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Explanation:

Refer to the Ideal Gas Mixtures section in the Thermodynamics chapter of the FE Reference Handbook.

The internal energy of a mixture of ideal gasses is just the mass-weighted sum of the internal energy of the constituent gasses, $$ u = \sum{y_iu_i} = y_{O_2}u_{O_2} + y_{CO_2}u_{CO_2} $$ where $y_i$ is the mass fraction of constituent $i$. In the problem statement, we are only given the partial pressures which means we can find the mole fraction, $$ x_i = \frac{P_i}{P} $$ or, $$ x_{O_2} = \frac{P_{O_2}}{P_{O_2} + P_{CO_2}} = \frac{80}{80 + 20} = 0.8\\ x_{CO_2} = \frac{P_{CO_2}}{P_{O_2} + P_{CO_2}} = 0.2 $$ Which we can then use to convert from mole fraction to mass fraction via the formula given in the FE Handbook, $$ y_i = \frac{x_i M_i}{\sum{x_i M_i}} $$ which for oxygen is $$ y_{O_2} = \frac{0.8 (\SI{32}{\kilo\gram\per\kilo\mole})}{\left[ 0.8 (\SI{32}{\kilo\gram\per\kilo\mole}) \right] + \left[ 0.2 (\SI{44}{\kilo\gram\per\kilo\mole}) \right]} = 0.744 $$ where $M_{O_2}$ and $M_{CO_2}$ can be found in the back of the Thermodynamics chapter in the Thermal and Physical Property Table. At this point, we could use the same formula to calculate the mole fraction of CO$_2$ but we can also use a shortcut because we know the total sum of mass fraction must equal 1. In this case we have only 2 constituent gases so we can straight away calculate $y_{CO_2} = 1 - 0.744 = 0.256$.

Finally we can evaluate for the total internal energy, $$ u = y_{O_2}u_{O_2} + y_{CO_2}u_{CO_2} = 0.744\cdot\SI{188.5}{\kilo\joule\per\kilo\gram} + 0.256\cdot\SI{151.1}{\kilo\joule\per\kilo\gram} \\ = \SI{178.9}{\kilo\joule\per\kilo\gram} $$

Explanation:

Refer to the Control Systems section in the Instrumentation, Measurement, and Control chapter of the FE Reference Handbook.

The controller plant, $G(s)$, is $\frac{25}{s(s+5)}$.

This is a unity feedback controller. It can be deduced that this is a unity feedback controller from looking at the classical model of a negative feedback controller, which is shown in the FE Reference Handbook as depicted in the image and comparing it to our system (noting that $G_2(s) = 1$, $H(s) = 1$ and $L(s) = 0$).

A unity feedback model is also shown explicitly in the section on Control Systems.

Thus, $H(s) = 1$ and the closed loop transfer function, $T(s)$, is $\frac{Y(s)}{R(s)}$ (output over the input) which is, $$T(s) = \frac{G(s)}{1+G(s)}$$ Substituting in $G(s)$, $$T(s) = \frac{\frac{25}{s(s+5)}}{1+\frac{25}{s(s+5)}}$$ $$T(s) = \frac{25}{s^2 + 5s + 25}$$ This is a second-order control system of the form shown in the handbook, $$\frac{Y(s)}{R(s)} = \frac{K\omega_n^2}{s^2 + 2 \zeta \omega_n s + \omega_n^2}$$ Where, $$\omega_n = \sqrt{25} = 5$$ $$2\zeta\omega_n = 5$$ Substituting $\omega_n$ and solving for $\zeta$ yields, $$\zeta = 0.5$$ Using the values for $\omega_n$ and $\zeta$, calculate the peak time from, $$t_p = \frac{\pi}{\omega_n\sqrt{1-\zeta^2}}$$ $$t_p = \frac{\pi}{5\sqrt{1-0.5^2}} = 0.726 \si{s}$$

Explanation:

Refer to the Chemical Reaction Engineering section in the Chemical Engineering chapter of the FE Reference Handbook.

