4.2 Convective Heat Transfer & Forced/Natural Convection Coefficients

Key Takeaways

  • Newton's Law of Cooling governs convection: q˙conv=hcA(Ts−T∞)=Ts−T∞Rconv\dot{q}_{conv} = h_c A (T_s - T_\infty) = \frac{T_s - T_\infty}{R_{conv}}, where hch_c is the convective heat transfer coefficient in Btu/(hr⋅ft2⋅∘F)\text{Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)} or W/(m2⋅K)\text{W/(m}^2\cdot\text{K)}.

  • The fundamental dimensionless numbers in convection are Reynolds (Re=ρVDμ\text{Re} = \frac{\rho V D}{\mu}), Prandtl (Pr=cpμk=να\text{Pr} = \frac{c_p \mu}{k} = \frac{\nu}{\alpha}), Nusselt (Nu=hDkf\text{Nu} = \frac{h D}{k_f}), Grashof (Gr=gβΔTL3ν2\text{Gr} = \frac{g \beta \Delta T L^3}{\nu^2}), and Rayleigh (Ra=Gr⋅Pr\text{Ra} = \text{Gr}\cdot\text{Pr}).

  • For fully developed turbulent flow in circular tubes (ReD≥10,000\text{Re}_D \ge 10,000, 0.6≤Pr≤1600.6 \le \text{Pr} \le 160), the Dittus-Boelter correlation gives NuD=0.023ReD0.8Prn\text{Nu}_D = 0.023 \text{Re}_D^{0.8} \text{Pr}^n with n=0.4n = 0.4 for heating the fluid and n=0.3n = 0.3 for cooling.

  • Non-circular ducts utilize hydraulic diameter Dh=4AcPwD_h = \frac{4 A_c}{P_w}; for rectangular ducts of width aa and height bb, Dh=2aba+bD_h = \frac{2 a b}{a + b}.

  • Natural convection is driven by buoyancy forces where Nu=CRam\text{Nu} = C \text{Ra}^m (m=1/4m = 1/4 for laminar boundary layers Ra<109\text{Ra} < 10^9, and m=1/3m = 1/3 for turbulent boundary layers Ra≥109\text{Ra} \ge 10^9).

Last updated: August 2026

4.2 Convective Heat Transfer & Forced/Natural Convection Coefficients

Convective heat transfer is the transport of energy between a solid boundary and an adjacent moving fluid (liquid or gas), combining microscopic molecular conduction and macroscopic bulk fluid advection. In HVAC and refrigeration engineering, convection occurs inside hydronic chiller/boiler tubes, across finned air-conditioning coils, through air supply ductwork, and along building architectural surfaces. Accurately determining convective heat transfer coefficients (hch_c) via dimensionless empirical correlations is critical for equipment sizing on the NCEES PE Mechanical exam.


1. Newton's Law of Cooling & Boundary Layer Theory

Convective heat transfer is governed by Newton's Law of Cooling:

q˙conv=hcA(Ts−T∞)\dot{q}_{conv} = h_c A (T_s - T_\infty)

Where:

  • q˙conv=convective heat transfer rate (Btu/hr or W)\dot{q}_{conv} = \text{convective heat transfer rate } (\text{Btu/hr or W})
  • hc=convective heat transfer coefficient (Btu/(hr⋅ft2⋅∘F)or W/(m2⋅K))h_c = \text{convective heat transfer coefficient } (\text{Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)} \text{or W/(m}^2\cdot\text{K)})
  • A=surface area in contact with fluid (ft2 or m2)A = \text{surface area in contact with fluid } (\text{ft}^2 \text{ or m}^2)
  • Ts=solid surface temperature (∘F or ∘C)T_s = \text{solid surface temperature } ({}^\circ\text{F or } {}^\circ\text{C})
  • T∞=free-stream bulk fluid temperature (∘F or ∘C)T_\infty = \text{free-stream bulk fluid temperature } ({}^\circ\text{F or } {}^\circ\text{C})

The equivalent convective thermal resistance is:

