Chemical Engineering - MCQ Practice Questions

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.

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.

Understanding:

We must identify the correct expression for the effective mean temperature difference used in shell-and-tube heat exchanger design when the arrangement is not purely countercurrent.

Formula:

The standard design equation for a heat exchanger is:

where the corrected mean temperature difference is:

Step 1: Role of the LMTD

The Log Mean Temperature Difference (LMTD) is first calculated assuming purely countercurrent flow between the same terminal temperatures.

Step 2: Role of the correction factor $F$

For multi-pass arrangements (e.g., 1-2, 2-4 exchangers), the actual mean driving force is less than the countercurrent LMTD. The dimensionless correction factor $F$ ($0<F\le 1$) accounts for the departure from pure countercurrent flow. It is obtained from standard charts as a function of two dimensionless ratios $P$ and $R$.

Step 3: Corrected mean temperature difference

The effective temperature difference used in the design equation is:

The LMTD is always calculated on the countercurrent basis regardless of the actual flow arrangement; $F$ then corrects it downward.

Answer:

The corrected mean temperature difference is $F$ multiplied by the countercurrent LMTD.

Quick Tip:

A value of $F<0.75$ is generally considered thermodynamically inefficient; the designer should reconsider the number of shell passes or flow arrangement.

Understanding:

We seek the condition that defines the optimum insulation thickness by minimising total annual cost.

Formula:

At the minimum of total cost:

Step 1: Differentiate total cost

Step 2: Interpret the condition

Since $C_i$ increases with $x$, $\frac{d{C}_i}{dx}>0$, and since $C_h$ decreases with $x$, $\frac{d{C}_h}{dx}<0$. At the optimum, these two slopes are equal in magnitude but opposite in sign, so their sum is zero. This is NOT the same as $C_i=C_h$ (which is a common misconception).

Step 3: Confirm it is a minimum

The second derivative test on $C_T$ confirms a minimum when the curvature is positive, which is the case for the typical cost curves in insulation problems.

Answer:

The optimum thickness satisfies the condition that the sum of the derivatives equals zero.

Quick Tip:

The condition $C_i=C_h$ gives the intersection of the two cost curves, not the minimum of their sum. Always differentiate $C_T$ to find the true optimum.

Understanding:

We must find the minimum height of a cylindrical tank that holds a given volume.

Formula:

Volume of a vertical cylinder:

Step 1: Solve for $H$

Step 2: Round to nearest 0.1 m

Verification:

Answer:

The minimum height of the cylindrical storage tank is approximately 7.9 m.

Quick Tip:

A common error is using diameter instead of radius in $\pi r^2$, which would give $H=100/(\pi \times 16)\approx 1.99\text{m}$ — always halve the diameter first.

Understanding:

We derive how the optimum economic pipe diameter scales with flow rate.

Formula:

For turbulent flow, pumping cost $\propto Q^a{D}^{-m}$ and capital cost $\propto D^n$. Total cost:

Step 1: Minimise with respect to $D$

Step 2: Solve for $D^{\ast }$

For turbulent flow in a smooth pipe, the friction factor gives $a=2$ in the pumping cost exponent (since power $\propto Q\cdot \Delta P\propto Q^3{D}^{-5}$, and capital cost $\propto D^n$), yielding:

Step 3: Interpretation

This is the standard result in plant design for economic pipe sizing. For typical values $m\approx 5$ and $n\approx 1$, $D^{\ast }\propto Q^{1/3}$, consistent with the widely used rule $D_{\text{opt}}\propto Q^{0.35\text{–}0.45}$.

Answer:

The optimum economic pipe diameter scales with flow rate as:

Quick Tip:

The Peters and Timmerhaus correlation for optimum economic diameter ($D_{\text{opt}}\approx 0.363Q^{0.45}{\rho }^{0.13}$ in SI) is derived from exactly this minimisation.

Understanding:

We must identify the correct thin-wall formula for hoop (circumferential) stress in a cylindrical pressure vessel.

Formula:

From equilibrium of a half-cylinder of unit length:

Step 1: Derive by equilibrium

Consider a free-body diagram of a unit length of a half-cylinder. The bursting force due to pressure acting on the projected area $2r\times 1$ is:

This is resisted by two wall cross-sections, each of area $t\times 1$:

Step 2: Equate forces

Step 3: Contrast with longitudinal stress

The longitudinal (axial) stress is half the hoop stress:

Hoop stress is therefore the governing stress for cylindrical vessel design, which is why the wall thickness is sized against it.

Answer:

The hoop stress in a thin-walled cylindrical pressure vessel is:

Quick Tip:

A common trap is confusing hoop stress $Pr/t$ with longitudinal stress $Pr/2t$. Remember: hoop stress is always twice the longitudinal stress in a cylinder, making it the design-limiting stress.

Understanding:

We must state the six-tenths rule for scaling equipment capital cost with capacity in plant design.

Formula:

The six-tenths (power-law or economy-of-scale) rule:

where the exponent $n=0.6$ for most chemical plant equipment.

Step 1: Origin of the exponent

Capital cost scales roughly as surface area (for vessels) while capacity scales as volume. Since area $\propto V^{2/3}\approx V^{0.667}$, the empirical exponent rounds to 0.6. Extensive plant data confirm this value.

Step 2: Apply the rule

For a capacity ratio $S_2/S_1$:

Step 3: Validity

The six-tenths rule is most reliable when $0.1\le S_2/S_1\le 10$. Outside this range, more detailed cost correlations (e.g., Guthrie or Ulrich method) are recommended.

Answer:

The six-tenths rule gives the scaled equipment cost as:

Quick Tip:

For gas compressors the exponent is closer to 0.82, and for furnaces it is about 0.77. The default exponent of 0.6 applies broadly to vessels, heat exchangers, and distillation columns.

Understanding:

We must identify the slope and orientation of the $q$-line on the McCabe–Thiele diagram for a saturated liquid (bubble-point) feed.

Formula:

The $q$-line equation is:

where $q$ is the liquid fraction parameter and $z_F$ is the feed composition.

Step 1: Value of $q$ for saturated liquid feed

For a feed that is a saturated liquid (bubble-point liquid), all the feed enters as liquid:

Step 2: Slope of the $q$-line

An infinite slope corresponds to a vertical line on the $x$–$y$ diagram.

Step 3: Summary of $q$ values

Answer:

For a saturated liquid feed, $q=1$, giving the $q$-line an infinite slope, i.e., the line is vertical passing through $x=z_F$.

Quick Tip:

Memorising the five feed conditions with their $q$ values and corresponding $q$-line orientations is essential for McCabe–Thiele problems in GATE Chemical Engineering.

Understanding:

We must calculate the payback period for a plant investment.

Formula:

Step 1: Calculate the denominator

Step 2: Calculate PBP

Verification:

An investment of \5\text{ M}$recoveredat$\$1\text{ M/yr}$ clearly takes exactly 5 years. $✓$

Answer:

The payback period for this plant is 5 years.

Quick Tip:

Depreciation is added back to net profit because it is a non-cash expense — the cash is still available to recover the investment. Using net profit alone (giving PBP $=6.25$ yr) is a very common exam trap.