Numericals on RAC-2
1 Answer

11. An air conditioned building has a sensible cooling load of 60 kW and latent load of 40 kW. The room is maintained at 24$^o$C (DBT) and 50% RH, while the outside design conditions are: 34$^o$C (DBT) and 40% RH. To satisfy the ventilation requirement, outdoor air is mixed with re-circulated air in the ratio of 1:3 (by mass). Since the latent load on the building is high, a reheat coil is used along with a cooling and dehumidifying coil. Air is supplied to the conditioned space at 14$^o$C (DBT). If the bypass factor of the cooling coil is 0.15 and the barometric pressure is 101.325 kPa, find: a) Mass flow rate of supply air, b) Required cooling capacity of the cooling coil and heating capacity of the reheat coil.


From psychrometric chart, the following properties are obtained:

Inside conditions: $_i =24^oC(DBT)$ and $RH_i = 50$%

$W_i = 0.0093$ kgw/kgda, $h_i = 47.66$ kJ/kgda

outside conditions: $t_o =34^oC(DBT)$ and $RH_o = 40$%

$W_o = 0.01335$ kgw/kgda, $h_i = 68.21$ kJ/kgda

Since the air is supplied to the room at 12$^o$C, the mass flow rate of supply air m$_3$ is obtained from sensible energy balance across the room, i.e.,

$m_3=\dfrac{Q_{s,r}}{C_{pm}(t_i-t_s)}=\dfrac{60}{1.0216(24-14)}=5.873kg/s$ (ans)

The moisture content of supply air is obtained from latent energy balance across the room as:

$W_s=W_i-\dfrac{Q_{l,r}}{m_3h_{fg}}=0.0093-\dfrac{40}{5.873\times2501}=0.0066$ kgw/kgda

Since 25% of the supply air is fresh air, the mass flow rates of fresh and re-circulated air are:

$m_o = 0.25 \times 5.873 = 1.468 kg/s$ and $m_{rc} = 0.75 \times 5.873 = 4.405 kg/s$ (Ans.)

b) From sensible energy balance for the mixing process of fresh air with re-circulated air (Fig.30.4), we obtain the mixed air conditions as:

$t_m = \dfrac{(m_o.t_o + m_{rc}.t_i)}{(m_o+m_{rc})} = 26.5^oC$

$W_m = \dfrac{(m_o.W_o + m_{rc}.W_i)}{(m_o+m_{rc})} = 0.0103$ kgw/kgda

$h_m = \dfrac{(m_o.h_o + m_{rc}.h_i)}{(m_o+m_{rc})} = 52.75$ kJ/kgda

Since heating in the reheat coil is a sensible heating process, the moisture content of air remains constant during this process. Then from Fig.30.4., writing the by-pass factor in terms of humidity ratios as:


From the above expression, the humidity ratio at coil ADP condition is found to be:

$W_{ADP}=\dfrac{(W_s – X.W_m)}{(1–X)} = \dfrac{(0.0066 – 0.15\times0.0103)}{(1.0–0.15)} = 0.00595$ kgw/kgda

The Coil ADP is the saturation temperature corresponding to a humidity ratio of WADP, hence, from psychrometric chart or using psychrometric equations, it is found to be:

$t_{ADP} = 6.38^oC$

Hence, the temperature of air at the exit of the cooling coil ($t_C$ in Fig.30.4) is obtained from the by-pass factor as:

$t_c = t_{ADP} + X (t_m – t_{ADP}) = 9.4^oC$

From $W_c (= W_s)$ and $t_c$, the enthalpy of air at the exit of the cooling coil is found from psychrometric chart as:

$h_c = 26.02$ kJ/kgda

Hence, from energy balance across cooling coil and reheater:

Required capacity of cooling coil, $Q_c = m_s(h_m – h_c) = 157.0 kW$ (Ans.)

Required capacity of reheat coil, $Q_{rh} = m_sc_{pm}(t_s–t_c) = 27.6 kW$ (Ans.)

12. A large warehouse located at an altitude of 1500 m has to be maintained at a DBT of 27$^o$C and a relative humidity of 50% using a direct evaporative cooling system. The outdoor conditions are 33$^o$C (DBT) and 15$^o$C (WBT). The cooling load on the warehouse is 352 kW. A supply fan located in the downstream of the evaporative cooler adds 15 kW of heat. Find the required mass flow rate of air. Assume the process in evaporative cooler to follow a constant WBT.


At 1500m, the barometric pressure is equal to 84.436 kPa.

