🎨 Add blkt pipes output

documentation/source/development/standards.md

@@ -259,6 +259,14 @@ This should be used for units of $\text{kg} \cdot \text{m}^{-2}\text{s}^{-1}$
 
 ---------------------
 
+##### Specific Heat Capacities
+
+- Specific heat capacities for materials $[\text{J/kg/K}]$ should start with the `heatcap_` prefix
+- Specific heat capacities at constant volume should start with the `heatcap_vol_` pefix
+- Specific heat capacities at constant pressure should start with the `heatcap_pres_` pefix
+
+---------------------
+
 ##### Pressures
 
 - Pressures should start with the `pres_` prefix

documentation/source/eng-models/blanket_overview.md

@@ -5,100 +5,3 @@
 
 --------------------
 
-### Coolant mechanical pumping power | `coolant_pumping_power()`
-
-To calculate the coolant pumping power we use the change in enthalpies of the coolant as it goes through the pump. 
-**We assume the pump is isentropic so the entropy change of the coolant is 0**. 
-
-The mechanical pumping power is defined as:
-
-$$
-P = \frac{\frac{\dot{m} \times \left(H_{\text{out}}-H_{\text{in}}\right)}{\eta}}{\left(1-fp\right)}
-$$
-
-where $\dot{m}$ is the coolant mass flow rate, $H$ is the coolant enthalpy, $\eta$ is the isentropic efficiency of the pump and $\gamma$ is the adiabatic index of the coolant.
-
-$$
-fp = \frac{T_{\text{pump,out}}\left(\frac{P_{\text{pump,out}}}{P_{\text{pump,in}}}\right)^{-\frac{\gamma -1}{\gamma}}}{\eta \left(T_{\text{pump,in}}-T_{\text{pump,out}}\right)}
-$$
-
-------------------
-
-### Coolant pressure drop | `coolant_friction_pressure_drop()`
-
-The pressure drop in the coolant is given by the [Darcy-Weisbach Equation](https://en.wikipedia.org/wiki/Darcy%E2%80%93Weisbach_equation)
-
-For a cylindrical pipe of uniform diameter the pressure loss due to viscous effects can be characterized by:
-
-$$
-\Delta P = L\left[f_{\text{D}}\frac{\rho}{2}\frac{\langle v \rangle^2}{D_{\text{H}}}\right]
-$$
-
-where $L$ is the pipe length, $f_{\text{D}}$ is the [Darcy friction factor](https://en.wikipedia.org/wiki/Darcy_friction_factor_formulae), $\rho$ is the coolant density, $\langle v \rangle$ is the mean flow coolant velocity and $D_{\text{H}}$ is the hydraulic diameter or the pipe diameter in this case.
-
-To find the Darcy friction factor we need to know the Reynolds number given by:
-
-$$
-\text{Re} = \frac{\rho v L}{\mu}
-$$
-
-here $L$ is the characteristic length which we set to be the pipe diameter and $\mu$ is the coolant dynamic viscosity.
-
-Using the Reynolds number we calculate the Darcy friction factor using the Haaland approximation calculated by [`darcy_friction_haaland()`](../eng-models/generic_methods/pumping.md#pumping-coolant-friction--darcy_friction_haaland).
-
-For the radius of the pipe bend we assume it to be 3 times the radius of the coolant channel.
-
-The elbow coefficients for the 90 and 180 degree bends $\left(f_{\text{90,elbow}}, f_{\text{180,elbow}}\right)$ are calculated via [`elbow_coeff()`](#pipe-bend-elbow-coefficient--elbow_coeff).
-
-The pressure drop for the straights along the entire pipe length is the same as above:
-
-$$
-\Delta P = L\left[f_{\text{D}}\frac{\rho}{2}\frac{\langle v \rangle^2}{D_{\text{H}}}\right]
-$$
-
-where we define $\frac{f_{\text{D}}L}{D_{\text{H}}}$ as our straight section coefficient.
-
-The pressure drop for the 90 and 180 degree bends are:
-
-$$
-\Delta P = N_{\text{90}} \left[f_{\text{90,elbow}} \frac{\rho \langle v \rangle^2}{2}\right]
-$$
-
-$$
-\Delta P = N_{\text{180}} \left[f_{\text{180,elbow}} \frac{\rho \langle v \rangle^2}{2}\right]
-$$
-
-where $N_{\text{90}}$ and $N_{\text{180}}$ are the number of 90 and 180 degree bends in the system.
-
-The total returned pressure drop is simply:
-
-$$
-\Delta P = L\left[f_{\text{D}}\frac{\rho}{2}\frac{\langle v \rangle^2}{D_{\text{H}}}\right] + N_{\text{90}} \left[f_{\text{90,elbow}} \frac{\rho \langle v \rangle^2}{2}\right] + N_{\text{180}} \left[f_{\text{180,elbow}} \frac{\rho \langle v \rangle^2}{2}\right]
-$$
-
--------------------
-
-### Pipe bend elbow coefficient | `elbow_coeff()`
-
-This function calculates the elbow bend coefficients for pressure drop calculations.
-
-$$
-a = 1.0 \quad \text{if} \ \theta = 90^{\circ} \\
-a = 0.9 \times \sin{\left(\frac{\theta \pi}{180^{\circ}}\right)} \quad \text{if} \ \theta < 70^{\circ} \\
-a = 0.7 + 0.35 \times \sin{\left(\frac{\theta}{90^{\circ}} \times \frac{\pi}{180^{\circ}}\right)} \quad \text{if} \ \theta > 90^{\circ} \\
-$$
-
-where $\theta$ is the angle of the pipe bend.
-
-$$
-b = \frac{0.21}{\sqrt{\frac{R_{\text{elbow}}}{D_{\text{pipe}}}}}\quad \text{if} \ \frac{R_{\text{elbow}}}{D_{\text{pipe}}} \ge 1 \\
-b = \frac{0.21}{\left(\frac{R_{\text{elbow}}}{D_{\text{pipe}}}\right)^{2.5}}\quad \text{if} \ \frac{R_{\text{elbow}}}{D_{\text{pipe}}} \le 1 \\
-\text{else} \quad b =0.21
-$$
-
-The elbow coefficient is given by:
-
-$$
-ab + \left( f_{\text{D}} \times \frac{R_{\text{elbow}}}{D_{\text{pipe}}}\right) \times \theta \times \left(\frac{\pi}{180^{\circ}}\right)
-$$
-

