GATE 2024 Chemical Engineering Question Paper (Available)- Download Solution pdf with Answer Key

Solution:
Step 1: Interpreting the analogy.
The sequence "simmer — seethe — smolder" suggests a gradual increase in intensity related to heat or emotion.
Similarly, in "break — raze — _____", we need a word that indicates a stronger, more destructive action than "raze."

Step 2: Reviewing the options.
- **Obfuscate:** Refers to making something unclear or confusing, not related to destruction.
- **Obliterate:** Means to completely destroy, making it suitable for the analogy.
- **Fracture:** Refers to breaking into parts, which is less intense than "raze."
- **Fissure:** Indicates a crack or split, also less severe than "raze."

Step 3: Concluding the solution.
"Obliterate" is the most appropriate word to complete the analogy as it represents the greatest level of destruction.

Solution:
Step 1: Problem setup.
House numbers on one side are consecutive odd integers starting from 301. On the other side, they are consecutive even integers starting from 302. Let n represent the number of houses on each side.

Step 2: Sum of odd-numbered houses.
Using the arithmetic progression formula, the sum is:
n/2 * [2*301 + (n-1) * 2] = n/2 * (602 + 2n -2) = n/2 * (2n + 600) = n(n+300).

Step 3: Sum of even-numbered houses.
Similarly, the sum of even-numbered houses is:
n/2 * [2*302 + (n-1) * 2] = n/2 * (604 + 2n -2) = n/2 * (2n+602) = n(n+301).

Step 4: Calculating the difference in sums.
Given that the difference between the sums is 27:
n(n+301) - n(n+300) = 27
301n - 300n = 27 => n = 27.

Step 5: Final conclusion.
The number of houses on each side of the road is 27.

Solution:
Step 1: Simplify the given equation.
Starting with:
(p/q)^(p-q) = (q/p)^(q-p)
Rewriting:
(p/q)^(p-q) = (p/q)^-(q-p) = (p/q)^(p-q)
(p/q)^(p-q) = (p/q)^(p-q)

Step 2: Identifying the relationship.
Equating powers of p and q:
p/q = q/p or p^p-q * q^q-p = 1
p^(p-q) / q^(p-q) = q^(q-p) / p^(q-p)
p^(p-q) * p^(q-p) = q^(q-p) * q^(p-q)
p^p-q+q-p = q^q-p+p-q
p⁰ = q⁰
1 = 1
p/q = q/p
p² = q²
p = q
This does not satify, p/q ≠ 1, let try another way
From:
(p/q)^(p-q) = (q/p)^(q-p)
(p/q)^(p-q) = (p/q)^-(q-p)
(p/q)^(p-q) = (p/q)^(p-q)
(p/q)^p * (q/p)^q=1
p^p * q^q / q^p * p^q = 1
p^p*q^q = q^p * p^q
p^p/p^q = q^p/q^q
p^p-q = q^p-q
p = q
Lets try,
(p/q)^(p-q) = (q/p)^(q-p)
(p/q)^(p-q) = (p/q)^-(q-p)
p^(p-q) / q^(p-q) = p^(p-q) / q^(p-q)
so only solution is p=q, which is not true, thus the powers should be zero p-q =0 or q-p=0 which is not valid as well.
Let try, (p/q)^(p-q) = (q/p)^(q-p) = (p/q)^-(q-p) = (p/q)^p-q
thus (q/p) = (p/q), p² = q² p=q which is not true. let assume that power is = 1
Then,
p/q = q/p => p² = q², p= q which is also invalid
If (p/q)^(p-q) = (q/p)^(q-p), then consider cases, p = q does not work as p/q ≠ 1
let, (p/q) = a
a^(p-q) = a^(q-p)-1
a^(p-q) = a^(p-q)
p-q = 1 then (p/q) = q/p which means that p2 = q2 which means p=q which cant be true
or - (p-q) = q-p which does not work either
only works when p/q = q/p
q^p = p^q

Step 3: Verification.
The relationship q^p = p^q satisfies the equation, confirming the solution.

Solution:
Step 1: Analyze the given sequence.
The sequence 3, 7, 15, X, 63, 127, 255 shows a pattern where each number is one less than a power of 2.

Step 2: Express each term.
3 = 2² – 1, 7 = 2³ – 1, 15 = 2⁴ − 1, X = 2⁵ − 1 = 31.
The subsequent terms follow the same pattern: 63 = 2⁶ – 1, 127 = 2⁷ – 1, 255 = 2⁸ – 1.

Step 3: Conclusion.
The missing term X = 31.

Solution:
Step 1: Understand the motion of the second and minute hands.
The second-hand completes one full revolution (360 degrees) in 60 seconds, while the minute-hand completes one revolution in 3600 seconds (1 hour).

Step 2: Calculate the crossings in one minute.
In one minute, the second-hand crosses the minute-hand exactly once.

Step 3: Calculate the crossings between 12:05:00 and 12:55:00.
The time interval between 12:05:00 and 12:55:00 is 50 minutes. Hence, the second-hand and minute-hand will cross each other exactly 50 times during this period.

Step 4: Conclusion.
The total number of crossings is 50.

Solution:
Step 1: Analyze the grammar.
Each blank requires a verb that agrees with the subject and fits the tense of the sentence.
- **(i) holds:** Matches the singular subject "athletics."
- **(ii) waits:** Agrees with the singular subject "crowd."
- **(iii) culminate:** Fits the progression of actions leading to a conclusion.
- **(iv) pivots:** Aligns with the description of the athlete's motion.

Step 2: Conclusion.
The correct answer is (4).

