Blanket

General blanket methods | BlanketLibrary

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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.

$$fp = \frac{T_{\text{pump,out}}\left(\frac{P_{\text{pump,out}}}{P_{\text{pump,in}}}\right)^{-\frac{\gamma -1}{\gamma}}}{\eta \left(T_{\text{pump,in}}-T_{\text{pump,out}}\right)}$$

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Coolant pressure drop | coolant_friction_pressure_drop()

The pressure drop in the coolant is given by the Darcy-Weisbach 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, $\rho$ 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().

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 clalculated via 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:

$$\Delta P = L\left[f_{\text{D}}\frac{\rho}{2}\frac{\langle v \rangle^2}{D_{\text{H}}}\right] + N_{\text{90}} \left[f_{\text{90,elbow}} \frac{\rho \langle v \rangle^2}{2}\right] + N_{\text{180}} \left[f_{\text{180,elbow}} \frac{\rho \langle v \rangle^2}{2}\right]$$

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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:

$$ab + \left( f_{\text{D}} \times \frac{R_{\text{elbow}}}{D_{\text{pipe}}}\right) \times \theta \times \left(\frac{\pi}{180^{\circ}}\right)$$