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 = 1 / 2 kmol/m³,
- C_A0 = 1 kmol/m³,
- t = 2 hours.
Substitute:
ln(1 / 2) = −2k.
Solve for k:
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:
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.

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 the 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).

Step 1: Eigenvalue equation for the matrix.The eigenvalues of a matrix A satisfy the characteristic equation:
det(A − λI) = 0,
where λ is the eigenvalue and I is the identity matrix.

Step 2: Write the characteristic equation for matrix A.Matrix A =[ -5 a ]
[ -2 -2 ]

The determinant of A − λI is:
det(A − λI) = det([−5 − λ a ] [−2 −2 − λ]).
det(A − λI) = (−5 − λ)(−2 − λ) − (−2)(a).
Simplify:
det(A − λI) = (−5 − λ)(−2 − λ) + 2a.
det(A − λI) = λ² + 7λ + 10 + 2a.

Step 3: Use the given eigenvalues.The eigenvalues are given as −1 and −6. The characteristic polynomial can also be written as:
det(A − λI) = (λ + 1)(λ + 6).
Expand:
det(A − λI) = λ² + 7λ + 6.

Step 4: Compare coefficients.From the expanded determinant, det(A − λI) = λ² + 7λ + 10 + 2a.
Comparing this with det(A − λI) = λ² + 7λ + 6, we get:
10 + 2a = 6.
Solve for a:
2a = 6 − 10.
a = −2.

Step 5: Conclusion.The value of a is −2.

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 of the reaction N₂ + 3H₂ → 2NH₃:
Moles of N₂ reacted = (1/3) × Moles of H₂ reacted
Moles of N₂ 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.
Outlet flow rate of N₂:
F_N2,out = F⁰_N2 − Moles of N₂ reacted
F_N2,out = 100 − 60 = 40 mol/s.

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

Step 1: Use the Gibbs-Duhem relation.The Gibbs-Duhem relation for partial molar properties in a binary mixture is:
(∂V̄_A/∂x_A)_T,P x_A + (∂V̄_B/∂x_A)_T,P (1 − x_A) = 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.

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.

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 = 48,000 W.
Convert to kilowatts:
Q = 48,000 W = 48 kW.

Step 4: Conclusion.The magnitude of the heat transfer rate is 48 kW.

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.

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:
u_Poiseuille(y) = ΔP H² / 2μL (1 − y² / H²).

2. Couette flow contribution:
u_Couette(y) = V/2 (y/H − 1).

Step 2: Combine the contributions.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).

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.
T₁ = 900 K, T₂ = 300 K.
T_s⁴ = (900⁴ + 300⁴) / 2.

Step 3: Calculate T_s⁴.
T_s⁴ = (6.561 × 10¹⁰ + 8.1 × 10⁸) / 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.

Step 1: Heat balance at steady state.Heat transfer by hot fluid = Heat gained by cold fluid.
100 × 2 × (110 − T_hot,out) = 50 × 4.2 × (70 − 30).

Step 2: Solve for T_hot,out.100 × 2 × (110 − T_hot,out) = 8400.
110 − T_hot,out = 8400 / 200 = 42.
T_hot,out = 110 − 42 = 68°C.

Step 3: Log Mean Temperature Difference (LMTD).ΔT₁ = T_hot,in − T_cold,out = 110 − 30 = 80.
ΔT₂ = T_hot,out − T_cold,in = 68 − 70 = −2.
ΔT_m = (80 − 38) / ln(80/38) ≈ 38.99°C.

Step 4: Heat transfer area calculation.Q = UAΔT_m, where Q = 8400 W.
8400 = 200 × A × 38.99.
A = 8400 / (200 × 38.99) ≈ 17.9 m².

Step 5: Conclusion.The heat transfer area is 17.9 m².

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 = (H₂² / H₁²) · τ₁ = (2H₁)² / H₁² · τ₁ = 4 · τ₁.

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

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 = 2G₀(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.