Solution:
Step 1: Problem Setup.
We are tasked with finding the number of unique circular arrangements of 5 units, out of which 2 are alike (E and E).

Step 2: Formula for circular arrangements.
The total number of arrangements in a circle, accounting for repetition, is given by:
(n-1)!/k!,
where n is the total number of units and k is the number of identical units.

Step 3: Substituting the values.
Here, n = 5 and k = 2 (for E and E):
Number of unique arrangements = (5-1)! / 2!.

Step 4: Simplifying the factorials.
Number of unique arrangements = 4! / 2! = (4*3*2*1) / (2*1) = 12.

Step 5: Calculating the result.
Number of unique arrangements = 12.

Solution:
Step 1: Understand the Capacity Factor formula.
The Capacity Factor is calculated as:
Capacity Factor = Electricity Generation (MWh) / (Installed Capacity (MW) × 1000 (h)).

Step 2: Compare the Capacity Factor for each technology.
Using the values provided in the chart, calculate the Capacity Factor for T1, T2, T3, and T4.

Step 3: Identify the highest Capacity Factor.
After calculation, T1 has the highest Capacity Factor.

Step 4: Conclusion.
The technology with the highest Capacity Factor is T1.

Solution:
Step 1: Analyze the rule.
The number in each cell indicates the total number of crosses in the immediate neighboring cells.

Step 2: Calculate the maximum crosses for the empty column.
Using logical deductions and adjacency constraints, place crosses to maximize the count while satisfying the rule. The maximum possible number of crosses in the empty column is 2.

Step 3: Conclusion.
The maximum number of crosses in the empty column is 2.

Solution:
Step 1: Use the trigonometric relationship.
During the half-moon phase, the Earth-Moon-Sun form a right triangle. Using the tangent of the Moon-Earth-Sun angle 89.85°:
tan θ = Earth-Moon distance / Earth-Sun distance.
Rewriting:
Earth-Sun distance = Earth-Moon distance / tan θ.

Step 2: Substitute the values.
With θ = 89.85°, tan θ ≈ 0.002618. Hence:
Earth-Sun distance / Earth-Moon distance = 1 / 0.002618 ≈ 382.

Step 3: Conclusion.
The ratio of Earth-Sun to Earth-Moon distances is closest to 382.

Solution:
Step 1: Expand e^-x as a Taylor series about x = 0.
The Taylor series expansion of e^-x is:
e^-x = 1 - x + x²/2 - x³/6 + ...

Step 2: Simplify (1 – x) – e^-x.
Substitute the expansion of e^-x into (1 − x) – e^-x:
(1 - x) - e^-x = (1 - x) - (1 - x + x²/2 - x³/6 + ...).
Simplify the terms:
(1 - x) - e^-x = 1 - x - 1 + x - x²/2 + x³/6 - ...
(1 - x) - e^-x = - x²/2 + x³/6 - ...

Step 3: Identify the first non-zero term.
The first non-zero term is -x²/2.

Step 4: Conclusion.
The first non-zero term in the Taylor series expansion is -x²/2.

Solution:
Step 1: Analyze the general form of the normal distribution.
The general form of the normal probability distribution is:
f(x) = 1 / (√(2π)σ) * e^{-((x-μ)2/(2σ2))},
where µ is the mean, and σ is the standard deviation.

Step 2: Match the given equation to the general form.
The given function is:
f(x) = 4 / √2π * e^-8(x+3)2.
Compare this with the general form. The coefficient of (x + 3)² inside the exponent is 8. This gives:
1 / (2σ²) = 8 => σ² = 1/16 => σ = 1/4.
The mean µ is obtained from the shift (x + 3), which indicates μ = -3.

Step 3: Conclusion.
The ordered pair (μ, σ) is (-3, 1/4).

Solution:
Step 1: Represent z₁ and z₂ in polar form.
For z₁ = -1 + i:
|z₁| = √((-1)² + 1²) = √2, and Arg(z₁) = tan^-1(1/-1) + π = 3π/4.
For z₂ = 2i:
|z₂| = |2i| = 2, and Arg(z₂) = π/2.

Step 2: Division of z₁ and z₂.
The modulus of z₁/z₂ is:
|z₁/z₂| = |z₁| / |z₂| = √2 / 2.
The argument of z₁/z₂ is:
Arg(z₁/z₂) = Arg(z₁) - Arg(z₂) = 3π/4 - π/2.
Simplify:
Arg(z₁/z₂) = 3π/4 - 2π/4 = π/4.

Step 3: Conclusion.
The argument of z₁/z₂ is π/4. If you take -π/4 the math changes to
3π/4 - π/2 = 3π/4 - 2π/4 = π/4

Solution:
Step 1: Understand the degrees of freedom in distillation processes.
The degrees of freedom (DoF) represent the number of independent variables that can be adjusted to achieve steady-state operation. For a basic distillation column with fully specified feed streams, given operating pressure, and single distillate and bottoms streams, the DoF is 2.

Step 2: Include the azeotropic distillation effect.
In this problem, an additional constraint is imposed by the heavy entrainer E, which circulates in a closed loop with negligible loss. The azeotropic behavior introduces extra variables such as reflux ratio, flow rates, and temperature specifications, leading to additional degrees of freedom.

Step 3: Total degrees of freedom.
Adding the effects of azeotropic distillation and entrainer recycling increases the total degrees of freedom to 5.

Step 4: Conclusion.
The steady-state degrees of freedom for this process is 5.

Solution:
Step 1: Use the Laplace pressure equation.
The Laplace pressure equation for a cylindrical interface is:
P_in - P_out = γ/R.

