen tetroxide (\(N_{2}O_{4}\)) or MON-3 oxidizer. Hypergolic propellants ignite spontaneously upon contact, making them easy to use in space without a separate ignition system.

Propulsion system: The propulsion system features a main engine for major orbital adjustments and 32 smaller thrusters for attitude control.

Mission tasks: The fuel is used to perform various maneuvers throughout the mission, including:Phasing and rendezvous with the International Space Station (ISS).Occasional reboosts of the ISS to counteract atmospheric drag.A final burn to de-orbit the spacecraft for a destructive reentry into Earth's atmosphere. 

Note on variants While the Standard Cygnus spacecraft had a similar fuel mass, information specifically points to the Enhanced version carrying roughly 800 kg of propellant. Later iterations, like those launched by the Falcon 9, have a larger overall launch mass and carry more propellant to support a heavier payload

We can estimate the weight of this vehicle by considering the Apollo Lunar Module (LM). As noted, our LEV will have a similar profile:
lightweight; intended to fly only in space and around the lunar surface, meaning it would not need a thick shell and heavy heat shield to protect it during re-entry into Earth’s atmosphere; and capable of carrying some
cargo, a crew of two, and life support for up to a few days.

The Apollo LM’s dry mass (its weight with crew and cargo but without fuel) was 5.2 metric tons. However, the LM carried two rocket engines and propulsion systems—one for descending from lunar orbit onto the Moon’s surface, another for ascending back to lunar orbit. The ascent portion (the “ascent stage”), which contained the crew cabin, crew, and life support equipment, is most similar to our purposes here. Its dry mass was 2.3 tons. If we used this figure for estimating the weight of our LEV, we would also need to add the weight of the landing legs, and make various other adjustments. But given a half-century of improvements in materials and avionics science and engineering, a LEV could surely make significant improvements in the weight. We will therefore estimate 2 tons for the LEV’s dry mass, again, including crew and cargo.

In Figure 2 (see page 37), we can see the mass requirements of our 2-ton LEV. In addition to the dry mass, about 6 tons of propellant are required for each mission that uses 6.1 km/s of delta-V. So the total mass (known as the “wet mass”), including ship, cargo, and propellant, is about 8 tons. Also, the required weight of the tanks and engines—which take up part of the 2-ton dry mass—still leaves 1.3 tons for the crew, crew cabin, and other cargo.

Table 1. Cargo Lander Mission (single stage)
Launcher Staging Orbit Propulsion Tank Length Payload Delivered
Falcon H LEO LOx/CH4 3.2 m 8.3 tons
Falcon H GTO LOx/CH4 1.05 8.3
Falcon H LEO LOx/H2 7.9 10.4
Falcon H GTO LOx/H2 2.5 9.6
New Glenn LEO LOx/CH4 1.12 6.0
New Glenn LEO ` LOx/H2 2.85 7.5
Vulcan LEO LOx/CH4 1.54 4.0
Vulcan LEO LOx/H2 3.8 5.0
SLS LEO LOx/CH4 1.9 12.0
SLS LEO LOx/H2 4.45 15.0
BFR LEO LOx/CH4 3.2 19.9
BFR TLI LOX/CH4 2.5 60.0

First stage Mass ('dry' without propellant) 22,200 kg (48,900 lb) Second stage Mass (without propellant) 4,000 kg (8,800 lb) First stage Mass ('wet' with propellant) 433,100 kg (954,800 lb) Second stage Mass (with propellant) 111,500 kg (245,800 lb)Dec 3, 2020

SECOND STAGE   
Height    13.8 m / 45.3 ft
Diameter    3.7 m / 12.1 ft
Empty Mass    3,900 kg / 8,598 lb