Rconv=1hcA[hr⋅∘FBtu]R_{conv} = \frac{1}{h_c A} \quad \left[\frac{\text{hr}\cdot{}^\circ\text{F}}{\text{Btu}}\right]
+-----------------------------------------------------------------------------------------+
| HYDRODYNAMIC AND THERMAL BOUNDARY LAYERS OVER A FLAT SURFACE                            |
+-----------------------------------------------------------------------------------------+
| Free Stream: V_inf, T_inf                                                               |
|      ====================>                                                              |
|                . - ~ ~ ~ - .  <--- Hydrodynamic Boundary Layer delta(x)                 |
|            . '               ` .    Velocity transitions from 0 at wall to V_inf        |
|        . ' - - - - - - - - - - - ` <--- Thermal Boundary Layer delta_t(x)               |
|    . '                              `   Temp transitions from T_s at wall to T_inf      |
| +-------------------------------------------------------------------------------------+ |
| | Solid Surface at Temperature T_s (No-Slip Condition: u(y=0) = 0)                   | |
+-----------------------------------------------------------------------------------------+

Boundary Layer Mechanics

When fluid flows past a stationary surface, viscous shear stresses create a hydrodynamic (velocity) boundary layer of thickness δ(x)\delta(x), where local velocity transitions from zero at the wall (the no-slip boundary condition) to 99%99\% of the free-stream velocity V∞V_\infty. Concurrently, a thermal boundary layer of thickness δt(x)\delta_t(x) develops where temperature transitions from TsT_s to T∞T_\infty. Because fluid velocity is zero at the solid interface, heat transfer across the interface occurs entirely by pure conduction through the fluid film:

q′′=−kf∂T∂y∣y=0=hc(Ts−T∞)  ⟹  hc=−kf∂T∂y∣y=0Ts−T∞q'' = -k_f \left.\frac{\partial T}{\partial y}\right|_{y=0} = h_c (T_s - T_\infty) \implies h_c = \frac{-k_f \left.\frac{\partial T}{\partial y}\right|_{y=0}}{T_s - T_\infty}

Where kfk_f is the fluid thermal conductivity. Thinner boundary layers produce steeper temperature gradients ∂T∂y\frac{\partial T}{\partial y}, resulting in substantially higher convection coefficients hch_c.


2. Fundamental Dimensionless Numbers in Convection

Dimensionless analysis reduces complex multi-variable fluid-thermal equations into standard governing parameters:

+-----------------------------------------------------------------------------------------+
| SUMMARY OF PRIMARY DIMENSIONLESS GROUPS                                                 |
+-----------------------------------------------------------------------------------------+
| Reynolds (Re):   Re = (rho * V * L) / mu = (V * L) / nu       [Inertia / Viscous]       |
| Prandtl (Pr):    Pr = nu / alpha = (c_p * mu) / k_f           [Momentum / Thermal Diff] |
| Nusselt (Nu):    Nu = (h * L) / k_f                           [Convection / Conduction] |
| Grashof (Gr):    Gr = (g * beta * Delta_T * L^3) / nu^2       [Buoyancy / Viscous]      |
| Rayleigh (Ra):   Ra = Gr * Pr = (g * beta * Delta_T * L^3)/(nu*alpha) [Natural Convect] |
+-----------------------------------------------------------------------------------------+

1. Reynolds Number (Re\text{Re})

Represents the ratio of dynamic inertial forces to viscous shear forces:

Re=ρVDμ=VDν\text{Re} = \frac{\rho V D}{\mu} = \frac{V D}{\nu}
  • Internal Flow in Circular Pipes:
    • Laminar Regime: Re<2,300\text{Re} < 2,300
    • Transition Regime: 2,300≤Re≤10,0002,300 \le \text{Re} \le 10,000
    • Fully Turbulent Regime: Re>10,000\text{Re} > 10,000

2. Prandtl Number (Pr\text{Pr})

Represents the ratio of momentum molecular diffusivity (ν=μ/ρ\nu = \mu/\rho) to thermal molecular diffusivity (α=k/(ρcp)\alpha = k/(\rho c_p)):

