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21. Calculate energy transmitted into a building at 3 P.M on July 21st due to solar radiation through a south facing window made of regular single glass. The dimensions of the window are height 2 m, width 1.5 m and the depth of inset 0.3 m. Find the energy transmitted if there is no overhang.

Answer:

From the above data the altitude angle $\beta$ and wall solar azimuth angle $\angle$ are found to be:

$\beta = 48.23^o, \alpha = 39.87^o$

Therefore area of the unshaded portion = x $\times$ y, where x and y are given by:

$x = W − d(\tan \alpha) =1.5 − 0.3(\tan39.87) =1.249 m$

$y=H-d\left(\dfrac{\tan\beta}{\cos\alpha}\right)=2.0-0.3\left(\dfrac{\tan48.23}{\cos39.87}\right)=1.562m$

$\therefore$ The heat transmission rate into the building through the unshaded portion $Q_{us}$ is given by:

$Q_{us} = (x.y).(SHGF_{max} ).SC = (1.249\times1.562)\times 230 \times1.0 = 448.7 W$

The heat transmission rate into the building through the unshaded portion $Q_{ss}$ is given by:

$Q_{ss} = (W.H − x.y).(SHGF_{max ,N} ).SC = (1.049) \times120 \times1.0 =125.9 W$

Hence the total amount of radiation transmitted into the building, $Q_{sg}$ is given by:

$Q_{sg} = Q_{us} + Q_{ss} = 574.6 W$ (Ans.)

Without overhang the heat transmission rate is:

$Q_{sg} = (W \times H)SHGF_{max} = 690 W$ (Ans.)

Thus there is a reduction of 115.4 W (16.7%) due to external shading. Of course, these values will be different for different periods.


22. A large air conditioned building with a total internal volume of 1,00,000 m$^3$ is maintained at 25$^o$C (DBT) and 50% RH, while the outside conditions are 35$^o$C and 45% RH. It has a design occupancy of 10,000 people, all nonsmoking. The infiltration rate through the building is equal to 1.0 ACH. Estimate the heat transfer rate due to ventilation and infiltration. Assume the barometric pressure to be 1 atm.

Answer:

From psychrometric chart:

For inside conditions: 24$^o$C (DBT) and 50% RH:

$W_i = 0.0093$ kgw/kgda, $h_i = 47.656$ kJ/kgda

For outside conditions: 35$^o$C (DBT) and 45% RH:

$W_o = 0.01594$ kgw/kgda, $h_o = 75.875$ kJ/kgda and $v_a = 0.89519 m^3$/kg

Heat transfer due to ventilation:

From Table 33.3, assume a ventilation requirement of 3.5 l/s/person. Hence the total OD air required is:

$V_{o,v} = 3.5 \times 10,000 = 35000 l/s = 35 m^3/s$

Hence the mass flow rate of ventilated air is:

$m_{o,v} = 35 /0.89519 = 39.1$ kg/s

Sensible heat transfer rate due to ventilation is given by:

$Q_{s,v} = m_{o,v}c_{pm}(t_o – t_i) = 39.1 \times 1.0216 \times (35 – 25) = 399.5$ kW

Latent heat transfer rate due to ventilation is given by:

$Q_{l,v} = m_{o,v}h_{fg}(W_o – W_i) = 39.1 \times 2501 \times (0.01594 – 0.0093) = 649.3$ kW

Hence total heat transfer rate due to ventilation is:

$Q_{t,v} = Q_{s,v} + Q_{l,v} = 1048.8$ kW (Ans.)

Heat transfer rate due to infiltration:

Infiltration rate, $V_{inf} = 1 ACH = 1,00,000/3600 = 27.78 m^3/s$

Hence mass flow rate of infiltrated air is:

$m_{inf} = V_{inf}/v_a = 27.78/0.89519 = 31 kg/s$

Hence using expressions similar to ventilation, the sensible, latent and total heat transfer rates due to infiltration are found to be:

$Q_{s,inf} = 316.7$ kW (Ans.)

