Scratch Build – In Progress Logic - V 01 - The cook book!

Hi I am almost back, Real Life called.

Oh well I started by checking on new hardware and ran into new ideas implemented on hardware at Computex 2023 and 2024.

IF you have seen design ideas like these below give me a like.

Havn HS420 - Youtube - Der Bauer

Montech. - Computex 2024 - Gear seekers - pull tap

Kins95 Radiator placement.

Antec Performance 1M - https://www.anandtech.com/show/21443/antecs-miniitx-chassis-can-house-a-geforce-rtx-4090

Noctua quiet air flow mod - Fans air layering with focus on sound volume - there is room for improvement.

Fractals ERA2 & MOOD - A unibody its beautiful.

Pogo pins in PC cases showed in 2023 there was 1 more but I cant remember them right now.

Inspiration and ideas are contagious its just small things seen on Computex 2023 and 2024 - Well I approve good ideas should multiply.

I will add some new ideas I noticed on Computex as seen on YouTube later as Picasso said Good Artists Copy; Great Artists Steal.

 

"Longevity? A back-connected motherboard is a wrench in the gearbox!

I thought my plan for ensuring the longevity of this case was nearly bulletproof, but it's not. The expectation was that by allowing for the insertion of a new motherboard tray, any challenge posed by a new motherboard design could be addressed, thus securing longevity through modularity. However, it's not that simple."

"The current design faces a significant issue that is nearly a dilemma. Typically, a problem offers multiple solutions, whereas a dilemma restricts us to two options, both of which are undesirable."

"Looking at the challenge shown in the pictures below, the problems with inserting a new type of motherboard are evident. The new cutout on the tray for a back-connected motherboard interferes with the airflow (recirculation and inter change between the 2 chambers), power supply unit (PSU) and access to wiring and cable routing, with cable routing being a significant challenge on its own."

looking at the options:

1-1: Shifting the PSU
2-0: changing the dimensions of the case to:
2-1: get room for a secondary motherboard tray,
2-2: a larger Water Cooling chamber for the PSU
2-3: or a wider Spine for cable routing inserted, see 2-0.

The pictures only show part of 1-1 and 2-2 essentially the problem no solutions.

Here's a refined and coherent presentation of my current status and the issues i'm facing with the motherboard tray design:

Present Status: Rear View



Revised Inspiration for Cutout Location



Cutouts Seen from the Water Cooling Rig Side (Backside of Motherboard Tray)



The blocking of two likely power insert areas is obvious. There will also be problems with the cables from the PSU interfering with access to the motherboard.

Attempt to Spin the PSU to the Side ![PSU Spun to the Side



First, I tried to spin the PSU to the side. This didn't fix the problem with access and introduced the next challenge with airflow.

Creating access to cables, and te next problem is obvious.





Creating access to cables introduces the next obvious problem. Interference with Fans on the Water Cooling Rig and Air flow

I have several options in mind, but all are quite challenging. I'll be back with more solutions.

After looking at the 2024 edition of Computex and companies like Fractal and others looking as if they have been inspired somewhere close to home, I decided to step up my game and work on the MK 02 that is more scifi inspired. Meaning work on this design is atm a secondary priority that will be updated when I relax. I learned a lot from this exercise and hopefully will learn more

 

This is part of the SAS4MC aka the Monitoring and control system.

To make this work I use the sensors previously described. The temperature sensors before and after the radiator and the flow meter that gives us the water speed an amount kg/s. This is a data based approach from 1 loop. This is a way around the lacking data on radiators ability to disperse energy during specific conditions.

On a personal note the companies that make quality radiators should test each radiator individually and use the data for marketing.

This approach can be used for both energy added and dissipated. Its not a beautiful scientific method / approach but it works and meets my wish of showing/knowing what is going on in the loops. It combines the effect of fan and liquid speed through the system on the temperature in system. the sensors precision are part of why I use this approach.

To calculate the heat transfer in a PC radiator loop, we can use the basic principles of thermodynamics. The heat transfer rate (Q) in this case is related to the change in temperature of the water as it flows through the radiator, the flow rate of the water, and the specific heat capacity of water.