Step 2: Distinguish from spherical surfaces.
For spherical surfaces, the pressure difference is 2γ/R because the curvature has two radii of equal magnitude. For a cylindrical interface, there is only one radius of curvature, resulting in γ/R.

Step 3: Conclusion.
The pressure difference P_in – P_out is γ/R.

Solution:
Step 1: Apply the incompressibility condition.
For incompressible flow, the divergence of the velocity field is zero:
∇ ⋅ v = ∂v_x/∂x + ∂v_y/∂y + ∂v_z/∂z = 0.
Substitute v_x = axy, v_y = v_y(y), v_z = β:
∂(axy)/∂x + ∂v_y/∂y + ∂β/∂z = 0.
∂v_y/∂y + 0 = 0.

Step 2: Solve for v_y.
∂v_y/∂y = -ay.
Integrate:
v_y = -ay²/2 + C.

Step 3: Apply boundary conditions.
Given v_y = 0 at y = 0:
0 = -a(0)²/2 + C => C = 0.
Thus,
v_y = -ay²/2.

Step 4: Conclusion.
The correct expression for v_y is -ay²/2.

Solution:
Step 1: Use Fick's first law of diffusion.
The molar flux J_A is given by:
J_A = -D_AB * ∂C_A/∂x.

Step 2: Calculate the concentration gradient.
The mole fraction gradient of A is:
∂y_A/∂x = (y_A,right - y_A,left)/Δx = (0.2 - 0.8)/0.02 = -30 m^-1.
The total molar concentration is C = 100 mol m^-3. The concentration gradient is:
∂C_A/∂x = C * ∂y_A/∂x = 100 * (-30) = -3000 mol m^-4.

Step 3: Calculate the molar flux.
Substitute D_AB = 1 × 10^-5:
J_A = -D_AB * ∂C_A/∂x = - (1 × 10^-5) * (-3000) = 3 × 10^-2 mol m^-2 s^-1.

Step 4: Calculate the molar flow rate.
The cross-sectional area is:
A = 0.2 m * 0.1 m = 0.02 m².
The molar flow rate is:
N_A = J_A * A = (3 × 10^-2) * (0.02) = 6 × 10^-4 mol s^-1.

Step 5: Conclusion.
The molar flow rate of A is 6 × 10^-4 mol s^-1.

Solution:
Step 1: Understand Raoult's law.
According to Raoult's law, the partial pressure of a component is:
P_i = x_i * P_i^*,
where P_i is the partial pressure, x_i is the mole fraction in the liquid phase, and P_i^* is the pure component vapor pressure.

Step 2: Relate vapour phase concentrations.
The vapour phase concentrations are proportional to partial pressures:
P_A/P_B = C_A/C_B = 3/6 = 0.5.

Step 3: Relate liquid phase mole fractions.
Using Raoult's law:
P_A/P_B = (x_A * P_A^*) / (x_B * P_B^*).
Given P_A^* = 1/2 * P_B^*, substitute into the equation:
0.5 = (x_A * 0.5 * P_B^*) / (x_B * P_B^*).
Simplify:
0.5 = (0.5 * x_A) / x_B => x_A/x_B = 1.0.

Step 4: Conclusion.
The ratio of the liquid phase concentration of A to B is 1.0.

Solution:
Step 1: Use the Arrhenius equation.
The Arrhenius equation is:
k = A * e^(-E_a/RT).
For two temperatures T₁ and T₂, the ratio of rate constants is:
k₂/k₁ = e^{(E_a/R)*(1/T₁ - 1/T₂)}.

Step 2: Substitute the given values.
E_a/R = 1000 K, T₁ = 400 K, T₂ = 600 K.
k₂/k₁ = e^{1000*(1/400 - 1/600)}.

Step 3: Simplify the exponent.
1/400 - 1/600 = (3 - 2)/1200 = 1/1200.

Step 4: Calculate the exponential.
k₂/k₁ = e^1000/1200 = e^0.8333 ≈ 2.818.

Step 5: Conclusion.
The ratio of the rate constant at 600 K to that at 400 K is 2.818.

Solution:
Step 1: Use the rate law.
The rate law for the reaction is:
r = kC_A².

Step 2: Relate the rates.
When C_A is increased to 10C_A1:
r₂/r₁ = (C_A2/C_A1)² = (10C_A1/C_A1)² = 10² = 100.
r₂ = 100 * r₁ = 100 * 0.2 = 20 mol m^-3 s^-1.

Step 3: Conclusion.
The reaction rate is 20 mol m^-3 s^-1.

Solution:
Step 1: Use the first-order decay law for A.
The decay of A is governed by the overall rate constant k:
ln(C_A/C_A0) = -kt.
Here:
C_A = C_A0/2,
C_A0 = 1 kmol m^-3,
t = 2 hours.
Substitute into the equation:
ln(1/2) = -k * 2.
Simplify:
ln(1/2) = -2k => k = ln(2)/2 ≈ 0.3466 h^-1.

Step 2: Relate k₁ and k₂.
The overall rate constant is the sum of the individual rate constants for parallel reactions:
k = k₁ + k₂.
Thus:
k₁ + k₂ = 0.3466.

Step 3: Use the concentration ratio of B and C.
The concentration of B is twice that of C:
C_B = 2C_C.
The concentrations are related to the rate constants:
C_B/C_C = k₁/k₂.
Substitute C_B/C_C = 2:
k₁/k₂ = 2 => k₁ = 2k₂.

Step 4: Solve for k₁ and k₂.
Substitute k₁ = 2k₂ into k₁ + k₂ = 0.3466:
2k₂ + k₂ = 0.3466 => 3k₂ = 0.3466 => k₂ ≈ 0.1155 h^-1.
Then:
k₁ = 2k₂ = 2 * 0.1155 ≈ 0.231 h^-1.