Use the expression to determine a general equation for the rate, $$r = \frac{\si{[NO]_2} - \si{[NO]_1}}{t_2-t_1} = \frac{\Delta[NO]}{\Delta t}$$ Now plug in the given values for the concentrations at times $t_2$ and $t_1$. $$r = \frac{(0.70-0.10) \si {M/liter}}{(5.0 - 1.0) \si{min}} =0.15 \text{ mole} \si{liter^{-1}}\si{min}^{-1}$$

Explanation:

Refer to the Angle of Repose section in the Chemical Engineering chapter of the FE Reference Handbook.

The pile of solids is formed in the shape of a cone by the depositing funnel. To calculate the angle of repose, we can drop a perpendicular from the top of the cone to the base to show the height and draw a bisector across the base to show the diameter.

The angle of repose is now the angle of a right triangle which is adjacent to half the length of the diameter, $d$, and opposite the length of the height $h$. In other words, the trigonometric tangent function. Calling the angle of repose $\theta$, we have, $$\tan{\theta} = \frac{h}{\frac{1}{2}d}$$ The radius is half the diameter by definition, so, $$\tan{\theta} = \frac{h}{r}$$ Solving for $\theta$, $$\theta = \tan^{-1}\frac{h}{r}$$ Plug in the given values of $r$ and $h$, $$\theta = \tan^{-1}\frac{100\si{cm}}{200\si{cm}} = 26.56^\circ$$ Thus, according to the table, this powder has an Excellent/Very Free Flow.

Explanation:

Refer to the Convection section in the Chemical Engineering chapter of the FE Reference Handbook.

Apply the equation that relates the individual given coefficients together. $$\frac{1}{K_L}= \frac{1}{Hk_G} + \frac{1}{k_L}$$ The value $H$ is Henry's Law constant where $p_{A_i} = HC_{AL}$ which is given as $0.25$. Plugging in these values yields, $$\frac{1}{K_L}= \frac{1}{(0.25)(0.25)} + \frac{1}{1.10} = 16.91$$ Taking the reciprocal to find $K_L$, $$K_L = 0.0592 \frac{\si{lb\cdot mole}}{\si{hr\cdot ft^2 \cdot \Delta P_A}}$$

Explanation:

Refer to the Ideal Gas section in the Thermodynamics chapter of the FE Reference Handbook.

Because we are given the pressure, volume, and temperature and wish to find mass, we will use the form of ideal gas law, $$ PV = mRT $$ where we must find the gas constant for this particular ideal gas. It is $$ R = \frac{\bar{R}}{mol. wt} = \frac{\SI{8.314}{\kilo\pascal\meter\cubed\per\kilo\mole\per\kelvin}}{\SI{24}{\kilo\gram\per\kilo\mole}} = \SI{0.346}{\kilo\pascal\meter\cubed\per\kilo\gram\per\kelvin} $$ Thus the mass is $$ m = \frac{PV}{RT} = \frac{(\SI{2050}{\kilo\pascal})\SI{0.6}{\meter\cubed}}{(\SI{0.346}{\kilo\pascal\meter\cubed\per\kilo\gram\per\kelvin})\cdot\SI{ (273+25)}{\kelvin}}=\SI{11.9}{\kilo\gram} $$

Explanation:

Refer to the Steam Tables section in the Thermodynamics chapter of the FE Reference Handbook.

In the steam tables, we can see for a liquid at $\SI{20}{\celsius}$ the saturate pressure is $\SI{2.339}{\kilo\pascal}$. This tells us that, at this temperature, water will remain a liquid until the pressure is reduced to $\SI{2.339}{\kilo\pascal}$. Our system pressure, $\SI{50}{\kilo\pascal}$, is clearly higher and thus we remain in the compressed liquid phase (this is also referred to by some as the sub-cooled liquid phase)

Explanation:

Refer to the Chemical Reaction Engineering section in the Chemical Engineering chapter of the FE Reference Handbook.