documentation/source/eng-models/generic_methods/pumping.md

@@ -1,5 +1,80 @@
 # Pumping Methods
 
+
+## Coolant mechanical pumping power | `coolant_pumping_power()`
+
+To calculate the coolant pumping power we use the change in enthalpies of the coolant as it goes through the pump. 
+**We assume the pump is isentropic so the entropy change of the coolant is 0**. 
+
+The mechanical pumping power is defined as:
+
+$$
+P = \frac{\frac{\dot{m} \times \left(H_{\text{out}}-H_{\text{in}}\right)}{\eta}}{\left(1-fp\right)}
+$$
+
+where $\dot{m}$ is the coolant mass flow rate, $H$ is the coolant enthalpy, $\eta$ is the isentropic efficiency of the pump and $\gamma$ is the adiabatic index of the coolant.
+
+$$
+fp = \frac{T_{\text{pump,out}}\left(\frac{P_{\text{pump,out}}}{P_{\text{pump,in}}}\right)^{-\frac{\gamma -1}{\gamma}}}{\eta \left(T_{\text{pump,in}}-T_{\text{pump,out}}\right)}
+$$
+
+------------------
+
+## Coolant pressure drop | `coolant_friction_pressure_drop()`
+
+The pressure drop in the coolant is given by the [Darcy-Weisbach Equation](https://en.wikipedia.org/wiki/Darcy%E2%80%93Weisbach_equation)
+
+For a cylindrical pipe of uniform diameter the pressure loss due to viscous effects can be characterized by:
+
+$$
+\Delta P = L\left[f_{\text{D}}\frac{\rho}{2}\frac{\langle v \rangle^2}{D_{\text{H}}}\right]
+$$
+
+where $L$ is the pipe length, $f_{\text{D}}$ is the [Darcy friction factor](https://en.wikipedia.org/wiki/Darcy_friction_factor_formulae), $\rho$ is the coolant density, $\langle v \rangle$ is the mean flow coolant velocity and $D_{\text{H}}$ is the hydraulic diameter or the pipe diameter in this case.
+
+To find the Darcy friction factor we need to know the Reynolds number given by:
+
+$$
+\text{Re} = \frac{\rho v L}{\mu}
+$$
+
+here $L$ is the characteristic length which we set to be the pipe diameter and $\mu$ is the coolant dynamic viscosity.
+
+Using the Reynolds number we calculate the Darcy friction factor using the Haaland approximation calculated by [`darcy_friction_haaland()`](../eng-models/generic_methods/pumping.md#pumping-coolant-friction--darcy_friction_haaland).
+
+For the radius of the pipe bend we assume it to be 3 times the radius of the coolant channel.
+
+The elbow coefficients for the 90 and 180 degree bends $\left(f_{\text{90,elbow}}, f_{\text{180,elbow}}\right)$ are calculated via [`elbow_coeff()`](#pipe-bend-elbow-coefficient--elbow_coeff).
+
+The pressure drop for the straights along the entire pipe length is the same as above:
+
+$$
+\Delta P = L\left[f_{\text{D}}\frac{\rho}{2}\frac{\langle v \rangle^2}{D_{\text{H}}}\right]
+$$
+
+where we define $\frac{f_{\text{D}}L}{D_{\text{H}}}$ as our straight section coefficient.
+
+The pressure drop for the 90 and 180 degree bends are:
+
+$$
+\Delta P = N_{\text{90}} \left[f_{\text{90,elbow}} \frac{\rho \langle v \rangle^2}{2}\right]
+$$
+
+$$
+\Delta P = N_{\text{180}} \left[f_{\text{180,elbow}} \frac{\rho \langle v \rangle^2}{2}\right]
+$$
+
+where $N_{\text{90}}$ and $N_{\text{180}}$ are the number of 90 and 180 degree bends in the system.
+
+The total returned pressure drop is simply:
+
+$$
+\Delta P = L\left[f_{\text{D}}\frac{\rho}{2}\frac{\langle v \rangle^2}{D_{\text{H}}}\right] + N_{\text{90}} \left[f_{\text{90,elbow}} \frac{\rho \langle v \rangle^2}{2}\right] + N_{\text{180}} \left[f_{\text{180,elbow}} \frac{\rho \langle v \rangle^2}{2}\right]
+$$
+
+-------------------
+
+
 ## Pumping coolant friction | `darcy_friction_haaland()`
 
