Sealant joint and heat loss calculations: Solutions

This assignment is based on a rectangular, brick-clad, 2-story building with a low-slope (essentially "flat") roof. The plan dimensions of the building are 60 feet x 120 feet, and the building height from grade is 24 feet. Vertical sealant joints in the brick veneer are spaced at the column lines, every 30 feet; there are no horizontal sealant joints. Use these dimensions in the calculations that follow.

Part I: Sealant joint calculation

Calculate the required width of a typical vertical sealant joint, assuming the following: sealant movement capability of 50%; coefficient of thermal expansion of brick = 0.0000036; temperature range from -15 degrees F to 105 degrees F; installation temperature range from 60 to 75 degrees F; moisture expansion coefficient of brick = 0.0002; maximum structural movement (due to wind or seismic forces) of 1/4"; construction tolerance = 1/8 inch. Round answer to nearest 1/8".

Remember to find the "worst-case" situation, either for maximum joint contraction or maximum joint expansion.

Solution:

The solution is best found by constructing a table with columns for maximum joint contraction (i.e., brick expansion), and maximum joint expansion (brick contraction). Note that 360" refers to the panel width, i.e., 30 feet.

	Max. joint contraction	Max. joint expansion
thermal expansion	(360")(0.0000036)(105-60) = 0.0583"	(360")(0.0000036)(75 - -15) = 0.1166"
moisture expansion	(360")(0.0002) = 0.072"	(360")(0) = 0"
structural deflection	0.25"	0.25"
total joint movement	0.3803"	0.3666"
required sealant width for movement: 100/50 = 2	2 x 0.3803" = 0.7607	2 x 0.3666" = 0.7332
additional sealant width for tolerance	0.125"	0.125"
Total sealant joint width	0.7607 + 0.125 = 0.8856"	0.7332 + 0.125 = 0.8582"

The answer is 0.8856" or, rounding to the nearest 1/8" increment, approximately 7/8".

Part II: Heat loss calculation

Calculate the (winter) heat loss from the building in BTU/hour units, using an indoor temperature of 70 degrees F and an outdoor temperature of -15 degrees F. Assume the following:

Window/door area is 25% of the total cladding surface (assume U-value for insulated, low-E windows/doors of 0.3)
For non-window/door wall surfaces, assume R-values for the typical wall section as follows: drywall=0.45; CMU=1.11; insulation=8.0; air space=1.0; brick=0.44; neglect air film contribution.
For roof surface, assume 4"-average thickness of rigid insulation with R=16 and drywall ceiling with R=0.45.
Neglect heat loss through the insulated slab-on-grade.

Provide the results in two parts:
A) The contribution to the total heat loss from

windows/doors;
other exterior wall surfaces;
roof;

B) The total heat loss (by adding the three component parts calculated in part "A").

Solution:

The temperature differential used in the following calculations is 70 - (-15) degrees, or 85 degrees F. The total exterior wall surface area (including windows/doors), and then the partial areas of solid wall, windows/doors, and roof, can be found by referring to the building image above:

total wall area = (120 + 120 + 60 + 60) x 24 = 8640 sq.ft.
total window/door area = 0.25(8640) = 2160 sq.ft.
total non-window/door wall area = 0.75(8640) = 6480 sq.ft.
total roof area = 60 x 120 = 7200 sq.ft.

A. Partial heat flow calculations.

Windows/doors: Q = (0.3)(2160)(85) = 55,080 BTU/hr.
Non-window/door walls: To find the U-value, first find the total R-value:
- R_t = 0.45 + 1.11 + 8.0 + 1.0 + 0.44 = 10.99
- U_t = 1 / 10.99 = 0.091
Q = (0.091)(6480)(85) = 50,118 BTU/hr.
Roof: To find the U-value, first find the total R-value:
- R_t = 16 + 0.45 = 16.45
- U_t = 1 / 16.45 = 0.0608
Q = (0.0608)(7200)(85) = 37,204 BTU/hr.

B. Total heat flow is the sum of the partial hat flows shown above: 142,402 BTU/hr..

Format

Submit on 8-1/1" x 11" paper. Show calculations in a neat, orderly, and clear format (in other words, do not simply submit your initial worksheet).

Academic Integrity

"Course Assignments. Students are encouraged to discuss the content of a course among themselves and to help each other to master it, but no student should receive help in doing a course assignment that is meant to test what he or she can do without help from others."

From the "Code of Academic Integrity," Policy Notebook for Cornell Community, Aug. 1998-99, p.78