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Numericals on RAC-1
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1. On a particular day the weather forecast states that the dry bulb temperature is 37$^o$C, while the relative humidity is 50% and the barometric pressure is 101.325 kPa. Find the humidity ratio, dew point temperature and enthalpy of moist air on this day.

Answer:

At 37$^o$C the saturation pressure (p$_s$) of water vapour is obtained from steam tables as 6.2795 kPa.

Since the relative humidity is 50%, the vapour pressure of water in air (p$_v$) is:

$p_v = 0.5 \times p_s = 0.5 \times 6.2795 = 3.13975$ kPa

the humidity ratio W is given by:

$W =\dfrac{ 0.622 \times p_v}{(p_t−p_v)} =\dfrac{ 0.622 \times 3.13975}{(101.325−3.13975)} = 0.01989$ kgw/kgda (Ans.)

The enthalpy of air (h) is given by the equation:

$h = 1.005t+W(2501+1.88t) = 1.005 \times 37+0.01989(2501+1.88 \times 37) = 88.31$ kJ/kgda (Ans.)


2. Will the moisture in the above air condense when it comes in contact with a cold surface whose surface temperature is 24$^o$C?

Ans.: Moisture in condense when it is cooled below its dew point temperature.

The dew point temperature of the air at 37$^o$C and 50 % relative humidity is equal to the saturation temperature of water at a vapour pressure of 3.13975 kPa.

From steam tables, the saturation temperature of water at 3.13975 Kpa is 24.8$^o$C, hence moisture in air will condense when it comes in contact with the cold surface whose temperature is lower than the dew point temperature. (Ans.)


3 Moist air at 1 atm. pressure has a dry bulb temperature of 32$^o$C and a wet bulb temperature of 26$^o$C. Calculate a) the partial pressure of water vapour, b) humidity ratio, c) relative humidity, d) dew point temperature, e) density of dry air in the mixture, f) density of water vapour in the mixture and g) enthalpy of moist air using perfect gas law model and psychrometric equations.

Answer:

a) Using modified Apjohn equation and the values of DBT, WBT and barometric pressure, the vapour pressure is found to be:

$p_v = 2.956$ kPa (Ans.)

b) The humidity ratio W is given by:

$W = \dfrac{0.622 \times 2.956}{(101.325-2.956)} = 0.0187$ kgw/kgda (Ans.)

c) Relative humidity RH is given by:

$RH = \left(\dfrac{p_v}{p_s}\right) \times 100 = \left(\dfrac{p_v}{\text{saturation pressure at }32^oC}\right) \times 100$

From steam tables, the saturation pressure of water at 32$^o$C is 4.7552 kPa, hence,

$RH = \left(\dfrac{2.956}{4.7552}\right) \times 100 = 62.16$% (Ans.)

d) Dew point temperature is the saturation temperature of steam at 2.956 kPa.

Hence using steam tables we find that:

$DPT = T_{sat}(2.956 kPa) = 23.8^oC$ (Ans.)

e) Density of dry air and water vapour

Applying perfect gas law to dry air:

Density of dry air $\rho_a =\dfrac{p_a}{R_aT}=\dfrac{(p_t−p_v)}{R_aT} = \dfrac{(101.325−2.956)}{(287.035\times 305)\times10^3}=1.1236 kg/m^3$ of dry air (Ans.)

f) Similarly the density of water vapour in air is obtained using perfect gas law as:

Density of water vapour $\rho_v = \dfrac{p_v}{R_vT} = \dfrac{2.956 \times 10^3}{(461.52 \times 305)} = 0.021 kg/m^3$ (Ans.)

g) Enthalpy of moist air is found from the equation:

$h = 1.005 \times t+W(2501+1.88\times t) = 1.005\times 32 + 0.0187(2501+1.88 \times 32)$

h= 80.05 kJ/kg of dry air (Ans.)


4. What is the required wattage of an electrical heater that heats 0.1 m$^3$/s of air from 15$^o$C and 80% RH to 55$^o$C? The barometric pressure is 101.325 kPa.

