4 MPa और 673 K पर भाप 60 m/s के वेग के साथ एक नोजल में स्थिरता से प्रवेश करती है और यह 2 MPa और 573 K पर बाहर निकलती है। नोजल का इनलेट क्षेत्र 50 cm2 है, और नोजल से  75 kJ/s की दर से ऊष्मा खो रही है। निकास वेग m/s में किसके द्वारा दिया जाता है?
(प्रविष्टि पर विशिष्ट आयतन 0.0734310 m3/kg के बराबर लें)

This question was previously asked in
BHEL ET Mechanical Held on May 2019
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  1. 62.425
  2. 58.899
  3. 624.25
  4. 581.723

Answer (Detailed Solution Below)

Option 4 : 581.723
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संकल्पना:

सातत्य समीकरण के अनुसार,हमारे पास है 

in = ṁout 

\(\dot m = \rho \times A \times v = \frac{{A \times velocity}}{{specific\;volume}}\;kg/s\)

अपरिवर्ती प्रवाह ऊर्जा समीकरण 

\(\dot m \times \left[ {{h_1} + \frac{{v_1^2}}{{2000}}} \right] - \dot Q = \dot m \times \left[ {{h_2} + \frac{{v_2^2}}{{2000}}} \right]\)

ṁ =नोजल में द्रव्यमान प्रवाह दर, h = विशिष्ट एन्थैल्पी, Q̇ = नोजल से खोई ऊष्मा, v = भाप का वेग, A =नोजल का क्षेत्र, ρ = घनत्व

गणना:

दिया गया है:

A = 50 cm2 = 50 × 10-4 m2, Q̇ = 75 kJ/s

नोजल इनलेट पर:

P1 = 4 MPa, T1 = 673 K, v1 = 60 m/s

नोजल के निकास(आउटलेट) पर:

P2 = 2 MPa, T2 = 573 K

\(m = \rho \times A \times v = \frac{{A \times velocity}}{{specific\;volume}}\;kg/s\)

\(̇̇ m = \frac{{50 \times {{10}^{ - 4}} \times 60}}{{0.077395}} = 3.877\;kg/s\)

\(\dot m \times \left[ {{h_1} + \frac{{v_1^2}}{{2000}}} \right] - \dot Q = \dot m \times \left[ {{h_2} + \frac{{v_2^2}}{{2000}}} \right]\)
\(3.877 \times \left[ {1.87 \times 673 + \frac{{{{60}^2}}}{{2000}}} \right] - 75 = \;3.877 \times \left[ {1.87 \times 573 + \frac{{v_2^2}}{{2000}}} \right]\)

v2 = 581.723 m/s.

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