Step 1: Understand the setup.SO₂ is diffusing through a stagnant air film of thickness 2 × 10⁻³ m. The steady-state flux of SO₂ is calculated using the given data:

Total pressure (P_T) = 1 bar = 10⁵ Pa
Temperature (T) = 30°C = 303 K
Diffusion coefficient (D_AB) = 1 × 10⁻⁵ m²/s
Partial pressures: P_A1 = 0.15 bar, P_A2 = 0.05 bar
P_B1 = P_T − P_A1 = 0.85 bar, P_B2 = P_T − P_A2 = 0.95 bar.

Step 2: Calculate the log mean partial pressure of B.P_{B, lm} = (P_B1 − P_B2) / ln(P_B1/P_B2)
P_{B, lm} = (0.85 − 0.95) / ln(0.95/0.85)
P_{B, lm} = 0.10 / ln(0.95/0.85).

Step 3: Use the flux equation.The steady-state flux of SO₂ is given by:
N_A|SO₂ = (D_ABP_T / RT) · ((P_A1 − P_A2) / (P_{B, lm}(z₂ − z₁))).

Substitute the given values:
N_A|SO₂ = (10⁻⁵ × 10⁵) / (8.314 × 303 × 2 × 10⁻³) × (0.10 / ln(0.95/0.85)).

Step 4: Simplify step-by-step.N_A|SO₂ ≈ 0.022 mol/m²·s.

Step 5: Conclusion.The steady-state flux of SO₂ is 0.022 mol/m²·s.

Step 1: Use equilibrium relationships for A and B.The equilibrium relationship for a binary mixture is:
y = (αx) / (1 + (α − 1)x),

where:
- α is the relative volatility of A with respect to B,
- x is the mole fraction in the liquid phase,
- y is the mole fraction in the vapor phase.

From the problem:
y₁ = 0.94, y₂ = 0.90, y₃ = 0.85.

Step 2: Solve for x₁ and x₂.For y₁ = 0.94:
0.94 = (2x₁) / (1 + x₁), solve for x₁:
x₁ = 0.94 / 1.06 = 0.8868.

For y₂ = 0.90:
0.90 = (2x₂) / (1 + x₂), solve for x₂:
x₂ = 0.90 / 1.10 = 0.8182.

Step 3: Apply the operating line equation.The operating line in the rectifying section is:
y = (R / (R + 1))x + (x_D / (R + 1)),

where the slope is:
Slope = R / (R + 1).

This slope is equal to the slope between the points (x₂, y₃) and (x₁, y₂):
R / (R + 1) = (y₃ − y₂) / (x₁ − x₂).

Step 4: Solve for R.R / (R + 1) = (0.90 − 0.85) / (0.8868 − 0.8182).
R / (R + 1) = 0.05 / 0.0686 = 0.729.
R = 0.729 / (1 − 0.729) = 2.68.

Step 5: Conclusion.The reflux ratio (R) is approximately 2.7.

Step 1: Constant-rate period drying time.In the constant-rate period, the drying rate is given as:
N_c = 2 kg/m²·h.
The drying time in the constant-rate period, t_c, is related to the change in moisture content (ΔX_c):
ΔX_c = X₀ − X_c,

where:
X₀ = 5 kg/kg dry solid (initial moisture content),
X_c = 2.5 kg/kg dry solid (final moisture content in constant-rate period).

The drying time is:
t_c = ΔX_c / (N_c · A),

where A = 0.5 m²/kg dry solid.
Substitute the values:
t_c = (5 − 2.5) / (2 × 0.5) = 2.5 hours.

Step 2: Falling-rate period.Since the moisture content reaches 2.5 kg/kg dry solid at the end of the constant-rate period, no falling-rate period is required to reduce the moisture content further for this question.

Step 3: Conclusion.The total drying time is t_total = t_c = 2.55 hours.

