1. One proposal for a combined heating and cooling system for a home is a so-called ice maker heat pump. It is shown in the sketch. The heat pump is operated using electricity during the heating season. It withdraws heat at 0oC from a pool of water, this is the low temperature heat source. It delivers heat at higher temperature to heat the house. As heat is transferred from the water, ice is formed. (In practice, there are series of pipes in the water. The pipes contain cold refrigerant; ice forms around the outside of the pipes.) Because of the rather low temperature of the ice and water the COP, ratio of heat delivered to electrical input, of the heat pump is only 2.5. To change phase from liquid water to ice, the internal energy change is approximately 334 kJ/kg. Neglect the change in volume for this process.
The home requires 60 x 106 kJ of heat during the winter and 30 x 106kJ of cooling (amount of heat transferred for cooling) for the summer. In the summer the heat pump is turned off and the melting ice is used to provide cooling for the home.
a) How much ice is formed over the heating season? Is it adequate to meet the cooling needs?
Answer:
For heat pump, COP=QH/W=2.5, and QH= W+ QC, so the QC=0.6Q QH=0.6*60x106 kJ=36 x106 kJ
The ice is formed over the heating season is:
∆Uwater-ice=Q-W, here W=0. ∆Uwater-ice=Mice*∆uwaterice=Q, so, Mice*=Q/∆uwater-ice=(36x106 kJ)/(334kJ/kg)=107784.4kg
In summer condition, the heat pump is turned off, and the cooling needs 30x106 kJ. The ice should be adequate to meet the cooling needs if the ice container’s insulation is good enough to keep ice from melting during spring. Additionally, the performance coefficient of the mechanical system used to transfer heat from home space to ice must be better than 83% (=30x106 kJ/36x106 kJ).
For any help regarding Architecture Assignment Help Visit :- https://www.architectureassignmenthelp.com/
Email :- support@architectureassignmenthelp.com call us/ WhatsApp :- +1 (315) 557-6473
b) Compare the electrical energy consumed for the ice maker heat pump system with a conventional system. In the latter the heat pump has a COP of 4.0, this is higher than the ice maker heat pump because the outside air is usually above the freezing point. In the conventional system the heat pump is also used as an air conditioner in the summer where the air conditioning COP is 3.0
Answer:
For the ice maker heat pump, COP=QH/W=2.5, and the electricity energy consumption : W=QH/2.5=24x106 kJ
For the conventional system, winter it needs electricity: W1= QH/4=15x106 kJ
In summer, it needs electricity: W2= QC/3=10x106 kJ The conventional system needs totally 25 x106 kJ.
From our calculation, it shows that the ice maker system will save 1 x106 kJ electricity energy.
For any help regarding Architecture Assignment Help Visit :- https://www.architectureassignmenthelp.com/
Email :- support@architectureassignmenthelp.com call us/ WhatsApp :- +1 (315) 557-6473
c) Comment on the feasibility of the ice –maker system for home heating and cooling.
Answer:
1. The container of ice will be huge. We can estimate the ice volume, and we need a 110m3 (such as 4.8mx4.8mx4.8m) size cubic tank to hold it. It at least needs a large scale storage space. Additionally, the ice container should be insulated so well in order to keep the ice from melting during spring period.
2. The mechanical system that is used in the ice maker system will be some complicated and highly cost. It needs to consider a lot of technical problems in order to in crease the system’s performance.
3. Actually, in summer condition, even if the heat pump is turned off, some electricity energy still needs to transfer heat between the house space and ice. So the ice maker system save energy comparing with conventional system may not happen. So, the ice maker system is not so feasible for general conditions. For example, it may not be used in a single house of which heating and cooling energy needs are not so big. This system will increase the first cost significantly. It may be applied in some special large buildings, which can afford the first and operation cost increasing.
For any help regarding Architecture Assignment Help Visit :- https://www.architectureassignmenthelp.com/ Email :- support@architectureassignmenthelp.com call us/ WhatsApp :- +1 (315) 557-6473
2. In summer, one person leaves her room with a opened cooker in which fully contains water. Some days later, she comes back home and found that some of the water in the cooker has been evaporated.
The room volume is 50m3, and current indoor air temperature is 35oC. Assume the room façade is insulated so well that there is no heat transfer through the façade. Also, the room is sealed well with no air leakage.
a). If the indoor air is saturated with vapor, how much vapor does the indoor air contain (kg)?
Answer:
From table K. in page 353, we find the specific volume of the saturated vapor at 35oC has v=25.22 m3/kg. Then the vapor contained in room air is:
Mv,1=(50m3)/(25.22m3/kg)=1.98kg
For any help regarding Architecture Assignment Help Visit :- https://www.architectureassignmenthelp.com/ Email :- support@architectureassignmenthelp.com call us/ WhatsApp :- +1 (315) 557-6473
b). The person feels that the indoor air temperature too hot. She turns on the air conditioner to cool the air. Two hours later, the indoor air temperature becomes 25oC. How much water condensed from air during the two hours? Please don’t consider the vapor generated by the person.
Answer:
From table K. in page 353, we find the specific volume of the saturated vapor at 25oC has v=43.36 m3/kg. Then now the vapor contained in room air is:
Mv,2=(50m3)/(43.36m3/kg)=1.15kg
The condensed water is:
Mwater,condensed= Mv,1 - Mv,2 =0.83kg
c). If the COP of the air conditioner is 3.0, how much electricity energy (kJ) was used to cool down the air from 35 oC to 25 oC? Assume that the person stayed in the room during the cooling two hours, and her heat amount that transferred to
For any help regarding Architecture Assignment Help Visit :- https://www.architectureassignmenthelp.com/
Email :- support@architectureassignmenthelp.com call us/ WhatsApp :- +1 (315) 557-6473
room air is 100W. Room air density between 35 oC to 25 oC assumes constant as 1.17 kg/m3, and specific heat of air Cv=700J/kg.K.
Answer:
We set the air at 35 oC as state condition 1, and the air at 25 oC as state condition 2. Take the room air (including the vapor) as our control system, then we have:
∆Uair=Q-W, here W=0.
Then, ∆Uair =Uair,2 - Uair,1 =Mair*Cv,air*(T2-T1)+ Mwater,condensed*(Uw,2-Uv,1)
=(1.17kg/m3)*(50m3)*(0.7kJ/kg.oC)*(250C 35oC) +(0.83kg)*(104.88kJ/kg2423.4kJ/kg)=- -2333.87kJ
The heat transferred into room air includes air conditioner’s cooling and person heat generation two parts:
Q=Qperson+Qcooling
Qperson=(100W)*(2*3600s)=720kJ
Then: Qcooling=∆Uair-Qperson=-3053.87kJ
That is to say, the air conditioner cooling heat |Qc|=3053.87kJ
COP=|Qc|/Welectr=3.0, so Welectr=|Qc|/3.0=1017.96kJ