(a) Discuss briefly the functional differences between a fan, a blower and a compressor. (10 marks) (b) Prove that shock cannot…

(a) A fan, a blower and a compressor all move gases, but their pressure-rise levels and thermodynamic behaviour differ.

• Fan: Produces a very small pressure rise, usually below about 1 kPa, giving a pressure ratio near 1. It moves large volumes of gas at low pressure. The gas density changes negligibly, so the flow is often treated as incompressible. Fans are used for ventilation, cooling, air conditioning, mine ventilation and draught in furnaces. Energy transfer is mainly kinetic; Bernoulli’s equation is often adequate.
• Blower: Produces a moderate pressure rise, roughly from a few kPa to about 0·1 MPa. Density change is small but not always negligible. Blowers are used for forced draught, exhaust systems, conveying light materials and industrial air supply. Both centrifugal and positive-displacement types are common.
• Compressor: Produces a high pressure rise, generally with pressure ratio greater than about 1·1 to 1·2, often much higher. Density and temperature rise significantly, so compressibility, enthalpy rise, cooling and thermodynamic irreversibility must be considered. Compressors are used in refrigeration, supercharging, gas turbines, pneumatic systems and chemical process plants. They require closer clearances and better sealing than fans and blowers.

Thus, the essential functional difference is the pressure rise and resulting density change: fan is low pressure, blower is medium pressure, and compressor is high pressure.

(b) Consider a steady, adiabatic normal shock in a perfect gas. Let state 1 be upstream and state 2 downstream. The basic conservation equations are:

• Continuity: ρ₁V₁ = ρ₂V₂
• Momentum: p₁ + ρ₁V₁² = p₂ + ρ₂V₂²
• Energy: h₁ + V₁²/2 = h₂ + V₂²/2

Combining these with the perfect-gas relations gives the normal-shock Mach-number relation:

M₂² = (1 + ((γ - 1)/2)M₁²)/(γM₁² - (γ - 1)/2)

The pressure ratio is:

p₂/p₁ = 1 + 2γ/(γ + 1)(M₁² - 1)

The entropy change for a perfect gas is:

s₂ - s₁ = Cₚ ln(T₂/T₁) - R ln(p₂/p₁)

Using the normal-shock relations, this becomes:

Δs/R = (1/(γ - 1)) ln A + (γ/(γ - 1)) ln B

where

A = (2γM₁² - (γ - 1))/(γ + 1)

and

B = ((γ - 1)M₁² + 2)/((γ + 1)M₁²)

Differentiating with respect to M₁²:

d(Δs/R)/d(M₁²) = 2γ(M₁² - 1)²/[M₁²(2γM₁² - (γ - 1))((γ - 1)M₁² + 2)] ≥ 0

At M₁ = 1, the shock is infinitely weak and Δs = 0. Since the derivative is non-negative, for M₁ < 1 we must have Δs < 0. But the second law of thermodynamics requires Δs ≥ 0 for an adiabatic shock. Therefore a normal shock cannot have an upstream subsonic Mach number. Hence shock cannot occur in subsonic flow; it must be supersonic upstream.

(c)(i) For steady one-dimensional conduction with constant k and no heat generation:

d²T/dx² = 0

Therefore:

T(x) = C₁x + C₂

At x = 0, T = T₁ = 80°C, so C₂ = 80°C.

At x = L, the conduction heat flux equals the convection heat loss:

-k dT/dx|ₓ₌ₗ = h(T(L) - T∞)

But dT/dx = C₁, and T(L) = 80 + C₁L. Thus:

-kC₁ = h(80 + C₁L - 15)

-kC₁ = h(65 + C₁L)

C₁(-k - hL) = 65h

C₁ = -65h/(k + hL)

Now h = 24 W/m²°C, L = 0·4 m, k = 2·3 W/m°C:

hL = 24 × 0·4 = 9·6 W/m°C

k + hL = 2·3 + 9·6 = 11·9 W/m°C

C₁ = -65 × 24/11·9 = -1560/11·9 = -15600/119 = -131·092 °C/m

Hence:

T(x) = 80 - (15600/119)x °C

or approximately:

T(x) = 80 - 131·092x °C

At the right face:

T(L) = 80 - (15600/119)(0·4) = 80 - 6240/119 = 3280/119 = 27·563°C

(c)(ii) The steady heat transfer rate through the wall is:

Q = -kA dT/dx = -kA C₁

Q = 2·3 × 20 × (15600/119)

Q = 717600/119 = 6030·25 W

Thus:

Q = 6·030 kW

Alternatively, using convection at the right surface:

Q = hA(T(L) - T∞) = 24 × 20 × (3280/119 - 15)

= 480 × 1495/119 = 717600/119 = 6030·25 W

Q = 6·030 kW

This result assumes steady state, constant thermal conductivity, no heat generation and negligible radiation.

(d) The proposed equation is:

h̄꜀ = 536.5(Tₛ - T∞)⁰·³³/T

with T = (Tₛ + T∞)/2. The problem states T is in °C, so at the film temperature 100°C, T = 100.

The dimensionless equation is:

h̄꜀L/K = C(Gr Pr)ᵐ

For the RHS to give the same dependence on L, the exponent must be m = 1/3, since Gr ∝ L³ and h is independent of L in the given correlation. The given exponent 0·33 is the rounded value of 1/3.

Take the film temperature as 100°C = 373·15 K for property evaluation.

Density of air:

ρ = P/(RT) = 101000/(287 × 373·15)

ρ = 0·9431 kg/m³

Kinematic viscosity:

ν = μ/ρ = 21·673 × 10⁻⁶/0·9431

ν = 22·98 × 10⁻⁶ m²/s

Prandtl number:

Pr = μCₚ/K = (21·673 × 10⁻⁶ × 1022)/0·0307

Pr = 0·7215

For an ideal gas:

β = 1/T = 1/373·15 = 2·680 × 10⁻³ K⁻¹

Now:

GrPr = gβΔT L³Pr/ν²

= (9·8 × 2·680 × 10⁻³ × 0·7215)/(22·98 × 10⁻⁶)² ΔT L³

= 3·588 × 10⁷ ΔT L³

Using m = 1/3:

hL/K = C(GrPr)¹/³

536·5 ΔT¹/³ L/(T K) = C(3·588 × 10⁷ ΔT L³)¹/³

536·5 ΔT¹/³ L/(100 × 0·0307) = C × 329·8 ΔT¹/³ L

C = 536·5/(100 × 0·0307 × 329·8)

C = 536·5/1012·5

C = 0·530

Therefore:

m = 1/3 = 0·333

C = 0·530

(e) Let the air-fuel ratio be A/F = 27. For 1 kg of air, fuel mass is 1/27 kg, and products mass is:

mₚ = 1 + 1/27 = 28/27 kg

Heat released per kg of products:

q = 43000/28 = 1535·714 kJ/kg

The specific heat at constant pressure is:

Cₚ = Cᵥ + R

Cₚ = (0·71 + 20 × 10⁻⁵T) + 0·287

Cₚ = 0·997 + 0·0002T kJ/kg K

At constant pressure, heat added equals enthalpy rise:

q = ∫ Cₚ dT from T₂ to T₃

1535·714 = ∫(0·997 + 0·0002T)dT from 870 K to T₃

1535·714 = 0·997(T₃ - 870) + 0·0001(T₃² - 870²)

1535·714 = 0·997T₃ - 867·39 + 0·0001T₃² - 75·69

0·0001T₃² + 0·997T₃ - 2478·794 = 0

Solving the quadratic:

T₃ = 2060·436 K

At constant pressure, using ideal gas law and including the added fuel mass:

V₃/V₂ = (mₚ/mₐ)(T₃/T₂)

V₃/V₂ = (28/27)(2060·436/870)

V₃/V₂ = 2·4560

The percentage of stroke at which combustion is completed is:

Percentage = (V₃ - V₂)/(V₁ - V₂) × 100

For compression ratio r = V₁/V₂ = 15:

V₁ - V₂ = 14V₂

Percentage = (V₃/V₂ - 1)/14 × 100

= (2·4560 - 1)/14 × 100

= 1·4560/14 × 100

= 10·40%

Combustion is completed at 10·40% of the stroke.