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Two-period discharge requirement calculator
Two-period discharge requirement
Integrate only positive load deficits when event charging is unavailable; spare chiller capacity during a lower-load interval cannot reduce prior required discharge.
Units IP (US) SI (metric)
C = max ( L 1 − Q , 0 ) × t 1 + max ( L 2 − Q , 0 ) × t 2 C = \max\left(L_{1} - Q,\ 0\right) \times t_{1} + \max\left(L_{2} - Q,\ 0\right) \times t_{2} C = max ( L 1 − Q , 0 ) × t 1 + max ( L 2 − Q , 0 ) × t 2
C = max ( L 1 − Q , 0 ) × t 1 + max ( L 2 − Q , 0 ) × t 2 C = \max\left(L_{1} - Q,\ 0\right) \times t_{1} + \max\left(L_{2} - Q,\ 0\right) \times t_{2} C = max ( L 1 − Q , 0 ) × t 1 + max ( L 2 − Q , 0 ) × t 2
Calculator
Enter the values you know and leave one box empty. Calculate finds the empty one, the result or any input.
Worked example
Given
L 1 L_{1} L 1 = 310 ton Q Q Q = 230 ton t 1 t_{1} t 1 = 4 h L 2 L_{2} L 2 = 170 ton t 2 t_{2} t 2 = 2 h
C = max ( 310 − 230 , 0 ) × 4 + max ( 170 − 230 , 0 ) × 2 = 320 C = \max\left(310 - 230,\ 0\right) \times 4 + \max\left(170 - 230,\ 0\right) \times 2 = 320 C = max ( 310 − 230 , 0 ) × 4 + max ( 170 − 230 , 0 ) × 2 = 320
Given
L 1 L_{1} L 1 = 1,050 kW Q Q Q = 700 kW t 1 t_{1} t 1 = 3 h L 2 L_{2} L 2 = 600 kW t 2 t_{2} t 2 = 2 h
C = max ( 1050 − 700 , 0 ) × 3 + max ( 600 − 700 , 0 ) × 2 = 1 , 050 C = \max\left(1050 - 700,\ 0\right) \times 3 + \max\left(600 - 700,\ 0\right) \times 2 = 1{,}050 C = max ( 1050 − 700 , 0 ) × 3 + max ( 600 − 700 , 0 ) × 2 = 1 , 050
Symbols and units
Symbol Meaning Unit
C C C required discharged cooling energy ton·h
L 1 L_{1} L 1 first cooling load ton
Q Q Q chiller capacity assigned to the load ton
t 1 t_{1} t 1 first duration h
L 2 L_{2} L 2 second cooling load ton
t 2 t_{2} t 2 second duration h
Symbol Meaning Unit
C C C required discharged cooling energy kWh
L 1 L_{1} L 1 first cooling load kW
Q Q Q chiller capacity assigned to the load kW
t 1 t_{1} t 1 first duration h
L 2 L_{2} L 2 second cooling load kW
t 2 t_{2} t 2 second duration h
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