Pulpul Exam Practice

GATE · ME Mechanical Engineering concept book

Heat Transfer

Dimensionless number relations in forced convection

Forced-convection correlations are written in dimensionless groups: the Nusselt number Nu = hL/k, the Reynolds number Re = \rho V L/\mu, the Prandtl number Pr = \mu c_p/k, and the Stanton number St = h/(\rho c_p V). Because the length, conductivity and viscosity cancel in the product, the Stanton number equals the Nusselt number divided by the product of the Reynolds and Prandtl numbers, St = Nu/(Re\,Pr), and this is the first relation to memorise. A common trap is to reverse the powers of Re and Pr, which gives a combination that no longer reduces to St. Dimensional consistency checks, such as confirming that every group is dimensionless, catch such errors quickly.

Correlations of the form Nu = f(Re, Pr) are the working tools of the subject. For fully developed laminar flow in a circular tube with a constant wall temperature, the Nusselt number is a constant, Nu = 3.66, independent of both Re and Pr. For laminar flow over a flat plate the average Nusselt number scales as Nu \propto Re_L^{1/2}; for fixed fluid and plate length this means Nu \propto u_\infty^{1/2}, so doubling the free-stream velocity multiplies Nu by \sqrt{2}. For turbulent flow with heating in a pipe, a Dittus–Boelter type correlation Nu = 0.023\,Re^{0.8}Pr^{0.4} is used, with Nu based on the diameter.

The Nusselt number also has a physical meaning. At the wall, Fourier's law and Newton's law of cooling give Nu = -\partial\theta^*/\partial y^* evaluated at the surface, where \theta^* is the non-dimensional temperature and y^* the non-dimensional normal coordinate, so Nu is the non-dimensional temperature gradient at the wall. For a uniform wall heat flux q'' on a tube of diameter D and length L, the energy balance q''\pi D L = \dot m c_p (T_{m,o} - T_{m,i}) fixes the length or the bulk temperature rise, and the outlet surface temperature follows from T_s = T_m + q''/h.

Radiation view factor between enclosed surfaces

The view factor F_{ij} is the fraction of the radiation leaving surface i that reaches surface j directly. Three rules govern view factors in an enclosure. Reciprocity gives A_i F_{ij} = A_j F_{ji}, so the view from a small body to a large surrounding surface can be found from the view of the large surface back to the small body. The summation rule requires \sum_j F_{ij} = 1 over all surfaces of a closed enclosure, which gives the self-view factor of a surface as one minus the fraction of its radiation that strikes other surfaces.

A convex surface or a plane surface cannot see itself, so F_{ii} = 0, whereas a concave surface can see part of itself. A convex body placed inside a larger enclosure sees only the enclosure, so its view factor to the enclosure is unity. In a hemispherical furnace with a flat base, the base is fully visible from the dome, so the dome-to-base view factor equals the ratio of the base area to the dome area, and the remainder is the dome's self-view factor. A flat bottom surface of a cylindrical enclosure has F_{13} = 1 - F_{12}, because the bottom sees only the top and the curved walls.

For black surfaces the net radiation exchange between surfaces i and j is Q_{ij} = A_i F_{ij}\sigma(T_i^4 - T_j^4), so once the view factors are known the problem reduces to multiplying area, view factor and the difference of fourth powers. For gray surfaces, a radiation network with surface resistances (1-\varepsilon)/(\varepsilon A) and space resistances 1/(A_iF_{ij}) is solved in series and parallel. A re-radiating (insulated) surface passes on all the radiation it receives, which makes it a path in parallel with the direct path to the surroundings.

Heat exchanger LMTD and NTU

The effectiveness \varepsilon of a heat exchanger is the ratio of the actual heat transfer to the maximum possible, \varepsilon = Q/Q_{max}, where Q_{max} = C_{min}(T_{h,in} - T_{c,in}) and C_{min} = \dot m c_p is the smaller heat capacity rate of the two fluids. Effectiveness is dimensionless and lies between 0 and 1, and once it is known the actual heat transfer is Q = \varepsilon C_{min}(T_{h,in} - T_{c,in}). When only partial data are given, first find each fluid's heat capacity rate from its own energy balance, C = Q/\Delta T, and then compare them to identify C_{min} before forming the effectiveness.