Inlet conditions to the evaporative cooling system are the outdoor conditions:

$t_o = 33^oC, WBT_o = 15^oC$

At these conditions and a barometric pressure of 84.436 kPa, the enthalpy of outdoor air is obtained using psychrometric equations1 as:

$h_o = 46.67$ kJ/kgda

The above system is shown on psychrometric chart in Fig.31.6

enter image description here

Assuming the evaporative process to follow a constant WBT and hence nearly a constant enthalpy line,

$h_o =h_{o’} = 46.67$ kJ/kgda

Note: Standard psychrometric chart cannot be used here as the barometric pressure is nit 1 atm.

Applying energy balance for the sensible heating process in the fan (process o’-s) and heating and humidification process through the conditioned space (process s-i), we obtain:

$m_s(h_s – h_{o’}) = 15 =$ sensible heat added due to fan (E.1)

$m_s(h_i – h_s) = 352 =$ cooling load on the room (E.2)

From psychrometric equations, for the inside condition of the warehouse (DBT=27$^o$C and RH = 50%), the enthalpy hi is found from psychrometric equations as:

$h_i = 61.38$ kJ/kgda

We have two unknowns (m$_s$ and h$_s$) and two equations (E.1 and E.2), hence solving the equations simultaneously yields:

$m_s = 24.94$ kJ/kg and $h_s = 47.27$ kJ/kgda (Ans.)

13. A winter air conditioning system maintains a building at 21$^o$C and 40% RH. The outdoor conditions are 0$^o$C (DBT) and 100% RH. The sensible load load on the building is 100 kW, while the latent heating load is 25 kW. In the air conditioning system, 50% of the outdoor air (by mass) is mixed with 50% of the room air. The mixed air is heated in a pre-heater to 25$^o$C and then required amount of dry saturated steam at 1 atm. pressure is added to the pre-heated air in a humidifier. The humidified air is then heated to supply temperature of 45$^o$C and is then supplied to the room. Find a) The required mass flow rate of supply air, b) Required amount of steam to be added, and c) Required heat input in pre-heater and re-heater. Barometric pressure = 1atm.


From psychrometric chart the following properties are obtained:

Outdoor conditions: 0$^o$C (DBT) and 100% RH

$W_o = 0.00377$ kgw/kgda, $h_o = 9.439$ kJ/kgda

Indoor conditions: $21^oC (DBT)$ and 40% RH

$W_i = 0.00617$ kgw/kgda, $h_i = 36.66$ kJ/kgda

Since equal amounts of outdoor and indoor air are mixed:

$t_m = 10.5^oC, W_m = 0.00497$ kgw/kgda, $h_m = 23.05$ kJ/kgda

From sensible energy balance across the room (Process s-i) in Fig.31.8:

a) Required mass flow rate of supply air is:

$m_s =\dfrac{ Q_s}{{c_{pm}(t_s – t_i)}} = \dfrac{100}{{1.0216(45 – 21)}} = 4.08$ kg/s (Ans.)

enter image description here

From latent energy balance for process s-i, the humidity ratio of supply air is found to be:

$W_s = W_i + Q_l/(h_{fg}.m_s) = 0.00617 + 25/(2501 \times 4.08) = 0.00862$ kgw/kgda

b) Required amount of steam to be added mw is obtained from mass balance across the humidifier (process r-h) as:

$m_w = m_s(W_s – W_m) = 4.08 \times (0.00862 – 0.00497) = 0.0149$ kg/s (Ans.)

c) Heat input to the pre-heater (process m-r) is obtained as:

$Q_{ph} = m_s.c_{pm}(t_r – t_m) = 60.44$ kW (Ans.)

Heat input to the re-heater (process h-s) is obtained as:

$Q_{rh} = m_s.c_{pm}(t_s – t_r) = 83.36$ kW (Ans.)

In the above example, it is assumed that during addition of steam, the dry bulb temperature of air remains constant. A simple check by using energy balance across the humidifier shows that this assumption is valid.

14. Calculate the local solar time and the corresponding hour angle at 9 A.M (local standard time, L.St.T) on October 21st, for the Indian city of Kolkata located at $22^\circ82'N$ and $88^\circ20'E$, the LSM for India is $82^\circ30'$. The EOT for Kolkata on October 21st is 15 minutes.


Local solar time, LST is given by the expression,

LST=LST+15+4(88.33-82.5)=9 hours 38.32 minutes A.M (ans.)

The corresponding hour angle is 324.6$^\circ$ (ans.)