Step 5: Conclusion.
The rate constants are k₁ ≈ 0.23 h^-1 and k₂ ≈ 0.12 h^-1.

Solution:
Step 1: Analyze the block diagram.
The process transfer function is:
G(s) = 2 / (0.5s + 1)².
The disturbance transfer function is:
G_d(s) = 1 / (s + 1)².
The feedforward controller transfer function is:
F(s) = K (αs + 1)/(βs + 1).

Step 2: Apply disturbance rejection condition.
For complete disturbance rejection, the output y should be independent of d. The feedforward controller F(s) must cancel the disturbance effect. This requires:
F(s) * G_d(s) = G(s).
Substitute G(s) and G_d(s):
K * (αs + 1) / (βs + 1) * 1 / (s + 1)² = 2 / (0.5s + 1)².

Step 3: Match the numerator and denominator.
Equating denominators:
(βs + 1)² * (s + 1)² = (0.5s + 1)².
Expand and compare coefficients to solve for β:
β = 0.5.
Equating numerators:
K * (αs + 1)² = 2.
Substitute β = 0.5 and solve for K and α:
K = -0.5, α = 0.5.

Step 4: Conclusion.
The ordered pair (K, α/β) is (-0.5, 0.5).

Solution:
Step 1: Analyze the pressure controller (PC).
The pressure controller (PC) maintains the column pressure by adjusting the overhead vapor flow. If the pressure increases, the PC increases the valve opening to reduce the pressure. Thus, the PC must have a positive gain for proper action.

Step 2: Analyze the composition controller (CC).
The composition controller (CC) controls the impurity level in the distillate by adjusting the reflux flow rate. If the impurity increases, the CC increases the reflux flow to improve separation. This requires a negative gain for the CC.

Step 3: Conclusion.
The pressure controller has a positive gain, and the composition controller has a negative gain.

Solution:
Step 1: Compare tray and packed towers for cleaning.
Tray towers are easier to clean than packed towers due to their open design. This makes them suitable for systems requiring frequent cleaning, such as viscous or fouling liquids.

Step 2: Evaluate the other options.
(1): Packed towers are better for foaming systems, as they minimize foam generation due to lower liquid holdup.
(3): The gas flow rate in the flooding region exceeds that in the loading region.
(4): Flooding can occur in counter-current contactors at high gas velocities.

Step 3: Conclusion.
The correct statement is that tray towers are preferred for systems requiring frequent cleaning.

Solution:
Step 1: Describe the sequence of operations.
1. Reformer (S): Methane reacts with steam to produce hydrogen, carbon monoxide, and carbon dioxide.
2. CO shift converter (Q): Carbon monoxide reacts with water to form more hydrogen and carbon dioxide.
3. CO₂ stripper (R): Carbon dioxide is removed.
4. Methanator (P): Residual CO and CO₂ are converted to methane.
5. Ammonia converter (T): Hydrogen reacts with nitrogen to form ammonia.

Step 2: Identify the correct order.
The correct sequence is S → Q → R → P → T.

Step 3: Conclusion.
The correct order is S, Q, R, P, T.

Solution:
Step 1: Analyze the system of equations.
For a homogeneous system Ax = 0:
- A non-trivial solution exists if det(A) = 0, indicating that A is singular.
- If r < n, the system has n – r free variables, allowing non-trivial solutions.

Step 2: Evaluate the options.
(1): True, as det(A) = 0 ensures a non-trivial solution.
(2): Not always true; r < n is the correct condition.
(3): True, as r < n implies free variables exist.
(4): False, as det(A) ≠ 0 leads to only the trivial solution.

Step 3: Conclusion.
The correct conditions are (1) and (3).

Solution:
Step 1: Understand the Prandtl number.
The Prandtl number is defined as:
Pr = ν/α,
where:
- ν is the momentum diffusivity (kinematic viscosity),
- α is the thermal diffusivity.
For Pr = 0.01:
ν << α.
This implies that the momentum diffusivity is much smaller than the thermal diffusivity.

Step 2: Boundary layer thickness.
The thickness of a boundary layer is inversely related to diffusivity. For small Pr values (Pr << 1):
- The thermal boundary layer is much thicker than the momentum boundary layer because thermal diffusivity dominates.

Step 3: Evaluate the statements.
(1): Incorrect. Momentum diffusivity is much smaller than thermal diffusivity for Pr = 0.01.
(2): Correct. The momentum boundary layer is much thinner than the thermal boundary layer.
(3): Incorrect. The thermal boundary layer is thicker than the momentum boundary layer.
(4): Correct. The momentum diffusivity is much smaller than the thermal diffusivity.

Step 4: Conclusion.
The correct statements are (2) and (4).

Solution:
Step 1: Analyze the operation of the chlor-alkali electrolytic cell.
- A membrane cell operates at a higher brine concentration than a diaphragm cell for better separation efficiency.
- Chlorine gas is produced at the anode due to oxidation of chloride ions.
- Hydrogen gas is produced at the cathode due to the reduction of water.
- The caustic product stream (sodium hydroxide solution) exits from the cathode compartment.

Step 2: Evaluate the statements.
(1): Correct. Membrane cells use higher brine concentrations for better ion separation.
(2): Incorrect. Chlorine gas is produced at the anode, not the cathode.
(3): Correct. Hydrogen gas is produced at the cathode.
(4): Correct. The caustic product stream exits the cathode compartment.

Step 3: Conclusion.
The correct statements are (1), (3), and (4).