$Q_{l,inf} = 514.8$ kW (Ans.)

$Q_{t,inf} = 831.5$ kW (Ans.)

It can be seen from the above example that the total load on the air conditioning system is very high ( = 1880.3 kW = 534.6 TR).


23. A building has to be maintained at 21$^o$C(dry bulb) and 50% relative humidity when the outside conditions are -30$^o$C(dry bulb) and 100% relative humidity. The inner and outer surface heat transfer coefficients are 8.3 W/m$^2$.K and 34.4 W/m$^2$.K, respectively. A designer chooses an insulated wall that has a thermal resistance (R-value) of 0.3 m$^2$.K/W. Find whether the wall insulation is sufficient to prevent condensation of moisture on the surface. If the chosen R-value of the wall can lead to condensation, what is the minimum thickness of additional insulation (thermal conductivity 0.036 W/m.K) required to prevent condensation. Take the barometric pressure as 101 kPa.

Answer:

From the psychrometric chart; for inside conditions of 21$^o$C and 50% RH:

Dew Point Temperature, $T_{DPT,i} = 10^\circ C$

The overall heat transfer coefficient for the wall U is given by:

$U=\left[R_{wall}+\left(\dfrac 1{h_i}\right)+\left(\dfrac1{h_o}\right)\right]^{-1}=\left[0.3+\left(\dfrac 1{8.3}\right)+\left(\dfrac1{34.4}\right)\right]^{-1}=2.224W/m^2.$K

Assuming steady state, the heat transfer rate through the wall is given by:

$q_w=U(T_i-T_o)=2.224\times(21-(-30))=113.424W/m^2$

The temperature of the inner surface of the wall, $T_{s,i}$ is obtained using the equation:

$q_w=h_i(T_i-T_{s,i})=113.424\Rightarrow T_{s,l}=7.33^\circ C$

Since, $T_{s,i}\lt T_{DPT,i}$

$\Rightarrow$ Condensation will take place on the inner surface of the wall. (ans)

To prevent condensation, the minimum allowable temperature of the inner surface is the DPT (10$^\circ$C).

Under this condition, the maximum allowable heat transfer rate is given by:

$q_{w, \text{allowable}}=h_i(T_i-T_{DPT,i})=8.3\times(21-10)=91.3W/m^2$

Hence the required $U_{req}$ value is:

$U_{req}=91.3/(T_i-T_o)=91.3/(21-(-30))=1.79W/m^2.$K

Hence the required resistance of the wall, $R_{w, req}$ is given by:

$R_{w,req} = (1/U_{req}) – (1/h_i) – (1/h_o) = 0.4091 m^2$.K/W

Hence the amount of additional insulation to be added is:

$R_{add} = (t_{add}/k_{add}) = 0.4091 – R_{wall} = 0.4091 – 0.3 = 0.1091 m^2.$K/W

$\Rightarrow$ Required insulation thickness, $t_{add} = 0.1091 \times 0.036 = 3.928\times10^{-3} m$ (Ans.)


24. A 4m $\times$ 5m wall consists of 3 glass windows of 1.5m $\times$ 1.0 m dimensions. The wall has thickness of 0.125 m and a thermal conductivity of 0.5 W/m.K, while the glass windows are 6 mm thick with a thermal conductivity of 1.24 W/m.K. The values of internal and external surface conductance for the wall (including glass) are 8.3 W/m$^2$.K and 34.4 W/m$^2$.K, respectively. The internal and external temperatures are 21$^o$C and –30$^o$C, respectively. Calculate the total heat transfer rate through the wall. What percentage of this heat transfer is through the windows?

