(a) A convergent-divergent nozzle is designed to expand air from a chamber in which the pressure is 800 kPa and temperature is 40°C to give Mach 2·5. The throat area of the nozzle is 0·0025 m². Find the following:

(i) The flow rate through the nozzl

Question

(a) A convergent-divergent nozzle is designed to expand air from a chamber in which the pressure is 800 kPa and temperature is 40°C to give Mach 2·5. The throat area of the nozzle is 0·0025 m². Find the following:

(i) The flow rate through the nozzle under design conditions.

(ii) The exit area of the nozzle.

(iii) The design back-pressure and the temperature of the air leaving the nozzle with this back-pressure.

(iv) The lowest back-pressure for which there is only subsonic flow in the nozzle.

(v) The back-pressure at which there is a normal shock wave on the exit plane of the nozzle.

(vi) The back-pressure below which there are no shock waves in the nozzle.

(vii) The back-pressure over which there are oblique shock waves in the exhaust from the nozzle.

(viii) The back-pressure over which there are expansion waves in the exhaust from the nozzle.

Use Isentropic and Shock tables attached at the end.

(20 marks)

(b)

A stainless steel plate of 1 m length, 1 m width and 10 mm thickness is kept horizontal. The thermal conductivity of the plate is 10 W/mK. Bottom surface of the plate is exposed to hot gases at 700°C with heat transfer coefficient at the bottom surface of 50 W/m²K. Top surface of the plate is cooled by air at 69°C and flowing parallel to the top surface. Any part of the plate should not exceed a maximum permissible temperature of 400°C to avoid failure. Find the minimum permissible velocity of the air required to ensure the plate does not get over-heated beyond the permissible limit. Neglect the heat loss from the side surfaces of the plate and assume one-dimensional heat transfer. Use the following correlation to solve the problem:

Nū_L = Pr⁰.333 [ 0.037 Re_L⁰.8 - 871 ]

Take the appropriate properties of the air from the table attached at the end.

20 marks

Nū_L = Pr⁰.333 [ 0.037 Re_L⁰.8 - 871 ]

Take the appropriate properties of the air from the table attached at the end.

(20 marks)

(c) (i) A vertical flat plate is maintained at a temperature of "T_w °C" and exposed to a stagnant atmospheric air "Tₐ °C". If T_w < Tₐ, show the shape of thermal and velocity boundary layers developed on the surface of the plate. Also show the variation "hₓ" along the vertical surface of the plate. Assume the flow is within the laminar region and consider only free convection.

(5 marks)

(ii) A fluid flows through a tube exposed to constant heat flux condition. The flow is in the turbulent flow regime for which the Dittus-Boelter correlation is applicable for determining the Nu number as given below:

Nu_d = (0·023) Re⁰.8 Pr⁰.4

In order to reduce the surface temperature of the tube, it is suggested to double the velocity of the flow. Find the percentage increase in the heat transfer coefficient due to increased velocity.

(5 marks)

