Chemical Engineering - MCQ Practice Questions

For adiabatic process, Q = 0, so ΔU = W (first law). Temperature increases during compression.

μ_JT = (∂T/∂P)_H. Negative coefficient means T increases with pressure drop (cooling requires very low T or high P).

Raoult's law: P_i = x_i × P_i° for ideal solutions. Both statements A and C are equivalent for ideal solutions.

First law: ΔU = Q - W = 500 - 300 = 200 J (using convention W = work by system).

At high pressures, molecular volume effect (b term) dominates, making Z > 1. At moderate pressures, Z < 1 due to intermolecular attractions.

At constant pressure: ΔH = Q_p (definition of enthalpy). From ΔU = Q - W and W = PΔV, we get Q = ΔH.

Entropy is a state function. ΔS is path-independent and same for all processes (reversible or irreversible) between fixed states.

For adiabatic process of ideal gas, PV^γ = constant where γ = Cp/Cv. This is the defining equation for adiabatic polytropic process.

Above inversion temperature (for methane ≈ 625 K), μ_JT < 0 (heating on expansion). Below it, μ_JT > 0 (cooling on expansion), enabling liquefaction.

ΔS_mix = -nR Σx_i ln x_i. For equal moles: ΔS_mix = -2R(0.5 ln 0.5 + 0.5 ln 0.5) = R ln 4 > 0, always positive.

Desalination (salt separation) is non-spontaneous: ΔG > 0. External work must be supplied to drive the process. This applies to RO and most separation processes.

Understanding:

We must identify the time constant $\tau$ and steady-state gain $K_p$ from the given transfer function.

Formula:

The standard first-order transfer function is:

where $K_p$ is the steady-state (process) gain and $\tau$ is the time constant.

Step 1: Compare with standard form.

Step 2: Read off parameters by direct comparison.

Step 3: Verify using the steady-state gain definition $G(0)$.

Answer:

The time constant is $2$ (time units) and the steady-state gain is $5$.

Quick Tip:

Always write the transfer function in the form $\frac{K_p}{\tau s+1}$ (denominator coefficient of $s^0$ equal to 1) before reading off $\tau$ and $K_p$. Never read $\tau$ and $K_p$ from a non-standard form.

Understanding:

We must identify the correct functional description of the derivative mode (${\tau }_{D}s$ term) in a PID controller.

Step 1: Review each mode's role.

The PID controller output $u(t)$ in time domain is:

Step 2: Identify each term.

Step 3: Match to options.

Eliminating steady-state offset is the role of the integral mode, not the derivative mode. The derivative mode acts on $\frac{de}{dt}$, providing a predictive (anticipatory) control action based on how fast the error is changing.

Answer:

The derivative mode responds to the rate of change of error, anticipating future trends and providing a damping (predictive) action.

Quick Tip:

A common exam trap: "eliminates offset" always refers to the integral (I) mode. The derivative (D) mode improves transient response and adds damping but cannot eliminate steady-state offset on its own.

Understanding:

We must find the critical gain $K$ beyond which the system becomes unstable using the Routh–Hurwitz criterion.

Formula:

The closed-loop characteristic equation is obtained from $1+G(s)=0$.

Step 1: Form the characteristic polynomial.

Step 2: Construct the Routh array.

Simplifying the $s^1$ row:

Step 3: Apply stability conditions.

All entries in the first column must be positive:

Step 4: Determine the critical (maximum) value of $K$.

At $K=48$, the $s^1$ element becomes zero, indicating the system is marginally stable (sustained oscillations).

Answer:

The system is stable for $0<K<48$, so the maximum value of $K$ for stability is:

Quick Tip:

For a third-order system $s^3+a{s}^2+bs+c=0$, the Routh stability condition simplifies to $ab>c$ and all coefficients positive. Here: $6\times 8=48>K$, giving $K<48$ directly.

Understanding:

We must find the output of a first-order system at $t=\tau$ for a unit step input.

Formula:

The step response of a first-order system is:

Step 1: Substitute $t=\tau =5$ and $K_p=1$.

Answer:

At $t=\tau$, the first-order system output reaches approximately $63.2\%$ of its final value.