Answer:

Air undergoes sensible heating as it flows through the electrical heater

From energy balance, the required heater wattage (W) is given by:

$W = m_a(h_e−h_i) \approx (V_a/ν_a).c_{pm}(T_e−T_i)$

Where $V_a$ is the volumetric flow rate of air in m$^3$/s and $ν_a$ is the specific volume of dry air. $T_e$ and $T_i$ are the exit and inlet temperatures of air and $c_{pm}$ is the average specific heat of moist air (≈1021.6 J/kg.K).

Using perfect gas model, the specific volume of dry air is found to be:

$ν_a = \dfrac{R_a.T}{P_a} = \dfrac{R_a.T}{( P_t −P_v)}$

At 15$^o$C and 80% RH, the vapour pressure pv is found to be 1.364 kPa using psychrometric chart or equations.

Substituting the values of $R_a$, T, $P_t$ and $P_v$ in the equation for specific volume, we find the value of specific volume to be 0.8274 m$^3$/kg

Therefore, Heater wattage, $W \approx (V_a/ν_a).c_{pm}(T_e−T_i)=(0.1/0.8274)\times1021.6(55-15) = 4938.8 W$ (ans.)


5. 0.2 kg/s of moist air at 45$^o$C (DBT) and 10% RH is mixed with 0.3 kg/s of moist air at 25$^o$C and a humidity ratio of 0.018 kgw/kgda in an adiabatic mixing chamber. After mixing, the mixed air is heated to a final temperature of 40$^o$C using a heater. Find the temperature and relative humidity of air after mixing. Find the heat transfer rate in the heater and relative humidity of air at the exit of heater. Assume the barometric pressure to be 1 atm.

Answer: Given:

Stream 1: mass flow rate, m$_1$ = 0.2 kg/s; $T_1 = 45^oC$ and RH = 10%.

Using psychrometric equations or psychrometric chart, the humidity ratio and enthalpy of stream 1 are found to be:

$W_1$ = 0.006 kgw/kgda & $h_1$ = 61.0 kJ/kgda

Stream 2: mass flow rate, m$_2$ = 0.3 kg/s; $T_2 = 45^oC$ and W$_2$ = 0.018 kgw/kgda

Using psychrometric equations or psychrometric chart, enthalpy of stream 2 is found to be:

$h_1$ = 71.0 kJ/kgda

For the adiabatic mixing process, from mass balance:

$W_3=\dfrac{m_{a,1}w_1+m_{a,2}w_2}{m_{a,1}+m_{a,2}}=\dfrac{0.2\times0.006+0.3\times0.018}{0.2+0.3}=0.0132$kgw/kgda

From energy balance (assuming the specific heat of moist air to remain constant):

$T_3=\dfrac{m_{a,1}T_1+m_{a,2}T_2}{m_{a,1}+m_{a,2}}=\dfrac{0.2\times45+0.3\times25}{0.2+0.3}=33^\circ C$ (ans)

From $T_3$ and $W_3$, the relative humidity of air after mixing is found to be:

$RH_3 = 41.8$% (ans.)

For the sensible heating process in the heater:

$Q_s = m_a(h_e−h_i)\approx m_a.c_{pm}(T_e−T_i) = 0.5\times1.0216(40-33) = 3.5756$ kW (ans.)

The relative humidity at the exit of heater is obtained from the values of DBT (40$^o$C) and humidity ratio (0.0132 kgw/kgda) using psychrometric chart/equations. This is found to be:

RH at 40$^o$C and 0.0132 kgw/kgda = 28.5 % (ans.)


6. A cooling tower is used for cooling the condenser water of a refrigeration system having a heat rejection rate of 100 kW. In the cooling tower air enters at 35$^o$C (DBT) and 24$^o$C (WBT) and leaves the cooling tower at a DBT of 26$^o$C relative humidity of 95%. What is the required flow rate of air at the inlet to the cooling tower in m$^3$/s. What is the amount of make-up water to be supplied? The temperature of make-up water is at 30$^o$C, at which its enthalpy (h$_w$) may be taken as 125.4 kJ/kg. Assume the barometric pressure to be 1 atm.