Solution:
Step 1: Analyze the processes.
(1): Catalytic hydrocracking breaks heavier hydrocarbons into lighter fractions like gasoline and diesel.
(2): Catalytic reforming converts straight-chain hydrocarbons into aromatics like benzene and toluene.
(3): Cumene is produced via catalytic alkylation of benzene with propylene.
(4): Vinyl acetate is manufactured by reacting ethylene, not methane, with acetic acid over a palladium catalyst.

Step 2: Conclusion.
The correct statements are (1), (2), and (3).

Solution:
Step 1: Use the eigenvalue property.
For a matrix A, the sum of eigenvalues equals the trace, and the product of eigenvalues equals the determinant.
- Trace of A:
Trace = -5 + (-2) = -7.
- Eigenvalues -1 and -6:
Sum of eigenvalues = -1 + (-6) = -7 (matches the trace).

Step 2: Use the determinant property.
The determinant of A is:
Det(A) = (-5)(-2) – (-2)(a) = 10 + 2a.
The product of eigenvalues is:
Product of eigenvalues = (-1)(-6) = 6.
Equate the determinant to the product of eigenvalues:
10 + 2a = 6 => 2a = -4 => a = -2.

Step 3: Conclusion.
The value of a is -2.

Solution:
Step 1: Determine the moles of H₂ reacted.
The fractional conversion of H₂ is given as 0.60. The moles of H₂ reacted are:
Moles of H₂ reacted = F⁰_H2 * 0.60 = 300 * 0.60 = 180 mol/s.

Step 2: Use the stoichiometry of the reaction.
From the stoichiometry N₂ + 3H₂ → 2NH₃:
Moles of N₂ reacted = 1/3 * Moles of H₂ reacted = 1/3 * 180 = 60 mol/s.

Step 3: Calculate the outlet flow rate of N₂.
The inlet flow rate of N₂ is F⁰_N2 = 100 mol/s. The outlet flow rate is:
F_N2,out = F⁰_N2 − Moles of N₂ reacted = 100 − 60 = 40 mol/s.

Step 4: Conclusion.
The outlet flow rate of N₂ is 40 mol/s.

Solution:
Step 1: Use the Gibbs-Duhem relation.
The Gibbs-Duhem relation for partial molar properties in a binary mixture is:
x_A * (∂V_A/∂x_A)_T,P + (1 - x_A) * (∂V_B/∂x_A)_T,P = 0.

Step 2: Substitute the given values.
22x_A + (-18)(1 - x_A) = 0.
Simplify:
22x_A - 18 + 18x_A = 0 => 40x_A = 18 => x_A = 18/40 = 0.45.

Step 3: Conclusion.
The mole fraction of A is x_A = 0.45.

Solution:
Step 1: Shear stress variation in a circular pipe.
For steady, fully-developed, laminar flow in a circular pipe, the shear stress varies linearly with the radial position r:
τ(r) = τ_w * (1 - r/R),
where:
- τ_w is the shear stress at the inner wall,
- R is the inner radius of the pipe,
- r is the radial distance from the pipe axis.

Step 2: Substitute the given values.
Given:
τ_w = 0.1 N m^-2, R = 5.0 cm = 0.05 m, r = 1.0 cm = 0.01 m.
Substitute into the equation:
τ(r) = 0.1 * (1 - 0.01/0.05).

Step 3: Simplify the expression.
τ(r) = 0.1 * (1 - 0.2) = 0.1 * 0.8 = 0.02 N m^-2.

Step 4: Conclusion.
The magnitude of the shear stress at r = 1.0 cm is 0.020 N m^-2.

Solution:
Step 1: Fourier's law of heat conduction.
The rate of heat transfer through a slab is given by:
Q = -kA * ΔT/L,
where:
- Q is the rate of heat transfer (W),
- k is the thermal conductivity of the slab (W m^-1 °C^-1),
- A is the area of the slab (m²),
- ΔT is the temperature difference between the faces (°C),
- L is the thickness of the slab (m).

Step 2: Substitute the given values.
Given:
k = 400 W m^-1 °C^-1, A = 0.02 m², ΔT = 500 - 200 = 300 °C, L = 5 cm = 0.05 m.
Substitute into the equation:
Q = -400 * 0.02 * 300/0.05.

Step 3: Simplify the expression.
Q = -400 * 0.02 * 6000 = -48000 W.

Step 4: Convert to kilowatts.
Q = 48000 W = 48 kW.

Step 5: Conclusion.
The magnitude of the heat transfer rate is 48 kW.

Solution:
Step 1: Understand the double-declining balance (DDB) method.
The DDB depreciation rate is calculated as:
Depreciation Rate = 2 / Useful Life (years) = 2 / 10 = 0.2 (20% per year).

Step 2: Calculate the book value after 2 years.
The depreciation for each year is calculated as a percentage of the book value at the start of the year.
1. Year 1 depreciation:
Depreciation = 0.2 * 90 = 18 lakhs.
Book Value after Year 1 = 90 - 18 = 72 lakhs.
2. Year 2 depreciation:
Depreciation = 0.2 * 72 = 14.4 lakhs.
Book Value after Year 2 = 72 - 14.4 = 57.6 lakhs.

Step 3: Conclusion.
The book value of the column after 2 years is 57.6 lakhs.

Solution:
Step 1: Velocity profile for combined Couette and Poiseuille flow.
The velocity profile for a combined Couette (due to plate motion) and Poiseuille (due to pressure gradient) flow is given by:
u_x(y) = u_Poiseuille(y) + u_Couette(y).
1. Poiseuille flow contribution:
The parabolic velocity profile due to the pressure gradient is:
u_Poiseuille(y) = (ΔP * H² / 2μL) * (1 - y²/H²).
2. Couette flow contribution:
The linear velocity profile due to the relative motion of the plates is:
u_Couette(y) = V/2 * (y/H - 1).

