Question

You need a rented apartment and you can see that the 3rd floors of each of the 2, 3 and 4 blocks of the buildings built in adjacent order as shown in the figure (front side) are for rent. You also learn that Block 2 is heated by a coal boiler operating at 75% efficiency, Block 3 is heated by a natural gas boiler operating at 92% efficiency, while Block 4 is heated by a heat pump with a COP of 4.5. All three blocks, as you know that the heating process takes place uninterruptedly for 150 days, the inside of the apartment is constantly constant at 22 ℃ for 150 days, the outside temperature is average ?∞ = 0 ℃ and the wind speed is average ∞ = 5 ? / ? Find the total heating cost for. Take the average heat transfer coefficient in the circle as ℎ?ç = 10 ? / ?2?, and the thermal resistance of the wall as ? wall = 1 ?2? / ?. Assume that heat is lost only from the front and back of the blocks. Also take the unit price of electricity as 0.71 TL / kWh, the unit price of natural gas as 0.14 TL / kWh, and the unit price of coal as 0.1 TL / kWh. (L1=15meter,L2=30 meter,L3=45 meter,L4=60 meter , 3.floor, H=4 meter

Kiralık bir daireye ihtiyacınız var ve şekilde gösterildiği gibi (ön yüzü) bitişik nizamda inşa edilmiş binalardan 2, 3 ve 4. blokların her birisinin 3. katlarının kiralık olduğunu görüyorsunuz. Ayrıca 2. Blok'un % 75 verimle çalışan kömür kazanı ile isıtıldığını, 3. Blok'un % 92 verimle çalışan doğalgaz kombisi ile Isıtıldığını, 4. Blok'un ise COP'si 4.5 olan isi pompası ile ile Isıtıldığını öğreniyorsunuz. Isitma işleminin 150 gün boyunca kesintisiz olarak gerçekleştiğini, bu 150 gün boyunca dairenin içinin sürekli olarak 22 °C'de sabit kaldığını, dış hava sıcaklığının ortalama To = 0 °C ve rüzgar hızının ortalama 2 = 5 m/s olduğunu bildiğinize göre her üç blok için toplam isitma maliyetini bulunuz. Daire içindeki ortalama isi taşınım katsayısını hiç = 10 W/m2K, duvarın ısıl direncini Rduvar = 1 m2K/W olarak alınız. Bloklann sadece ön ve arka yüzünden ısı kaybı olduğunu kabul ediniz. Ayrıca elektrik birim fiyatını 0.71 TL/KWh, doğalgaz birim fiyatını 0.14 TL/KWh, kömürün birim fiyatını 0.1 TL/KWh olarak alınız. NOT: Çözümde eğer soruda verilmeyen bir kabul yapacaksınız bunu açık bir şekilde ifade ediniz. Rüzgar 3. kat H=4 metre u = 5 m/s T = 0°C 1. Blok 2. Blok 3. Blok 4. Blok 5. Blok = 15 m L2 = 30 m La = 45 m LA = 60 m

Answer 1

Firstly, we shall calculate the average heat transfer coefficient of outside air.

It is given that only 3 blocks add heat to the outside air, therefore, the total length of blocks through which heat is added to the outside air is 15 + 15 + 15 = 45 m

This can be considered as the vertical plate subjected to forced convection.

Air > u = 5 m/s T(air) = 0 C > Heating blocks 2. 3 4 5 Blocks 1

So, first, we will calculate the Reynold's number of the flow.

ᏙᎠ Re V

where

• V = Velocity of flow = 5 m/s
• D = Length of heat conducting blocks = 45 m
• $\nu$ = kinematic viscosity of air at 0 C = 13.28e-6 m2/s

ᏙᎠ Re V

Re 5 x 45 13.28e - 6 16942771.08

The FLOW IS TURBULENT.

From the heat transfer data book, the appropriate Nusselt's Number relation is as follows:

Nu = 0.037 0,037 X Re0.8 x P7033

where

• Re = 16942771.08
• Pr = Prandtl number for air at 0 C = 0.7362 (from EES software)

Nu = 0.037 0,037 X Re0.8 x P7033

0.037 x 16942771.080.8 x 0.73620.3 20300.087

Now, we know,

hD Nu =

where

• D = 45 m
• k = thermal conductivity of air at 0 C = 0.02364 W/mK (from EES software)

hD Nu =

20300.087 h x 45 0.02364 10.664 W/m²K

Now, for all the heat producing blocks, the following thermal resistances occur in series:

1. internal convectiion (R1)
2. conduction through wall (R2)
3. External convection (R3)

-----

Internal convection (R1)

Thermal resistance =

R1 hA

where

• h = internal convection coeff = 10 W/m2K
• A = Area of a single wall = 4 x 15 = 60 m2

Thermal resistance =

R1 1.667e - 3 K/W 10 x 60

-----

Conduction through wall (R2)

Thermal resistance =

R R2

where

• R = resistance = 1 m2K/W (given)
• A = 60 m2

Thermal resistance =

R2 0.01667 K/W 60

-----

External convection (R3)

Thermal resistance =

R3 = hA

where

• h = outside heat transfer coeff = 10.664 W/m2K
• A = 60 m2

Thermal resistance =

R3 1.5629e - 3 K/W 10.664 x 60

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So, heat required from one wall =

- Tair Q Tin R1 + R2 + R3

where

• T(in) = 22 C
• T(air) = 0 C

- Tair Q Tin R1 + R2 + R3

Q= 22 – 0 1.667e – 3 + 0.01667 +1.5629e – 3 1105.534 W

Now, the building has 3 front facing walls and 3 rear facing walls.

Therefore, net heat lost from ONE floor of ONE block is 1105.534 x 2 = 2211.067 W

Now, the following table applies:

	Block 2	Block 3	Block 4
Net heat loss rate	2211.067 W	2211.067 W	2211.067 W
Heating medium	Coal Boiler	Natural gas	Heat pump
Efficiency/COP	0.75	0.92	4.5
Actual energy input	2211.067 / 0.75 = 2948.089 W	2211.067 / 0.92 = 2403.333 W	2211.067 / 4.5 = 491.348 W
No. of days working	150	150	150
Total energy consumption	3.8207e10 J	3.1147e10 J	6367872960 J
No of kWh	10613.055	8651.944	1768.854
Unit cost	0.1$ / kWh	0.14 $ / kWh	0.71 $ / kWh
Net cost	1061.306 $	1211.273 $	1255.89 $
TOTAL COST	$ 3528.465 / 150 DAYS

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