2 - RTS Example 8-14

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RTS Method Example 8-14 (Textbook) (cooling load from exterior walls) Fall 2023 BLDG 471 - Dr. M. Zaheeruddin 1 Solution Method: ❑ Step-1: Calculate the conduction heat gains from walls ❑ Step-2: Using the conduction heat gains, calculate the cooling loads

RTS Method - Example 8-14 (Text Book) The Wall-1 (Table 8-16) is exposed to the sol-air temperatures shown in Table 8- 15. Determine the conduction heat gains per unit area for hour 15:00. The constant indoor air temperature is 72 0 F. Solution: ?′ ???? , 𝑗 , 𝜃 = 𝑌 ?? (t ? , 𝑗 , 𝜃−?𝛿 − 𝑡 ?? ) 23 ? =0 Conduction Heat gains for the hour 15:00 can be obtained by using Equation 8-64 as follows: ?′ ???? , 𝑗 ,15 = 𝑌 ? 0 (t ? , 𝑗 ,15 − 72) + 𝑌 ? 1 (t ? , 𝑗 ,14 − 72) + 𝑌 ? 2 (t ? , 𝑗 ,13 − 72) + ⋯ + 𝑌 ? 23 (t ? , 𝑗 ,16 − 72) Substitute Y pn from Table 8-18 and t e θ from Table 8-15 in the above equation Fall 2023 BLDG 471 - Dr. M. Zaheeruddin 2 (8-64)

RTS Method - Example 8-14 Fall 2023 BLDG 471 - Dr. M. Zaheeruddin 3

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RTS Method - Example 8-14 ?′ ???? , 𝑗 ,15 = 𝑌 ? 0 (t ? , 𝑗 ,15 − 72) + 𝑌 ? 1 (t ? , 𝑗 ,14 − 72) + 𝑌 ? 2 (t ? , 𝑗 ,13 − 72) + ⋯ + 𝑌 ? 23 (t ? , 𝑗 ,16 − 72) Substitute Y pn from Table 8-18 and t e θ from Table 8-15 in the above equation Fall 2023 BLDG 471 - Dr. M. Zaheeruddin 4

RTS Method - Example 8-14 q cond (15) = Y p ( 0) [t e (15) – t rc ] + Y p ( 1) [t e (14) – t rc ] + Y p ( 2) [t e (13) – t rc ] 0.000156 [(151.2-72)] + 0.00560 [(138.1-72)] + 0.014795 [(120.3-72)] Y p ( 3) [t e (12) – t rc ] + Y p ( 4) [t e (11) – t rc ] + Y p ( 5) [t e (10) – t rc ] 0.014441 [(108.9-72)] + 0.009628 [(103.6-72)] + 0.005414 [(98.3-72)] Y p ( 6) [t e ( 9) – t rc ] + Y p ( 7) [t e ( 8) – t rc ] + Y p ( 8) [t e ( 7) – t rc ] 0.002786[(92.9-72)] + 0.001363 [(87.1-72)] + 0.000647 [(80.7-72)] Y p ( 9) [t e ( 6) – t rc ] + Y p (10) [t e ( 5) – t rc ] + Y p (11) [t e ( 4) – t rc ] 0.000301 [(78.4-72)] + 0.000139 [(78.0-72)] + 0.000063 [(78.2-72)] Y p (12) [t e ( 3) – t rc ] + Y p (13) [t e ( 2) – t rc ] + Y p (14) [t e ( 1) – t rc ] 0.000029 [(78.8-72)] + 0.000013 [(79.7-72)] + 0.000006[(80.7-72)] Y p (15) [t e (24) – t rc ] + Y p (16) [t e (23) – t rc ] + Y p (17) [t e (22) – t rc ] 0.000003 [(81.8-72)] + 0.000001 [(83.0-72)] + 0.000001 [(84.7-72)] Y p (18) [t e (21) – t rc ] + Y p (19) [t e (20) – t rc ] + Y p (20) [t e (19) – t rc ] 0.000000 [(86.8-72)] + 0.000000 [(101.4-72)] + 0.000000 [(131.0-72)] Y p (21) [t e (18) – t rc ] + Y p (22) [t e (17) – t rc ] + Y p (23) [t e (16) – t rc ] 0.000000 [(148.6-72)] + 0.000000[(157.3-72)] + 0.000000 [(157.8-72)] q cond (15) = 2.1646 BH/ft2 Fall 2023 BLDG 471 - Dr. M. Zaheeruddin 5

RTS Method - Example 8-14 (continued) Conversion of Conduction heat gains to cooling loads By using the conduction heat gains for the south wall obtained in the previous Example 8-14, Calculate the cooling load for the south facing wall at 15:00hrs. Assume a zone type MW2 (Table 8-21). Fall 2023 BLDG 471 - Dr. M. Zaheeruddin 6

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RTS Method - Example 8-14 (continued) Conversion of Conduction heat gains to cooling loads Split the conduction heat gains into radiative and convective parts Using Table 8-20 (for walls: radiative fraction = 0.63, convective fraction=0.37) The cooling load based on the current and past values of radiative heat gains : Q θ,cl = 0.63*{r 0 q θ + r 1 q θ - δ + r 2 q θ - 2δ + r 3 q θ - 3δ + . . . + r 23 q θ - 23δ } For the hour 15:00, Q 15, cl = 0.63*{r 0 q 15 + r 1 q 14 + r 2 q 13 + r 3 q 12 + . . . + r 23 q 16 } Substitute the values of r from Table (8-21) and heat flux q from Table (8-19) in the above equation Fall 2023 BLDG 471 - Dr. M. Zaheeruddin 7

RTS Method - Example 8-14 (continued) Conversion of Conduction heat gains to cooling loads Q 15,cl = 0.63*{r 0 q 15 + r 1 q 14 + r 2 q 13 + r 3 q 12 + r 4 q 11 + r 5 q 10 0.25509*2.165 + 0.11396*1.711 + 0.06959*1.385 + 0.05133*1.1 + 0.04259*0.832 + 0.03771*0.602 r 6 q 9 + r 7 q 8 + r 8 q 7 + r 9 q 6 + r 10 q 5 + r 11 q 4 0.03461*0.442 + 0.03241*0.377 + 0.03071*0.379 + 0.02931*0.413 + 0.02809* 0.468 + 0.02700*0.545 r 12 q 3 + r 13 q 2 + r 14 q 1 + r 15 q 24 + r 16 q 23 + r 17 q 22 0.02598*0.652 + 0.02504*0.814 + 0.02414*1.071 + 0.02328*1.49 + 0.02246*2.135 + 0.02167*2.984 r 18 q 21 + r 19 q 20 + r 20 q 19 + r 21 q 18 + r 22 q 17 + r 23 q 16 } 0.02091*3.792 + 0.02018*4.244 + 0.01948*4.277+ 0.01888*3.973 + 0.01815*3.429 + 0.01751*2.774 Q 15,cl = 1.060 BH/ft2 (from radiative heat gain) Total Cooling load at 15:00 = {1.060 (radiative)+2.165*0.37(convective)} = 1.861BH/ft2 Fall 2023 BLDG 471 - Dr. M. Zaheeruddin 8