🎨 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 +1,82 @@
 # 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:
+
+$$
+P = \frac{\frac{\dot{m} \times \left(H_{\text{out}}-H_{\text{in}}\right)}{\eta}}{\left(1-f_p\right)}
+$$
+
+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.
+
+$f_p$ is a seocondary correction accounting for the fact that some of the pump's mechanical work reheats the coolant, and that reheat is already implicitly present in the prescribed $\left(T_{\text{pump,in}}-T_{\text{pump,out}}\right)$ used elsewhere in the plant energy balance, without this correction that portion of the pump work would be double counted. The term uses the classical isentropic ideal-gas relation $T_2/T_1 = (P_2/P_1)^{(\gamma-1)/\gamma}$
+
+$$
+f_p = \frac{T_{\text{pump,out}}\left(\frac{P_{\text{pump,out}}}{P_{\text{pump,in}}}\right)^{-\frac{\gamma -1}{\gamma}}}{\eta \left(T_{\text{pump,in}}-T_{\text{pump,out}}\right)}
+$$
+
+------------------
+
+## 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 +138,62 @@ 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 because
+    cooling loops are open, steady-flow systems: coolant flows continuously through pipes, heat
+    exchangers and pumps at a system pressure that is actively maintained by the pump's operating
+    point and any pressure-control components (e.g. pressurisers, control valves), rather than
+    varying freely with the coolant's volume.
+
+    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.

process/core/io/plot/summary.py

@@ -91,6 +91,7 @@
     PlasmaProfileShapeType,
     calculate_profile_shell_contributions,
 )
+from process.models.power import PumpingPowerModelTypes
 from process.models.pulse import PulseTimings
 from process.models.superconductors import SuperconductorModel
 from process.models.tfcoil.base import (
@@ -16813,6 +16814,55 @@ def plot_cumulative_plasma_thermal_energy_profiles(axis, m_file: MFile, scan: in
     )
 
 
+def plot_blanket_coolant_properties(fig: plt.Figure, m_file: MFile, scan: int):
+    """Combined plot of blanket coolant channel structure and properties."""
+    for side, x_position in (("inboard", 0.1), ("outboard", 0.5)):
+
+        def get(variable: str):
+            return m_file.get(variable, scan=scan)
+
+        text = (
+            f"$\\mathbf{{{side.capitalize()} \\ blanket:}}$\n \n"
+            f"Radius of blanket channel: {m_file.get('radius_blkt_channel', scan=scan):.4f} m\n"
+            f"Channel roughness ($\\epsilon$): {m_file.get('roughness_fw_channel', scan=scan):.4e} m\n\n"
+            f"Radial coolant channel length: {get(f'len_blkt_{side}_coolant_channel_radial'):.4f} m\n"
+            f"Poloidal coolant channel length: {get(f'len_blkt_{side}_segment_poloidal'):.4f} m\n"
+            f"Number of radial channels: {get(f'n_blkt_{side}_module_coolant_sections_radial')}\n"
+            f"Number of poloidal channels: {get(f'n_blkt_{side}_module_coolant_sections_poloidal')}\n"
+            f"Total length of coolant channel straight sections: {get(f'len_blkt_{side}_channel_total'):.4f} m\n\n"
+            f"Pressure drop for straight sections: {get(f'dpres_blkt_{side}_coolant_channel_straight_total'):,.2f} Pa\n"
+            f"Pressure drop for 90° bends: {get(f'dpres_blkt_{side}_coolant_channel_90_bend'):,.2f} Pa\n"
+            f"Total pressure drop for 90° bends: {get(f'dpres_blkt_{side}_coolant_channel_90_bends_total'):,.2f} Pa\n"
+            f"Pressure drop for 180° bends: {get(f'dpres_blkt_{side}_coolant_channel_180_bend'):,.2f} Pa\n"
+            f"Total pressure drop for 180° bends: {get(f'dpres_blkt_{side}_coolant_channel_180_bends_total'):,.2f} Pa\n"
+            f"Total pressure drop for all bends: {get(f'dpres_blkt_{side}_bends_total'):,.2f} Pa\n\n"
+            f"Reynolds number ($Re$): {get(f'reynolds_blkt_{side}_coolant'):,.4f}\n"
+            f"Darcy Friction factor ($f$): {get(f'darcy_frict_blkt_{side}_coolant'):.4f}\n\n"
+            f"Friction drop coefficient for straight sections: {get(f'f_straight_blkt_{side}_coolant'):.4f}\n"
+            f"Friction drop coefficient for 90° bends: {get(f'f_elbow_blkt_{side}_90_bend'):.4f}\n"
+            f"Friction drop coefficient for 180° bends: {get(f'f_elbow_blkt_{side}_180_bend'):.4f}\n\n"
+            f"Total coolant mass flow rate: {get(f'mflow_blkt_{side}_coolant'):.4f} kg/s\n"
+            f"Coolant mass flow rate in single channel: {get(f'mflow_blkt_{side}_coolant_channel'):.4f} kg/s\n"
+            f"Coolant velocity in single channel: {get(f'vel_blkt_{side}_coolant'):.4f} m/s"
+        )
+
+        fig.text(
+            x_position,
+            0.5,
+            text,
+            fontsize=9,
+            verticalalignment="top",
+            horizontalalignment="left",
+            transform=fig.transFigure,
+            bbox={
+                "boxstyle": "round",
+                "facecolor": "wheat",
+                "alpha": 1.0,
+                "linewidth": 2,
+            },
+        )
+
+
 def main_plot(
     m_file: MFile,
     scan: int,
@@ -17449,17 +17499,18 @@ def _add_page(name: str | None = None):
     )
     plot_fw_90_deg_pipe_bend(pages["fw_td_cross_section"].add_subplot(337), m_file, scan)
 
-    plot_blkt_pipe_bends(_add_page("blkt_pipe_bends"), m_file, scan)
-    ax_blanket = pages["blkt_pipe_bends"].add_subplot(122, aspect="equal")
+    ax_blanket = _add_page("blkt_structure").add_subplot(122, aspect="equal")
     plot_blkt_structure(
-        ax_blanket,
-        pages["blkt_pipe_bends"],
-        m_file,
-        scan,
-        radial_build,
-        colour_scheme,
+        ax_blanket, pages["blkt_structure"], m_file, scan, radial_build, colour_scheme
     )
 
+    plot_blkt_pipe_bends(_add_page("blkt_cooling"), m_file, scan)
+    if (
+        m_file.get("i_p_coolant_pumping", scan=scan)
+        == PumpingPowerModelTypes.CALCULATE_PRESSURE_DROP
+    ):
+        plot_blanket_coolant_properties(pages["blkt_cooling"], m_file, scan)
+
     plot_main_power_flow(
         _add_page("main_power_flow").add_subplot(111, aspect="equal"),
         m_file,
@@ -17578,8 +17629,6 @@ def add_page_footer(
             color="dimgray",
         )
 
-    # create main plot
-    # Increase range when adding new page
     # run main_plot
     mfile_obj = MFile(mfile) if mfile else MFile("MFILE.DAT")
     run_label = f"{mfile_obj.get('fileprefix', scan=-1)} | scan {scan or -1} | {mfile_obj.get('date', scan=-1)} {mfile_obj.get('time', scan=-1)} | {mfile_obj.get('tagno', scan=-1)} | Branch: {mfile_obj.get('branch_name', scan=-1)}  "