Paper II — Q4
(a) A single-stage impulse turbine rotor has a mean blade ring diameter of 500 mm and rotates at a speed of 10000 r.p.m. The…
A single-stage impulse turbine rotor has a mean blade ring diameter of 500 mm and rotates at a speed of 10000 r.p.m. The nozzle angle is 20° and the steam leaves the nozzles with a velocity of 900 m/s. The blades are equiangular and the blade friction factor is 0·85. Construct velocity diagrams for the blades and determine the inlet angle of the blades for shockless entry of steam. Also, determine (i) the diagram power for a steam flow of 750 kg/hr, (ii) the diagram efficiency, (iii) the axial thrust and (iv) the loss of kinetic energy due to friction.
20 marks
Explain the effect of impeller blade shape on the performance of a centrifugal compressor with the help of an exit velocity diagram and pressure ratio-mass flow rate curve.
Discuss the phenomena of surging and choking in centrifugal compressors.
20 marks
A shell and tube heat exchanger operates with two shell passes and four tube passes. The shell side fluid is ethylene glycol, which enters at 140 °C and leaves at 80 °C with a flow rate of 4500 kg/hr. Water flows in the tubes, entering at 35 °C and leaving at 85 °C. The overall heat transfer coefficient for this arrangement is 850 W/m²-°C. Calculate the flow rate of water required and the area of the heat exchanger. The specific heat of ethylene glycol may be taken as 2·742 J/g-°C and the specific heat of water may be taken as 4·175 J/g-°C. For NTU relations, the following figure may be used.
10 marks
हिंदी में प्रश्न पढ़ें
औसत फलक (ब्लेड) वलय व्यास 500 mm वाली एक एकल-चरण आवेगी टर्बाइन का रोटर 10000 r.p.m. की गति से घूमता है। नोजल कोण 20° है तथा नोजल से भाप 900 m/s के वेग से बाहर निकलती है। फलक समानकोणिक है तथा फलक घर्षण गुणांक 0·85 है। फलकों के लिये वेग आरेख बनाइये तथा भाप की प्रयात्-रहित प्रविष्टि के लिये फलकों पर प्रवेश का कोण मालूम कीजिये। यह भी मालूम कीजिये (i) भाप प्रवाह 750 kg/hr के लिये आरेख शक्ति, (ii) आरेख दक्षता, (iii) अक्षीय प्रयोद और (iv) गतिज ऊर्जा की घर्षण के कारण हानि।
(20 अंक)
एक अपकेन्द्री समीपदक के निष्पादन पर प्ररोदक के फलक (ब्लेड) की आकृति के प्रभाव को, एक निर्गम वेग आरेख तथा दाब अनुपात-मात्रा प्रवाह दर वक्र की सहायता से समझाइये।
अपकेन्द्री समीपदकों में प्रोल्क्षण व प्रोधन घटनाओं को समझाइये।
(20 अंक)
एक कोष तथा नलिका उष्मा विनिमयित्र दो कोष पथ तथा चार नलिका पथ के साथ कार्यरत है। कोष की ओर का द्रव एथिलीन ग्लाइकॉल है, जो 140 °C पर प्रविष्ट होता है तथा 80 °C पर 4500 kg/hr की प्रवाह दर से बाहर निकलता है। नलिकाओं में प्रवाहित जल, 35 °C पर प्रविष्ट हो रहा है तथा 85 °C पर बाहर निकल रहा है। इस व्यवस्था के लिये समग्र उष्मा अन्तरण गुणांक 850 W/m²-°C है। वांछित जल-प्रवाह दर की तथा उष्मा विनिमयित्र के क्षेत्रफल की गणना कीजिये। एथिलीन ग्लाइकॉल की विशिष्ट उष्मा 2·742 J/g-°C तथा जल की विशिष्ट उष्मा 4·175 J/g-°C ली जा सकती है। एन० टी० यू० सम्बन्धों के लिये निम्न आरेख उपयोग में लिया जा सकता है।
(10 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.