15. Find the maximum altitude angle for Kolkata ($l = 22^o82’N$) on June 21st.


On June 21st, the declination angle,d is 23.5$^o$.

The maximum altitude angle occurs at solar noon at which the hour angle is zero.

Hence, Maximum altitude angle, $\beta_{max}$ is given by:


16. Find the sunrise, sunset and total sunshine hours at Kolkata (≈22$^o$N) on September 9th.


On September 9th, N = 252, hence the declination, d is equal to 4.62$^o$.

The hour angle at sunrise and sunset is given by,

$h_0=\cos^{-1}(− \tan 22.\tan 4.62)=91.87^\circ$

Since each 15$^o$ is equal to 1 hour, 91.87$^o$ is equal to 6 hours and 8 minutes. Hence,

Sunrise takes place at (12.00 - 6.08) = 5.52 A.M (solar time) (Ans.)

Sunset takes place at (12.00+6.08) = 6.08 P.M. (solar time)

Total sunshine hours are 2 $\times$ 6.08 = 12 hours and 16 minutes

17. What is the angle of incidence at 3 P.M. (solar time) of a north-facing roof that is tilted at an angle of 15$^\circ$ with respect to the horizontal. Location:22$^\circ$N, and date September 9$^{th}$.


Given: Latitude, l=22$^\circ$ (N)

Solar time = 3 P.M $\Rightarrow $ hour angle, h=45$^\circ$

Date = September 9th $\Rightarrow$ declination, d=4.62$^\circ$ (from earlier example)

Altitude angle, $\beta=\sin^{-1}(\cos l.\cos d.\cos h+\sin l.\sin d)=43.13^\circ$

Since the roof is north facing, the surface azimuth angle $\xi$ is equal to 180$^\circ$.

THe solar azimuth angle $\gamma$ is given by:

$\gamma=\sin^{-1}\left(\dfrac{\cos d.\sin h}{\cos\beta}\right)=\sin^{-1}\left(\dfrac{\cos 4.62\sin 45}{\cos 43.13}\right)=74.96^\circ$

The wall solar azimuth angle, $\alpha=180-(\gamma+\xi)=180-(74.96-180)=285^\circ$

Hence the angle of incidence is

$\theta=\cos^{-1}(\sin 43.13\cos15-\cos43.13.\cos 285.\sin 15)=52.3^\circ$

18. Find the direct normal radiation at Kolkata on June 21st at solar noon?


From the earlier example, on June 21st at solar noon, the altitude angle for Kolkata is 89.3$^\circ$. Hence the direct solar radiation is given by:

$I_{DN}=A.\exp\left(-\dfrac B{\sin\beta}\right)=1080.\exp\left(-\dfrac{0.21}{\sin89.3}\right)=875.4W/m^2$ (ans)

19. Find the diffuse and total solar radiation incident on a horizontal surface located at Kolkata on June 21st at solar noon?


From the earlier example, on June 21st at solar noon, the direct solar radiation is equal to 875.4$W/m^2$. Since the surface is horizontal, the view factor for diffuse radiation, $F_{WS}$ is equal to 1, whereas it is 0 for reflected radiation.

Hence, the diffuse solar radiation is given by:

$I_d=C.I_{DN}.F_{ws}=0.135\times875.4=118.18W/m^2$ (ans)

Since the surface is horizontal, the reflected solar radiation is zero. The angle of incidence is given by:


Hence the total incident solar radiation is given by:

$I_t=I_{DN}.\cos(\theta)+I_d=875.4\times\cos(0.7)+118.18=993.5W/m^2$ (ans)

20. Calculate the maximum heat transfer rate through a 1.5 m$^2$ area, unshaded, regular double glass facing south during the months of June and December without internal shading and with internal shading consisting of light venetian blinds. Location 32$^o$N


For the month of June the SHGF_{max} from Table 33.1 is 190 W/m$^2$. Using the values of shading coefficients from Table 33.2, the heat transfer rate is:

Without internal shading (SC = 0.9):

$Q_{sg} = A.(SHGF_{max} ).(SC) =1.5 \times190 \times0.9 = 256.5 W$ (Ans.)

With internal shading (SC = 0.51):

$Q_{sg} = A.(SHGF_{max} ).(SC) =1.5 \times190 \times0.51=145.35 W$ (Ans.)

These values for the month of December (SHGF$_{max}$ = 795 W/m$^2$) are:

Without internal shading: $Q_{sg} = 1073.25 W$ (Ans.)

With internal shading: $Q_{sg} = 608.175 W$ (Ans.)

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