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2. 1) The attic of a single family house has 8 inches of fiberglass insulation ( k = 0.028
BTU/hr ft F) on the attic floor. The homeowner has added 2 inches of foam insulation (k
= 0.015 Btu/he ft F) on the underside of the roof. The outside air temperature is 40 F and
the temperature of the occupied space beneath the attic floor is 70 F. The outside
environment and the indoor environment can be looked as black body, which has same
temperature as the outside and inside air temperature respectively. The foam and
fiberglass insulation also can be viewed as black body. Assume that the air in the attic is
at a uniform temperature.
a) If the attic is sealed so that no outside air can leak in, set up but do not solve, the
electrical analogy for the heat transfer from the interior of the home to the outside
air through the attic. Estimate the numerical value of each resistance element in the
electrical analogy in terms of the attic floor area, Afloor.
b) If the outside air flows into the attic at a flow rate of Mo, set up, but do not solve,
the equations that will allow prediction of the attic air temperature. Clearly identify
the definition of all terms in your equations. Assume the air in the attic is well
mixed at a uniform temperature.
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3. 2) After he has conquered most of the architectural challenges throughout the world, it
might be appropriate for the Stata Center architect to help NASA design the future Space
based Stata Center. As usual, the design will be over the top requiring substantial electrical
power for many years of operation in space. It is proposed to use a reversible engine that
receives heat from an isotope heat source. The isotope provides heat to the engine at
1000K, T1
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4. On the low temperature side of the engine, heat is rejected by radiation heat transfer to
outer space (effectively at 0 K) by use of a black flat panel “radiator” at temperature T2.
Only one side of the panel acts as the radiator. Most of the power plant system weight is
due to the radiator panel, whose weight varies linearly with its area A2. The engine must
produce a continuous power of 1 MW. Note, in this problem the large temperature
difference between T2 and outer space precludes the use of the linearized form of the
radiation heat transfer.
a) If T2 is set at 300 K how large an area A2 is required for the system to generate
1 MW of power?
b) Consider different possible values for T2. Sketch a graph showing how q2, the
heat rejected by radiator, varies with T2 for a fixed power output and fixed T1.
Now sketch the ratio of A2/q2 as it varies with T2.
c) Indicate graphically how you would use the results of part b to find the value of
T2 that minimizes the radiator area A2 for a fixed power output and a fixed
value of the upper temperature T1.
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5. 3) In winter, a MIT conference room has 100 people gathered together for a meeting.
The current indoor air temperature is Tin=28o C, RHin=90%. Because there are heat
sources generated from people and equipment in the room, this conference room may
become hotter and hotter if we don’t supply cooling air into it.
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6. (a). Some people suggest supplying outside air directly into the space via diffusers
without any heating or moisture addition. The current outside air temperature is 0o C,
RH=80%. Is this strategy feasible? If yes, please give its advantages. What are its
drawbacks, and how would you improve it?
(b). The indoor air temperature becomes Tin=25o C, RHin =60% after one hour. In
order to keep this condition steady, we need to supply a constant flow rate of air into
this conference room. If the heat generated by each person is 80W, the vapor generated
by each person is 180g/hr, the heat generated by equipments is 5000W, the heat loss
through the room’s façade is 2000W, how much air should be supplied into the room if
we keep the supply air temperature as Ts=18o C? What should the relative humidity of
the supply air be to maintain the room RHin at 60%? Assume that the exhaust air has
same state as indoor air (Tin=25o C, RHin=60%). Assume that there isn’t any air
leakage or other heat transfer into the room. Specific heat of air, Cp=1.01kJ/kg.o C, is
constant between temperature 0o C to 30o C.
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7. 4 In a large air conditioner the refrigerant in the condenser transfers heat to cooling water
(see the picture). The cooling water flows to a cooling tower where it is sprayed as very
small water drops into the top of the cooling tower. Ambient air will be induced into the
cooling tower from the bottom by a fan installed on the top of the cooling tower. There
will be some water evaporation from the droplets and there will also be heat transfer
between the air and water in the cooling tower. Then the cooled water will be returned to
the air conditioner’s condenser.
We assume:
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8. a. The water pipe between the condenser and cooling tower is insulated so that there
isn’t any heat exchange with the outside environment.
b. The cooling tower façade is insulated very well and there isn’t any heat exchange with
outside environment.
c. The condenser is insulated so well that there isn’t any heat exchange between the
condenser and ambient environment. That is to say, the refrigerant only exchanges
heat with water in the condenser.