Assumptions:

The system is steady-state.
The water is in-compressible.
No phase change occurs in the water (it remains liquid).

Equation:

The heat transfer rate Q can be calculated using the following equation:

Q=m˙⋅cp⋅ΔT

Where:

Q is the heat transfer rate (in watts, W).
m˙ is the mass flow rate of the water (in kg/s).
cp is the specific heat capacity of water (approximately 4184 J/(kg·°C) at room temperature and sea level). 4184 J/(kg·°C)
ΔT is the temperature difference between the inlet and outlet of the radiator (in °C or K).

Steps to Calculate Heat Transfer:

1. Determine the Mass Flow Rate (m˙\dot{m}m˙): This is the mass of water flowing through the radiator per second. It's usually given in kg/s. 0,1 kg/s

2. Measure the Inlet and Outlet Temperatures:

1. Tin is the temperature of the water before it enters the radiator. Tin 40 ∘C
Tout is the temperature of the water after it exits the radiator. Tout 35 ∘C

3. Calculate the Temperature Difference (ΔT):

ΔT=Tin−Tout ΔT 5 ∘C

This gives the temperature drop of the water as it passes through the radiator.

4. Plug the Values into the Heat Transfer Equation: Once you have}m˙, cp, and ΔT, you can calculate Q.

Example Calculation:

Let's assume:

The mass flow rate m˙= 0.1kg/s.
The inlet temperature Tin= 40∘
The outlet temperature Tout=35∘C
The specific heat capacity of water cp=4184 J/(kg*°C)

Then:


2. Using the heat transfer equation: Q= 0.1 kg/s * 4184  J/(kg·°C)* 5°C 2092 W

So, the heat transfer rate is 2092 W

This means that the radiator is dissipating approximately 2092 watts of heat from the water in this example.

This approach can be used for both energy added and dissipated. Its not the approach I would have preferred but it works and meets my wish of showing/knowing what is going on in the loops. It combines the effect of fan and liquid speed through the system on the temperature in system. It also makes the Water Cooling Rig able to stand alone and be used to compare with parts of other systems.

i will add a few more articles to show the way I approach the data in the SAS4MC. If you got questions post a reply.

The specific heat capacity (cp) of a liquid refers to the amount of heat energy required to raise the temperature of a unit mass of the liquid by one degree Celsius (or one Kelvin). The specific heat capacity depends on the type of liquid used in the system. Will be used in the Cooling Loops.

Water Water is commonly used in cooling systems due to its high specific heat capacity, which allows it to absorb a significant amount of heat before its temperature rises substantially. 4184 J/kg\cdotpK

Water-Glycol Mixture a 50/50 water-ethylene glycol mixture The specific heat capacity of the mixture is lower than pure water, which means it will absorb less heat per kilogram than water alone. However, glycol mixtures are preferred in certain situations for their lower freezing points and corrosion protection. 3300 J/kg\cdotpK

Ethylene Glycol (pure) Pure ethylene glycol has a lower specific heat capacity compared to water, meaning it heats up more quickly for the same amount of absorbed heat. 2300 J/kg\cdotpK

Propylene Glycol Propylene glycol is similar to ethylene glycol but is less toxic and often used in applications where safety is a concern. 2400 J/kg\cdotpK

 

Next I will add some info on Initial conditions: Ambient since the entire System relies on environmental conditions, making it a dynamic system dependent on external factors such as air temperature, moisture, and pressure. This will be a detailed breakdown of my thoughts, math and physics as far as I understand it.

To be posted soon, i need to make it understandable first.

 

! A short comment on why I think that Air density is important. My initial idea was to be able to control the environment inside the case knowing how the ambient conditions are, how the parts of the cooling system work and are controlled (Fans, Radiators and Pumps). After considering the many variables I have come to the conclusion that it needs to be calculated and manipulated on the Go, the system reacts on input, so not predictive but reactive. This is my thought on this for now.
If I ever get to a point where I think Machine Learning "AI" is the solution I might add this but for now this is it.

