Question 5.18: (a) Justify the statement ‘‘Nozzles are more efficient than ...
(a) Justify the statement ‘‘Nozzles are more efficient than diffusers’’.
(b) Derive the following expression for nozzle flow :
\frac{dA}{A} = \frac{1}{γ} \frac{dp}{p} (\frac{1 – M²}{M²}), the symbols having usual meanings.
(c) It is proposed to design steam nozzles for the following data :
Initial pressure = 30 bar
Initial temperature = 450°C
Back pressure = 6 bar
Nozzle efficiency = 90%
Initial steam velocity = 60 m/s
Mass flow rate = 2 kg/s
Assuming circular cross-section, calculate the inlet, throat and exit diameter of nozzle. (AMIE)
(a) Nozzles are more efficient than diffusers because in nozzles the flow is in the direction of decreasing pressure (favourable pressure gradient). Hence the boundary layer is thin and the frictional losses are less. In diffusers the boundary layers are thick due to adverse pressure gradients. This has greater frictional effects. Further, the supersonic diffusers have invariably a normal shock at entry, even when operating under design conditions. This further reduces the efficiency.
(b) From continuity equation, \dot{m} = ρ A C = constant = \frac{AC}{v}
Taking logarithms, and the differentials, we have
\frac{dA}{A} + \frac{dC}{C} – \frac{dv}{v} = 0
or \frac{dA}{A} = \frac{dv}{v} – \frac{dC}{C} …(i)
For isentropic flow, from first law of thermodynamics,
Tds = dh – vdp = 0
or dh = vdp …(ii)
From steady flow energy equation
h + \frac{C²}{2} = constant
or dh = – C dC …(iii)
From (ii) and (iii), we have
dC = – \frac{v d p}{C} …(iv)
For an ideal gas, isentropic relation,
p v ^{γ} = constant
or \frac{dp}{p} + γ \frac{dv}{v} = 0
or \frac{dv}{v} = – \frac{1}{ γ } \frac{dp}{p} …(v)
Substituting from (iv) and (v) in (i), we get
\frac{dA}{A} = – \frac{1}{ γ } \frac{dp}{p} + \frac{v d p}{C²} = \frac{1}{ γ } \frac{dp}{p} (\frac{ p v ^{γ}}{C²} – 1)= \frac{1}{ γ } \frac{dp}{p} ( \frac{ γR T}{C²} – 1) , as pv = RT
Further, noting that sonic velocity C_{sonic} = \sqrt { γRT} , and M = Mach Number =\frac{C}{ C_{sonic} } , the above relation reduces to
\frac{dA}{A} = \frac{1}{ γ } \frac{dp}{p} [\frac {C²_{sonic} }{C²} – 1] = \frac{1}{ γ } \frac{dp}{p} [ \frac{1}{M²} – 1]or \frac{dA}{A} = \frac{1}{ γ } \frac{dp}{p} [\frac{1 – M²}{M²}] , the required derivation.
(c) Refer Fig. 22.
p_{1} = 30 bar, p_{3} = 6 bar, p_{2} = throat pressure, C_{1} = 60 m/s, \dot{m} = 2 kg/s.
Inlet : From Mollier chart :
h_{1} = 3343 kJ/kg, v_{1} = 0.1078 m³/kg
h_{01} = h_{1} + \frac{C_{1}² }{ 2 } = 3343 + \frac{60² }{ 2 × 1000 }
= 3344.8 kJ/kg ≈ h_{1} [P_{01} ≈ p_{1} = 30 bar]
Also, \dot{m} = \frac{A_{1} C_{1}}{ v_{1} }
or A_{1} = \frac{\dot{m} v_{1}}{ C_{1} } = \frac{2 × 0.1078 }{60 } = 0.003593 m²
or \frac{π}{4 } D_{1}² = 0.003593
∴ D_{1} = ( \frac{0.003593 × 4 }{π} )^{ 1/ 2} = 0.0676 m = 67.6 mm.
Throat : Assume that frictional losses occur in diverging part only ; i.e., flow upto throat is frictionless.
\frac{p_{2} }{ p_{1}} = 0.546
∴ p_{2} = 30 × 0.546 = 16.38 bar [For superheat steam n = 1.3
\frac{p_{2} }{ p_{1}} = (\frac{2}{n + 1}) ^{\frac{n}{n – 1}} = (\frac{2}{1.3 + 1}) ^{\frac{1.3}{1.3 – 1}} = 0.546 ]
From Mollier chart :
h_{2} = 3155 kJ/kg, v_{2} = 0.19 m³/kg
C_{2} = 44.72 \sqrt {(3344.8 – 3155 )} = 616 m / s
∴ A_{2} = \frac{\dot{m}v_{2}}{ C_{2} } = \frac{2 × 0.19 }{616 } = 0.000617 m²
or \frac{π}{4 }D_{2}² = 0.000617
∴ D_{2} = ( \frac{0.000617 × 4 }{π} )^{ 1/ 2} = 0.028 m or 28 mm.
Exit : From Mollier chart : h_{3} = 2905 kJ/kg, v_{3}′ = 0.4 m³/kg
h_{3}′ = h_{01} – η_{nozzle} (h_{01} – h_{3})
= 3344.8 – 0.9(3344.8 – 2905) = 2949 kJ/kg
∴ C_{3}′ = 44.72 \sqrt {(3344.8 – 2949)} \simeq 890 m / s
∴ A_{3} = \frac{\dot{m} v_{3}′ }{ C_{3}′ } = \frac{2 × 0.4 }{890 } = 0.000899 m²
or \frac{π}{4 } D_{3}² = 0.000899
or D_{3} = ( \frac{0.000899 × 4 }{π} )^{ 1/ 2} = 0.0338 m or 33.8 mm.