Step 1: Intraphase diffusion-controlled reaction.For intraphase diffusion-controlled reactions, the observed reaction rate is proportional to the concentration and inversely proportional to the particle diameter. The relationship is:

r_obs ∝ C / D_p,

where C is the concentration and D_p is the particle diameter.

Step 2: Relate the observed rates.Let the observed rates for the 3 mm and 6 mm particles be r₁ and r₂, respectively. The relationship is:

r₂ / r₁ = (C₂ / C₁) × (D_p1 / D_p2).

Substitute the given values:
C₁ = 0.3 mol/L,
C₂ = 0.1 mol/L,
D_p1 = 3 mm,
D_p2 = 6 mm,
r₁ = 0.2 mol/s·L⁻¹ catalyst volume.

r₂ = r₁ × (C₂ / C₁) × (D_p1 / D_p2)
r₂ = 0.2 × (0.1 / 0.3) × (3 / 6).

Step 3: Simplify the calculation.r₂ = 0.2 × (1 / 3) × (1 / 2) = 0.033 mol/s·L⁻¹ catalyst volume.

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

Step 1: Overall conversion in the PFR system.The overall conversion is calculated as:
X = 1 − (C_A2 / C_A1) = 1 − (0.2 / 1) = 0.8.

Step 2: Rate constant for the PFRs.For a first-order reaction in a PFR:
C_A = C_A0e^−kτ,

where τ = V / ν, with V = 1 m³ and ν = 10 m³/h, so τ = 0.1 h.
For the overall reaction in two PFRs:
C_A = C_A0e^{−k(τ₁+τ₂)}, where τ₁ = τ₂ = 0.1 h.
Substitute values:
0.2 = 1e^−k·0.2.
ln(0.2) = −0.2k.
k = −ln(0.2) / 0.2 = 8.047 h⁻¹.

Step 3: Replace PFRs with CSTRs.For a CSTR:
C_A2 / C_A1 = 1 / (1 + kτ).
Since both CSTRs have equal volume:
C_A2 / C_A0 = 1 / (1 + kτ)².
Substitute values:
0.2 = 1 / (1 + 8.047τ)².
Take the square root:
√0.2 = 1 / (1 + 8.047τ).
1 + 8.047τ = 1 / √0.2.
8.047τ = (1 / √0.2) − 1.
τ = ((1 / √0.2) − 1) / 8.047 ≈ 0.15316 h.

Step 4: Calculate the volume of each CSTR.V = τν = 0.15316 × 10 = 1.5316 m³.

Step 5: Conclusion.The volume of each CSTR is approximately 1.54 m³.

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,
- k = 0.2 min⁻¹,
- E(t) is the residence time distribution function.

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

Step 3: Substitute E(t) into the integral.C_out = C_in ∫₃⁵ (1/2)e^−0.2t dt.
C_out = 2 × (1/2) ∫₃⁵ e^−0.2t dt.
C_out = ∫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 from t = 3 to t = 5:
∫₃⁵ e^−0.2t dt = (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 0.905 mol/L.

Step 1: Formula for area in polar coordinates.The area A enclosed by a curve in polar coordinates is given by:
A = (1/2) ∫₀²π r² dθ.

Step 2: Substitute r = a(1 − cos θ).The square of r is:
r² = [a(1 − cos θ)]² = a²(1 − 2 cos 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) ∫₀²π a²[(3/2) − 2 cos θ + cos(2θ)/2] dθ.
Factor out a²:
A = (a²/2) ∫₀²π [(3/2) − 2 cos θ + cos(2θ)/2] dθ.

Step 5: Evaluate each term of the integral.1. The integral of (3/2):
∫₀²π (3/2) dθ = (3/2) × 2π = 3π.
2. The integral of −2 cos θ:
∫₀²π −2 cos θ dθ = −2[sin θ]₀²π = −2(0 − 0) = 0.
3. The integral of cos(2θ)/2:
∫₀²π (cos(2θ)/2) dθ = (1/2)[sin(2θ)/2]₀²π = (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.