UPSC Answer Check · Accepted Answer

(a) Take air as an ideal gas, γ=1.4, R=287 J/kgK. Stagnation state: T₀=313.15 K, P₀=800 kPa, throat area A*=0.0025 m². Pressures are absolute; the flow is isentropic until a shock appears. The design flow is choked because Me=2.5>1. (i) Choked-throat mass-flow theorem: m = A*P₀/√T₀ × √(γ/R) × 1.2⁻³. For γ=1.4, 1.2⁻³=0.578704. √313.15=17.70 and √(1.4/287)=0.06984. m = 0.0025×800000/17.70 × 0.06984 × 0.578704 = 4.57 kg/s. (ii) Isentropic area–Mach relation: A/A* = (1/M)[(2/(γ+1))(1+(γ-1)M²/2)]³. At Me=2.5, 1+0.2×6.25=2.25 and (5/6)×2.25=1.875. Ae/A* = (1/2.5)×1.875³ = 675/256 = 2.6367. Ae = 2.6367×0.0025 = 0.00659 m². (iii) Isentropic pressure and temperature relations: Pe/P₀ = [1+0.2Me²]⁻³·⁵ = 2.25⁻³·⁵ = 128/2187 = 0.058528. The design back-pressure is the static exit pressure, so pb = Pe = 800×0.058528 = 46.82 kPa. Te/T₀ = 1/2.25 = 4/9, so Te = 313.15×4/9 = 139.18 K = -134.0 °C. (iv) The lowest pb for subsonic-only flow is the limiting case with the throat just sonic and the exit subsonic. Interpolating the subsonic branch for A/A*=2.6367 gives Me≈0.23 and Pe/P₀≈0.964. Hence pb≈800×0.964 = 771 kPa. Below this, the diverging section becomes supersonic and shocks or exhaust waves appear. (v) Normal shock on the exit plane: M₁=2.5 and P₁=46.82 kPa. Normal-shock pressure ratio: P₂/P₁ = 1 + [2γ/(γ+1)](M₁²-1) = 1 + (7/6)×5.25 = 7.125. Since the shock is on the exit plane, pb=P₂=7.125×46.82 = 333.6 kPa. (vi) No shock wave is inside the nozzle when pb is below the exit-plane normal-shock value: pb < 333.6 kPa. At 333.6 kPa the shock is on the plane; below it the interior is shock-free, although exhaust waves may exist. (vii) Oblique shock waves in the exhaust occur when the jet is over-expanded, i.e. pb is above the design value but below the exit normal-shock value: 46.82 kPa < pb < 333.6 kPa. (viii) Expansion waves in the exhaust occur when the jet is under-expanded: pb < 46.82 kPa. (b) Steady one-dimensional heat balance: q'' = hb(Tg-Tsb) = (kp/Lp)(Tsb-Tst) = ht(Tst-Ta). Side losses are neglected, so the same q'' passes through the plate. The maximum plate temperature is at the bottom surface, so set Tsb=400 °C for the minimum-velocity limit. q'' = 50×(700-400) = 15000 W/m². Plate temperature drop ΔTp = q''Lp/kp = 15000×0.010/10 = 15 K, so Tst=385 °C. Required top-surface coefficient: ht = q''/(Tst-Ta) = 15000/(385-69) = 47.47 W/m²K. Film temperature Tf=(385+69)/2=227 °C≈500 K. Air properties at 500 K: k=0.04038 W/mK, ν=37.90×10⁻⁶ m²/s, Pr=0.680. Average Nusselt number: Nū_L = htL/k = 47.47×1/0.04038 = 1175.5, with L=1 m. Given correlation: Nū_L = Pr⁰·³³³[0.037Re_L⁰·⁸ - 871]. Pr⁰·³³³=0.680⁰·³³³=0.8794. Thus 0.037Re_L⁰·⁸ - 871 = 1175.5/0.8794 = 1336.8, so Re_L⁰·⁸ = (1336.8+871)/0.037 = 5.967×10⁴. Re_L = (5.967×10⁴)¹·²⁵ = 9.33×10⁵. Since Re_L > 5×10⁵, the turbulent flat-plate correlation is applicable. Minimum velocity: V = Re_Lν/L = 9.33×10⁵×37.90×10⁻⁶/1 = 35.3 m/s. (c) (i) For T_wδ. A sketch would show both layers widening downward and hₓ starting high at the top and decreasing smoothly. The local coefficient is largest at the top edge because the boundary layers are thinnest there. In laminar free convection, hₓ∝x⁻⁰·²⁵. (ii) Dittus-Boelter: Nu_d = 0.023Re⁰·⁸Pr⁰·⁴. With the same fluid, tube diameter and Pr, h∝Nu∝Re⁰·⁸∝V⁰·⁸. Doubling velocity gives h₂/h₁=2⁰·⁸=1.741. Percentage increase = (1.741-1)×100 = 74.1%.

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