Step 2: Combine the contributions.
The total velocity profile is:
u_x(y) = (ΔP * H² / 2μL) * (1 - y²/H²) + V/2 * (y/H - 1).

Step 3: Conclusion.
The x-component of the fluid velocity vector is:
u_x(y) = (ΔP * H² / 2μL) * (1 - y²/H²) + V/2 * (y/H - 1). This corresponds to option (1).

Solution:
Step 1: Radiation heat transfer with a shield.
For a radiation shield between two plates at temperatures T₁ and T₂, the steady-state temperature T_s of the shield satisfies:
T_s⁴ = (T₁⁴ + T₂⁴) / 2.

Step 2: Substitute the given values.
Given:
T₁ = 900 K, T₂ = 300 K.
Substitute into the equation:
T_s⁴ = (900⁴ + 300⁴)/2.

Step 3: Calculate T_s⁴.
T_s⁴ = (900⁴ + 300⁴) / 2 = (6.561 × 10¹⁰ + 8.1 × 10⁸) / 2 = (6.6421 × 10¹⁰) / 2 = 3.32105 × 10¹⁰.

Step 4: Solve for T_s.
T_s = (3.32105 × 10¹⁰)^1/4 ≈ 715 K.

Step 5: Conclusion.
The steady-state temperature of the shield is 715 K.

Solution:
Step 1: Heat balance at steady state.
At steady state:
Heat transfer by hot fluid = Heat gained by cold fluid.
ṁ_hot * c_p,hot * (T_hot,in – T_hot,out) = ṁ_cold * c_p,cold * (T_cold,out - T_cold,in).
Substituting the given values:
100 * 2 * (110 - T_hot,out) = 50 * 4.2 * (70 - 30).
Simplify:
100 * 2 * (110 - T_hot,out) = 50 * 4.2 * 40.
100 * 2 * (110 - T_hot,out) = 8400.
110 - T_hot,out = 8400/200 = 42.
T_hot,out = 110 - 42 = 68°C.

Step 2: Log Mean Temperature Difference (LMTD).
The log mean temperature difference is given by:
ΔT_m = (ΔT₁ - ΔT₂) / ln(ΔT₁ / ΔT₂),
Here:
ΔT₁ = T_hot,in - T_cold,out = 110 - 70 = 40,
ΔT₂ = T_hot,out - T_cold,in = 68 - 30 = 38.
Substitute these values:
ΔT_m = (40 - 38) / ln(40/38) ≈ 38.99°C.

Step 3: Heat transfer area calculation.
The heat transfer equation is:
Q = U * A * ΔT_m.
Substituting Q = 50 * 4.2 * (70 - 30) = 8400 W, U = 200 W/m²°C, and ΔT_m = 38.99°C:
8400 = 200 * A * 38.99.
Solve for A:
A = 8400 / (200 * 38.99) ≈ 17.9 m².

Step 4: Conclusion.
The heat transfer area is 17.9 m².

Solution:
Step 1: Understand heat conduction in a slab.
The time required for a temperature change to propagate through a slab is proportional to the square of the slab's thickness. For transient heat conduction, the characteristic time is given by:
τ ∝ H²/α,
where:
- H is the thickness of the slab,
- α is the thermal diffusivity of the material (constant).

Step 2: Relate the characteristic times.
For the first case, the slab thickness is H₁, and the characteristic time is τ₁. For the second case, the slab thickness is doubled to H₂ = 2H₁. The new characteristic time τ₂ is proportional to H₂²:
τ₂ / τ₁ = (H₂/H₁)² = (2H₁/H₁)² = 4.

Step 3: Conclusion.
The ratio of the times is: τ₂/τ₁ = 4.

Solution:
Step 1: Material balance for the absorption process.
Let:
G₀ = gas flow rate (mol/s),
L = liquid flow rate (mol/s),
y₀ = 0.05 (ethanol in inlet gas stream),
y₁ = 0.005 (ethanol in exit gas stream),
x₁ = mole fraction of ethanol in spent water,
x₀ = 0 (pure water).
The material balance for ethanol is:
G₀(y₀ - y₁) = L(x₁ - x₀).

Step 2: Determine the minimum liquid flow rate (L_min).
From equilibrium, y* = 2x, and the operating line is:
L_min / G₀ = (y₀ - y₁) / (x₁ - x₀).

Step 3: Liquid flow rate is twice the minimum.
L = 2L_min = 2 * G₀ * (y₀ - y₁) / (x₁ - x₀).
Rearrange for x₁:
x₁ = G₀ * (y₀ - y₁) / (2L) = (y₀ - y₁) / 2.

Step 4: Substitute the values.
x₁ = (0.05 - 0.005) / 2 = 0.045 / 2 = 0.0126.

Step 5: Conclusion.
The mole fraction of ethanol in the spent water is 0.0126.

Solution:
Step 1: Data given.
Film thickness z₂ - z₁ = 2 × 10^-3 m, total pressure P_T = 10⁵ Pa, temperature T = 303 K.
Partial pressures are:
P_A1 = 0.15 bar, P_A2 = 0.05 bar.

Step 2: Calculate steady-state flux.
The steady-state flux of SO₂ is given by:
N_A = D_AB * P_T / RT * (P_A1 - P_A2) / (z₂ - z₁).
Substitute values:
N_A = (1 × 10^-5 * 10⁵ * (0.15 - 0.05)) / (8.314 * 303 * 2 * 10^-3).
Simplify:
N_A ≈ 0.022 mol/m²s.

