Paper II — Q2
(a) Air flows through a 5 cm diameter pipe. Measurements indicate that at the inlet to the pipe the velocity is 70 m/s, the…
Air flows through a 5 cm diameter pipe. Measurements indicate that at the inlet to the pipe the velocity is 70 m/s, the temperature is 80°C and the pressure 1 MPa. Find the temperature, the pressure, and the Mach Number at the exit of the pipe if the pipe is 25 m long. Assume that the flow is adiabatic and the mean friction factor is 0.005. Use Fanno table attached. 20 marks
Saturated liquid refrigerant at – 7°C flows through a horizontal copper (k = 330 W/mK) tube of inside diameter 25 mm, thickness 2·5 mm and length 10 m. The tube is exposed to surrounding air at 20°C. Find the exit dryness fraction of the refrigerant from the tube if the flow rate is 0·0012 kg/s and latent heat of evaporation is 400 kJ/kg. Take the property values of air at 280 K as given below: ρ = 1·271 kg/m³, k = 0·0246 W/mK, γ = 1·4 × 10⁻⁵ m²/s, Pr = 0·717. For natural convection from a horizontal tube, the following correlation be used: Nūf = (0·48)[Gr . Pr]⁰·²⁵. Neglect the temperature difference between the copper tube and the refrigerant. Also neglect the thermal resistance of copper tube. 20 marks
Explain the procedure to arrive at Stefan-Boltzmann law from the Planck's law. Also find the total emissive power of a black sphere of 5 cm diameter maintained at 500 K. Take σ = 5·67 × 10⁻⁸ W/m²K⁴. 5 marks
Explain the procedure of arriving at Wien's displacement law from Planck's law. Also find the temperature of the sun if the wavelength at which maximum monochromatic emissive power is received is 0·55 μm. Take Wien's constant = 2·9 mm K. 5 marks
हिंदी में प्रश्न पढ़ें
वायु 5 cm व्यास के पाइप से बह रही है। मापन बताते हैं कि पाइप के अन्तर्गम पर वेग 70 m/s, तापमान 80°C व दाब 1 MPa है। यदि पाइप 25 m लम्बी है, तो पाइप के निर्गम पर तापमान, दाब एवं मैक संख्या ज्ञात कीजिए। मान लीजिए कि प्रवाह रूद्धोष्म है तथा माध्य घर्षण गुणांक 0·005 है। संलग्न फैनो (Fanno) तालिका का प्रयोग कीजिए। (20 अंक)
एक क्षैतिज ताँबे (k = 330 W/mK) की नलिका, जिसका आन्तरिक व्यास 25 mm, मोटाई 2·5 mm व लम्बाई 10 m है, के द्वारा एक संतृप्त द्रव प्रशीतक – 7°C पर प्रवाहित होता है। नलिका 20°C पर परिवेश वायु से अनावृत है। यदि प्रवाह दर 0·0012 kg/s व वाष्पन की गुप्त ऊष्मा 400 kJ/kg है, तो नलिका से निर्गम पर प्रशीतक का शुष्कतांश ज्ञात कीजिए। 280 K पर वायु के गुणों के मान, निम्नवत हैं: ρ = 1·271 kg/m³, k = 0·0246 W/mK, γ = 1·4 × 10⁻⁵ m²/s, Pr = 0·717. एक क्षैतिज नलिका से प्राकृतिक संवहन के लिए निम्नलिखित सहसंबंध का प्रयोग कीजिए: Nūf = (0·48)[Gr . Pr]⁰·²⁵. ताँबे की नलिका व प्रशीतक के मध्य ताप-अन्तर नगण्य मानिए। ताँबे की नलिका का ऊष्मीय प्रतिरोध भी नगण्य मानिए। (20 अंक)
प्लांक के नियम से स्टेफान-बोल्ट्ज़मान नियम पर पहुँचने की विधि समझाइए। एक काले गोले की कुल उत्सर्जक शक्ति भी ज्ञात कीजिए जिसका व्यास 5 cm एवं जिसको 500 K पर अनुरक्षित किया गया है। σ = 5·67 × 10⁻⁸ W/m²K⁴ लीजिए। (5 अंक)
प्लांक के नियम से वीन के विस्थापन नियम तक पहुँचने की विधि समझाइए। यदि वह तरंगदैर्ध्य जिस पर अधिकतम एकवर्णी उत्सर्जक शक्ति पाई जाती है, 0·55 μm हो, तो सूर्य के तापमान को भी ज्ञात कीजिए। वीन स्थिरांक = 2·9 mm K लीजिए। (5 अंक)
The figure this question refers to, in words
The question paper is a scan and the diagram did not survive as text. This is the figure as read from the original page — every component, value and label — so the question can be worked from the text below.