Step 1: Identify the transfer function relationship.The transfer function G_d = y₂(s) / d(s) relates the disturbance d(s) to the output y₂(s). This requires analyzing the block diagram and accounting for feedback paths.

Step 2: Determine the effect of d(s).1. The disturbance d(s) directly affects the first process block G_p1, producing:
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 loops.1. The first feedback loop has a feedback gain of G_c1G_c2G_p1, resulting in a closed-loop transfer function for this segment:
1 / (1 + G_c1G_c2G_p1).
2. The second feedback loop has a gain of G_c2G_p2. Combining the loops yields:
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).

Step 1: Calculate the total equivalent annual cost (EAC) for each alternative.The formula for EAC is:
EAC = (C × i) / (1 − (1 + i)⁻ⁿ) + Annual Maintenance Cost,
where:
C = Installed cost,
i = Interest rate = 0.10,
n = Equipment life (years).

Step 2: Calculate EAC for each alternative.1. **Alternative P:**
C = 15, n = 3, Annual Maintenance Cost = 4.
EAC = (15 × 0.10) / (1 − (1 + 0.10)⁻³) + 4.
EAC ≈ (1.5) / (1 − 0.7513) + 4 = (1.5) / 0.2487 + 4 ≈ 6.03 + 4 = 10.03.

2. **Alternative Q:**
C = 25, n = 5, Annual Maintenance Cost = 3.
EAC = (25 × 0.10) / (1 − (1 + 0.10)⁻⁵) + 3.
EAC ≈ (2.5) / (1 − 0.6209) + 3 = (2.5) / 0.3791 + 3 ≈ 6.59 + 3 = 9.59.

3. **Alternative R:**
C = 35, n = 7, Annual Maintenance Cost = 2.
EAC = (35 × 0.10) / (1 − (1 + 0.10)⁻⁷) + 2.
EAC ≈ (3.5) / (1 − 0.5132) + 2 = (3.5) / 0.4868 + 2 ≈ 7.19 + 2 = 9.19.

Step 3: Compare EAC values.EAC_R = 9.19,
EAC_Q = 9.59,
EAC_P = 10.03.
The order from least to most expensive is:
R, Q, P.

Step 4: Conclusion.The correct option is R, Q, P.

Step 1: Given function and initial guess.f(x) = e^x − 5x, with x⁽⁰⁾ = 1.

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

Step 3: Newton-Raphson iterative formula.The formula for the next iterate is:
x⁽¹⁾ = x⁽⁰⁾ − f(x⁽⁰⁾) / f'(x⁽⁰⁾).

Step 4: Substitute x⁽⁰⁾ = 1.f(1) = e¹ − 5 × 1 = e − 5.
f'(1) = e¹ − 5 = e − 5.
x⁽¹⁾ = 1 − (e − 5) / (e − 5) = 1 − 1 = 0.

Step 5: Conclusion.The value of the next iterate x⁽¹⁾ is 0.00.

Step 1: Given parameters.F(r) = xi + yj + zk.
Path C is defined as r(t) = cos(t)i + sin(t)j + tk.

Step 2: Compute F(r) · dr.F(r) = cos(t)i + sin(t)j + tk.
dr = (-sin(t)dt)i + (cos(t)dt)j + dt k.
F(r) · dr = cos(t)(-sin(t)dt) + sin(t)(cos(t)dt) + t(dt).
F(r) · dr = -cos(t)sin(t)dt + sin(t)cos(t)dt + t(dt).
F(r) · dr = t(dt).

Step 3: Set up the integral.∫_C F(r) · dr = ∫₀^π t dt.

Step 4: Solve the integral.∫₀^π t dt = [t² / 2]₀^π = π² / 2 − 0 = π² / 2 ≈ 4.93.

Step 5: Conclusion.The value of the integral is 4.93.