Pr=να=cpμkf\text{Pr} = \frac{\nu}{\alpha} = \frac{c_p \mu}{k_f}
  • Pr≈0.70−0.72\text{Pr} \approx 0.70 - 0.72 for air and common gases at atmospheric conditions
  • Pr≈2.0−10.0\text{Pr} \approx 2.0 - 10.0 for liquid water (40∘F40^\circ\text{F} to 200∘F200^\circ\text{F})
  • Pr≈50−10,000\text{Pr} \approx 50 - 10,000 for heavy oils and glycols
  • Pr≪1\text{Pr} \ll 1 ( 0.01~0.01) for liquid metals

3. Nusselt Number (Nu\text{Nu})

Represents the ratio of convective heat transfer to pure molecular conduction across a fluid layer of characteristic length LL:

Nu=hcLkf\text{Nu} = \frac{h_c L}{k_f}

Once Nu\text{Nu} is calculated from an empirical correlation, the convective coefficient is determined directly:

hc=Nu⋅kfLh_c = \frac{\text{Nu} \cdot k_f}{L}

4. Grashof Number (Gr\text{Gr}) and Rayleigh Number (Ra\text{Ra})

In natural (free) convection, fluid motion is induced by buoyancy forces resulting from temperature-driven density differences:

Gr=gβ(Ts−T∞)L3ν2\text{Gr} = \frac{g \beta (T_s - T_\infty) L^3}{\nu^2} Ra=Gr⋅Pr=gβ(Ts−T∞)L3να\text{Ra} = \text{Gr} \cdot \text{Pr} = \frac{g \beta (T_s - T_\infty) L^3}{\nu \alpha}

Where:

  • g=gravitational acceleration (32.174 ft/s2 or 9.807 m/s2)g = \text{gravitational acceleration } (32.174\text{ ft/s}^2 \text{ or } 9.807\text{ m/s}^2)
  • β=volumetric thermal expansion coefficient (∘R−1 or K−1)\beta = \text{volumetric thermal expansion coefficient } ({}^\circ\text{R}^{-1} \text{ or } \text{K}^{-1}); for ideal gases, β=1Tfilm=1Tf (in ∘R)\beta = \frac{1}{T_{film}} = \frac{1}{T_f\text{ (in }{}^\circ\text{R)}}
  • Tf=Ts+T∞2=film temperature at which fluid properties are evaluatedT_f = \frac{T_s + T_\infty}{2} = \text{film temperature at which fluid properties are evaluated}

3. Forced Convection Correlations (Internal Flow)

Fully Developed Turbulent Flow in Smooth Pipes

For turbulent flow in circular tubes (ReD≥10,000\text{Re}_D \ge 10,000, 0.6≤Pr≤1600.6 \le \text{Pr} \le 160, and LD≥10\frac{L}{D} \ge 10), the Dittus-Boelter equation is the standard relation in the NCEES Reference Handbook:

NuD=0.023ReD0.8Prn\text{Nu}_D = 0.023 \text{Re}_D^{0.8} \text{Pr}^n

Where the exponent nn depends on the direction of heat transfer:

  • n=0.4n = 0.4 when the fluid is being heated (Ts>TbT_s > T_b, e.g., heating coils, boiler tubes)
  • n=0.3n = 0.3 when the fluid is being cooled (Ts<TbT_s < T_b, e.g., chiller evaporator tubes, cooling coils)

All fluid properties are evaluated at the bulk mean fluid temperature TbT_b.

Gnielinski Correlation (Higher Accuracy for 3,000≤ReD≤5×1063,000 \le \text{Re}_D \le 5 \times 10^6)

When greater precision is required across the transition-to-turbulent region (0.5≤Pr≤20000.5 \le \text{Pr} \le 2000):

NuD=(f/8)(ReD−1000)Pr1+12.7(f/8)1/2(Pr2/3−1)\text{Nu}_D = \frac{(f/8)(\text{Re}_D - 1000)\text{Pr}}{1 + 12.7(f/8)^{1/2}(\text{Pr}^{2/3} - 1)}

Where ff is the Darcy friction factor for smooth tubes: f=(0.790ln⁡(ReD)−1.64)−2f = (0.790 \ln(\text{Re}_D) - 1.64)^{-2}.