Answer:

The total heat transfer rate through the wall is given by:

$Q_{total} = U_oA_{total}(T_i – T_o)$

The value of $U_oA_{total}$ is given by:

$U_oA_{total} = U_{wall}A_{wall} + U_{glass}A_{glass}$

The U values for the wall and glass are obtained from their individual resistance values as:

$U_{wall}=[(0.125/0.5)+(1/8.3)+(1/34.4)]^{-1}=2.503E/m^2.$K

$U_{wall} = [(0.125/0.5) + (1/8.3) + (1/34.4)]^{-1} = 2.503 W/m^2$.K

$U_{glass} = [(0.006/1.24) + (1/8.3) + (1/34.4)]^{-1} = 6.48 W/m^2$.K

The area of glass, $A_{glass} = 3 \times 1.5 \times 1.0 = 4.5 m^2$

The area of wall, $A_{wall} = 4 \times 5 - 4.5 = 15.5 m^2$

Hence, $U_oA_{total} = U_{wall}A_{wall} + U_{glass}A_{glass} = 2.503 \times 15.5 + 6.48 \times 4.5 = 67.96$ W/K

Hence, $Q_{total} = U_{wall} A_{total} (T_i – T_o) = 67.96 (21+30) = 3465.96 W$ (Ans.)

% of heat transfer rate through glass = ${U_{glass}A_{glass}(T_i – T_o)/Q_{total}}\times100 = 42.9$% (ans.)


25. A multi-layered wall consists (from inside to outside) 6mm thick plywood, 125 mm thick common brick, 2.1 mm thick air space, 125 mm thick common brick and 6 mm thick cement plaster. The values of internal and external surface conductance for the wall are 8.3 W/m$^2$.K and 34.4 W/m$^2$.K, respectively. Find the overall heat transfer coefficient of the wall. What is the value of U, if the air space is replaced by 20 mm thick EPS board? Assume the temperature difference across the air space to be 10 K.

Answer:

For the composite wall, the overall heat transfer coefficient U is given by:

$\left(\dfrac 1U\right)=R_{tot}=\left(\dfrac1{h_i}\right)+\sum_{i=1}^{N} \left(\dfrac{\Delta x_i}{k_{w,i}}\right)+\sum_{j=1}^{M} \left(\dfrac1{C_j}\right)+\left(\dfrac1{h_o}\right)$

Substituting the values of individual resistances using the input values of wall thickness and thermal conductivity and thermal conductance (From Tables 34.2 and 34.3), the overall heat transfer coefficient is given by:

$\left(\dfrac 1U\right)=\left(\dfrac 1{8.3}\right)+\left(\dfrac {0.006}{0.1}\right)+\left(\dfrac {0.125}{0.77}\right)+\left(\dfrac 1{5.8}\right)+\left(\dfrac {0.125}{0.77}\right)+\left(\dfrac {0.006}{8.65}\right)+\left(\dfrac 1{34.4}\right)=0.7073m^2$K/W

$\Rightarrow U=1.414W/m^2$.K (ans)

If the air space is replaced by 20 mm EPS (k = 0.037 W/m.K), then the new U-value is:

$U_{EPS}=[(1/U)-(1/5.8)+(0.02/0.037)]^{-1}=0.93W/m^2.$K (ans)

Thus replacing the air gap with EPS leads to a decrease in the U-value by about 34 percent.


26. Determine the sol-air temerature for a flat roof if the direct radiation normal to the sun’s rays ($I_{DN}$) is 893 W/m$^2$ and the intensity of scattered radiation normal to the roof ($I_d$ ) is 112 W/m$^2$. Take the absorptivity of the roof for direct and scattered radiation as 0.9, the heat transfer coefficient of the outside surface as 34.4 W/m$^2$, the outside air temperature as 37$^o$C and the solar altitude angle as 80$^o$. If the time lag of the roof structure is zero and its decrement factor is unity, calculate the heat gain to the room beneath the roof if the U-value of the roof is 0.5 W/m2.K and the room temperature is 25$^o$C.

