# How do I determine the temperature inside a transformer vault?

I am trying to determine the temperature of an underground concrete vault buried 10m underground. The vault has the following dimensions: 2640mm x 5995mm x 2640mm with a wall thickness of 250mm. The transformer has the following dimensions: 2083mm by 1702mm by 1257mm. It has heat losses of 2080W to 11000W. How can I determine the temperature inside the vault? How would adding mechanical ventilation change the temperature inside this vault?

• What is the temperature of the ground at a 10m depth? Does it vary during the year? Jul 31 '19 at 12:35
• I am not sure, but I wanted to do the same calculation for a range between 10C and 30C. Jul 31 '19 at 12:38
• I know, for where I live, that the annual ground temperature varies between about 5 to 8 degrees C... Because I planned the incoming air for the ventilation to come in underground for 30 meters... Jul 31 '19 at 12:40
• I think that due to the vents that underground transformer vaults have, the temperature may fluctuate more. Jul 31 '19 at 12:41

# Assumptions

Assume steady state and neglect all radiation terms. Assume the surroundings to the concrete wall is an infinite heat sink at a fixed (constant) temperature). Finally, neglect the heat transfer that occurs at the corners of the concrete box (set the internal and external areas of the box to be equal as far as heat transfer is concerned).

# Formulations

The energy balance from the transformer to the inside air of the concrete wall is

$$P = h_i A_T (T_T - T_{ai})$$

In this, $$P$$ is the power supplied, $$T_T, T_{ai}$$ are the transformer and internal air temperatures, $$h_i$$ is the convection coefficient inside the box, and $$A_T$$ is the area of the transformer. This equation says that power generated by the transformer is convected away to the inside air of the container. The convection coefficient $$h_i$$ depends on how fast the air is moving (stagnate/free convection to forced convection). This gives you one equation with two unknowns (the temperatures).

The energy balance from the air in the box to the concrete wall is

$$P = h_i A_T (T_T - T_{ai}) = h_i A_w (T_{ai} - T_{wi})$$

In this, $$T_{wi}$$ is the temperature of the internal wall and $$A_w$$ is the area of the wall. This equation says that power generated by the transformer is convected away to the inside air of the container that is then convected to the box wall. It is one equation with two unknowns (the temperatures).

The energy balance on the concrete wall without vents is

$$k \frac{(T_{wi} - T_g)}{w} = h_i (T_{ai} - T_{wi})$$

In this, $$k$$ is the thermal conductivity of concrete, $$T_g$$ is the (constant) ground temperature, and $$w$$ is the wall thickness. This equation says that heat flow to the inside walls from the air is conducted away through the walls. It gives one more equation with the same two unknowns (the temperatures).

The energy balance on the concrete wall with vents is

$$k (1 - f_v) \frac{(T_{wi} - T_g)}{w} + h_v f_v (T_{wi} - T_g) = h_i (T_T - T_{wi})$$

In this, $$f_v$$ is the fractional area of the wall that contains vents and $$h_v$$ is a pseudo coefficient for convection-type heat transfer through the vents. It also gives one more equation with the same two unknowns (the temperatures).

# Solution

In all cases without or with vents, you can combine the system to obtain two equations and two unknowns. The system can be solved.

A starting point is to solve the flux equation without vents to find $$T_{wi}$$.

$$P = k A_w \frac{(T_{wi} - T_g)}{w}$$

This says that the energy from the transformer goes through the walls by conduction.

Use the value of $$T_{wi}$$ and the equation $$P = h_i A_w(T_{ai} - T_{wi})$$ to find $$T_{ai}$$. Continue with the heat transfer from transformer to air to find $$T_T$$. Finally, use these first guesses in the equation with vents to re-calculate the temperatures.

A comparable problem is given at this link.

• I included the fact that you likely have air inside the box between the transformer and the concrete walls. This does not change the fact that this problem can be solved. Aug 1 '19 at 0:32
• Thank you very much, I was able to determine the temperature this way. As a follow-up question, how would having airflow change the temperature in the room? Does the current method assume that there is no airflow inside of the vault? Aug 1 '19 at 12:06
• I also think there is a mistake in the final equation. You advise to solve for Tai however, the equation only has Twi as an unknown. Aug 1 '19 at 12:24
• @Lechuga I've updated the post to give answers for both of your questions. Aug 1 '19 at 15:50
• Does this assume that the power dissipated happens in one hour? How could I calculate the temperature if many more hours occurred? Aug 28 '19 at 12:44