Agricultural and Biosystems Engineers Licensure Examination Heat & Mass Transfer: practice questions

C. 1/(hA)

Newton's law Q = hA ΔT has the form Q = ΔT/R, so R_conv = 1/(hA) in K/W. hA is the convective conductance, the inverse of the resistance.

D. first increases, reaches a maximum at the critical radius, then decreases

Total resistance ln(r/r1)/(2πk) + 1/(2πrh) is a minimum at r = k/h, so heat loss is a maximum there. Below r_c the gain in outer area outweighs the added conduction resistance. "First decreases, reaches a minimum" inverts the result: the critical radius gives maximum loss, not minimum.

A. 281 W

Small body in a large enclosure: Q = εAσ(T1⁴ − T2⁴), with A = πDL = 0.3142 m². Q = 0.9 × 0.3142 × 5.67 × 10⁻⁸ × (400⁴ − 300⁴) = 281 W. Ignoring emissivity gives 312; using °C gives 4.16; taking A = 2πDL (using D as a radius) doubles the answer to 561 W.

D. transmits short-wave solar radiation but blocks the long-wave radiation emitted by the crop and floor

Solar radiation peaks near 0.5 µm and passes through glazing; surfaces inside at about 300 K emit near 10 µm, where glass and most films are nearly opaque. Reduced convective exchange with outside air also contributes. Calling the glazing transparent to long-wave and opaque to short-wave radiation reverses the spectral behaviour.

C. that the Biot number is below 0.1

Heisler charts exist precisely for cases where Bi is not small, so internal gradients matter; they cover a wide range of 1/Bi. They do assume one-dimensional conduction, constant properties and h, a sudden step in fluid temperature and uniform initial temperature, and the one-term form is accurate for Fo > 0.2.

D. 390 W/m²·K

1/U = 1/500 + 0.001/16 + 1/2000 = 0.002 + 0.0000625 + 0.0005 = 0.0025625, so U = 390 W/m²·K. Omitting the wall gives 400; averaging or adding the coefficients treats series resistances incorrectly. U is always less than the smaller h, so improving the 500 W/m²·K side helps most. Averaging gives 1,250 W/m²·K and adding gives 2,500 W/m²·K.

A. 0.374 kg/h

ṁ = h_m A(ρ_v,s − ρ_v,∞) = 0.01 × 1 × (0.0173 − 0.0069) = 1.04 × 10⁻⁴ kg/s = 0.374 kg/h. Ignoring the vapour already in the air gives 0.623 kg/h; the kg/s value labelled as kg/h is a time-unit slip; adding the densities has no basis. Adding the densities gives 0.871 kg/h.

B. 27.2 °C

T_out = T_db − η(T_db − T_wb) = 40 − 0.8 × 16 = 27.2 °C. 24 °C is the wet-bulb limit, reached only by an ideal (100%) pad; using (1 − η) gives 36.8 °C; 12.8 °C is the temperature drop, not the outlet temperature.

C. where the constant-rate period ends

At the critical moisture content the surface can no longer be kept wet by internal moisture movement, so the rate starts to fall. The moisture below which no further drying is possible in given air is the equilibrium moisture content. The surface is at the wet-bulb temperature during the constant-rate period, not at its end. That is also where the falling-rate period begins.

D. the insulation layer

In a series circuit the temperature drop across each layer is proportional to its resistance. Insulation has by far the largest L/k, so most of the drop occurs across it. If a structural layer or film took the largest drop, the insulation would be too thin or wet.