Step 1: Solve the differential equation.The given equation is a Cauchy-Euler equation:
x² d²y/dx² − x dy/dx − 3y = 0.
Assume a solution of the form y = x^m. Substituting this into the equation gives:
m(m − 1) − m − 3 = 0.
Simplify:
m² − 2m − 3 = 0.
Factorize:
(m − 3)(m + 1) = 0.
The roots are m = 3 and m = −1.

Step 2: General solution.The general solution is:
y(x) = C₁x⁻¹ + C₂x³.

Step 3: Apply boundary conditions.1. At x = 1, y(1) = 2:
2 = C₁(1)⁻¹ + C₂(1)³ = C₁ + C₂.
C₁ + C₂ = 2. . . (i)
2. At x = 2, y(2) = 17/2:
17/2 = C₁(2)⁻¹ + C₂(2)³ = C₁/2 + 8C₂.
Multiply through by 2:
17 = C₁ + 16C₂. . . (ii).

Step 4: Solve for C₁ and C₂.From equation (i):
C₁ = 2 − C₂.
Substitute into equation (ii):
17 = (2 − C₂) + 16C₂.
17 = 2 + 15C₂.
15C₂ = 15.
C₂ = 1, C₁ = 1.

Step 5: Compute y at x = 3/2.y(3/2) = C₁(3/2)⁻¹ + C₂(3/2)³.
y(3/2) = 1/(3/2) + (3/2)³.
y(3/2) = 2/3 + 27/8.
y(3/2) = 0.6667 + 3.375 = 4.0417 ≈ 4.06.

Step 6: Conclusion.The solution y(x) at x = 3/2 is 4.06.

Step 1: Compute the gradient of the function f(x, y, z).
The gradient is given by:
∇f = (∂f/∂x)i + (∂f/∂y)j + (∂f/∂z)k.
f(x, y, z) = x⁴ + 2y³ + z².
∂f/∂x = 4x³, ∂f/∂y = 6y², ∂f/∂z = 2z.
∇f = 4x³i + 6y²j + 2zk.

Step 2: Evaluate the gradient at P(−1, −1, −1).
∇f|_P = 4(−1)³i + 6(−1)²j + 2(−1)k.
∇f|_P = −4i + 6j − 2k.

Step 3: Normalize the direction vector a = i + j.
a = i + j.
|a| = √(1² + 1²) = √2.
Unit vector: â = (i + j) / √2.

Step 4: Compute the directional derivative.
The directional derivative is:
DD = ∇f · â.
DD = (−4i + 6j − 2k) · ((i + j) / √2).
DD = (−4/√2) + (6/√2).
DD = (−4 + 6) / √2 = 2 / √2 = √2 ≈ 1.41.

Step 5: Conclusion.
The directional derivative of the function at P(−1, −1, −1) along (i + j) is 1.41.

Step 1: Define variables.Let F = feed flow rate of A, P = purge flow rate, R = recycle flow rate.
The relationship between recycle and purge streams is:
R / P = 10 ⇒ R = 10P.

Step 2: Material balance on A.The total flow of A entering the reactor is:
F + R = F + 10P.
Since the single-pass conversion is 50%, unreacted A is:
Unreacted A = 0.5(F + 10P).
The unreacted A splits into recycle and purge streams:
R + P = 0.5(F + 10P).
Substitute R = 10P:
10P + P = 0.5(F + 10P).
11P = 0.5F + 5P.
6P = 0.5F.
F = 12P.

Step 3: Solve for P.The total product rate B is 100 kmol/h, and the feed is 90% A:
F × 0.9 = 100 ⇒ F = 100 / 0.9 = 111.1 kmol/h.
Substitute F = 12P:
111.1 = 12P.
P = 111.1 / 12 ≈ 18.2 kmol/h.

Step 4: Conclusion.The flow rate of A in the purge stream is 18.2 kmol/h.

