inputs/givens 1.volume of ice (3.5 gal) 2.density of ice (736 kg/m 3 ) 3.latent heat of ice, h sf...

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Inputs/Givens 1. Volume of Ice (3.5 gal) 2. Density of Ice (736 kg/m 3 ) 3. Latent Heat of Ice, h sf (333.6 KJ/kg) 4. Melt time of 1 hour (3600 s) Constraints and Assumptions 1. Steady State 2. Ice can be melted in 1 hour Output 1. Cooling Load (900 watts) Governing Equations Efficiency Estimation Functional Diagram: Part 1 Cooling Load Required ( ) 900 sfi ce out melt Vh Q W t

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Page 1: Inputs/Givens 1.Volume of Ice (3.5 gal) 2.Density of Ice (736 kg/m 3 ) 3.Latent Heat of Ice, h sf (333.6 KJ/kg) 4.Melt time of 1 hour (3600 s) Constraints

Inputs/Givens1. Volume of Ice

(3.5 gal)2. Density of Ice

(736 kg/m3)3. Latent Heat of Ice, hsf

(333.6 KJ/kg)4. Melt time of 1 hour

(3600 s)

Constraints and Assumptions1. Steady State2. Ice can be melted in 1 hour

Output1. Cooling Load

(900 watts)

Governing Equations

Efficiency Estimation Functional Diagram: Part 1Cooling Load Required

( )900sf ice

outmelt

VhQ W

t

Page 2: Inputs/Givens 1.Volume of Ice (3.5 gal) 2.Density of Ice (736 kg/m 3 ) 3.Latent Heat of Ice, h sf (333.6 KJ/kg) 4.Melt time of 1 hour (3600 s) Constraints

Inputs/Givens1. Heat Flux (900W)2. Fluid properties of air

and water

Constraints and Assumptions1. Ideal gas2. Incompressible flow3. Constant Pressure (Cp)4. Uniform Flow5. Steady State6. Ambient air Temp of 22 C and

output temp of 13 C7. Coolant temp of 0 C from ice box8. Ice can be melted in 1 hour

Output1. Air Flow Rate

(0.12 m3/s)2. Coolant Flow Rate

(1 GPM -> at least 0.5)

Governing Equations

Efficiency Estimation Functional Diagram: Part 2Fan/Pump sizing

900 ( )p airW Qc T

39000.12

( )ap air

W mQ sc T

900 ( )p coolantW Qc T

max

9000.5

( )p coolant

WQ GPM

c T

Page 3: Inputs/Givens 1.Volume of Ice (3.5 gal) 2.Density of Ice (736 kg/m 3 ) 3.Latent Heat of Ice, h sf (333.6 KJ/kg) 4.Melt time of 1 hour (3600 s) Constraints

Input data1. Cooling Load

(900 watts)2. Coolant Flow rate

(1 GPM -> at least 0.5)

Constants and Knowns1. Density of Water

(1000 kg/m3)

Constraints and Assumptions1. No pumping losses2. 65% pump efficiency (low)3. Fan at 100% power4. Steady State5. 2x calculated pump power to

accommodate losses6. z (H) of coolant in pumping loop

equal to 1m (would be less in actual unit)

OutputCOP = 10

Governing Equations

Efficiency Estimation Functional Diagram: Part 3COP calculation

in

gQHW

in

CCCOP

W

Page 4: Inputs/Givens 1.Volume of Ice (3.5 gal) 2.Density of Ice (736 kg/m 3 ) 3.Latent Heat of Ice, h sf (333.6 KJ/kg) 4.Melt time of 1 hour (3600 s) Constraints

Measured data1. T of water in and out

of radiator2. Win from “plug power

meter”3. Coolant Flow Rate

Constants and givens1. Area, A, of air flow2. Fluid properties of air

(density, Cp)

Constraints and Assumptions- Ideal gas- Incompressible flow- Constant Pressure (Cp)- Uniform Flow

OutputFinal/Overall COP of unit

Governing Equations

Final Efficiency Functional Diagram(Final Testing)

outCC Q VACp T

( )avg testoverall

in

CC tCOP

W

( )avgCC mean CC

Page 5: Inputs/Givens 1.Volume of Ice (3.5 gal) 2.Density of Ice (736 kg/m 3 ) 3.Latent Heat of Ice, h sf (333.6 KJ/kg) 4.Melt time of 1 hour (3600 s) Constraints

Measured data1. T of water in and out

of radiator2. Win from “plug power

meter”3. Coolant Flow Rate

Constants and givens1. Area, A, of air flow2. Fluid properties of air

(density, Cp)

Constraints and Assumptions- Ideal gas- Incompressible flow- Constant Pressure (Cp)- Uniform Flow

OutputFinal/Overall COP of unit

Governing Equations

Final Efficiency Functional Diagram(Final Testing)

outCC Q VACp T

( )avg testoverall

in

CC tCOP

W

( )avgCC mean CC