Answer:

At the inlet to cooling tower: DBT = 35$^o$C and WBT = 24$^o$C

From psychrometric chart/equations the following values are obtained for the inlet:

Humidity ratio, $W_i$ = 0.01426 kgw/kgda

Enthalpy, $h_i$ = 71.565 kJ/kgda

Sp. volume, $ν_i$ = 0.89284 m$^3$/kgda

At the outlet to cooling tower: DBT = 26$^o$C and RH = 95%

From psychrometric chart/equations the following values are obtained for the outlet:

Humidity ratio, $W_o$ = 0.02025 kgw/kgda

Enthalpy, $h_i$ = 77.588 kJ/kgda

From mass and energy balance across the cooling tower:

$Q_c = m_a{(h_o−h_i) − (W_o−W_i)h_w} = 100 kW$

Substituting the values of enthalpy and humidity ratio at the inlet and outlet of cooling tower and enthalpy of make-up water in the above expression, we obtain:

m$_a$ = 18.97 kg/s,

hence, the volumetric flow rate, $V_i = m_a \times ν_i = 16.94 m^3/s$ (ans.)

Amount of make-up water required mw is obtained from mass balance as:

$m_w = m_a(W_o - W_i) = 18.97(0.02025 − 0.01426) = 0.1136 kg/s = 113.6$ grams/s (ans.)


7. In an air conditioning system air at a flow rate of 2 kg/s enters the cooling coil at 25°C and 50% RH and leaves the cooling coil at 11$^o$C and 90% RH. The apparatus dewpoint of the cooling coil is 7°C. Find a) The required cooling capacity of the coil, b) Sensible Heat Factor for the process, and c) By-pass factor of the cooling coil. Assume the barometric pressure to be 1 atm. Assume the condensate water to leave the coil at ADP (hw = 29.26 kJ/kg)

Answer:

At the inlet to the cooling coil; $T_i = 25^oC$ and RH = 50%

From psychrometric chart; Wi = 0.00988 kgw/kgda and hi = 50.155 kJ/kgda

At the outlet of the cooling coil; $T_o = 11^oC$ and RH = 90%

From psychrometric chart; $W_o$ = 0.00734 kgw/kgda and $h_o$ = 29.496 kJ/kgda

a) From mass balance across the cooling coil, the condesate rate, $m_w$ is:

$m_w = m_a(W_i − W_o) = 2.0(0.00988 − 0.00734) = 0.00508 kg/s$

From energy balance across the cooling tower, the required capacity of the cooling coil, $Q_c$ is given by:;

$Q_c = m_a(h_i − h_o) – m_w.h_w = 2.0(50.155 − 29.496) − 0.00508 \times 29.26 = 41.17$ kW (ans.)

b) The sensible heat transfer rate, $Q_s$ is given by:

$Q_s = m_ac_{pm}(T_i − T_o) = 2.0 \times 1.0216 \times (25 − 11) = 28.605$ kW

The latent heat transfer rate, $Q_l$ is given by:

$Q_s = m_ah_{fg}(W_i − W_o) = 2.0 \times 2501.0 x (0.00988 − 0.00734) = 12.705$ kW

The Sensible Heat Factor (SHF) is given by:

$SHF = \dfrac{Q_s}{(Q_s + Q_l)} =\dfrac{ 28.605}{(28.605 + 12.705)} = 0.692$ (ans.)

c) From its definition, the by-pass factor of the coil, BPF is given by:

$BPF = \dfrac{(T_o − ADP)}{(T_i − ADP)} = \dfrac{(11 − 7)}{(25 − 7)} = 0.222$


8. A 1.8 meter tall human being with a body mass of 60 kg performs light work (activity = 1.2 met) in an indoor environment. The indoor conditions are: DBT of 30$^o$C, mean radiant temperature of 32$^o$C, air velocity of 0.2 m/s. Assuming an average surface temperature of 34$^o$C for the surface of the human being and light clothing, find the amount of evaporative heat transfer required so that the human being is at neutral equilibrium.