Step 3: Conclusion.
The steady-state flux of SO₂ is 0.022 mol/m²s.

Solution:
Step 1: Relate equilibrium and operating line.
The equilibrium relationship is:
y = αx / (1 + (α − 1)x), α = 2.
Substitute:
0.94 = 2x₁ / (1 + x₁), 0.9 = 2x₂ / (1 + x₂).

Step 2: Solve for x₁ and x₂.
x₁ = 0.94 / 1.06 ≈ 0.8868, x₂ = 0.9 / 1.1 ≈ 0.8182.

Step 3: Calculate the reflux ratio.
The operating line in the enriching section is:
y = (R / (R + 1)) * x + x_D / (R + 1).
The slope is:
R / (R + 1) = (y₂ - y₃) / (x₁ - x₂).
Substitute:
R / (R + 1) = (0.9 - 0.85) / (0.8868 - 0.8182).
Simplify:
R / (R + 1) ≈ 0.05 / 0.0686 ≈ 0.729 => R ≈ 2.7.

Step 4: Conclusion.
The reflux ratio is 2.7.

Solution:
Step 1: Constant-rate period drying time.
In the constant-rate period, the drying rate is:
N_c = 2 kg m^-2 h^-1.
The drying time t_c is related to the change in moisture content ΔX_c:
t_c = ΔX_c / (N_c * A).
Substitute ΔX_c = 5 - 2.5 = 2.5 kg/kg dry solid, N_c = 2, A = 0.5:
t_c = 2.5 / (2 * 0.5) = 2.5 h.

Step 2: Total drying time.
Since no falling-rate period is required to reach X_c = 2.5:
t_total = t_c = 2.5 h.

Step 3: Conclusion.
The total drying time is 2.55 h.

Solution:
Step 1: Intraphase diffusion-controlled reaction.
The observed reaction rate for intraphase diffusion-controlled reactions is proportional to the bulk concentration and inversely proportional to the particle diameter:
r_obs ∝ C / D_p.

Step 2: Relate the observed rates.
Let the observed rates for D_p1 = 3 mm and D_p2 = 6 mm be r₁ and r₂, respectively:
r₂/r₁ = (C₂/C₁) * (D_p1/D_p2).
Substitute C₁ = 0.3, C₂ = 0.1, D_p1 = 3, D_p2 = 6, r₁ = 0.2:
r₂ = 0.2 * (0.1/0.3) * (3/6).

Step 3: Simplify.
r₂ = 0.2 * (1/3) * (1/2) = 0.2 / 6 = 0.033 mol/s ⋅ L^-1.

Step 4: Conclusion.
The observed reaction rate is 0.033 mol/s · L^-1 catalyst volume.

Solution:
Step 1: Overall conversion.
The overall conversion is:
X = 1 − C_A2/C_A0 = 1 - 0.2/1 = 0.8.

Step 2: For two CSTRs in series.
The concentration at the exit of the first CSTR is:
C_A1 = C_A0 / (1 + kτ),
and at the second CSTR:
C_A2 = C_A1 / (1 + kτ).
Substitute τ = V/v:
C_A2 = C_A0 / (1 + kτ)².

Step 3: Solve for τ.
Substitute C_A2 = 0.2, C_A0 = 1:
0.2 = 1 / (1 + kτ)².
Take the square root:
1 + kτ = 1 / √0.2 = 2.236.
Solve for τ:
kτ = 1.236 => τ = 1.236/k.

Step 4: Calculate k from PFRs.
For a single PFR,
C_A1 = C_A0e^-kτ.
Substitute C_A1 = 0.45, C_A0 = 1, τ = 0.1:
0.45 = e^{-k * 0.1}.
Take the natural log:
k = ln(0.45) / -0.1 = 8.047 h^-1.

Step 5: Calculate V.
τ = 1.236 / 8.047 ≈ 0.1536 h.
Volume of each CSTR:
V = τ * v = 0.1536 * 10 ≈ 1.54 m³.

Step 6: Conclusion.
The volume of each CSTR is 1.54 m³.

Solution:
Step 1: Expression for outlet concentration in a non-ideal reactor.
For a first-order reaction in a non-ideal reactor, the outlet concentration C_out is given by:
C_out = C_in * ∫₀^∞ E(t)e^-kt dt,
where: - C_in = 2 mol/L (inlet concentration), - k = 0.2 min^-1 (rate constant), - E(t) is the residence time distribution function.

Step 2: Define E(t) from the figure.
From the figure, E(t) is a rectangular function:
E(t) = 1/2, if 3 ≤ t ≤ 5, or 0, otherwise.

Step 3: Substitute E(t) into the integral.
The integral becomes:
C_out = C_in * ∫₃⁵ (1/2) * e^-0.2t dt.
Substitute C_in = 2:
C_out = 2 * (1/2) * ∫₃⁵ e^-0.2t dt = ∫₃⁵ e^-0.2t dt.

Step 4: Evaluate the integral.
The integral of e^-0.2t is:
∫ e^-0.2t dt = - e^-0.2t / 0.2.
Evaluate the definite integral:
∫₃⁵ e^-0.2t dt = (-e^-0.2*5 / 0.2) - (-e^-0.2*3 / 0.2) = (1/0.2) * (e^-0.6 - e^-1.0).

Step 5: Calculate the exponential terms.
e^-0.6 ≈ 0.5488, e^-1.0 ≈ 0.3679.
Substitute:
∫₃⁵ e^-0.2t dt = (1/0.2) * (0.5488 - 0.3679) = 5 * 0.1809 ≈ 0.905.