Laminar Flow in Circular Tubes (ReD<2,300\text{Re}_D < 2,300)

For fully developed laminar internal flow, Nu\text{Nu} is independent of Reynolds and Prandtl numbers:

  • Constant Surface Heat Flux (q′′=constantq'' = \text{constant}): NuD=hDkf=4.364\text{Nu}_D = \frac{h D}{k_f} = 4.364
  • Constant Surface Temperature (Ts=constantT_s = \text{constant}): NuD=hDkf=3.656\text{Nu}_D = \frac{h D}{k_f} = 3.656

Non-Circular Ducts & Hydraulic Diameter (DhD_h)

For non-circular ducts (rectangular HVAC ducts, annular passages), the characteristic dimension is the hydraulic diameter:

Dh=4AcPwD_h = \frac{4 A_c}{P_w}

Where:

  • Ac=cross-sectional flow area (ft2)A_c = \text{cross-sectional flow area } (\text{ft}^2)
  • Pw=wetted perimeter in contact with fluid (ft)P_w = \text{wetted perimeter in contact with fluid } (\text{ft})

For a rectangular duct of cross-sectional width aa and height bb:

Dh=4(ab)2(a+b)=2aba+bD_h = \frac{4 (a b)}{2(a + b)} = \frac{2 a b}{a + b}

For an annular pipe (inner pipe OD = D1D_1, outer pipe ID = D2D_2):

Dh=4[π4(D22−D12)]π(D2+D1)=D2−D1D_h = \frac{4 \left[\frac{\pi}{4}(D_2^2 - D_1^2)\right]}{\pi (D_2 + D_1)} = D_2 - D_1

4. Forced Convection Correlations (External Flow)

External Flow Over Flat Surfaces

For fluid flowing parallel to a flat plate of length LL:

  • Laminar Boundary Layer (Rex<5×105\text{Re}_x < 5 \times 10^5): Nux=0.332Rex1/2Pr1/3  ⟹  Nu‾L=0.664ReL1/2Pr1/3\text{Nu}_x = 0.332 \text{Re}_x^{1/2} \text{Pr}^{1/3} \quad \implies \quad \overline{\text{Nu}}_L = 0.664 \text{Re}_L^{1/2} \text{Pr}^{1/3}
  • Turbulent Boundary Layer (Rex≥5×105\text{Re}_x \ge 5 \times 10^5): Nux=0.0296Rex4/5Pr1/3  ⟹  Nu‾L=0.037ReL4/5Pr1/3\text{Nu}_x = 0.0296 \text{Re}_x^{4/5} \text{Pr}^{1/3} \quad \implies \quad \overline{\text{Nu}}_L = 0.037 \text{Re}_L^{4/5} \text{Pr}^{1/3}

Cross-Flow Across Single Cylinders & Tube Banks

For cross-flow across a single tube of outer diameter DD (Hilpert correlation):

Nu‾D=CReDmPr1/3\overline{\text{Nu}}_D = C \text{Re}_D^m \text{Pr}^{1/3}

In finned-tube HVAC cooling and heating coils, air flows across tube banks arranged in in-line or staggered configurations. Staggered tube banks produce higher turbulence and convective heat transfer coefficients for equivalent air face velocities at the expense of higher fan static pressure drop.

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Tube Bank Layouts: In-Line vs Staggered Coil Geometries

5. Natural (Free) Convection Correlations

Natural convection correlations take the general power-law form:

Nu‾L=hˉLkf=CRaLm=C(GrL⋅Pr)m\overline{\text{Nu}}_L = \frac{\bar{h} L}{k_f} = C \text{Ra}_L^m = C (\text{Gr}_L \cdot \text{Pr})^m

Where the constants CC and mm depend on geometry, orientation, and flow regime:

  • Laminar Flow (Ra<109\text{Ra} < 10^9): m=1/4m = 1/4 (0.250.25)
  • Turbulent Flow (Ra≥109\text{Ra} \ge 10^9): m=1/3m = 1/3 (0.3330.333)

Summary of Natural Convection Geometries (ASHRAE / NCEES Standards)