The Stand Alone System for Monitoring and Control (SAS4MC) involves the Hardware Chamber and the Water Cooling System (WCR) that relies on environmental conditions, making it a dynamic system dependent on external factors such as air temperature, moisture, and pressure. This is a detailed breakdown of my thoughts, math and physics as far as I understand it.

The system incorporates sensors BME280 in all chambers including the 3 tech chambers on the latest case iteration.

BME280 is using the I2C communication protocol and delivers temperature, humidity, pressure, and estimates altitude. These 5 sensors are part of the base delivering data for calibration, ΔT, ΔRH%, ΔPa. The difference between the Ambient and single parts of the case is the key to get things on point.

Initial conditions, Ambient:

Altitude (from sensors): Above sea level manually or from sensor = 0 meter
Pressure = 101,325 Pa (standard atmospheric pressure at sea level) Manually added = 101,325 Pa
Temperature (From sensors): = 25 °C
Relative Humidity (RH%) (From sensors): (RH%) = 60%

Basics - Constants:

0 °C in Kelvin = 273,15 K
liters per cubic meter = 1000 liters (l) or kg at sea level.

Specific gas constant for dry air (Rd): (Rd): = 287,05 J/(kg·K)
Specific gas constant for water vapor (Rv): = 461,5 J/(kg·K)

H2O Temperature from Sensors in Reservoir, L1 and L2. equal to air temperature after longer inactivity 25 °C - More on this later!

Air Density (Ideal Gas Law): ρ=(P/(R⋅T))
P air density (kg/m³)
P air pressure (Pa)
R specific gas constant for air (J/kg*K)
T absolute temperature (K)

This Table shows common maximum saturation vapor pressure values at certain temperatures (in °C) at sea level, these values are used to calculate the needed values:


Temperature (°C) Saturation Vapor Pressure (KPa)

10°C 1,227 KPa

20°C 2,338 KPa

25°C 3,169 KPa

30°C 4,246 KPa

35°C 5,622 KPa

40°C 7,375 KPa

50°C 12,336 KPa

The air density directly affects the cooling performance of a radiator. When air density increases, the radiator becomes more efficient at transferring heat from the liquid to the air, enhancing cooling performance. Conversely, reduced air density decreases the heat transfer efficiency, reducing the cooling performance, even if the liquid flow rate remains constant.

When working with weather data think of a layer cake. The situation at the bottom defines the weather above, lets just say its all connected.

Saturation vapor pressure at 25°C (from a reference table): from Table manually inserted need to program 3,169 Kpa

Step 1: Convert Temperature to Kelvin T(K)=25+273.15=298.15K

Step 2: Calculate the Partial Pressure of Water Vapor(Pv)
Pv=Relative Humidity×Saturation Vapor Pressur Pv=0.60×3,169Pa=1,9014 KPa

Step 3: Calculate the Partial Pressure of Dry Air (Pd)

The total pressure is the sum of the partial pressures of dry air and water vapor. Therefore, the partial pressure of dry air is: Pd=P−Pv Pd=101,325 Pa−1,901.4 Pa=99,423.6 Pa 99,4236 kPa

Step 4: Calculate Air Density (ρ)

Now, using the formula that accounts for both dry air and water vapor: ρ=((Pd/(Rd⋅T))+(Pv/(Rv⋅T)))

1. Dry air contribution: ρd=(99,423.67 Pa/(287.05 J/(kg\cdotpK)×298.15 K))*1000≈ 1,16170837 kg/m³

2. Water vapor contribution: ρv=1,901.4 Pa/(461.5 J/(kg\cdotpK)×298.15 K)*1000≈ 0,01382 kg/m³

Finally, add the two contributions: ρ= ρd+ρv = 1.163 kg/m3+0.0138 kg/m3 ≈ 1,17553 kg/m3

*This part was for fun and to see if I could. The Air Density impacts you fans and radiators performance in both the hardware and the Water cooling part. To make this data usable and easy accessible I am still working on a part that chooses / combines the best cooling option when looking at Fan - Water Pump / Air and water speed. The article above of watt dissipated is a good start to reach the goal.

