# Conductive Heat Transfer calculation over time

I am trying to heat the oil in a 1/2 inch thick rectangular steel box by using an Insulated heated blanket covered on all 6 faces of the box, the working conditions of the operation are at -50 Celsius, the temperature of the oil is required to be at 15 Celcius during the operation, so I need to determine how much heat/power is required by the blanket and how much time will it take for the temperature of the oil to change from -50 Celsius to 15 Celsius?

one face resistive diagram, Rb = resistance by the blanket thickness of 1 inch, Rw = rectangular steel box wall

• Is the oil well-mixed? Will you regulate the power to the heater as a constant or will you regulate the temperature of the heater as a constant? – Jeffrey J Weimer Nov 5 '18 at 1:56
• What time frame? 1 hour, 10 hours, 1 day, 1 week? If you define the time then that will give the power required, if you specify the time then that will give the power needed. – Solar Mike Nov 5 '18 at 7:06
• You seem to be ignoring convection from the surface of the insulation to the air - this will have a major effect. Will this be during the day or night? Black sky is much cooler... – Jonathan R Swift Nov 5 '18 at 8:45
• The time frame will work itself out in a dimensionless equation. Convection is considered in the network side as a resistance to air. I doubt that we will have any interest in radiation at these temperatures. The OP needs first to address the two key questions of mixing and power/temperature regulation. All else is secondary. – Jeffrey J Weimer Nov 5 '18 at 14:04

# Foundations

The governing equation for the heat flow to the tank is

$$k\nabla^2 T + \rho \tilde{C}_p \frac{\partial T}{\partial t} = \hat{\dot{q}}_o$$

where temperature, thermal conductivity, density, and specific heat refer to the oil. The heat flow is per unit volume.

--> Assume the oil is well-mixed. The internal gradients disappear, giving

$$\rho \tilde{C}_p V \frac{d T}{d t} = \dot{q}_o$$

The governing equation for the heat flow to the air is

$$(T_h - T_a) = \dot{q}_a R_a$$

using the temperatures of the heater and air as well as the thermal resistance to the air. Allow that $$\dot{q}_o$$ is also defined by a thermal resistance equation as $$(T_h - T) = \dot{q}_o R_o$$. The total power taken is $$\dot{q}_T = \dot{q}_o + \dot{q}_a$$.

The system has three unknowns, $$T$$, $$T_h$$, and $$\dot{q}_T$$.

# Solutions

## General

Define $$\Theta = (T_h - T) / (T_a - T_o) = (T_h - T) / \Delta T_{ao}$$, where $$T_o$$ is the initial temperature of the oil. At time $$t = 0$$, we can set a boundary condition $$\Theta = 1$$. This means, the heater as the same temperature as the air to start. The option to set $$\Theta(0) = 0$$ (the heater the same as the oil) is more difficult to resolve. We can combine and rewrite the first governing equations as

$$-\rho \tilde{C}_p V \frac{d \Theta}{d t} = \frac{\Theta}{R_o} \Rightarrow - \frac{d \Theta}{d \tau} = \Theta$$

with $$\tau = t / \rho \tilde{C}_p V R_o$$. We find the solution as

$$\Theta = \exp(-\tau)$$

This gives an equation for the temperature difference between the heater and the oil as a function of time.

Write the thermal resistance equations as

$$\dot{q}_o R_o = \Theta \Delta T_{ao}$$ $$\dot{q}_a R_a = - \Phi \Delta T_{ao}$$

with $$\Phi = (T_a - T_h) / \Delta T_{ao}$$. The boundary condition is that $$\Phi = 0$$ at $$t = 0$$. We end with one equation.

$$\dot{q}_T = \left(\frac{\Theta}{R_o} - \frac{\Phi}{R_a}\right)\Delta T_{ao}$$

# Solutions

Solve the equation for $$\Theta = \exp(-\tau)$$ to obtain $$T_h - T$$ versus $$t$$.

## Constant Total Power

Solve the equation for $$\dot{q}_T$$ to obtain $$\Phi(\tau)$$.

$$\Phi(\tau) = \frac{R_a}{R_o}\exp(-\tau) - \frac{R_a \dot{q}_T}{\Delta T_{ao}}$$

Solve this for $$T_h(\tau)$$ from $$T_h(\tau) = T_a - \Phi(\tau) \Delta T_{ao}$$. Plug the equation for $$T_h(\tau)$$ into the equation $$T = T_h(\tau) - \exp(-\tau)\Delta T_{ao}$$ to find the temperature change of the oil with time.

$$\frac{T_a - T}{\Delta T_{ao}} = \left(1 + \frac{R_a}{R_o}\right)\exp(-\tau) - \frac{R_a \dot{q}_T}{\Delta T_{ao}}$$

## Constant Heater Temperature

Solve the equation $$\Theta(\tau)$$ directly for $$T$$ versus $$t$$. The value of $$\Phi$$ is a constant. Plug both $$\Theta$$ and $$\Phi$$ into the equation for $$\dot{q}_T$$ to find $$\dot{q}_T$$ versus $$t$$.

## Translation to Picture

The translation to the picture occurs as

$$R_o = R_{CL} + R_w + R_{B1}$$ $$R_a = R_{CA} + R_{B2}$$ $$T = T_{Lubricant} \ \ \ T_h = T_s \ \ \ T_a = T_\infty$$

## Example

Here is an example of a plot of $$\frac{T_a - T}{\Delta T_{ao}}$$ versus $$\tau$$ generated using a python/Jupyter code.

# Summary

Take the starting point that you regulate the temperature of the heater jacket to obtain a desired heating time. Allow the power to the heater to vary. The power will start high and fall off over time (as the oil heats). Estimate the initial power using values at $$t = 0$$. This will give you an easy estimate to the starting point for the level power that you need when time is a constraint.

When you start with the initial power and hold it constant, you will heat the oil faster. The temperature of the jacket will increase with time. Find the time when you have to cutoff or reduce the power to avoid burning out the heater.

When time is critical, you will do better to insulate (increase $$R_a/R_o$$) than to increase power. Notice that $$(1 + R_a/R_o)$$ and $$R_a$$ are multipliers, the former especially to an exponential of time, while power is just an additive term.

• I edited this to fix a mistake in the boundary condition on the heater at t=0 and a missing negative sign in the conversion of the governing equation to a dimensionless form. The solution now carries a (-) sign in the $\exp(-\tau)$ expression. – Jeffrey J Weimer Nov 6 '18 at 13:49