(c) The figure consists of two parts: a schematic diagram of a heat exchanger and a performance chart. The schematic shows a shell-and-tube heat exchanger with two shell passes and four tube passes. The shell fluid enters at the top right and exits at the bottom right, labeled with (mc)_s = C_s. The tube fluid enters at the bottom left and exits at the top left, labeled with (mc)_t = C_t. The chart below has a vertical axis labeled 'Effectiveness, epsilon, %' ranging from 0 to 100 with grid lines every 20 units. The horizontal axis is labeled 'Number of transfer units, NTU_max = AU/C_min' ranging from 0 to 5 with grid lines every 1 unit. The chart contains a family of curves representing different values of the ratio C_min/C_max. The curves are labeled with values 0, 0.25, 0.50, 0.75, and 1.00. The curve for C_min/C_max = 0 is the highest, starting at 0 and rising steeply to approach 100%. The curve for C_min/C_max = 1.00 is the lowest, rising more gradually. All curves start at the origin (0,0).
Model answer
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(a) Single-stage impulse turbine
u = πDN/60 = π × 0.5 × 10000/60 = 261.8 m/s. Inlet: Cw₁ = 900 cos20° = 845.7 m/s, Cf₁ = 900 sin20° = 307.8 m/s. Blade inlet angle for shockless entry: tanβ₁ = Cf₁/(Cw₁ − u) = 307.8/583.9 = 0.527 ⟹ β₁ = 27.8° (= β₂, equiangular blades). Cr₁ = Cf₁/sinβ₁ = 660 m/s, Cr₂ = 0.85 × 660 = 561 m/s. Outlet: Cf₂ = 561 sin27.8° = 261.6 m/s, Cw₂ = 561 cos27.8° − 261.8 = 234.5 m/s (opposite to blade motion). ṁ = 750/3600 = 0.2083 kg/s, ΣCw = 845.7 + 234.5 = 1080.2 m/s.
(i) Diagram power P = ṁ u ΣCw = 0.2083 × 261.8 × 1080.2 = 58.9 kW. (ii) Diagram efficiency η_d = 2uΣCw/C₁² = 2 × 261.8 × 1080.2/900² = 69.8%. (iii) Axial thrust F_a = ṁ(Cf₁ − Cf₂) = 0.2083 × 46.2 = 9.6 N. (iv) Friction loss = ṁ(Cr₁² − Cr₂²)/2 = 0.2083 × (660² − 561²)/2 = 12.6 kW.
(b)(i) Effect of impeller blade shape
The exit blade angle β₂ fixes the exit whirl Cw₂ and hence the ideal work u₂Cw₂ and pressure rise.
- Backward-curved (β₂ below 90°): Cw₂ less than u₂, and Cw₂ falls as flow rises, so the pressure ratio falls steadily with mass flow. This negative slope gives stable operation, a wide range and the highest efficiency, though pressure ratio per stage is lower.
- Radial (β₂ = 90°): Cw₂ ≈ u₂ (less slip), the ideal head is independent of flow, and the curve is nearly flat. Radial blades are strong, simple to make and suit high tip speeds, so most aircraft and high-speed compressors use them.
- Forward-curved (β₂ above 90°): Cw₂ greater than u₂, giving the largest theoretical pressure ratio, but it rises with flow, so operation is unstable and prone to surge, with high exit kinetic energy that the diffuser recovers poorly. They are rarely used.
(b)(ii) Surging and choking
Surging: at low flow, on the rising (positive-slope) part of the characteristic, the pressure the compressor can deliver drops below the downstream pressure. Flow momentarily reverses, the delivery pressure falls, forward flow resumes and the cycle repeats. The result is violent pressure and flow oscillation, noise, vibration, overheating and possible blade damage. It is avoided by operating to the right of the surge line, with blow-off or recirculation valves, variable inlet guide vanes or diffuser vanes.