The Arrhenius equation can be applied directly from the givens, $$k = Ae^{\frac{-E_a}{\overline R T}}$$ where $$\begin{align} A & = \text{pre-exponential factor} \\ E_a &= \text{activation energy} (\si{J/mol, cal/mol)} \\ T &= \text{temperature } ({\si K}) \\ \overline R &= \text{gas law constant} = 8.314 \si {J/(mol \cdot K)} \end{align}$$ Rearrange the equation to solve for $E_a$, $$E_a =- \ln\frac{k}{A} (\overline RT)$$ Plug in the values, $$E_a = \ln \left ( \frac{12}{15} \right ) (8.314 \times 22) = 40.81 \si{J/mol}$$

Explanation:

Refer to the Chemical Reaction Engineering section in the Chemical Engineering chapter of the FE Reference Handbook.

The reaction rate equation will have the form, $$r = kC_A^xC_B^y$$ Where $C_A$ and $C_B$ are the reactants and $x$ and $y$ are the respective orders. Thus, we know the expression needs to be of the form, $$r = k \si{[NO]}^x\si{[H_2]}^y$$ Next, we can use the data from the table to determine the reaction order. In experiments 1 and 2, the concentration of $\si{H_2}$ is maintained constant. Thus we can write the ratio, $$\frac{r_1}{r_2} = \frac{k\si{[NO]_1^x[H_2]_1^y}}{k\si{[NO]_2^x [H_2]_2^y}}$$ The rates of reaction for $\si {NO}$ can be deduced from the table as, $$r_1 = \frac{\Delta A_1}{\Delta t} = \frac{1-0.8}{1 \si {hr}} = 0.2 \si{M/hr} \\ r_2 = \frac{\Delta A_2}{\Delta t} = \frac{2-1.2}{1 \si {hr}} = 0.8 \si{M/hr} $$ Plugging in the values (noting that $k$ and the concentration of $\si H_2$ are cancelled out), $$\frac{0.2}{0.8} = \frac{1.0^x}{2.0^x}$$ Solve for $x$, $$\ln (\frac{0.2}{0.8})= x \ln \frac{1}{2} \\ x = 2$$ Apply the same logic to solve for $y$, this time using experiments 1 and 3 where the initial concentrations of $\si {NO}$ are held constant. $$r_1 = \frac{\Delta B_1}{\Delta t} = \frac{1-0.8}{1 \si {hr}} = 0.2 \si{M/hr} \\ r_2 = \frac{\Delta B_2}{\Delta t} = \frac{2-1.6}{1 \si {hr}} = 0.4 \si{M/hr} $$ Using the ratio, $$\frac{0.2}{0.4} = \frac{1.00^y}{2.00^y}$$ Solve for $y$, $$\ln(\frac{0.2}{0.4}) = y\ln\frac{1}{2} \\ y = 1$$ Thus, the final rate equation for this reaction is therefore, $$r = k \si{[NO]^2[H_2]}$$

Explanation:

Refer to the Fate and Transport section in the Environmental Engineering chapter of the FE Reference Handbook.

The control volume is the CSTR itself and there is a single intake and outtake. The question asks for the concentration of the outlet only and all values are constants. Thus, it is a steady-state problem and we can begin by writing an expression for the flow, $$\frac{dm}{dt} = \frac{dm_{in}}{dt} - \frac{dm_{out}}{dt} + \frac{dm_{r}}{dt}$$ $$ 0 = QC_{in} - QC - kCV$$ Solving for the unknown $C$, $$C = C_{in} \cdot \frac{Q}{Q +kV}$$ Substituting in the known values, $$C = 100 \frac{\si{mg}}{\si{L}} \cdot \frac{50 \frac{\si {m^3}}{\si {day}}} {50 \frac{\si {m^3}}{\si {day}}+(0.2 \si {day}^{-1})(500 \si m^3)} = 33.3\frac{\si{mg}}{\si{L}} \approx 33 \frac{\si{mg}}{\si{L}} $$