Chemical Engineering - MCQ Practice Questions

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.

Understanding:

We must verify the units of ${\text{HTU}}_{OG}$ from the given defining equation.

Formula:

Step 1: Write out the units

Step 2: Cancel common units

Step 3: Physical meaning

${\text{HTU}}_{OG}$ is a length — specifically, the height of packing required to achieve one transfer unit of separation. The total packing height is:

where ${\text{NTU}}_{OG}$ is dimensionless.

Answer:

The units of ${\text{HTU}}_{OG}$ are metres.

Quick Tip:

Always verify HTU units via dimensional analysis before using correlations — this quickly catches errors in which phase (gas or liquid) the coefficient and flow rate correspond to.

Understanding:

We must calculate the annual straight-line depreciation for a heat exchanger.

Formula:

Straight-line depreciation:

Step 1: Compute depreciable amount

Step 2: Divide by service life

Verification:

Over 10 years, total depreciation = 10 \times 10{,}000 = \100{,}000$,whichexactlyrecovers$C_p - C_s$.$\checkmark$

Answer:

The annual straight-line depreciation charge is \10{,}000$ per year.

Quick Tip:

A common mistake is dividing the full purchase price $C_p$ (not $C_p-C_s$) by $n$, which would give \12{,}000/\text{yr}$. Always subtract the salvage value before dividing.