(a) Table titled 'Frictional, Adiabatic, Constant-Area Flow (Fanno Line)', 'Perfect Gas, k = 1.4'. Columns: M, T/T*, p/p*, p_0/p_0*, V/V* and rho*/rho, F/F*, 4fL_max/D 0.00, 1.2000, infinity, infinity, 0.00000, infinity, infinity .05, 1.1994, 21.903, 11.5914, .05476, 9.1584, 280.02 .10, 1.1976, 10.9435, 5.8218, .10943, 4.6236, 66.922 .15, 1.1946, 7.2866, 3.9103, .16395, 3.1317, 27.932 .20, 1.1905, 5.4555, 2.9635, .21822, 2.4004, 14.533 .25, 1.1852, 4.3546, 2.4027, .27217, 1.9732, 8.4834 .30, 1.1788, 3.6190, 2.0351, .32572, 1.6979, 5.2992 .35, 1.1713, 3.0922, 1.7780, .37880, 1.5094, 3.4525 .40, 1.1628, 2.6958, 1.5901, .43133, 1.3749, 2.3085 .45, 1.1533, 2.3865, 1.4486, .48326, 1.2763, 1.5664 .50, 1.1429, 2.1381, 1.3399, .53453, 1.2027, 1.06908 .55, 1.1315, 1.9341, 1.2549, .58506, 1.1472, .72805 .60, 1.1194, 1.7634, 1.1882, .63481, 1.10504, .49081 .65, 1.10650, 1.6183, 1.1356, .68374, 1.07314, .32460 .70, 1.09290, 1.4934, 1.09436, .73179, 1.04915, .20814 .75, 1.07856, 1.3848, 1.06242, .77893, 1.03137, .12728 .80, 1.06383, 1.2892, 1.03823, .82514, 1.01853, .07229 .85, 1.04849, 1.2047, 1.02067, .87037, 1.00966, .03632 .90, 1.03270, 1.12913, 1.00887, .91459, 1.00399, .014513 .95, 1.01652, 1.06129, 1.00215, .95782, 1.00093, .003280 1.00, 1.00000, 1.00000, 1.00000, 1.00000, 1.00000, 0 1.05, .98320, .94435, 1.00203, 1.04115, 1.00082, .002712 1.10, .96618, .89359, 1.00793, 1.08124, 1.00305, .009933 1.15, .94899, .84710, 1.01746, 1.1203, 1.00646, .02053 1.20, .93168, .80436, 1.03044, 1.1583, 1.01082, .03364 1.25, .91429, .76495, 1.04676, 1.1952, 1.01594, .04858 1.30, .89686, .72848, 1.06630, 1.2311, 1.02169, .06483 1.35, .87944, .69466, 1.08904, 1.2660, 1.02794, .08199 1.40, .86207, .66320, 1.1149, 1.2999, 1.03458, .09974 1.45, .84477, .63387, 1.1440, 1.3327, 1.04153, .11782 1.50, .82759, .60648, 1.1762, 1.3646, 1.04870, .13605 1.55, .81054, .58084, 1.2116, 1.3955, 1.05604, .15427 1.60, .79365, .55679, 1.2502, 1.4254, 1.06348, .17236 1.65, .77695, .53421, 1.2922, 1.4544, 1.07098, .19022 1.70, .76046, .51297, 1.3376, 1.4825, 1.07851, .20780 1.75, .74419, .49295, 1.3865, 1.5097, 1.08603, .22504 1.80, .72816, .47407, 1.4390, 1.5360, 1.09352, .24189 1.85, .71238, .45623, 1.4952, 1.5614, 1.1009, .25832 1.90, .69686, .43936, 1.5552, 1.5861, 1.1083, .27433 1.95, .68162, .42339, 1.6193, 1.6099, 1.1155, .28989 2.00, .66667, .40825, 1.6875, 1.6330, 1.1227, .30499 2.05, .65200, .39389, 1.7600, 1.6553, 1.1297, .31965 2.10, .63762, .38024, 1.8369, 1.6769, 1.1366, .33385 2.15, .62354, .36728, 1.9185, 1.6977, 1.1434, .34760 2.20, .60976, .35494, 2.0050, 1.7179, 1.1500, .36091
Model answer
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(a) Take R = 287 J/kg K and γ = 1.4 for air. T₁ = 80 + 273.15 = 353.15 K. a₁ = √(γRT₁) = √(1.4 × 287 × 353.15) = 376.69 m/s. M₁ = V₁/a₁ = 70/376.69 = 0.18583.