Choking: at high flow the velocity at the impeller eye, throat or diffuser reaches sonic speed. Mass flow then cannot increase at that speed however low the back pressure, efficiency and pressure ratio fall steeply, and the speed lines become vertical. It sets the maximum capacity and is delayed by larger flow passages and suitable inlet design.
(c) Shell-and-tube heat exchanger (2 shell passes, 4 tube passes)
Heat duty from glycol: Q = (4500/3600) × 2742 × (140 − 80) = 205.65 kW. Water flow: ṁ_w = 205650/(4175 × 50) = 0.985 kg/s (≈ 3546 kg/h).
C_h = 1.25 × 2742 = 3427.5 W/K = C_min, C_c = 0.985 × 4175 = 4113 W/K, Cr = 0.833. ε = Q/[C_min(140 − 35)] = 205650/(3427.5 × 105) = 0.571. From the two-shell-pass ε–NTU relation (or chart), NTU ≈ 1.27. A = NTU × C_min/U = 1.27 × 3427.5/850 ≈ 5.1 m². (Check by LMTD: ΔT_lm,cf = 49.8 °C, giving F ≈ 0.95, consistent.)
How this answer will be evaluated
What earns full marks: All parts fully answered with correct calculations, clear diagrams, and physical interpretation.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Velocity diagrams, blade angle, power, efficiency, thrust, and friction loss. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Calculate blade velocity (u) from diameter and speed
- Construct inlet and outlet velocity triangles
- Determine blade inlet angle for shockless entry
- β₁ ≈ 27.8°, P ≈ 58.9 kW, η ≈ 69.8%, thrust ≈ 9.6 N, friction loss ≈ 12.6 kW
Loses marks
- Missing velocity triangles
- Incorrect blade angle calculation
Earns more
- Correct application of blade friction factor
- Consistent units in power and energy calculations
Extra mark
- Clear, labelled velocity diagrams
- (b)(i) Effect of impeller blade shape on compressor performance. 10 marks
explain— definition/context → points in order → small example → short close
Must cover
- Describe effect of blade shape on exit velocity diagram
- Relate blade shape to pressure ratio-mass flow curve
Loses marks
- No link between blade shape and performance curves
Earns more
- Mention specific blade shapes (radial, backward, forward)
Extra mark
- Sketch of exit velocity diagram
- (b)(ii) Phenomena of surging and choking in centrifugal compressors. 10 marks
discuss— intro → 3-4 dimensions → example → balanced close
Must cover
- Define surging and its causes
- Define choking and its causes
- Explain effects on compressor operation
Loses marks
- Confusing surging with choking
- No explanation of causes or effects
Earns more
- Mention methods to prevent surging or choking
Extra mark
- Sketch of performance curve showing surge and choke points
- (c) Water flow rate and heat exchanger area. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Calculate heat duty from ethylene glycol data
- Determine water flow rate from heat duty
- ε = 0.571, Cr = 0.833, NTU ≈ 1.27, area ≈ 5.1 m² with water flow ≈ 0.985 kg/s
Loses marks
- Incorrect heat duty calculation
- Wrong application of LMTD or NTU method
Earns more
- Correct use of specific heat values
- Appropriate application of NTU chart
Extra mark
- Clear statement of assumptions
What "Construct" is asking you to do
Build the required object — a velocity diagram, a sequential test, a control chart, a geometrical figure — step by step, so the sequence is visible on the page. The steps are marked, not only the finished thing.
Structure that answers it
Data and requirement → scale or basis chosen, stated → construction steps in order → the finished construction, labelled → quantities read off, or the result it yields
Where marks are lost
A diagram drawn without a stated scale, so nothing can be scaled off it and the quantities that follow lose their support. In statistics, writing down the procedure without fixing its defining constants — stopping bounds in terms of the two error probabilities, or the control limits — leaves it unmarkable.
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