Quick Tip:

This is a universal result: any first-order system reaches $63.2\%$ of its final (steady-state) value at exactly $t=\tau$, regardless of the numerical value of the time constant. At $t=2\tau$: $86.5\%$; at $t=3\tau$: $95\%$; at $t=5\tau$: $99.3\%$ (considered fully settled).

Understanding:

We must compute the gain margin in decibels given the open-loop gain magnitude at the phase crossover frequency.

Formula:

The gain margin is:

In decibels:

Step 1: Compute GM as a ratio.

Step 2: Convert to decibels.

Answer:

A gain margin of $12\text{dB}$ means the system can tolerate a $4\times$ increase in open-loop gain before becoming unstable.

Quick Tip:

Remember: $20{\log }_{10}(2)\approx 6\text{dB}$ and $20{\log }_{10}(4)=20{\log }_{10}(2^2)=2\times 6=12\text{dB}$. These conversions appear frequently in Bode plot problems.

Understanding:

We must recall the Ziegler–Nichols continuous cycling (ultimate gain) tuning rules for a PID controller.

Formula:

The Ziegler–Nichols PID tuning rules (ultimate gain method) are:

Step 1: Identify the integral time parameter.

From the table above, the integral time for a PID controller is:

Step 2: Verify against the other options.

Answer:

According to the Ziegler–Nichols ultimate gain method for a PID controller, the integral time is:

Quick Tip:

For the Ziegler–Nichols PID rules, a useful mnemonic: $K_c=0.6K_u$, ${\tau }_I=P_u/2$, ${\tau }_D=P_u/8$. For a PI controller: $K_c=0.45K_u$, ${\tau }_I=P_u/1.2$.

Understanding:

We must find the phase angle contributed by a pure dead-time element at a given frequency.

Formula:

For a pure dead-time element $G(s)=e^{-\theta s}$, the frequency response is:

The magnitude is $|G(j\omega )|=1$ for all $\omega$, and the phase angle is:

Converting to degrees:

Step 1: Compute phase in radians.

Step 2: Convert to degrees.

Answer:

The phase lag introduced by the dead-time element at $\omega =1\text{rad/min}$ is:

Quick Tip:

Dead time contributes no change in amplitude (gain $=1$ always) but adds an ever-increasing phase lag proportional to frequency. This makes dead time particularly detrimental to closed-loop stability at higher frequencies.

Understanding:

We must derive the standard closed-loop transfer function (CLTF) for a unity negative feedback system.

Formula:

For a negative feedback system with forward gain $G(s)$ and feedback gain $H(s)$, the closed-loop transfer function is:

Step 1: Apply unity feedback condition $H(s)=1$.

Step 2: Interpret the result.

The error signal is $E(s)=R(s)-C(s)$, and $C(s)=G(s)E(s)$. Substituting:

This confirms the result above.

Answer:

The closed-loop transfer function for a unity negative feedback system is:

Quick Tip:

For a non-unity feedback system: $\frac{C}{R}=\frac{G}{1+GH}$. The characteristic equation (denominator set to zero) $1+G(s)H(s)=0$ governs the stability of the closed-loop system.

Understanding:

We must identify the correct characteristic of an integral (I) only controller in a feedback control system.

Step 1: Recall the integral controller action.

The integral controller output is:

In the Laplace domain:

The $1/s$ term adds a pole at the origin to the open-loop transfer function.

Step 2: Analyse steady-state behaviour.

Adding an integrator in the forward path increases the system type by one. A Type 1 system has zero steady-state error for a step input. Therefore, the integral mode eliminates steady-state offset entirely.

Step 3: Analyse stability and speed effects.

The additional pole at the origin from the integrator increases the order of the system, which introduces additional phase lag. This reduces phase margin, making the system more prone to oscillation and potentially sluggish in its response.

Step 4: Eliminate wrong options.

Answer:

The integral controller eliminates steady-state offset by accumulating error over time, but its added phase lag can cause sluggish response and oscillation tendency.

Quick Tip:

Integral windup is a practical concern with I and PI controllers: if the error persists for a long time (e.g., during actuator saturation), the integral term can grow excessively large, causing a large overshoot when the error finally reduces. Anti-windup mechanisms are used to prevent this.