FE Mechanical Practice Exam

For an isentropic process of an ideal gas with constant specific heats, temperature and pressure are related by

This follows from combining $Pv^k = \text{constant}$ with the ideal-gas law $Pv = RT$ to eliminate $v$.

The pressure ratio is $800/100 = 8$ and the exponent is $(1.4 - 1)/1.4 = 0.2857$. Absolute temperatures must be used because the relation is a ratio.

Evaluate the power with logarithms: $\ln 8 = 2.079$, so $0.2857\times 2.079 = 0.594$ and $e^{0.594} = 1.811$.

The exit temperature is well above the inlet temperature, as expected for compression, and the rise is moderate for a pressure ratio of 8, which is consistent with an exponent less than 1.

T₂ ≈ 543 K (option B)

The most tempting wrong answer is C (689 K), which uses the exponent $k - 1 = 0.4$ instead of $(k-1)/k$; that exponent belongs to the temperature-volume relation $T_2/T_1 = (v_1/v_2)^{k-1}$, not the temperature-pressure one. Option A (322 K) applies the ratio to 27°C and converts to kelvin afterwards; ratios of temperature only hold on an absolute scale. Option D uses the exponent $1/k$, which belongs to the volume-pressure relation.

At steady state the same heat flux passes through the inside convection film, the wall and the outside film, so the three resistances act in series between the two air temperatures. Per unit area, the resistances are $1/h$ for each convection film and $L/k$ for conduction through the wall.

Total thermal resistance per unit area:

The conduction term dominates, so the wall itself is the main barrier to heat flow.

Heat flux is the overall temperature difference divided by the total resistance, with the difference taken between the two air temperatures (the surface temperatures are unknown):

The flux is positive from inside to outside, as expected since the room is warmer. The inside surface temperature follows from the inside film alone: $T_{s,i} = 22 - 76.9/10 = 14.3$°C.

q″ ≈ 77 W/m², directed from the room to the outdoors

(a) Torque. Power transmitted by a rotating shaft is $P = T\omega$, so the torque is the power divided by the angular speed in rad/s:

(b) Minimum diameter. The maximum shear stress in a solid circular shaft occurs at the outer surface: $\tau_{\max} = Tc/J$ with $c = d/2$ and $J = \pi d^4/32$, which simplifies to $\tau_{\max} = 16T/(\pi d^3)$. Set it equal to the allowable stress and solve for $d$:

Any larger diameter keeps the stress below 40 MPa, so 26 mm is an acceptable standard size.

(c) Angle of twist. For the 26 mm shaft, the polar moment of inertia is

The angle of twist over length $L$ is $\theta = TL/(GJ)$, in radians, with all quantities in base SI units:

Convert to degrees: $0.0355\times 180/\pi = 2.03°$.

T ≈ 127.3 N·m; $d_{\min}$ ≈ 25.3 mm; θ ≈ 0.0355 rad ≈ 2.03° per metre of shaft