The number of transfer units is NTU = UA/C_{min}, and it measures the size of the exchanger relative to the fluid capacity. A fluid that undergoes condensation or boiling keeps a constant temperature, so its heat capacity rate is effectively infinite, the capacity ratio is zero, and the effectiveness reduces to \varepsilon = 1 - e^{-NTU}. In a condenser, the coolant is the only fluid with a finite capacity rate, so its outlet temperature follows from T_{out} = T_{in} + \varepsilon(T_{sat} - T_{in}), and the effectiveness equals the coolant temperature rise divided by the difference between the steam and coolant inlet temperatures.

The heat released by condensation is \dot Q = \dot m_s h_{fg}, so when this heat is known and the effectiveness is given, the coolant mass flow is found from \dot Q = \varepsilon C_{min}\Delta T_{max}. A counterflow arrangement reaches a higher effectiveness than parallel flow for the same NTU, and the LMTD method relates Q to UA through Q = UA\,\Delta T_{lm}.

Transient conduction and Heisler chart

Heisler charts give the transient centre temperature of a plane wall, an infinite cylinder or a sphere, and they are plotted against two dimensionless groups: the Biot number Bi = hL_c/k and the Fourier number Fo = \alpha t/L_c^2. The centre temperature ratio is therefore read from the chart using Bi and Fo, whereas the Reynolds, Prandtl, Froude, Nusselt and Grashof numbers belong to flow, convection or free-surface problems and do not enter transient conduction inside a solid.

The lumped capacitance method is valid only when the Biot number is small, typically Bi = hL_c/k < 0.1, where the characteristic length is L_c = V/A_s. For a plate convected on both faces L_c = d/2, and for a sphere L_c = r/3. Checking the criterion for each combination of h, k and thickness shows which plates may be treated as lumped. A thin, highly conducting plate with modest h qualifies, whereas a thick, poorly conducting plate with large h does not, because the internal temperature gradient is then not negligible.

When the method applies, the temperature decays exponentially, (T - T_\infty)/(T_i - T_\infty) = \exp(-t/\tau), with time constant \tau = \rho c V/(hA_s). For a sphere this becomes \tau = \rho c r/(3h), and the time to reach a target temperature is t = \tau\ln[(T_i - T_\infty)/(T - T_\infty)].

Fin heat transfer

For a very long fin of uniform cross-section, the heat rate is \dot Q = \sqrt{hPkA_c}\,\theta_b, where P is the perimeter, A_c the cross-sectional area and \theta_b the base excess temperature. The heat rate therefore scales as the square root of the conductivity and of the product PA_c. For an infinitely long fin, doubling the conductivity multiplies the heat transfer by \sqrt{2} while the other quantities stay fixed.

When two fins of different shape are compared with the same material and the same convection coefficient, the ratio of their heat rates is the ratio of \sqrt{P_1A_1} to \sqrt{P_2A_2}, because h, k and \theta_b cancel. The geometry alone determines the comparison, so the perimeter and area of each cross-section must be computed before forming the ratio.

For a rod whose lateral surface is insulated and whose end is exposed to convection, the net internal heat generation must leave through the tip. The tip condition is -k\,dT/dx = h(T - T_\infty), and integrating the generation over the whole length gives the total heat leaving the tip. If this integral is zero, the tip heat flux vanishes and the tip temperature equals the ambient, so the temperature field no longer depends on the convection coefficient h.

Wall heat flux from temperature profile

At a solid–fluid boundary the wall heat flux follows from Fourier's law, q'' = -k\,(\partial T/\partial y)_{y=0}, so the local gradient taken from a given temperature profile, multiplied by the conductivity of the medium, gives the flux directly. At an interface, heat flux is continuous, so k_{fluid}(\partial T/\partial y)_{fluid} = k_{solid}(\partial T/\partial y)_{solid}. A solid of high conductivity therefore carries a much smaller temperature gradient than the adjacent fluid for the same heat flux.

What comes next

Sign in to read the full text free, up to 2 units a day.

Read this unit in the app

Exams that include this subject: GATE ME · Mechanical Engineering