Consider moist air with absolute humidity of 0.02 (kg moisture)/(kg dry air) at 1 bar pressure. The vapor pressure of water is given by the equation: \[ \ln P_{{sat}} = 12 - \frac{4000}{T - 40} \] where \( P_{{sat}} \) is in bar and \( T \) is in K. The molecular weight of water and dry air are 18 kg/kmol and 29 kg/kmol, respectively. The dew temperature of the moist air is ____________ ℃ (rounded off to the nearest integer).
\[ \ln P_{\text{sat}} = 12 - \frac{4000}{T - 40} \]
Step 2: Calculate the partial pressure of the water vapor:\[ y_{H_2O} = \frac{\text{absolute humidity} \times 1000}{M_{\text{water}}} \times \frac{M_{\text{air}}}{1000} \] \[ y_{H_2O} = \frac{0.02 \times 1000}{18} \times \frac{29}{1000} = 0.0322 \]
The partial pressure of water vapor is:
\[ P_{H_2O} = y_{H_2O} \times P_{\text{total}} = 0.0322 \times 1 = 0.0322 \, \text{bar} \]
Step 3: Solve for the dew temperature \( T \):\[ P_{\text{sat}} = P_{H_2O} = 0.0322 \, \text{bar} \] \[ \ln 0.0322 = 12 - \frac{4000}{T - 40} \] \[ -3.442 = 12 - \frac{4000}{T - 40} \] \[ \frac{4000}{T - 40} = 15.442 \Rightarrow T - 40 = \frac{4000}{15.442} = 259.4 \Rightarrow T = 259.4 + 40 = 299.4 \, K \] \[ T_{\text{dew}} = 299.4 - 273.15 = 26.25^\circ C \]
Therefore, the dew temperature is approximately 26°C.
Is there any good show __________ television tonight? Select the most appropriate option to complete the above sentence.
Consider a process with transfer function: \[ G_p = \frac{2e^{-s}}{(5s + 1)^2} \] A first-order plus dead time (FOPDT) model is to be fitted to the unit step process reaction curve (PRC) by applying the maximum slope method. Let \( \tau_m \) and \( \theta_m \) denote the time constant and dead time, respectively, of the fitted FOPDT model. The value of \( \frac{\tau_m}{\theta_m} \) is __________ (rounded off to 2 decimal places).
Given: For \( G = \frac{1}{(\tau s + 1)^2} \), the unit step output response is: \[ y(t) = 1 - \left(1 + \frac{t}{\tau}\right)e^{-t/\tau} \] The first and second derivatives of \( y(t) \) are: \[ \frac{dy(t)}{dt} = \frac{t}{\tau^2} e^{-t/\tau} \] \[ \frac{d^2y(t)}{dt^2} = \frac{1}{\tau^2} \left(1 - \frac{t}{\tau}\right) e^{-t/\tau} \]