Question
Download Solution PDF4 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)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
सातत्य समीकरण के अनुसार,हमारे पास है
ṁin = 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.
Last updated on May 20, 2025
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