Specific heat of dry air Cpa 1.006 kJ/kgK;
Specific heat of water vapour Cpv- 1.87 kJ/kgK
Specific heat of moist air: c_{p} = 1.02kJ / k * gK
Density of dry air rho = 1.2kg / (m ^ 3)
Latent heat of vaporisation of water at 0°C hf. 0 2501 kJ/kgK
Gas constant of water vapor R_{v} = 0.4615kJ / k * gK
Gas constant of dry air R_{a} = 0.2871kJ / k * gK
Q3. (Answer the following in short answer)
(i) On an arbitrary psychometric chart, show the air conditioning process of:
a) cooling and dehumidifying
b) evaporative cooling
c) heating and humidifying
d) sensible heating
(12 marks)
(ii) The atmospheric pressure of air in a room is 101.325kPa. It’s dry bulb and wet bulb. temperature are 25 °C and 20°C, respectively. Using Table Q3, calculate
a) the humidity ratio
b) the partial pressure of water vapor
c) the density of moist air
d) the dew point temperature
The Correct Answer and Explanation is :
address your queries, we’ll first illustrate the specified air conditioning processes on a psychrometric chart and then perform calculations based on the given room conditions. (i) Air Conditioning Processes on a Psychrometric Chart:**
- Cooling and Dehumidifying:
- Process Description: Thiprocess involves lowering the air’s dry-bulb temperature and reducing its moisture content. It’mmonly achieved by passing air over a cooling coil, where the air is cooled below its dew point, causing moisture to condense.
- epresentation on the Chart:** On the pchrometric chart, this is depicted as a line moving downward and to the left, indicating a decrease in both temperature and humidity ratio.
- Evrative Cooling:
- Process Description: In this pross, air is cooled by evaporating water into it without adding external heat. The sensiblat of the air is converted into latent heat, increasing the humidity ratio while decreasing the dry-bulb temperature.
- Represation on the Chart: This is shown as line moving upward and to the left along a constant wet-bulb temperature line, indicating a decrease in dry-bulb temperature and an increase in humidity ratio.
- Heating anumidifying:
- Process Description: This process involv increasing both the air’s temperature and its moisture content. It’s typically achi by adding steam or spraying water into the air while simultaneously heating it.
- Representation the Chart: On the psychrometric cha, this is depicted as a line moving upward and to the right, indicating increases in both dry-bulb temperature and humidity ratio.
- *Sensible Heating: – Process Description: This process increases the r’s dry-bulb temperature without changing its moisture content. It’s commonly achieved by png air over a heating coil.
- Representation on the rt: This is shown as a horizontal li moving to the right, indicating an increase in dry-bulb temperature while the humidity ratio remains constant.
(ii) Calculations Based on RoConditions:
Given:
- Atmospheric pressure (P): 101.325 k
- Dry-bulb temperature (T_db): 25°CWebulb temperature (T_wb): 20°C
- Hidity Ratio (W):**
The humiditytio is calculated using the fmula: here \( P_{v} \) is the partial pressure of wrapor. First, we need to determine the saturation vapor sre at the wet-bulb temperature (T_wb). Using the Tetens equation: \[ P_{ws} = 0.61078 \timexp\left( \frac{17.27 \timTwb}}{T_{wb} + 237.3} \right) \] Plugging in T_wb = 20°C: \[ P_{ws} = 0.61078 \times \expf \frac{17.27 \times 20} 237.3} \right) \] \[ P_{ws} \approx 2.338 \, \text{kPa} \] Next, we calculate thcal vapor pressure (P_v) using the formu [ P_{v} = P_{ws} – P \times A \times (T_{db} – T_{wb}) \] Where A ihpsychrometric constant, approximately 0.00066°C⁻¹. \[ P_ 2.338 – 101.325 \times 0.00066 \times (25 – 20) \] \[ P_{v} = 28 0.3344 \] \[ P_{v} \approx 2.0036 \, \text{kPa} \] Now, stuting \( P_{v} \) into the ity ratio formula: \[ W = 0.622 \times a2.0036}{101.325 – 2.0036} \] \[ W = 0.622 \times \frac{2.00393214} \] \[ W \approx 0.01254 \, \text{kg water vapogry air} \] **b) Partial Pressure of Water VrP_v):** As calculated above, the partial pressure of wateapor is approximately 2.0036 kPa. **c) Density Moist Air (ρ):** The density of moist air can be calculated using the formula: \[ho = \frac{P_d}{R_d \times T} + \fr{P_v}{R_v \times T} \] Where: – \( P_d \) = Partial pressurfry air = \( P – P_v \) – \( R_d \) = Specific gas constant for r = 071J/(kg·K) – \( R_v \) = Specific gas constant for waterpo= 0.4615 kJ/(kg·K) – \( T \) = Absolute temperature in Kelvin =db 273.15 Calculating each component: \[ P_d = 101.325 – 2.0036 = 921\, \text{kPa