Step 1: Calculate the stoichiometric air-to-methane ratio r_s.The combustion reaction requires 2 moles of O₂ per mole of CH₄. Since air contains 20% O₂:
r_s = 2 mol O₂ / 0.2 mol O₂/mol air = 10.

Step 2: Determine methane conversion X.For r = 1.05r_s:
r = 1.05 × 10 = 10.5.
Substitute r into the conversion equation:
X = 1 − 0.1e^{−2(10.5 − 10)} = 1 − 0.1e^−2(0.5).
e⁻¹ ≈ 0.3679.
X = 1 − 0.1 × 0.3679 = 1 − 0.03679 = 0.9632.

Step 3: Energy balance for the flue gas.The total heat released by combustion is:
Q = X × ΔH_r = 0.9632 × (−880) = −847.6 kJ/mol.
The molar heat capacity of the flue gases is given as c_p = 40 J/mol·°C.
The temperature rise of the flue gas is:
ΔT = −Q / c_p.
Substitute Q = −847.6 kJ/mol = −847600 J/mol and c_p = 40 J/mol·°C:
ΔT = 847600 / 40 = 21190 °C.
The exit flue gas temperature is:
T_exit = T_inlet + ΔT = 50 + 21190 = 1727 °C.

Step 4: Conclusion.The exit flue gas temperature is 1727.0 °C.

Step 1: Apply the first law of thermodynamics.For an isolated system, the change in internal energy (ΔU) equals the work done (ΔW):
ΔU = ΔW, and ΔQ = 0 (no heat transfer).

Step 2: Calculate the initial number of moles of gas.Using the ideal gas law, P × V = n × R × T:
P = 5 × 10⁵ Pa, V = 0.1 m³, R = 400 J/mol·K, T = 400 K.
n = (P × V) / (R × T) = (5 × 10⁵ × 0.1) / (400 × 400) = 125 mol.

Step 3: Work done by the spring.The work done by the spring is:
W = −(k × x²) / 2, where k = 0.3 N/m and x = 0.2 m.
W = −(0.3 × 0.2²) / 2 = −0.03 J.

Step 4: Relating internal energy to temperature.For a monatomic ideal gas, ΔU = (3/2) n R ΔT.
Using ΔU = W:
(3/2) × 125 × 400 × (T_f − 400) = −0.03.
Solve for T_f:
T_f = 400 − (0.03 / (3/2 × 125 × 400)).
T_f = 400 − 0.0001 ≈ 397 K.

Step 5: Conclusion.The final equilibrium temperature is 397.0 K.

Step 1: Relation between Z and B.The virial equation of state is:
Z = P V / RT = 1 + B P / RT.
The slope of Z with respect to P is:
∂Z/∂P|_T = B / RT.

Step 2: Solve for B.Given ∂Z/∂P|_T = −3.538 × 10⁻³ bar⁻¹, T = 340 K, R = 83.14 bar·cm³/mol·K:
B = (∂Z/∂P|_T) × RT.
B = −3.538 × 10⁻³ × (83.14 × 340).
B ≈ −99.0 cm³/mol.

Step 3: Relation to molar residual Gibbs free energy.∂G_R/∂P|_T = B.
Thus, ∂G_R/∂P|_T = −99.0 cm³/mol.

Step 4: Conclusion.The value of ∂G_R/∂P|_T is −99.0 cm³/mol.

Step 1: Force balance for terminal velocity.At terminal velocity, the gravitational force is balanced by the drag force:
(π/6) d³ρ_pg = (1/2) C_Dρ_gA v_t².
Substitute A = (π/4)d²:
(π/6) d³ρ_pg = (1/2) C_Dρ_g(π/4)d²v_t².

Step 2: Solve for terminal velocity.v_t² = (4dρ_pg) / (3C_Dρ_g).
Substitute the given values:
d = 0.001 m, ρ_p = 7001 kg/m³, ρ_g = 1 kg/m³, g = 9.8 m/s², C_D = 0.44:
v_t² = (4 × 0.001 × 7001 × 9.8) / (3 × 0.44 × 1).
v_t² ≈ 207.91.
v_t ≈ √207.91 ≈ 14.50 m/s.