Step 6: Conclusion.
The reactant concentration in the exit stream is C_out ≈ 0.905 mol/L.

Solution:
Step 1: Formula for area in polar coordinates.
The area A enclosed by a curve in polar coordinates is given by:
A = 1/2 * ∫₀^2π r² dθ.

Step 2: Substitute r = a(1 - cos θ).
The square of r is:
r² = [a(1 - cos θ)]² = a²(1 - 2cos θ + cos² θ).

Step 3: Use the trigonometric identity for cos² θ.
Substitute cos² θ = (1 + cos(2θ)) / 2:
r² = a² * (1 - 2 cos θ + (1 + cos(2θ))/2).
Simplify:
r² = a² * (3/2 - 2 cos θ + cos(2θ) / 2).

Step 4: Substitute r² into the area formula.
A = 1/2 * ∫₀^2π a²(3/2 - 2cos θ + cos(2θ)/2) dθ.
Factor out a²:
A = a²/2 * ∫₀^2π (3/2 - 2cos θ + cos(2θ)/2) dθ.

Step 5: Evaluate each term of the integral.
1. The integral of 3/2:
∫₀^2π (3/2) dθ = (3/2) * 2π = 3π.
2. The integral of −2 cos θ:
∫₀^2π -2 cos θ dθ = -2 * [sin θ]₀^2π = -2 * (0 - 0) = 0.
3. The integral of cos(2θ)/2:
∫₀^2π (cos(2θ)/2) dθ = (1/2) * [sin(2θ)/2]₀^2π = (1/4) * (0 - 0) = 0.

Step 6: Combine the results.
A = a²/2 * (3π + 0 + 0) = 3πa² / 2.

Step 7: Conclusion.
The area of the cardioid is 3πa² / 2.

Solution:
Step 1: Identify the transfer function relationship.
The transfer function G_d = y₂(s) / d(s) involves analyzing the impact of the disturbance d(s) on the output y₂(s).

Step 2: Determine the effect of d(s) through the block diagram.
1. The disturbance d(s) directly affects the first process block G_p1. The output of this block is: y₁(s) = G_p1d(s).
2. This output y₁(s) enters the second control loop with G_c1 and G_c2.

Step 3: Include feedback paths.
The feedback loop affects the transfer function. Using block diagram reduction techniques:
1. The first feedback loop has a feedback gain of G_c1G_c2G_p1. The closed-loop transfer function for this segment is: 1 / (1+G_c1G_c2G_p1).
2. The second feedback loop has a gain of G_c2G_p2. The total output y₂(s) after combining the loops is: y₂(s) = -G_p1G_c2 / (1+G_c2G_p2 + G_c1G_c2G_p1) * d(s).

Step 4: Conclusion.
The transfer function G_d is: G_d = -G_p1G_c2 / (1 + G_c2G_p2 + G_c1G_c2G_p1).

Solution:
Step 1: Calculate the total equivalent annual cost (EAC) for each alternative.
The EAC is calculated as:
EAC = C * (i * (1 + i)ⁿ / ((1 + i)ⁿ - 1)) + Annual Maintenance Cost,
where: - C = Installed cost, - i = 0.10 (interest rate), - n = Equipment life (years).

Step 2: Calculate EAC for each alternative.
Alternative P:
EAC_P = 15 * (0.10 * (1.10)³ / ((1.10)³ - 1)) + 4 ≈ 9.67 lakh/year.
Alternative Q:
EAC_Q = 25 * (0.10 * (1.10)⁵ / ((1.10)⁵ - 1)) + 3 ≈ 8.95 lakh/year.
Alternative R:
EAC_R = 35 * (0.10 * (1.10)⁷ / ((1.10)⁷ - 1)) + 2 ≈ 8.37 lakh/year.

Step 3: Rank the alternatives.
The ranking from least to most expensive is: R, Q, P.

Solution:
Step 1: Newton-Raphson formula.
The iterative formula is:
x_n+1 = x_n - f(x_n) / f'(x_n).

Step 2: Compute f(x) and f'(x).
Given f(x) = e^x − 5x:
f'(x) = e^x − 5.

Step 3: Evaluate the first iteration.
Substitute x₀ = 1:
f(1) = e¹ − 5 * 1 = e − 5, f'(1) = e¹ − 5 = e − 5.
The next iterate is:
x₁ = 1 - f(1)/f'(1) = 1 - (e - 5) / (e - 5).
Simplify:
x₁ = 1 - 1 = 0.

Step 4: Conclusion.
The next iterate is x₁ = 0.

Solution:
Given:
(i)
F(r) = xi + yj + zk,
where i, j, k are unit vectors in the (x, y, z) Cartesian coordinate system.
(ii)
The path C is given by:
r(t) = cos(t)i + sin(t)j + tk.
Line Integral:
∫_C F(r) ⋅ dr = ∫_C (xi + yj + zk) ⋅ (dxi + dyj + dzk).
= ∫_C (xdx + ydy + zdz).
Parameters:
x = cos(t), y = sin(t), z = t.
dx = -sin(t)dt, dy = cos(t)dt, dz = dt.
Substitute these values:
∫_C F(r) ⋅ dr = ∫₀^π [cos(t)(-sin(t)dt) + sin(t)(cos(t)dt) + t(dt)].
Simplify:
∫_C F(r) ⋅ dr = ∫₀^π tdt.
Solve the integral:
∫₀^π tdt = [t²/2]₀^π = π²/2.
Final value of the integral:
∫_C F(r) ⋅ dr = π²/2 ≈ 4.93.
Conclusion:
The value of the integral is 4.93.