Surface Geometry & OrientationCharacteristic Length (LL)Rayleigh Number RangeCorrelation Equation (Nu‾L\overline{\text{Nu}}_L)
Vertical Plate / Vertical WallHeight HH104≤RaL≤10910^4 \le \text{Ra}_L \le 10^9 (Laminar)Nu‾L=0.59RaL1/4\overline{\text{Nu}}_L = 0.59 \text{Ra}_L^{1/4}
Vertical Plate / Vertical WallHeight HH109≤RaL≤101310^9 \le \text{Ra}_L \le 10^{13} (Turbulent)Nu‾L=0.10RaL1/3\overline{\text{Nu}}_L = 0.10 \text{Ra}_L^{1/3}
Horizontal Cylinder (Pipes)Outer Diameter DD104≤RaD≤10910^4 \le \text{Ra}_D \le 10^9Nu‾D=0.53RaD1/4\overline{\text{Nu}}_D = 0.53 \text{Ra}_D^{1/4}
Horizontal Cylinder (Pipes)Outer Diameter DD109≤RaD≤101210^9 \le \text{Ra}_D \le 10^{12}Nu‾D=0.13RaD1/3\overline{\text{Nu}}_D = 0.13 \text{Ra}_D^{1/3}
Hot Horizontal Plate Facing UpLc=AsPL_c = \frac{A_s}{P}104≤RaL≤10710^4 \le \text{Ra}_L \le 10^7 (Laminar)Nu‾L=0.54RaL1/4\overline{\text{Nu}}_L = 0.54 \text{Ra}_L^{1/4}
Hot Horizontal Plate Facing UpLc=AsPL_c = \frac{A_s}{P}107≤RaL≤101110^7 \le \text{Ra}_L \le 10^{11} (Turbulent)Nu‾L=0.15RaL1/3\overline{\text{Nu}}_L = 0.15 \text{Ra}_L^{1/3}
Hot Horizontal Plate Facing DownLc=AsPL_c = \frac{A_s}{P}104≤RaL≤10910^4 \le \text{Ra}_L \le 10^9Nu‾L=0.27RaL1/4\overline{\text{Nu}}_L = 0.27 \text{Ra}_L^{1/4}

6. Phase Change Convection: Boiling & Condensation Fundamentals

Phase-change convective processes provide the highest heat transfer coefficients in HVAC and refrigeration engineering (h≈500 to 10,000+ Btu/(hr⋅ft2⋅∘F)h \approx 500 \text{ to } 10,000+\text{ Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)}).

The Pool Boiling Curve

When a submerged heating surface operates at temperature TsT_s above the fluid saturation temperature TsatT_{sat} (excess temperature ΔTe=Ts−Tsat\Delta T_e = T_s - T_{sat}):

  1. Natural Convection Regime (ΔTe<9∘F\Delta T_e < 9^\circ\text{F} / 5∘C5^\circ\text{C}): Liquid motion is driven purely by natural convection without vapor bubble formation.
  2. Nucleate Boiling Regime (9∘F<ΔTe<54∘F9^\circ\text{F} < \Delta T_e < 54^\circ\text{F} / 5∘C<ΔTe<30∘C5^\circ\text{C} < \Delta T_e < 30^\circ\text{C}): Vapor bubbles nucleate at microscopic surface cavities, detach rapidly, and create vigorous fluid agitation. This regime yields maximum heat transfer efficiency and governs direct-expansion (DX) and flooded refrigeration evaporators.
  3. Critical Heat Flux (qmax′′q''_{max} / Departure from Nucleate Boiling - DNB): The peak heat flux point. Exceeding qmax′′q''_{max} causes vapor bubbles to coalesce into an insulating blanket.
  4. Film Boiling Regime (Large ΔTe\Delta T_e): A continuous vapor film completely blankets the heater surface, dropping hh drastically and causing severe surface temperature spikes (burnout).

Condensation: Film vs. Dropwise

  • Film Condensation (Nusselt Film Theory): Liquid condensate forms a continuous laminar liquid film that flows downward under gravity. The liquid film introduces a conductive thermal barrier, reducing hh.
  • Dropwise Condensation: Condensate forms discrete droplets that break away, keeping the cold surface directly exposed. Dropwise condensation achieves heat transfer coefficients 4 to 8 times higher than film condensation.