Step 3: Conclusion.The terminal settling velocity is 14.50 m/s.

Step 1: Apply Bernoulli’s equation with pump head.The energy equation between Reservoir 1 and Reservoir 2 is:
h_p = h_L + (z₂ − z₁),
where:
- h_p = pump head,
- h_L = head loss in the pipe,
- z₂ − z₁ = 50 m (elevation difference).

Step 2: Calculate head loss h_L.The head loss in the pipe is:
h_L = (f L v²) / (D 2g),
where:
- f = 0.06,
- L = 100 m,
- D = 0.1 m,
- v = Q / A = (3.14 × 10⁻²) / [π (0.1/2)²] ≈ 4 m/s.
Substitute values:
h_L = (0.06 × 100 × 4²) / (0.1 × 2 × 9.8).
h_L = 96 / 1.96 ≈ 49 m.

Step 3: Total pump head h_p.h_p = h_L + (z₂ − z₁) = 49 + 50 = 99 m.

Step 4: Pump power.The pump power is:
P = ρ g Q h_p,
where:
- ρ = 1000 kg/m³,
- g = 9.8 m/s²,
- Q = 3.14 × 10⁻² m³/s,
- h_p = 99 m.
Substitute values:
P = 1000 × 9.8 × (3.14 × 10⁻²) × 99 ≈ 30,506 W = 30.50 kW.

Step 5: Conclusion.The pump power is 30.50 kW.

Step 1: Relation between flow rate and pressure difference.The flow rate through a Venturi meter is given by:
Q = C_dA √(2ΔP / ρ),
where:
- C_d = discharge coefficient,
- A = cross-sectional area,
- ΔP = pressure drop,
- ρ = fluid density.
Since ΔP ∝ Q², the manometer reading Δh (which represents ΔP) is also proportional to Q²:
Δh ∝ Q².

Step 2: Ratio of manometer readings.If the flow rate is doubled (Q₂ = 2Q₁):
Δh₂ / Δh₁ = (Q₂ / Q₁)² = (2Q₁ / Q₁)² = 2² = 4.

Step 3: Conclusion.The ratio of the manometer readings Δh₂/Δh₁ is 4.

Step 1: Carnot efficiency for each stage.The Carnot efficiency for a heat engine is:
η_Carnot = 1 − (T_cold / T_hot).
1. For Stage 1:
η₁ = 1 − (360 / 400) = 0.1 (10%).
2. For Stage 2:
η₂ = 1 − (300 / 360) = 0.167 (16.7%).

Step 2: Effective Carnot efficiency of the process.The overall Carnot efficiency is:
η_overall = η₁ + (1 − η₁) · η₂.
Substitute values:
η_overall = 0.1 + (1 − 0.1) · 0.167 = 0.1 + 0.9 · 0.167 = 0.1 + 0.1503 = 0.2503 (25.03%).

Step 3: Actual efficiency.The actual process efficiency is 50% of the Carnot efficiency:
η_actual = 0.5 × 0.2503 = 0.12515 (12.52%).

Step 4: Power delivered by the process.The power delivered by the process is:
P = η_actual × Heat input.
Substitute η_actual = 0.12515 and Heat input = 2 kW:
P = 0.12515 × 2 = 0.25 kW.

Step 5: Conclusion.The power delivered by the process is 0.25 kW.

Step 1: Calculate the dilution rate D.The dilution rate is defined as:
D = F / V,
where F = 75 L/h and V = 200 L.
D = 75 / 200 = 0.375 h⁻¹.