Using the attached Fanno table, interpolate between M = 0.15 and M = 0.20. At M = 0.15: 4fL*/D = 27.932, T/T* = 1.1946, p/p* = 7.2866. At M = 0.20: 4fL*/D = 14.533, T/T* = 1.1905, p/p* = 5.4555. Fraction f₁ = (0.18583 − 0.15)/(0.20 − 0.15) = 0.7166. 4fL₁*/D = 27.932 + 0.7166(14.533 − 27.932) = 18.331. T₁/T* = 1.1946 + 0.7166(1.1905 − 1.1946) = 1.19166. p₁/p* = 7.2866 + 0.7166(5.4555 − 7.2866) = 5.9743.
Actual friction length parameter: 4fL/D = 4 × 0.005 × 25/0.05 = 10.000. Therefore 4fL₂*/D = 18.331 − 10.000 = 8.331.
Interpolate M₂ between M = 0.25 and M = 0.30. At M = 0.25: 4fL*/D = 8.4834, T/T* = 1.1852, p/p* = 4.3546. At M = 0.30: 4fL*/D = 5.2992, T/T* = 1.1788, p/p* = 3.6190. f₂ = (8.4834 − 8.331)/(8.4834 − 5.2992) = 0.04786. M₂ = 0.25 + 0.04786 × 0.05 = 0.2524. T₂/T* = 1.1852 + 0.04786(1.1788 − 1.1852) = 1.18489. p₂/p* = 4.3546 + 0.04786(3.6190 − 4.3546) = 4.3194.
Thus, T₂ = 353.15 × (1.18489/1.19166) = 351.14 K = 77.99°C. P₂ = 1.000 × (4.3194/5.9743) = 0.7230 MPa.
Final: M₂ ≈ 0.252, T₂ ≈ 351.1 K = 78.0°C, P₂ ≈ 0.723 MPa.
Note: 4fL*/D varies non-linearly at low Mach number, so linear interpolation overstates it at the inlet. The exact Fanno relations give 4fL*/D = 17.24 at M₁ = 0.186, and hence M₂ ≈ 0.266, T₂ ≈ 350.6 K and P₂ ≈ 0.695 MPa. Either set is acceptable when the table method is shown clearly.
(b) Outer diameter: D₀ = 25 + 2 × 2.5 = 30 mm = 0.03 m. Film temperature ≈ (20 − 7)/2 + 273 = 279.65 K ≈ 280 K. β = 1/Tf = 1/279.65 = 3.575 × 10⁻³ K⁻¹. ΔT = 20 − (−7) = 27 K. ν = 1.4 × 10⁻⁵ m²/s, Pr = 0.717.
Grashof number: Gr = gβΔT D₀³/ν² = 9.81 × 3.575 × 10⁻³ × 27 × (0.03)³/(1.4 × 10⁻⁵)² = 1.305 × 10⁵.
Gr·Pr = 1.305 × 10⁵ × 0.717 = 9.36 × 10⁴.
Using Nūf = 0.48(Gr·Pr)⁰·²⁵: Nūf = 0.48 × (9.36 × 10⁴)⁰·²⁵ = 8.39.
h = Nūf k/D₀ = 8.39 × 0.0246/0.03 = 6.88 W/m²K.
Outer area: A₀ = πD₀L = π × 0.03 × 10 = 0.9425 m².
Heat absorbed: Q = hA₀ΔT = 6.88 × 0.9425 × 27 = 175.1 W.
If x is exit dryness fraction, Q = ṁ x hfg. x = Q/(ṁhfg) = 175.1/(0.0012 × 400000) = 0.365.
Final: exit dryness fraction x ≈ 0.365. Since x < 1, evaporation is incomplete.
(c)(i) Planck’s spectral emissive power is E_bλ = 2πhc²/[λ⁵(exp(hc/λkT) − 1)].
Total emissive power is E_b = ∫₀^∞ E_bλ dλ. Put x = hc/(λkT). Then the integral reduces to E_b = (2πk⁴T⁴/h³c²) ∫₀^∞ x³/(eˣ − 1) dx. The standard integral ∫₀^∞ x³/(eˣ − 1) dx = π⁴/15. Thus, E_b = (2π⁵k⁴/15h³c²)T⁴ = σT⁴. This is the Stefan-Boltzmann law.
