# How to decrease step from 15 mm for Surface node temperature [closed]

Hello Everybody! I am currently working on DesignBuilder and trying to obtain surface node temperature of the wall. I was able to switch this function on in Outputs tab (Outputs/Simulation Output Options/Miscellaneous Outputs -> Surface node temperature...). I know that when you use this option, it gives wall temperature for every 15 mm of the wall thickness. Problem is that I need a temperature for a specific layer which is only 5 mm thick and located between two nodes. Is there any option to decrease the node step from 15 mm to 5 mm? Thank you in advance!

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### Closed for the following reason the question is answered, right answer was accepted by Almas close date 2019-09-26 05:29:37.006406

1

You'll have to change the global Timestep altogether, setting it to Timestep, 12; will give you 5min timesteps.

( 2019-09-24 12:19:32 -0500 )edit

Thanks, You are right)

( 2019-09-26 05:05:09 -0500 )edit

Actually, guys, I've across the paper of Paulo Tabares-Velasco and Brent Griffith called "Diagnostic test cases for verifying surface heat transfer algorithms and boundary conditions in building energy simulation programs". They have explained how DesignBuilder (DB) calculates step for surface temperature calculation. They use the formula delta(x) = sqrt(c * alfa * delta(t)), where c - space discretization constant (standard value for DB is 3, can be modifyed in "Edit calculation Options"-> "Options" -> Advanced -> General Solution -> Finite Difference Setting)...

( 2019-09-26 05:13:08 -0500 )edit

delta(t) - time step can be modified in "Edit calculation Option", and alfa is a thermal diffusivity of material and calculated by the formula of alfa=thermal conductivity/(density*specific heat). The values of thermal conductivity, density, and specific heat can be found in material properties in DB. Knowing c, alfa, delta(t), we can calculate delta(x) which is a nodal step, then knowing the thickness of the layer we can exactly calculate how many nodes will be there and their location.

( 2019-09-26 05:17:06 -0500 )edit

For example: I have a layer of cement of 0.2m. Thermal conductivity, k = 0.8. specific heat = 840. density 1600. Time step is 2 minutes = 120 sec. c=3. thermal diffusivity, alfa= 0.8/8401600= 5.952e-7 delta(x)= sqrt(5.952e-73*120)=0.0146 m number of nodes = 0.2/0.0146=13.69 which means we will have 13 nodes for this layer (programm takes smallest round number). new delta(x)= 0.2/13=0.01538 m. For 0.2 m thick cement DB will show 14 nodal surface temperature (one for the zero point)

( 2019-09-26 05:23:35 -0500 )edit