Step 2: Use the Monod equation to solve for C_S.The specific growth rate is equal to the dilution rate at steady state:
μ_g = μ_m C_S / (K_S + C_S).
Substitute the given values: μ_g = 0.375, μ_m = 0.25, K_S = 1:
0.375 = 0.25 C_S / (1 + C_S).
Rearrange:
1.5 = C_S / (1 + C_S).
1.5 (1 + C_S) = C_S.
1.5 + 1.5C_S = C_S.
1.5 = −0.5C_S.
C_S = 3 g/L.

Step 3: Conclusion.The glucose concentration in the recycle stream C_S1 is 3.0 g/L.

Step 1: Define the dynamics of the system.
The rate of change of hold-up V in the surge drum is:
dV/dt = F_in − F_out,
where:
F_out = K_c(V − V̅) + F_out.

Step 2: Initial steady state.
At steady state, F_in = F_out = 1 m³/h and V = V̅ = 5 m³.

Step 3: Maximum allowable deviation in hold-up.
The maximum allowable hold-up is V_high = 0.8V_full. Thus, the maximum deviation is:
ΔV = V_high − V̅ = 0.8V_full − 0.5V_full = 0.3V_full.

Step 4: Relation between K_c and ΔV.
The maximum deviation in hold-up ΔV is related to the step increase in F_in by:
ΔV = ΔF_in / K_c,
where ΔF_in = F_in,new − F_in,old = 2 − 1 = 1 m³/h.
Rearrange for K_c:
K_c ≥ ΔF_in / ΔV.
Substitute ΔF_in = 1 m³/h and ΔV = 0.3V_full:
K_c ≥ 1 / (0.3 × 10) = 0.33 h⁻¹.

Step 5: Conclusion.
The minimum controller gain K_c^min is 0.33 h⁻¹.

Step 1: Write the characteristic equation for the closed-loop system.The characteristic equation is given by:
1 + G_cG_p = 0.
Substitute G_c and G_p:
1 + [K_c (τ_Ds + 1) / (τ_Ds)] × [1 / ((s − 1)(10s + 1))] = 0.
Simplify:
1 + K_c (τ_Ds + 1) / [τ_Ds (s − 1)(10s + 1)] = 0.
Multiply through by the denominator:
(10s³) + s²(10 − τ_DK_c) + s(1 − K_cτ_D) + K_c = 0.

Step 2: Apply the Routh-Hurwitz stability criterion.For stability, all coefficients in the Routh array must be positive. The characteristic polynomial is:
10s³ + s²(10 − τ_DK_c) + s(1 − K_cτ_D) + K_c.
The Routh array is constructed as follows:
s³: 10, (1 − K_cτ_D)
s²: (10 − τ_DK_c), K_c
s¹: [(10 − τ_DK_c)K_c − 10(1 − K_cτ_D)] / (10 − τ_DK_c), 0
s⁰: K_c.

Step 3: Solve for τ_D from stability conditions.The first condition for stability is:
(10 − τ_DK_c) > 0.
Solve for τ_D:
τ_D < 10 / K_c.
The second condition comes from the s¹ row, ensuring the numerator is positive. Simplify and solve to find the maximum τ_D for stability:
τ_D = 22.2 minutes.

Step 4: Conclusion.The maximum feasible value of τ_D is 22.2 minutes.

Step 1: Calculate the column cross-sectional area.The cross-sectional area of the column is:
A_column = π(D/2)²,
where D = 120 cm.
A_column = π × (120/2)² = π × 60² = π × 3600 ≈ 11310 cm².

Step 2: Calculate the total area occupied by downcomers.Assuming there are two downcomers, the total area occupied is:
A_downcomers = 2 × 575 = 1150 cm².

Step 3: Calculate the percentage of area not available for vapor flow.The percentage area occupied by the downcomers is:
% Area = (A_downcomers / A_column) × 100.
Substitute A_downcomers = 1150 cm² and A_column = 11310 cm²:
% Area = (1150 / 11310) × 100 ≈ 10.2%.

Step 4: Conclusion.The percentage column cross-sectional area not available for vapor flow due to the downcomers is 10.2%.