We know:
a. The water temperature before entering the cooling tower is Tw,1=35o C, and the
water flow rate is 60kg/s. Specific heat of water Cv=4.18kJ/kg.o C.
b. The air temperature before entering the cooling tower is Ta,1=25o C, RH=60%, and
its leaving temperature Ta,2=28o C, RH=100%. The flow rate of the air is 25kg/s.
c. The COP of the air conditioner is 3.0.
Questions:
a. What is the rate of water evaporation in the cooling tower (kg/s)?
b. Calculate the value of the temperature (Tw,2) of the water returning from the cooling
tower, before it enters the condenser.
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9. c. If the air conditioner is used in steady state to cool a room space, what is the rate of
cooling (Qc)?
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10. Problem 1:
We name the heat transfer coefficients as the following figure labels:
(a). The electrical analogy for the heat transfer from the interior of the home to the
outside air through the attic can be drawn as:
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11. Estimation of the numerical values of the resistance elements is shown in Table 1.
From Table 1, we see that the dominant resistance of the heat transfer through the attic
lies at the two conduction resistances of foam board and fiberglass insulation. This
finding may help us to understand question (b)’s solution.
Table 1. Estimation of the resistance elements (Unit: 1/[(Btu/hr.ft 2 .F).Afloor])
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12. b). We can choose the attic space (including the air in the attic) as our control system
(see the following figure). Then we can write the energy conservation equations as:
The equation that will allow prediction of the attic air temperature can be derived as:
where Rin and Rout are shown in part (a) solution.
Note: Two comparisons should be mentioned here:
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13. 1. When air goes through the attic, the convection heat transfer coefficients h2 and h3
will become close to “forced convection” heat transfer coefficients, which will be
much larger than the radiation coefficient h3r. Hence, the radiation h3r can be
neglected comparing with the “forced convection” because of their parallel
connection.
2. Comparing with Rin and Rout, the resistances of “forced convection” (1/h2(1.2Afloor)
and1/ h3Afloor) can be neglected because of their serial connection.
Finally, we can reach the above solutions for question (b).
Problem 2:
(a). Assume this power generation process is ideal. We have:
so the heat transfer through the flat black panel equals:
q2=0.3/0.7W=0.43 (MW)
This heat transfer q2 also equals: then
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14. (b). From (a), we get: then we can derive:
Considering the heat transfer through the flat black panel’s radiation, we also can get:
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15. (c) Because there has a relationship of A2=q2*A2/q2, we can get the figure of A2
which shows as Fig. 3. From the graph, we may get the minimum value of A2 at
fixed power output (W).
From another point view, we also can use numerical analysis to check the minimum area
A2. Combining equation (1) and (2), we can get:
A2 will reach its minimum value when
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16. Fig. 3 Getting the minimum value of area A2 from graph
Problem 3:
(a) This strategy is infeasible although it can save energy. The two main reasons are
listed as
1. If we supply the outside air (0 C, RH=80%) directly into the room, there will have
vapor condensation at diffuser or the places close to the diffuser, which w produce water
drops or frog in front of the diffuser. We can draw line between the indoor air state point
and outside air state point in the psychometric chart for mixing. It is obvious that this
mixing line crosses the 100% relative humidity line.
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17. 2. Supplying this so cold air directly into the space will generate ‘draft’ problems.
Especially for the people who stay under the diffusers, they will feel too cold and may
complain.
In order to improve this strategy, we can add a heat exchanger between the exhaust
air and the supply air. If we properly design the amount of energy and meet the
comfort requirement as well.
(b).
The indoor sensible heat sources equal to: 80W*100+5000W= 13 KW
Then the supply air should carry heat load as: Q=13kW-2000W=11 kW
For supplying air, we have:
From other side, the indoor moisture load equals to: mw=180g/hr*100=180kg/hr=5g/s.
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18. We know two parameters of the supply air, w and T. We can find the supply air state in
psychometric chart and read the RH around 69%.
Problem 4:
First, let’s read the air state parameters from psychometric chart:
Air before entering the cooling tower:
Ta,1=25o C, RH=60%, humidity ratio w1=12 g/kgair, enthalpy h1=56 kJ/kgair.
Air leaving the cooling tower:
Ta,1=28o C, RH=100%, humidity ratio w2=24.2 g/kgair, enthalpy h2=90 kJ/kgair.
(a). The water evaporation rate equals the vapor increasing rate in the air:
So the water evaporation rate in the cooling tower is 0.305kg/s.
(b). Choose the cooling tower as our controlled system, then we can write the energy
conservation equation as:
That is to say,
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19. Note: The water evaporation rate (0.305kg/s) is so small comparing with the air and
water flow te that the mass flow rate changes of the air and water flow are
negligible.
(c). The heat exchange between water and refrigerant in the condenser QH equals:
The air conditioning system’s COP=30, then:
So, we can calculate the cooling rate as:
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