Answer:

For a flat roof, the angle of incidence $\theta$ is given by:

$\theta=(\pi/2)-\beta=(\pi/2)-80=10^\circ$

where $\beta$ is the altitude angle

Total solar irradiation on the flat roof It is given by:

$I_t = I_{DN}.\cos (\theta) + I_d = 893 \times \cos (10) + 112 = 991.43 W/m^2$

Hence the sol-air temperature is given by:

$T_{sol-air}=T_o+\left(\dfrac{\alpha_DI_D+\alpha_dI_d-R}{h_0}\right)=37+\dfrac{0.9\times991.43}{34.4}=62.94^\circ$ (ans)

Since the time lag is 0 and decrement factor is 1.0 for the roof, the heat transfer rate through the roof is given by:

$q=U(T_{sol-air}-T_i)=18.97W/m^2$ (ans)


27. A building has its north, west facing walls and the roof exposed to sun. The dimensions of the building are 12 m $\times$ 12 m $\times$ 5 m ($W\times L\times H$). The U-value of the walls are 0.5 W/m$^2$.K, while it is 0.4 W/m$^2$.K for the roof. There are no windows on north and west walls, and the other two walls are exposed to air conditioned spaces. The outside design temperature is 41$^o$C while the indoor is maintained at 25$^o$C, while the average temperature for the design day is 31$^o$C. Calculate heat transfer rate to the building at 5 P.M., 6 P.M and & 7 P.M. Assume the walls are of D-Type and the roof is of Type 5.

Answer:

Since the average outside temperature is different from 29$^o$C, adjustments have to be made to the values obtained from the CLTD tables.

$CLTD_{adj} = CLTD_{Table} + (T_{av} – 29) = CLTD_{Table} + 2$

a) Heat transfer rate through the roof:

From the Table of CLTD values for roof (Table 34.5), the CLTD values at 5 P.M., 6 P.M.and 7 P.M. are 29$^o$C, 30$^o$C and 29$^o$C, respectively

$\therefore Q_{roof,5 P.M.} = U_{roof}A_{roof}CLTD_{adj,5 P.M.} = 0.4 \times 144 \times (29 + 2) = 1785.6 W$

$Q_{roof, 6.P.M.} = U_{roof}A_{roof}CLTD_{adj,6 P.M.} = 0.4 \times 144 \times 32 = 1843.2 W$

$Q_{roof, 7 P.M.} = Q_{roof, 5 P.M.} = 1785.6 W$ (as the CLTD values are same)

b) Heat transfer rate through north facing wall:

Table 34.6 is used for obtaining CLTD values for the walls

$Q_{north, 5 P.M.} = U_{wall}A_{wall}CLTD_{adj, 5 P.M.} = 0.5 \times60 \times 10 = 300 W$

$Q_{north, 6. P.M.} = U_{wall}A_{wall}CLTD_{adj, 6 P.M.} = 0.5\times 60 \times 11 = 330 W$

$Q_{north,7 P.M.} = U_{wall}A_{wall}CLTD_{adj, 7 P.M.} = 0.5 \times 60 \times 12 = 360 W$

c) Heat transfer rate through the west facing wall:

Similar to the north facing wall, the heat transfer rates through the west facing walls are found to be:

$Q_{west,5 P.M} = 450 W$

$Q_{west,6 P.M} = 570 W$

$Q_{west,7 P.M} = 660 W$

Total heat transfer through the building is:

$Q_{total, 5 P.M.} = 1785.6 + 300 + 450 = 2535.6 W$ (Ans.)

$Q_{total, 6 P.M. }= 1843.2 + 330 + 570 = 2743.2 W$ (Ans.)

$Q_{total, 7 P.M.} = 1785.6 + 360 + 660 = 2805.6 W$ (Ans.)