7. Step-by-Step Worked Example: Internal Forced Convection Sizing

Problem Statement

Water at a bulk mean temperature of Tb=140.0∘FT_b = 140.0^\circ\text{F} flows through a 1-inch1\text{-inch} nominal Type L copper pipe with an inside diameter of Di=1.025 in.=0.08542 ftD_i = 1.025\text{ in.} = 0.08542\text{ ft} at a volumetric flow rate of 5.0 GPM5.0\text{ GPM}. The pipe wall is maintained at Ts=160.0∘FT_s = 160.0^\circ\text{F} by an external heating source (heating fluid, n=0.4n = 0.4).

From NCEES water property tables at 140∘F140^\circ\text{F} (60∘C60^\circ\text{C}):

  • Density: ρ=61.38 lbm/ft3\rho = 61.38\text{ lbm/ft}^3
  • Dynamic Viscosity: μ=1.14×10−3 lbm/(ft⋅s)=4.104 lbm/(ft⋅hr)\mu = 1.14 \times 10^{-3}\text{ lbm/(ft}\cdot\text{s)} = 4.104\text{ lbm/(ft}\cdot\text{hr)}
  • Kinematic Viscosity: ν=1.857×10−5 ft2/s\nu = 1.857 \times 10^{-5}\text{ ft}^2/\text{s}
  • Thermal Conductivity: kf=0.376 Btu/(hr⋅ft⋅∘F)k_f = 0.376\text{ Btu/(hr}\cdot\text{ft}\cdot{}^\circ\text{F)}
  • Specific Heat: cp=1.000 Btu/(lbm⋅∘F)c_p = 1.000\text{ Btu/(lbm}\cdot{}^\circ\text{F)}
  • Prandtl Number: Pr=3.01\text{Pr} = 3.01

Calculate:

  1. Cross-sectional flow area AcA_c (ft2\text{ft}^2) and mean flow velocity VV (ft/s\text{ft/s})
  2. Reynolds number ReD\text{Re}_D and flow regime
  3. Nusselt number NuD\text{Nu}_D via the Dittus-Boelter equation
  4. Internal convective heat transfer coefficient hih_i (Btu/(hr⋅ft2⋅∘F)\text{Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)})
  5. Convective heat gain rate per linear foot of pipe length q˙′\dot{q}' (Btu/(hr⋅ft)\text{Btu/(hr}\cdot\text{ft)})

Solution Steps

Step 1: Calculate flow area and fluid velocity

Ac=π4Di2=π4(0.08542 ft)2=0.005731 ft2A_c = \frac{\pi}{4} D_i^2 = \frac{\pi}{4} (0.08542\text{ ft})^2 = 0.005731\text{ ft}^2 Q=5.0 GPM×(0.13368 ft31 gal)×(1 min60 s)=0.01114 ft3/sQ = 5.0\text{ GPM} \times \left(\frac{0.13368\text{ ft}^3}{1\text{ gal}}\right) \times \left(\frac{1\text{ min}}{60\text{ s}}\right) = 0.01114\text{ ft}^3/\text{s} V=QAc=0.01114 ft3/s0.005731 ft2=1.944 ft/sV = \frac{Q}{A_c} = \frac{0.01114\text{ ft}^3/\text{s}}{0.005731\text{ ft}^2} = 1.944\text{ ft/s}

Step 2: Calculate Reynolds number

ReD=VDiν=1.944 ft/s×0.08542 ft1.857×10−5 ft2/s=0.166061.857×10−5=8942\text{Re}_D = \frac{V D_i}{\nu} = \frac{1.944\text{ ft/s} \times 0.08542\text{ ft}}{1.857 \times 10^{-5}\text{ ft}^2/\text{s}} = \frac{0.16606}{1.857 \times 10^{-5}} = 8942

Alternatively, using mass flow rate m˙=ρQ=61.38×(0.01114×3600)=2461.4 lbm/hr\dot{m} = \rho Q = 61.38 \times (0.01114 \times 3600) = 2461.4\text{ lbm/hr}:

ReD=4m˙πDiμ=4×2461.4π×0.08542×4.104=9845.61.1014=8939≈8940\text{Re}_D = \frac{4 \dot{m}}{\pi D_i \mu} = \frac{4 \times 2461.4}{\pi \times 0.08542 \times 4.104} = \frac{9845.6}{1.1014} = 8939 \approx 8940

Because ReD≈8940\text{Re}_D \approx 8940 is within the turbulent range ( ≥10,000\,\ge 10,000 for standard handbook Dittus-Boelter application; transition-turbulent), apply Dittus-Boelter:

Step 3: Calculate Nusselt number via Dittus-Boelter Since fluid is being heated (Ts>TbT_s > T_b), n=0.4n = 0.4:

NuD=0.023ReD0.8Pr0.4=0.023×(8940)0.8×(3.01)0.4\text{Nu}_D = 0.023 \text{Re}_D^{0.8} \text{Pr}^{0.4} = 0.023 \times (8940)^{0.8} \times (3.01)^{0.4} (8940)0.8=1445.0(8940)^{0.8} = 1445.0 (3.01)0.4=1.5532(3.01)^{0.4} = 1.5532 NuD=0.023×1445.0×1.5532=51.62\text{Nu}_D = 0.023 \times 1445.0 \times 1.5532 = 51.62

Step 4: Calculate convective heat transfer coefficient hih_i

hi=NuD⋅kfDi=51.62×0.376 Btu/(hr⋅ft⋅∘F)0.08542 ft=19.4090.08542=227.2 Btu/(hr⋅ft2⋅∘F)h_i = \frac{\text{Nu}_D \cdot k_f}{D_i} = \frac{51.62 \times 0.376\text{ Btu/(hr}\cdot\text{ft}\cdot{}^\circ\text{F)}}{0.08542\text{ ft}} = \frac{19.409}{0.08542} = 227.2\text{ Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)}

Step 5: Calculate convective heat transfer rate per linear foot Inside surface area per linear foot (L=1 ftL = 1\text{ ft}):

Ai′=πDiL=π×0.08542 ft×1.0 ft=0.26836 ft2/ftA_i' = \pi D_i L = \pi \times 0.08542\text{ ft} \times 1.0\text{ ft} = 0.26836\text{ ft}^2/\text{ft} q˙′=hiAi′(Ts−Tb)=227.2×0.26836×(160.0−140.0)=227.2×0.26836×20.0=1219.4 Btu/(hr⋅ft)\dot{q}' = h_i A_i' (T_s - T_b) = 227.2 \times 0.26836 \times (160.0 - 140.0) = 227.2 \times 0.26836 \times 20.0 = 1219.4\text{ Btu/(hr}\cdot\text{ft)}
Test Your Knowledge

Water at 60°F flows through a circular tube in fully developed turbulent flow at a velocity of 4 ft/s, resulting in a convective heat transfer coefficient of h_1 = 600 Btu/(hr·ft2·°F). If the water flow velocity is increased to 8 ft/s while keeping tube diameter and fluid properties constant, what is the new convective heat transfer coefficient h_2 according to the Dittus-Boelter correlation?

A

792 Btu/(hr·ft2·°F)

B

1,200 Btu/(hr·ft2·°F)

C

1,045 Btu/(hr·ft2·°F)

D

1,584 Btu/(hr·ft2·°F)

Test Your Knowledge

A rectangular air distribution duct measures 24 inches wide by 12 inches high. What is the hydraulic diameter (D_h) of this duct for use in Reynolds and Nusselt number calculations?

A

18.0 inches (1.500 ft)

B

20.0 inches (1.667 ft)

C

24.0 inches (2.000 ft)

D

16.0 inches (1.333 ft)

Test Your Knowledge

Which of the following dimensionless numbers represents the ratio of thermal molecular diffusivity to momentum molecular diffusivity in fluid heat transfer?

A

Inverse of Prandtl Number (1 / Pr)

B

Reynolds Number (Re)

C

Grashof Number (Gr)

D

Nusselt Number (Nu)

Test Your Knowledge

In refrigeration evaporators, which boiling regime provides the highest heat transfer coefficients and optimal thermodynamic performance without risking surface burnout?

A

Natural Convection Boiling

B

Nucleate Boiling

C

Transition Boiling

D

Film Boiling

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