Answer:

Using Du Bois equation, the surface area of the human being A$_s$ is:

$A_{Du} = 0.202m^{0.425} h^{0.725} = 0.202 \times 60^{0.425} \times 1.8^{0.725} =1.7625 m^2$

Hence the total heat generation rate from the body, $Q_g$ is:

$Q_g = A_s \times (\text{Activity level in met}) \times 58.2 = 1.7625 \times 1.2 \times 58.2 = 123.1 W$

Using Belding & Hatch equations, the convective and radiative heat losses from the surface of the body are found as:

$Q_c=14.8V^{0.5}(t_b-t)=14.8\times0.2^{0.5}(34-30)=26.48W$

$Q_r=11.603(t_b-t_s)=11.603(34-32)=23.2W$

For neutral equilibrium,

$Q_g = Qc_ + Q_r + Q_e = \Rightarrow Q_e = Q_g – (Q_c + Q_r)$

Substituting the values of $Q_g, Q_c$ and $Q_r$ in the above expression, we find that the required amount of evaporative heat transfer $Q_e$ is equal to:

$Q_e = 123.1 – (26.48+23.2) = 73.42 W$ (Ans.)


9. A 100% outdoor summer air conditioning system has a room sensible heat load of 400 kW and a room latent heat load of 100 kW. The required inside conditions are 24$^o$C and 50% RH, and the outdoor design conditions are 34$^o$C and 40% RH. The air is supplied to the room at a dry bulb temperature of 14$^o$C. Find a) the required mass flow rate of air b) moisture content of supply air, c) Sensible, latent heat loads on the coil, and d) The required cooling capacity of the coil, Coil Sensible Heat Factor and coil ADP if the by-pass factor of the coil is 0.2. Barometric pressure = 1 atm. Comment on the results.

Answer:

The psychrometric process for this system is shown in Fig.30.5.

The psychrometric properties at inside and outside conditions are:

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

From psychrometric chart or using psychrometric equations; the moisture content and enthalpy of inside air are:

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

enter image description here

outside conditions: $t_i =34^oC(DBT)$ and $RH_i = 40$%

From psychrometric chart or using psychrometric equations; the moisture content and enthalpy of inside air are:

$W_o = 0.01335$ kgw/kgda, $h_1 = 68.21$ kJ/kgda

a) From sensible energy balance equation for the room, we find the required mass flow rate of air as:

$m_s=\dfrac{Q_{s,r}}{C_{pm}(t_i-t_s)}=\dfrac{400}{1.0216(24-14)}=39.154kg/s$ (ans.)

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

$W_s=W_i-\dfrac{Q_{l,r}}{m_sh_{fg}}=0.0093-\dfrac{100}{39.154\times2501}=0.0083$ kgw/kgda (ans.)

c) From energy balance, the sensible and latent loads on the coil are obtained as:

$Q_{s,c}= m_sC_{pm}(t_o-t_s)=39.154\times1.0216\times(34-14)=800$ kW

$Q_{l,c}= m_sh_{fg}(W_o-W_s)=39.154\times2501\times(0.01335-0.0083)=494.5$ kW (ans.)

d) The required cooling capacity of the coil is equal to the total load on the coil, $Q_{t,c}$:

$Q_{t,c} = Q_{s,c} + Q_{l,c} = 800 + 494.5 = 1294.5$ kW (Ans.)

Coil Sensible Heat Factor, $CSHF = \dfrac{Q_{s,c}}{Q_{t,c}} = 0.618$ (Ans.)

Coil ADP is obtained by using the definition of by-pass factor (X) as:

$t_{ADP}(1 − X) = t_s − X.t_o$

$\Rightarrow t_{ADP} = \dfrac{(t_s − X.t_o)}{(1-X)} =\dfrac{ (14 − 0.2 \times 34)}{(1 − 0.2)} = 9^oC$ (Ans.)

Comments:

  1. It is seen that with 100% outdoor air, the load on the coil (or required cooling capacity of the coil) is much higher compared to the cooling load on the building (Required coil capacity = 1294.5 kW whereas the total load on the room is 500 kW). Since 100% outdoor air is used, the relatively cold and dry indoor air is exhausted without re-circulation and the hot and humid air is conditioned using the coil coil. Thus the required cooling capacity is very high as the cooling coil has to cool and dehumidify outdoor air.