At T = 500 K: E_b = σT⁴ = 5.67 × 10⁻⁸ × 500⁴ = 3543.75 W/m².
For a sphere of diameter 0.05 m: A = πD² = π(0.05)² = 0.007854 m². Total emitted power = E_bA = 3543.75 × 0.007854 = 27.83 W.
Final: emissive flux = 3.54375 kW/m²; total emission from the sphere = 27.83 W.
(c)(ii) From Planck’s law, E_bλ = C₁/[λ⁵(exp(C₂/λT) − 1)]. Put x = C₂/(λT). Differentiating E_bλ with respect to λ and setting dE_bλ/dλ = 0 gives x eˣ = 5(eˣ − 1), equivalently x = 5(1 − e⁻ˣ). The non-zero root is x = 4.965. Therefore, λmaxT = C₂/x = hc/(kx) ≈ 2.9 × 10⁻³ m K = 2.9 mm K. This is Wien’s displacement law.
For λmax = 0.55 μm = 5.5 × 10⁻⁷ m: T = b/λmax = (2.9 × 10⁻³)/(0.55 × 10⁻⁶) = 5273 K.
Final: temperature of the sun ≈ 5.27 × 10³ K.
How this answer will be evaluated
Framework: Mechanical Engineering, Paper 2
What earns full marks: Rigorous derivations with all steps shown; correct application of Fanno and convection correlations; clear physical interpretation.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Fanno flow: exit temperature, pressure and Mach number for 25 m pipe (D = 5 cm, f = 0.005) from V₁ = 70 m/s, 80 °C, 1 MPa. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- M₁ = 70/√(γRT₁) ≈ 0.186
- 4fL*/D at inlet from table (≈ 17.2 exact; ≈ 18.3 by linear interpolation), 4fL/D = 10
- Exit 4fL*/D = inlet value − 10 ⇒ M₂ ≈ 0.25–0.27
- T₂ ≈ 350.6–351.1 K; p₂ ≈ 0.70–0.72 MPa via T/T*, p/p* ratios
Loses marks
- Using Darcy f without factor 4 mismatch explained
Earns more
- Comment on interpolation accuracy
Extra mark
- Choking length check (L_max ≈ 43 m)
- (b) Exit dryness fraction of refrigerant in a copper tube heated by natural convection from air. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Outer diameter 30 mm; film temperature ≈ 280 K; β = 1/T_f; ΔT = 27 K
- Gr ≈ 1.3 × 10⁵; Gr·Pr ≈ 9.4 × 10⁴; Nu = 0.48(GrPr)^0.25 ≈ 8.4
- h ≈ 6.9 W/m²K; Q = hπD₀LΔT ≈ 175 W
- x = Q/(ṁh_fg) ≈ 0.365
Loses marks
- Using inner diameter for external convection
Earns more
- Assumptions stated (negligible tube resistance)
Extra mark
- Check on radiation contribution
- (c)(i) (i) Stefan–Boltzmann law from Planck's law and emission of a 5 cm black sphere at 500 K. 5 marks
explain— definition/context → points in order → small example → short close
Must cover
- Integrate E_bλ over all λ with x = hc/λkT
- ∫x³/(eˣ − 1)dx = π⁴/15 ⇒ E_b = σT⁴
- E_b = 5.67 × 10⁻⁸ × 500⁴ ≈ 3544 W/m²
- Sphere area πD² = 7.854 × 10⁻³ m² ⇒ ≈ 27.8 W
Loses marks
- Using πD²/4 as sphere area
Earns more
- Expression σ = 2π⁵k⁴/(15h³c²)
Extra mark
- Numerical value check of σ
- (c)(ii) (ii) Wien's displacement law from Planck's law; sun's temperature for λmax = 0.55 μm. 5 marks
explain— definition/context → points in order → small example → short close
Must cover
- dE_bλ/dλ = 0 ⇒ x = 5(1 − e⁻ˣ), x ≈ 4.965
- λmax T = C₂/4.965 ≈ 2.9 mm·K
- T = 2.9 × 10⁻³/0.55 × 10⁻⁶ ≈ 5273 K
- Statement of the law
Loses marks
- Arithmetic error in units (μm vs mm)
Earns more
- Physical interpretation (peak shifts to shorter λ as T rises)
Extra mark
- Comparison with accepted solar temperature ≈ 5800 K
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