EFPLAB2: Heat Exchanger

EFPLAB2 Lab Experiment: Heat Exchanger

T,A - Diagram

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T,Q̇ - Diagram

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Test Rig Animation

Bei Bedarf Zusatzinformationen Einblenden.

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Variation of the input parameters

In some exceptional cases, the heat exchanger cannot be simulated and therefore not be displayed correctly. This applies in particular to the counter current heat exchanger with the settings where the temperature of the fluids is completely equalized.

Fluid 1:

Mass flow ${\dot{m}}_{1}$ = kg/s

Inlet temperature $ϑ_{1 α}$ = °C

Heat exchanger Overall heat transfer coefficient k = 1300 W/(m²K)

Flow direction of the fluids in the heat exchanger

Cocurrent

Countercurrent

Heat transfer surface $A$ = m²

Fluid 2:

Mass flow ${\dot{m}}_{2}$ = kg/s

Inlet temperature $ϑ_{2 α}$ = °C

Calculation Table

In the calculation table, equations and results are displayed depending on the input parameters. If the temperature difference α or β drops to 0 K, the calculations are suspended.

Temperature difference A

Δ T_{A} = ϑ_{1 α} - ϑ_{2 ω} = 20.25 K

Δ T_{A} = ϑ_{1 α} - ϑ_{2 α} = 20.25 K

Temperature difference B

Δ T_{B} = ϑ_{1 ω} - ϑ_{2 α} = 20.25 K

Δ T_{B} = ϑ_{1 ω} - ϑ_{2 ω} = 20.25 K

log oder arith

Δ T_{m} = Δ ϑ_{l n} = \frac{Δ ϑ_{α} - Δ ϑ_{ω}}{l n (\frac{Δ ϑ_{α}}{Δ ϑ_{ω}})} = 20.45 K

Δ T = Δ ϑ_{α} = Δ ϑ_{ω} = 20.45 K

Transferred heat flow in a heat exchanger

\dot{Q} = k \cdot A \cdot Δ T_{m} = 10000 W

\dot{Q} = k \cdot A \cdot Δ T = 10000 W

Enthalpy flow difference fluid 1

{Δ \dot{H}}_{F l u i d, 1} = {\dot{m}}_{1} \cdot c_{p 1} \cdot (ϑ_{1 ω} - ϑ_{1 α}) = 10000 W

Enthalpy flow difference fluid 2

{Δ \dot{H}}_{F l u i d, 2} = {\dot{m}}_{2} \cdot c_{p 2} \cdot (ϑ_{2 ω} - ϑ_{2 α}) = 10000 W

Heat recovery number 1

P_{1} = \frac{ϑ_{1 α} - ϑ_{1 ω}}{ϑ_{1 α} - ϑ_{2 α}} = 2.34

Heat recovery number 2

P_{2} = \frac{ϑ_{2 ω} - ϑ_{2 α}}{ϑ_{1 α} - ϑ_{2 α}} = 2.34