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MrRissso [65]
3 years ago
15

Assignment Overview

Engineering
1 answer:
Elanso [62]3 years ago
4 0

Answer:HI! Just listen to the explanation

Explanation:

So, to create the storyboard, you need to make boxes that resemble what you see in comics and fill them in with process of how the shipment would work.

All you need to do is describe how the shipment works! :) If it seems hard to get the word count in, just ask your supervisor to shorten the word count slightly to make it as easy as possible!!! I hope this really helped and made it easier on you! Good luck to everyone working on this assignment!!! ;)

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Answer:

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The structure of a house is such that it loses heat at a rate of 4500kJ/h per °C difference between the indoors and outdoors. A
adelina 88 [10]

Answer:

15.24°C

Explanation:

The quality of any heat pump pumping heat from cold to hot place is determined by its coefficient of performance (COP) defined as

COP=\frac{Q_{in}}{W}

Where Q_{in} is heat delivered into the hot place, in this case, the house, and W is the work used to pump heat

You can think of this quantity as similar to heat engine's efficiency

In our case, the COP of our heater is

COP_{heater} = \frac{\frac{4500\ kJ}{3600\ s} *(T_{house}-T_{out})}{4\ kW}

Where T_{house} = 24°C and T_{out} is temperature outside

To achieve maximum heating, we will have to use the most efficient heat pump, and, according to the second law of thermodynamics, nothing is more efficient that Carnot Heat Pump

Which has COP of:

COP_{carnot}=\frac{T_{house}}{T_{house}-T_{out}}

So we equate the COP of our heater with COP of Carnot heater

\frac{1.25 *(T_{house}-T_{out})}{4}=\frac{T_{house}}{T_{house}-T_{out}}

Rearrange the equation

\frac{1.25}{4}(24-T_{out})^2-24=0

Solve this simple quadratic equation, and you should get that the lowest outdoor temperature that could still allow heat to be pumped into your house would be

15.24°C

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Answer:

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   y=0

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       if i not in ["U","L","D","R"]:

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