Comments:

  1. The difference in design dry bulb temperature between outdoor and indoor is 17$^o$C, it is observed that the CLTD value ranges between 31 to 32$^o$C for the roof, 10 to 12$^o$C for the north facing the wall and 15 to 22$^o$C for the west facing wall. The difference between CLTD values and $(T_o – T_i)_{design}$ is due to varying outdoor temperatures, varying solar radiation and finally due to the thermal capacity of the walls.

  2. It is seen that the maximum amount of heat transfer rate is through the roof, hence, putting additional insulation on the roof will reduce the cooling load.

  3. Due to the thermal lag effect of the building, the peak heat transfer takes place not during sunshine, but after sunset.


28. A building has a U-value of 0.5 W/m$^2$.K and a total exposed surface area of 384m$^2$. The building is subjected to an external load (only sensible) of 2 kW and an internal load of 1.2 kW (sensible). If the required internal temperature is 25$^o$C, state whether a cooling system is required or a heating system is required when the external temperature is 3°C. How the results will change, if the U-value of the building is reduced to 0.36 W/m.K?

Answer:

From the energy balance,

$T_{out, bal}=T_{in}-\dfrac{(Q_{solar}+Q_{int})_{sensible}}{UA}=25-\dfrac{(2+1.2)\times 1000}{0.5\times 384}=8.33^\circ C$

Since the outdoor temperature at balance point is greater than the external temperature ($T_{ext} \lt T_{out,bal}$);

the building requires heating (Ans.)

When the U-value of the building is reduced to 0.36 W/m.K, the new balanced outdoor temperature is given by:

$T_{out, bal}=T_{in}-\dfrac{(Q_{solar}+Q_{int})_{sensible}}{UA}=25-\dfrac{(2+1.2)\times 1000}{0.36\times 384}=1.85^\circ C$

Since now the outdoor temperature at balance point is smaller than the external temperature ($T_{ext}\gt T_{out,bal}$);

the building now requires cooling (Ans.)

The above example shows that adding more insulation to a building extends the cooling season and reduces the heating season.


29. An air conditioned room that stands on a well ventilated basement measures 3 m wide, 3 m high and 6 m deep. One of the two 3 m walls faces west and contains a double glazed glass window of size 1.5 m by 1.5 m, mounted flush with the wall with no external shading. There are no heat gains through the walls other than the one facing west. Calculate the sensible, latent and total heat gains on the room, room sensible heat factor from the following information. What is the required cooling capacity?

Inside conditions : 25$^o$C dry bulb, 50 percent RH

Outside conditions : 43$^o$C dry bulb, 24$^o$C wet bulb

U-value for wall : 1.78 W/m$^2$.K

U-value for roof : 1.316 W/m$^2$.K

U-value for floor : 1.2 W/m$^2$.K

Effective Temp. Difference (ETD) for wall: 25$^o$C

Effective Temp. Difference (ETD) for roof: 30$^o$C

U-value for glass ; 3.12 W/m$^2$.K

Solar Heat Gain (SHG) of glass ; 300 W/m$^2$

Internal Shading Coefficient (SC) of glass: 0.86

Occupancy : 4 (90 W sensible heat/person) (40 W latent heat/person)

Lighting load : 33 W/m$^2$ of floor area

Appliance load : 600 W (Sensible) + 300 W(latent)

Infiltration : 0.5 Air Changes per Hour

Barometric pressure : 101 kPa

Answer:

From psychrometric chart,

For the inside conditions of 25$^o$C dry bulb, 50 percent RH:

$W_i = 9,9167 \times 10^{-3}$ kgw/kgda

For the outside conditions of 43$^o$C dry bulb, 24$^o$C wet bulb:

$W_o = 0.0107$ kgw/kgda, density of dry air = 1.095 kg/m$^3$

External loads:

a) Heat transfer rate through the walls: Since only west wall measuring 3m $\times$ 3m with a glass windows of 1.5m $\times$ 1.5m is exposed; the heat transfer rate through this wall is given by:

$Q_{wall} = U_{wall}A_{wall}ETD_{wall} = 1.78 \times (9-2.25) \times 25 = 300.38 W$ (Sensible)

b) Heat transfer rate through roof:

$Q_{roof} = U_{roof}A_{roof}ETD_{roof} = 1.316 \times 18 \times 30 = 710.6 W$ (Sensible)

c) Heat transfer rate through floor: Since the room stands on a well-ventilated basement, we can assume the conditions in the basement to be same as that of the outside (i.e., 43$^o$C dry bulb and 24$^o$C wet bulb), since the floor is not exposed to solar radiation, the driving temperature difference for the roof is the temperature difference between the outdoor and indoor, hence:

$Q_{floor} = U_{floor}A_{floor}ETD_{floor} = 1.2 \times 18 \times 18 = 388.8 W $(Sensible)

d) Heat transfer rate through glass: This consists of the radiative as well as conductive components. Since no information is available on the value of CLF, it is taken as 1.0. Hence the total heat transfer rate through the glass window is given by:

$Q_{glass} = A_{glass} [U_{glass}(T_o−T_i)+SHGF_{max}SC] = 2.25[3.12 \times 18 + 300 \times 0.86] = 706.9 W$ (Sensible)

e) Heat transfer due to infiltration: The infiltration rate is 0.5 ACH, converting this into mass flow rate, the infiltration rate in kg/s is given by:

$m_{inf} =\text{ density of air }\times (ACH \times\text{ volume of the room})/3600 = 1.095 \times (0.5 \times 3\times3\times6)/3600$

$m_{inf} = 8.2125 \times 10^{-3} kg/s$

Sensible heat transfer rate due to infiltration,$Q_{s,inf}$;

$Q_{s,inf} = m_{inf}c_{pm}(T_o−T_i) = 8.2125 \times 10^{-3}\times 1021.6 \times (43 – 25) = 151 W$ (Sensible)

Latent heat transfer rate due to infiltration, $Q_{l,inf}$:

$Q_{l,inf} = m_{inf}h_{fg}(W_o−W_i) = 8.8125\times10^{-3}\times 2501\times10^3(0.0107−0.0099)=16.4 W$ (sensible)

Internal loads:

a) Load due to occupants: The sensible and latent load due to occupants are:

$Q_{s,occ} =\text{ no.of occupants }\times SHG = 4 \times 90 = 360 W$

$Q_{l,occ} =\text{ no.of occupants }\times LHG = 4\times 40 = 160 W$

b) Load due to lighting: Assuming a CLF value of 1.0, the load due to lighting is:

$Q_{lights} = 33\times\text{ floor area }= 33 \times 18 = 594 W$ (Sensible)

c) Load due to appliance:

$Q_{s,app} = 600 W$ (Sensible)

$Q_{l,app} = 300 W$ (Latent)

Total sensible and latent loads are obtained by summing-up all the sensible and latent load components (both external as well as internal) as:

$Q_{s,total} = 300.38+710.6+388.8+706.9+151+360+594+600 = 3811.68 W$ (Ans.)

$Q_{l,total} = 16.4+160+300 = 476.4 W$ (Ans.)

Total load on the building is:

$Q_{total} = Q_{s,total} + Q_{l,total} = 3811.68 + 476.4 = 4288.08 W$ (Ans.)

Room Sensible Heat Factor (RSHF) is given by:

$RSHF = \dfrac{Q_{s,total}}{Q_{total}} = \dfrac{3811.68}{4288.08} = 0.889$ (Ans.)

To calculate the required cooling capacity, one has to know the losses in return air ducts. Ventilation may be neglected as the infiltration can take care of the small ventilation requirement. Hence using a safety factor of 1.25, the required cooling capacity is:

Required cooling capacity = 4288.08$\times$ 1.25 = 5360.1 W ≈ 1.5 TR (Ans.)

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