11127-10 exercises lecture 07 solutions

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					Exercises on Domestic Hot Water Systems Lecture 07
Course 11127 Sustainable Heating and Cooling of Buildings                                       03.2010

BYG.DTU - 11127 Sustainable Heating and Cooling of Buildings

           Solutions to Exercises on DHW Systems (lecture 07)
Scenario 1:
1) If a flow through heat exchanger is used instead of the hot water tank to provide hot water,
determine the effective power of the heat exchanger (heat loss from the heat exchanger and the
distribution pipes can be neglected).
 Answer:
The effective power of the heat exchanger is determined by the power needed to heat up the cold
water to hot water temperature at the maximum flow rate.

Peff  4.2  qV  TV  TK 
qv in l/s, Tv and Tk in ºC.
Peff  4.2  6 / 60  55  10  =18.9      [kW]

Scenario 2:
2) Determine the necessary tank volume.
Answer:
Energy balance of the tank must be analyzed after each hot water draw-off. The following equation
must be observed.
E1+(Peff ·Δt- Pheat loss ·Δt) ≥ Etap
E1≥ Etap - (Peff ·Δt- Pheat loss ·Δt)

E1 is the effective energy content of the tank at the start of the tapping program. [kWh]
Peff, effective charging power of the district heating net, [kW].
Pheat loss, total heat loss from the hot water tank and the distribution pipes, [kW].
Δt, duration from the start of the tapping program to the end of the hot water draw-off, [h].

In order to determine E1, the heat loss from the hot water tank and the distribution pipes must be
calculated.
Heat loss from the top of the tank:
           
               d   y     es 
                                2

U t op  4
               et         1
                    
                        atop
Heat loss from the side of the tank:
                                        
U side                                     H
         1               d y  2es
                          1        1
            ln         
        2       dy      aside d y  2es
Heat loss from the bottom of the tank:
               
                    d   y    es 
                                    2

U bottom  4
                 1  eb
                         
             abot
dy: outer diameter of the tank.
H : height of the tank.

                                                   -1-
Exercises on Domestic Hot Water Systems Lecture 07
Course 11127 Sustainable Heating and Cooling of Buildings                                        03.2010

αtop, αbot, αside: heat transfer coefficient of the top (10 W/m2K) , the bottom 5.88 W/m2K and the side
surface of the tank (7.69 W/m2K) respectively.
λ, Thermal conductivity of the insulation materials, 0.04 W/m/K.
et, es, eb, Thickness of insulation materials at the top, the side and the bottom of the tank respectively
(0.05 m).
Volume of the tank has to be known before the heat loss of the tank can be calculated, but volume
of the tank is unknown. It will be determined by energy balance analysis. An iteration process is
therefore needed. It is assumed that the tank volume is 124 l.
The H/ dy ratio is 2.8.
Volume of the tank v=H*(3.14/4* dy2)=2.8*D(3.14/4* dy 2)=2.198* dy 3
dy =(0.124/2.198)1/3=0.384 m
Heat loss from the top of the tank:
                                    
             d   y     et 
                             2
                                         0.384  0.052
U t op  4                       =4         =0.109                   W/K
             et 1           0.05 1
                                 
              atop         0.04 10
Heat loss from the side of the tank:
                         
U side                                     H=
          1    d y  2es     1        1
            ln            
         2        dy       aside d y  2es
                         
                                                               * 2.8 * 0.384 =1.068 [W/K]
    1       0.384  2 * 0.05    1          1
         ln                  
2 * 0.04        0.384          7.69 0.384  2 * 0.05
Heat loss from the bottom of the tank:
                                        
                  d   y    es 
                                 2
                                             0.384  0.052
U bottom  4          =4               =0.104                        [W/K]
                 1eb      0.05    1
                              
             abot        0.04 5.88
U=0.109+1.068+0.104=1.28              [W/K]
Heat loss from the tank=1.28*(40-20)=25.6                            [W]
                          = 0.0256                                   [kW]

Heat loss from the distribution pipes can be obtained from exercise of lecture 06.
QW , dis ,ls  1.19  14 .06  15 .25                                [MJ/day]
               =15.25*1000/(24*60*60)=0.177                          [kW]
Pheat loss=0.203          [kW]

Energy balance of the tank must be analyzed after each hot water draw-off. As example, the energy
balance of the tank is analyzed at the end of the first draw-off, see the tapping program No.3.
The first draw-off has a tapped energy of 0.105 kWh, the equivalent volume at a temperature of
55C can be calculated:
     T  V             (55  10)V
 Q           0.105 
       860                    860
 v=2.01       [l]
The hot water tapping flow rate for small tapping is 3 l/min. Duration from the start of the tapping
program to the end of the hot water draw-off, Δt, is 2.01/3=0.67 [min]=0.0112 [h]
E1≥ Etap - (Peff ·Δt- Pheat loss ·Δt)
                                                               -2-
Exercises on Domestic Hot Water Systems Lecture 07
Course 11127 Sustainable Heating and Cooling of Buildings                                  03.2010

E1≥ 0.105 - (0.8 ·0.0112- 0.203 ·0.0112)=0.10           [kWh]

The energy balance is then analyzed at the end of each draw-off and the effective energy content of
the tank E1 is calculated, see the following table. The effective energy content of the tank is the
max. E1 otherwise the tapping program can not be fulfilled.
E1=4.63                       [kWh]
           860            860
Veff  E      = 4.63             =88.48 [l]
           T           (55  10 )
V = 1.4  Veff =88.48*1.4=124 [l]

Since the calculated tank volume is the same as the assumed, the necessary hot tank volume is 124
litres.