**Depending on the parts CPU water cooler, GPU water cooler or Radiator. The lack of data on heat transfer at different water speeds/amounts kg/s makes it a jungle of the lucky and why do I think that? Because the weakest link sets the pace and performance. Its not by default a bad thing to be slow if you deliver or dissipate more energy from the liquid in the same time frame. The key is the ΔT (Difference in temperature) in the loop vs the ambient temperature. More time in a radiator for a given amount of liquid might make more energy dissipate, the only disadvantage might be a 5 degree C rise in CPU or GPU temperature but that might be within the norm, this part depends on your hardware specs and attitude towards noise that could see a significant reduction.

-- This next part was to satisfy my curiosity and to see what impact a change in altitude might have on a system (air density 2).

Altitude (Air Pressure) impact on the above.
Step 1: Calculate Temperature at Altitude

The temperature at a given altitude can be calculated using the standard lapse rate for the troposphere (up to 11,000 meters), which is approximately: Lapse rate= −6.5∘C/km =−0.0065K/m
Starting with the sea level temperature of See above°C ditto here K), the temperature at altitude h (in meters) can be calculated as: T(h)=T0−L⋅h (Here comes the Layer Cake)

Where:

T0= (sea level standard temperature) °K 273,15 °K
L=0.0065 K/m (temperature lapse rate) 0,0065 K/m
h is the 3000 meter

Temperature at altitude -At -20°C, the saturation vapor pressure over ice (since water is typically frozen at this temperature) is approximately: T(h)=T0−L⋅h 253,65 °K

Step 2: Calculate Pressure at Altitude

The pressure at a given altitude can be calculated using the barometric formula: P(h)=P0⋅(T0T(h))^(R⋅Lg⋅M)

Where:
P0=101325Pa (sea level standard pressure) P0 = 101325 Pa
g=9.80665m/s2 (acceleration due to gravity) g = 9,8066500 m/s2
Molar mass of Earth's air M= 0,0289644 kg/mol
Universal gas constant R = 8,3144598 J/(mol\c*K)
n = (R⋅Lg⋅M) 5,2557877

Pressure at a given altitude of 3000 meters P(h)=P0⋅(T0T(h))^(R⋅Lg⋅M) 68,653 Pa

*** We are now at 3000 meter

now re-running the 1st part

Air density considering humidity = 1,17553 kg/m³ at 0 meter vs at 3000 meter 0,95914 kg/m³

The efficiency of the fans/radiator because of the lower Air density is now only 81,5924 % or lets put it like this your fans need to run and be 18,4 % more effective same goes for your radiator. This is where the measuring of Watt comes in and combines both in a more efficient form.

Next part should be about Newton’s Law of Cooling used on the Reservoir. This is slightly more complex, why you might ask! Because of the impact slow cooling liquid has on the entire system when doing a Statistical Analysis or Calibration of the sensors thus when booting. Knowing what is in the system and how and why it works as it does gives me the option/ability to force a result to the systems advantage.

At present I am looking at Linear Regression: Using Excel's LINEST function to fit a linear trend to the temperature data. Exponential Trendline: On a graph of the temperature data, we can add an exponential trendline and display the equation on the chart. This can help model the cooling behavior. Forecasting: Use Excel’s FORECAST.ETS function to predict future values based on your temperature data series. This is a might add not a necessity.

I like to keep my options open. So next Newton’s Law of Cooling .

 

I did briefly comment on the Reservoirs role above, just posting the calculations will add comments/explanations later.

Newton’s Law of Cooling: To model how the reservoir loses heat to its surroundings (typically through radiation and convection to the ambient air):

dT/dt=−k(T−Tambient)

Where:

T is the temperature of the reservoir.

Tambient is the ambient temperature.

k is a constant that depends on the heat transfer characteristics of the system.

Suppose a reservoir initially at 40°C is cooling down to ambient air at 25°C in 21 h 20 minutes Reservoir running temperature 40 °C

Ambient air at 25 °C

Time to dissipate heat from 40 to 25 °C t 1280 min

Cooling constant -k: -0,0030706 min⁻¹
Contestant e is equal to: 2,71828183

After 10 minutes, the temperature drops to approximately 39,546 °C.

