(a) Heat is generated in a stainless steel plate (thermal conductivity = 22 W/m-K) of thickness 1 cm, at a uniform rate of 600…

(a) Temperature distribution and heat flux in a plate with heat generation

Take x from the left face (0 ≤ x ≤ L, L = 0.01 m). For steady one-dimensional conduction with uniform generation q̇ = 600 × 10⁶ W/m³ and k = 22 W/m·K: k d²T/dx² + q̇ = 0. Integrating twice, T = −q̇x²/(2k) + C₁x + C₂. Boundary conditions: T(0) = 200°C gives C₂ = 200. T(L) = 100°C gives C₁ = [−100 + q̇L²/(2k)]/L. q̇L²/(2k) = 600 × 10⁶ × 10⁻⁴/44 = 1363.6 K, so C₁ = (1363.6 − 100)/0.01 = 1.2636 × 10⁵ K/m.

(i) Temperature distribution: T(x) = 200 + 1.2636 × 10⁵ x − 1.3636 × 10⁷ x² (°C, x in m). It is a downward-opening parabola.

(ii) The maximum is where dT/dx = 0: x_max = C₁k/q̇ = 1.2636 × 10⁵/(2.727 × 10⁷) = 4.63 mm from the left face. T_max = 200 + C₁x_max/2 = 200 + 1.2636 × 10⁵ × 4.633 × 10⁻³/2 = 492.7°C.

(iii) Heat flux q = −k dT/dx, positive to the right. At x = 0: q = −22 × 1.2636 × 10⁵ = −2.78 × 10⁶ W/m², so 2.78 MW/m² leaves through the left face. At x = L: dT/dx = 1.2636 × 10⁵ − 2.727 × 10⁷ × 0.01 = −1.4636 × 10⁵ K/m, so q = +3.22 × 10⁶ W/m², and 3.22 MW/m² leaves through the right face. Check: 2.78 + 3.22 = 6.00 MW/m² = q̇L. The right side takes more because the maximum lies closer to the left face and the right face is colder.

(b) Heat engine driving a heat pump from waste energy

Temperatures: T_waste = 323.15 K, T_L = 303.15 K, T_H = 423.15 K. Reversible heat engine: Ẇ = Q̇_W1(1 − T_L/T_waste) = Q̇_W1 × 20/323.15. Reversible heat pump: Q̇_H/T_H = Q̇_W2/T_waste, and Ẇ = Q̇_H − Q̇_W2 = Q̇_W2(T_H/T_waste − 1) = Q̇_W2 × 100/323.15. Equating the work: Q̇_W1 × 20 = Q̇_W2 × 100, so Q̇_W1 = 5Q̇_W2. With Q̇_W1 + Q̇_W2 = 5 MW: Q̇_W2 = 5/6 MW and Q̇_W1 = 25/6 MW. Q̇_H = Q̇_W2 × T_H/T_waste = (5/6)(423.15/323.15) = 1.09 MW.

(c) Centrifugal compressor impeller type and velocity triangles

N = 6000 rpm = 100 rev/s. Blade speeds: U₁ = π(0.5)(100) = 157.1 m/s, U₂ = π(1.0)(100) = 314.2 m/s. With no pre-whirl C_w1 = 0. Power input factor ψ = 1.04, so P = ψṁU₂C_w2 and C_w2 = 50 000/(1.04 × 1.25 × 314.2) = 122.4 m/s. Check: the shaft work is 50/1.25 = 40 kJ/kg, and ψU₂C_w2 = 1.04 × 38.46 = 40 kJ/kg.

A slip factor of unity means no slip, that is, the whirl equals that of the blade-guided flow. Since C_w2 = 122.4 m/s is less than U₂ = 314.2 m/s, the relative outlet velocity must have a component against the direction of rotation: the blades are backward curved. (A radial impeller would give C_w2 = U₂ and a power of about 128 kW.)

Velocity triangles: Inlet: U₁ = 157.1 m/s tangential, C₁ = C_m1 purely radial (no whirl), and the relative velocity W₁ closes the triangle. Outlet: U₂ = 314.2 m/s, C_w2 = 122.4 m/s, and the relative velocity W₂ has a tangential component U₂ − C_w2 = 191.7 m/s opposite to U₂. The radial component C_m2 is not given, so the outlet blade angle follows from tan β₂ = C_m2/191.7 once C_m2 is known.