                                Tapped       Tapping      Time from
            Type       Time     Energy      flow rate        start     Etap         E1       Veff
 hold          -      hh:mm      kWh          l/min          min       kWh         kWh       liter
   1        small      07:00     0.105          3             0.7      0.105       0.10       1.9
   2       shower      07:05      1.4           6             9.5      1.505       1.41      26.9
   3        small      07:30     0.105          3            30.7      1.61        1.30      24.9
   4        small      07:45     0.105          3            45.7      1.715       1.26      24.1
   5        bath       08:05     3.605          6            76.7      5.32        4.56      87.1
   6        small      08:25     0.105          3            85.7      5.425       4.57      87.4
   7        small      08:30     0.105          3            90.7      5.53        4.63      88.4
   8        small      08:45     0.105          3           105.7      5.635       4.58      87.6
   9        small      09:00     0.105          3           120.7      5.74        4.54      86.7
  10        small      09:30     0.105          3           150.7      5.845       4.34      83.0
            floor
 11       cleaning    10:30      0.105          3           210.7      5.95        3.85      73.6
 12         small     11:30      0.105          3           270.7      6.055       3.36      64.2
 13         small     11:45      0.105          3           285.7      6.16        3.31      63.3
 14     dishwashing   12:45      0.315          4           346.5      6.475       3.02      57.8
 15         small     14:30      0.105          3           450.7      6.58        2.09      39.9
 16         small     15:30      0.105          3           510.7      6.685       1.60      30.5
 17         small     16:30      0.105          3           570.7      6.79        1.10      21.1
 18         small     18:00      0.105          3           660.7      6.895       0.31       6.0
         household
 19       cleaning    18:15      0.105          3           675.7        7         0.27       5.1
         household
 20       cleaning    18:30      0.105          3           690.7      7.105        0.22      4.3
 21         small     19:00      0.105          3           720.7       7.21        0.03      0.5
 22     dishwashing   20:30      0.735          4           813.6      7.945       -0.16     -3.1
 23         bath      21:00      3.605          6           851.7      11.55        3.06     58.5
 24         small     21:30      0.105          3           870.7      11.655       2.98     56.9

Energy balance for a 24 h period is analyzed,
Peff ≥ (Etap +Pheat loss ·24)/ 24
Etap is the tapped energy in a 24 h period, 11.655 [kWh].

                                                -3-
Exercises on Domestic Hot Water Systems Lecture 07
Course 11127 Sustainable Heating and Cooling of Buildings                                       03.2010

Peff, effective charging power of the district heating net, 0.8 [kW].
 Pheat loss, total heat loss from the hot water tank and the distribution pipes, 0.203 [kW].
Δt, 24 [h].
 Peff=0.8 > (11.655+0.203*24)/24=0.69
It means that the effective charging power of the district heating net is large enough.

3) Determine the daily energy output from the district heating net to the house.

The daily energy output is calculated using Eq. 1 in EN 15316-3-3,
QW , gen,out  QW  QW ,dis ,ls  QW , st ,ls  QW , p ,ls
QW,gen,out      Total generation output [MJ/day]
QW              Domestic hot water requirement calculated in EN 15316-3-1, [MJ/day].
The daily tapped energy is 11.655 [kWh/day] = 42 [MJ/day]
QW,dis,ls       Thermal loss from domestic hot water distribution system [MJ/day].
From the exercise of lecture 06, the daily thermal loss from the distribution system is 15.25
[MJ/day].
QW,st,ls        Thermal loss from the domestic hot water storage vessel [MJ/day].
From 7:00 to 21:30, water in the hot water tank has an average temperature of 40ºC, the heat loss
from the tank is 0.0256 kW.
The heat loss of the tank is 0.0256*1000*14.5*60*60/106=1.34 [MJ/day]
For the rest of the day, the tank is on standby. The average water temperature in the tank is 55ºC.
Heat loss from the tank=1.28*(55-20)=44.8                              [W]
The heat loss of the tank is 0.0448*1000*(24-14.5)*60*60/106=1.53 [MJ/day]
There is no need to consider thermal loss from primary pipes.
QW , gen,out  QW  QW ,dis ,ls  QW ,st ,ls  QW , p ,ls  42  15 .25  1.34  1.53  60 .1 [MJ/day]




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