  The pressure drop is based on the Darcy friction factor, using the [Haaland equation](https://en.wikipedia.org/wiki/Darcy_friction_factor_formulae#Haaland_equation), an approximation to the implicit Colebrook–White equation. 
@@ -61,4 +136,58 @@ where $\rho$ is the coolant density and $\mu$ is the coolant viscosity.
 
     $$
     h = \frac{\mathrm{Nu_D}k}{2r_{\text{channel}}}
-    $$
+    $$
+
+-------------------------
+
+## Pipe bend elbow coefficient | `elbow_coeff()`
+
+This function calculates the elbow bend coefficients for pressure drop calculations.
+
+$$
+a = 1.0 \quad \text{if} \ \theta = 90^{\circ} \\
+a = 0.9 \times \sin{\left(\frac{\theta \pi}{180^{\circ}}\right)} \quad \text{if} \ \theta < 70^{\circ} \\
+a = 0.7 + 0.35 \times \sin{\left(\frac{\theta}{90^{\circ}} \times \frac{\pi}{180^{\circ}}\right)} \quad \text{if} \ \theta > 90^{\circ} \\
+$$
+
+where $\theta$ is the angle of the pipe bend.
+
+$$
+b = \frac{0.21}{\sqrt{\frac{R_{\text{elbow}}}{D_{\text{pipe}}}}}\quad \text{if} \ \frac{R_{\text{elbow}}}{D_{\text{pipe}}} \ge 1 \\
+b = \frac{0.21}{\left(\frac{R_{\text{elbow}}}{D_{\text{pipe}}}\right)^{2.5}}\quad \text{if} \ \frac{R_{\text{elbow}}}{D_{\text{pipe}}} \le 1 \\
+\text{else} \quad b =0.21
+$$
+
+The elbow coefficient is given by:
+
+$$
+ab + \left( f_{\text{D}} \times \frac{R_{\text{elbow}}}{D_{\text{pipe}}}\right) \times \theta \times \left(\frac{\pi}{180^{\circ}}\right)
+$$
+
+--------------
+
+## Required mass flow rate | `calculate_required_mass_flow_rate()`
+
+The required mass flow rate of a coolant is given simply by the fundamental heat transfer equation:
+
+$$
+\dot{m} = \frac{P}{c_{\text{p}}(T)\times \Delta T}
+$$
+
+where $\dot{m}$ is the required mass flow rate in, $P$ is the heating power to be removed, $c_{\text{p}}$ is the coolant specific heat capacity for constant pressure and $\Delta T$ is the temperature change in the coolant.
+
+!!! note "Variation specific heat capacity"
+
+    The heat capacity itself is a function of temperature. Therefore it is common to use the heat capacity value at the simple average between the initial and final temperature.
+    This however assumes a linear relationship. Ideally the equation should be solves as:
+
+    $$
+    \dot{m} = \frac{P}{\int_{T_{\text{in}}}^{T_{\text{in}}}c_{\text{p}}(T) dT}
+    $$
+
+
+!!! info "Choice of specific heat capacity"
+
+    For pumping, the specific heat capacity for constant pressure $(c_{\text{p}})$ is used as cooling loops are open-flow systems where the fluid moves continuously through pipes, heat exchangers, and pumps. As the coolant heats up, it expands freely along the loop. Because it is free to expand, the local pressure remains relatively constant while the volume changes.
+
+    You would only use the specific heat capacity for constant volume $(c_{\text{v}})$ if the coolant was completely sealed inside a rigid, unyielding container with zero flow, where heating it would cause the pressure to spike but the volume to stay exactly the same.