THEORY EXAMINATION (SEM–IV) 2016-17 HEAT TRANSFER

Course: B.Tech (Chemical Engineering)
Subject Code: CH401
Subject Title: Heat Transfer
Exam Type: Theory (Semester IV, 2016–17)
Duration: 3 Hours
Maximum Marks: 100

Note: Be concise. Assume missing data for numerical problems.

SECTION – A (10 × 2 = 20 Marks)

Short conceptual questions testing theoretical clarity and definitions.

Topics:

Newton’s Law of Cooling:
Rate of heat transfer ∝ temperature difference between surface and fluid.

1. Q=hA(Ts−T∞)Q = hA(T_s - T_\infty)Q=hA(Ts​−T∞​)

Fin Efficiency and Effectiveness:

Efficiency: Ratio of actual heat transferred by fin to heat transferred if the entire fin were at base temperature.

Effectiveness: Ratio of heat transferred by fin to heat transferred without fin.

Biot Number (Bi):

1. Bi=hLckBi = \frac{hL_c}{k}Bi=khLc​​

Indicates the ratio of internal to surface thermal resistance.

Black Body vs Gray Body:

Black body: Absorbs all incident radiation.

Gray body: Absorptivity < 1 and independent of wavelength.

Natural vs Forced Convection:

Natural: Fluid motion due to density difference (buoyancy).

Forced: External devices (fan/pump) cause motion.

Boiling and Condensation:

Boiling: Liquid → vapor phase transition due to heat addition.

Condensation: Vapor → liquid phase change due to cooling.

Compact Heat Exchanger:
Large surface area per unit volume (>700 m²/m³), e.g., car radiators, air heaters.

Fouling Factor:
Represents additional thermal resistance due to deposits on heat transfer surfaces.

Emissive Power & Monochromatic Emissive Power:

Emissive Power (E): Total radiant energy emitted per unit area.

Monochromatic Eλ: Emission per wavelength interval.

Kirchhoff’s Law:
For a body in thermal equilibrium, emissivity = absorptivity at a given wavelength.

SECTION – B (Attempt Any Five – 5 × 10 = 50 Marks)

This section mixes numerical and theoretical questions covering conduction, convection, heat exchangers, and radiation.

(a) Composite Wall Heat Loss Calculation

A wall has 7.5 cm fire plate + 0.65 cm mild steel plate.
Inside gas = 650°C, outside air = 27°C.
Given:

hi=60 W/m2Kh_i = 60\, W/m^2Khi​=60W/m2K, ho=8 W/m2Kh_o = 8\, W/m^2Kho​=8W/m2K

k1=1.035 W/mKk_1 = 1.035\, W/mKk1​=1.035W/mK, k2=53.6 W/mKk_2 = 53.6\, W/mKk2​=53.6W/mK
Find:

Heat loss (W/m²)

Outside surface temperature.

(b) Insulated Pipe Heat Loss

Steel tube (inner dia 5.08 cm, outer 7.62 cm, k = 43.26 W/mK) + insulation (2.5 cm thick, k = 0.208 W/mK).
Inside gas = 316°C, hi=28 W/m2Kh_i = 28\, W/m^2Khi​=28W/m2K, outside air = 30°C, ho=17 W/m2Kh_o = 17\, W/m^2Kho​=17W/m2K.
→ Calculate heat loss for 3 m length.

(c) Parallel vs Counter-Flow Heat Exchanger Efficiency

Oil (120°C → 90°C) heats water (30°C → 60°C).
Both have same U.
→ Find A1/A2A_1/A_2A1​/A2​ for co-current vs counter-current flow.

(d) Double Pipe Heat Exchanger Design

Cold fluid: 4 kg/s (20°C → 40°C)
Hot fluid: 1 kg/s (160°C), same Cp, U=640W/m2KU = 640 W/m^2KU=640W/m2K
→ Find area ratio for co-current and counter-current arrangements.

(e) Radiation Exchange Between Plates

Two parallel iron plates (3 m × 3 m, 25 mm apart) at 100°C and 40°C.
Find net radiant heat transfer if emissivity (ε) is same for both.

(f) Hydrodynamic and Thermal Boundary Layers

Explain with a diagram:

Development of boundary layers.

Distinguish viscous sub-layer (laminar) and buffer layer (transition zone).

(g) Critical Thickness of Insulation

Derive condition for optimum insulation thickness in a cylinder.

rc=khr_c = \frac{k}{h}rc​=hk​

Explain physical significance.

(h) Triple-Effect Evaporators

Draw and explain forward, backward, and mixed feed systems.
Discuss relative merits & demerits (e.g., steam economy vs complexity).

SECTION – C (Attempt Any Two – 2 × 15 = 30 Marks)

Analytical and conceptual derivations or design-based problems.

3. Pool Boiling & Heat Exchanger Redesign

Explain pool boiling regimes: natural convection, nucleate, transition, and film boiling.

Numerical:
Oil cooled from 450K → 410K by water (300K → 350K).
Redesign to cool oil to 390K while keeping other parameters constant.
→ Find new tube length required.

4. LMTD Derivation

Derive the Log Mean Temperature Difference (LMTD) for a parallel-flow heat exchanger:

ΔTm=(ΔT1−ΔT2)ln⁡(ΔT1/ΔT2)\Delta T_m = \frac{(\Delta T_1 - \Delta T_2)}{\ln(\Delta T_1 / \Delta T_2)}ΔTm​=ln(ΔT1​/ΔT2​)(ΔT1​−ΔT2​)​

State assumptions (steady state, no phase change, constant U).
Explain why counter-flow exchangers are more efficient.

5. Heat Exchanger Classification & NTU Method

Classify exchangers:

Direct Contact, Indirect Contact, Recuperators, Regenerators.

Draw temperature distribution in condenser & evaporator.

Derive effectiveness (ε) of a parallel-flow exchanger using NTU method:

• ε=1−e−NTU(1+Cr)1+Cr\varepsilon = \frac{1 - e^{-NTU(1 + C_r)}}{1 + C_r}ε=1+Cr​1−e−NTU(1+Cr​)​

Summary

This paper comprehensively tests knowledge of:

Modes of Heat Transfer (conduction, convection, radiation).

Heat Exchanger Design (LMTD & NTU methods).

Critical insulation, boundary layers, and boiling regimes.