Ambient Temperature = 25 °C

Initial Temperature Difference = 15 °C

Exponent Value (-k * t) = -0,9825791


=EXP(-0,9825791) = 0,37434437


= 25 °C + ( 15 °C * 0,37434437 ) (Calculates final temperature) 30,6151655

After (t) minutes, the temperature would drop to approximately t °C.
10 - 39,546
20 - 39,106
40 - 38,266
80 - 36,733
160 - 34,178
320 - 30,615 - 5 h 20 min
640 - 27,102 - 10 h 40 min
1280 - 25,295 - 21 h 20 min

Assumptions and Parameters for calculations with Reservoir of acrylic:

Thermal conductivity of Acrylic ( kacrylick): 0.185 W/m*K (average).
Link https://en.wikipedia.org/wiki/List_of_thermal_conductivities

Thickness of the Acrylic Wall ( d): 0,01 meters (10 mm).
Surface Area of the Reservoir (A): 0.0864 square meters. 120*120*6 mm = 0,0864 m2
Specific Heat Capacity of Water (cp): 4184 J/kg*K.
Mass of Water (m): 1 kg (1 liter). (m) 1,0368 kg (liter at sea level).
Ambient Temperature: 25°C.
Initial Reservoir Temperature: 40°C.
Temperature Difference: 15°C - (40°C - 25°C).

1. Heat Transfer Rate Q:

The heat transfer rate is given by:

Q= (kacrylic*A*(Treservoir−Tambient))/d ((0,185*0,0864*15)/0,01) 23,976 Watt

2. Cooling Constant k:

Now, using the formula for the cooling constant k:

k= Q/(m⋅cp⋅(Treservoir−Tambient) k= (23,976/(1,0368*4184*15)) = 0,00036847 min−1

Final Result:

The new cooling constant k with a wall thickness of 10 mm (0.01 meters) is approximately 0,00036847 min⁻¹.

Update: Let me start by stating that the simple solution to the extreme long time for the water in the Reservoir to cool down to ambient temperature (20+ hours) can be mitigated by the cooling loop, that will keep running for a bit of time after the system is shut down. I could calculate this but the I have a temperature sensor in the Reservoir to make this easy

Why then do this calculation? My reason is the rest of system and sensors that will get incorrect data for an extended period, now I know what to look for and for how long.

 

The Next calculations should be for the radiator and Fans but the calculations are to imprecise. I will have to make long term observations and calculate what the real values for the fan and radiator efficiency is, this is the part that has a lot of variables and is to easy to ignore. Why easy to ignore because there are other ways (see the 1st part of the calculations) to get the options into the SAS4MC.

This data below is only some of the data displayed on 3 OLEDs of 7-15 OLED.

Data Displayed:

Reservoir:

1-1. MPR12 – Water Level: 80 %
1-2. H2O Temperature: 22.21 *C
1-3. H2O Quality MPR12 or TDS Meter w Sensor: 60 ppm – Data over time
1-4. LED/RGB RGB Color

Loop 1:

1. Reservoir *
2-2. Pump PWM: 1200 rpm - 2.1 lpm - 20% Load
2-3. CPU: 49 C - 59 % Load
2-4. H2O Temperature: 24 C diff 1.8 C
-

3-5. Radiator fan: 1400 rpm
3-6. H2O Temperature: 22 C Diff 2 C
3-7. Flow Meter: 2.1 lpm - INS-FM14 Coolant Flow Meter
1. Reservoir *

This data changes with a constant interval or if something NOT NOMINAL happens.

This part i logically showing the placement of sensors on the Reservoir and in Loop 1. The data from the sensors is used for the calculations to be sure that all is working as expected. If its not Nominal OLEDs, RGB and sound is used to get attention. This is before or during a hard shut down. This is a minor part of the SAS4MC there are more sensors in the case then on the Water Cooling rig (Reservoir and cooling Loops). A lot of work has gone into making sure that the data shown on the OLED is intuitive, still the work is not finished, more options keep popping-up.