  2. It is observed that the CSHF (0.618) is much smaller compared to the room SHF (0.8), hence, the coil ADP is much smaller than the room ADP.


10. A room is air conditioned by a system that maintains 25$^o$C dry bulb and 50 % RH inside, when the outside conditions are 34$^o$C dry bulb and 40% RH. The room sensible and latent heat gains are 60 kW and 12 kW respectively. As shown in the figure below, The outside fresh air first flows over a first cooler coil and is reduced to state 1 of 10$^o$C dry bulb and a relative humidity of 85%. It is then mixed with re-circulated air, the mixture (state 2) being handled by a fan, passed over a second cooler coil and sensibly cooled to 12$^o$C dry bulb (state 3). The air is then delivered to the room. If the outside fresh air is used for dealing with the whole of the room latent heat gain and if the effects of fan power and duct heat gains are ignored, find: a) mass flow rates of outside fresh air and supply air; b) DBT and enthalpy of the air handled by the fan (state 2); and c) required cooling capacity of first cooler coil and second sensible cooler coil.

enter image description here

Answer:

From psychrometric chart, the following properties are obtained:

Inside conditions: $t_i =24^oC(DBT)$ 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

At state 1: $t_1 = 10^oC(DBT)$ and$ RH_1 = 85$%

$W_1$ = 0.00647 kgw/kgda, $h_1$ = 26.31 kJ/kgda

a) 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.,

$m_3=\dfrac{Q_{s,r}}{C_{pm}(t_i-t_3)}=\dfrac{60}{1.0216(24-12)}=4.894$kg/s (ans.)

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

$W_s=W_i-\dfrac{Q_{l,r}}{m_3h_{fg}}=0.0093-\dfrac{12}{4.894\times2501}=0.0083$kgw/kgda

Since the fresh air takes care of the entire latent load, the heat transfer across coil 2 is only sensible heat transfer. This implies that:

$W_2 = W_3 = 0.0083$ kgw/kgda

Applying mass balance across the mixing of re-circulated and fresh air (1-2), we obtain:

$m_1W_1+(m_2-m_1)W_i = m_2.W_2$

From the above equation, we get $m_1$ as:

$m_1= m_2\dfrac{(W_i –W_2)}{(Wi – W1)} = 1.73 kg/s$

Hence the mass flow rate of re-circulated air is:

$m_{rc} = m_2 – m_1 = (4.894 - 1.73) = 3.164 kg/s$

b) From energy balance across the mixing process 1-2, assuming the variation in cpm to be negligible, the temperature of mixed air at 2 is given by:

$t_2 = \dfrac{(m_1t_1 + m_{rc}t_i)}{m_2} = 19.05^oC$ (Ans.)

From total enthalpy balance for the mixing process, the enthalpy of mixed air at 2 is:

$h_2 =\dfrac{ (m_1h_1 + m_{rc}h_i)}{m_2} = 40.11$ kJ/kgda (Ans.)

c) From energy balance, cooling capacity of 1st cooler coil is given by:

$Q_{c,1} = m_1(h_o – h_1) = 1.73 \times (68.21 – 26.31) = 72.49$ kW (Ans.)

From energy balance across the 2nd cooler coil, the cooling capacity of the second coil is given by:

$Q_{c,2} = m_2.c_{pm}(t_2 – t_3) = 4.894 \times 1.0216 \times (19.05 – 12.0) = 35.25$ kW (Ans.)

Comment: It can be seen that the combined cooling capacity (72.49 + 35.25 =107.74 kW) is larger than the total cooling load on the building (60 + 12 = 72 kW).The difference between these two quantities (107.74 – 72 = 35.74 kW) is equal to the cooling capacity required to reduce the enthalpy of the fresh air from outdoor conditions to the required indoor conditions. This is the penalty one has to pay for providing fresh air to the conditioned space. Larger the fresh air requirement, larger will be the required cooling capacity.

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800m³ / min of recalculated air at 22 degrees C DBT and 10 degrees C DPT is to be mixed with 300m³ / mIn of fresh air at 30 degrees C DBT and 50% RH. Determine the enthalpy, specific volume, humidity ratio and dew point temperature of the mixture.

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