An NACA 4415 airfoil is mounted in a high-speed subsonic wind tunnel. The lift coefficient is measured as 0.85. If the test-section Mach number is 0.7, at what angle of attack is the airfoil?<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine the angle of attack (\u03b1)<\/strong> of an NACA 4415 airfoil given the following conditions:<\/p>\n\n\n\n The NACA 4415<\/strong> airfoil has:<\/p>\n\n\n\n For subsonic conditions, the lift coefficient can be approximated by:<\/p>\n\n\n\n [ Where:<\/p>\n\n\n\n For thin airfoils in compressible subsonic flow:<\/p>\n\n\n\n [ [ For NACA 4415, zero-lift angle ( \\alpha_0 \\approx -2^\\circ )<\/strong> = (-0.035) radians.<\/p>\n\n\n\n [ [ [ [ Convert to degrees:<\/p>\n\n\n\n [ Angle of attack ( \\alpha \\approx 3.5^\\circ )<\/strong><\/p>\n\n\n\n The angle of attack at which an airfoil generates a given lift coefficient depends on its aerodynamic shape, flow conditions, and compressibility effects. For a NACA 4415 airfoil, which has moderate camber and thickness, the lift characteristics are predictable using classical thin airfoil theory modified for compressibility.<\/p>\n\n\n\n In subsonic compressible flows (Mach < 1), the lift curve slope increases compared to incompressible flow. This effect is corrected by the Prandtl-Glauert<\/strong> factor, where the lift slope increases by ( 1\/\\sqrt{1 – M^2} ). At a Mach number of 0.7, this results in a higher lift curve slope, roughly 8.8 per radian<\/strong> instead of the incompressible ( 2\\pi \\approx 6.28 ).<\/p>\n\n\n\n Given a lift coefficient of 0.85, and knowing that the NACA 4415 has a zero-lift angle<\/strong> around \u22122 degrees<\/strong>, we solve the linear lift equation:<\/p>\n\n\n\n [ After plugging in the known values and solving, we find the angle of attack is approximately 3.5 degrees<\/strong>. This indicates the airfoil must be pitched slightly above the chord line to counteract its negative zero-lift angle and produce the required lift.<\/p>\n\n\n\n This method assumes linear behavior and is valid up to moderate angles of attack where flow remains mostly attached. It’s a good approximation in wind tunnel tests and early design phases of aircraft performance analysis.<\/p>\n","protected":false},"excerpt":{"rendered":" An NACA 4415 airfoil is mounted in a high-speed subsonic wind tunnel. The lift coefficient is measured as 0.85. If the test-section Mach number is 0.7, at what angle of attack is the airfoil? The correct answer and explanation is : To determine the angle of attack (\u03b1) of an NACA 4415 airfoil given the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-210352","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/210352","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=210352"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/210352\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=210352"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=210352"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=210352"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
\n\n\n\nStep 1: Understand the airfoil characteristics<\/h3>\n\n\n\n
\n
C_L = C_{L_\\alpha} (\\alpha – \\alpha_0)
]<\/p>\n\n\n\n\n
\n\n\n\nStep 2: Estimate lift curve slope<\/h3>\n\n\n\n
C_{L_\\alpha} = \\frac{2\\pi}{\\sqrt{1 – M^2}} \\quad \\text{(per rad)}
]<\/p>\n\n\n\n
C_{L_\\alpha} = \\frac{2\\pi}{\\sqrt{1 – 0.7^2}} = \\frac{2\\pi}{\\sqrt{1 – 0.49}} = \\frac{2\\pi}{\\sqrt{0.51}} \\approx \\frac{6.283}{0.714} \\approx 8.8 \\, \\text{per rad}
]<\/p>\n\n\n\n
\n\n\n\nStep 3: Use known zero-lift angle for NACA 4415<\/h3>\n\n\n\n
\n\n\n\nStep 4: Solve for angle of attack<\/h3>\n\n\n\n
0.85 = 8.8(\\alpha – (-0.035))
]<\/p>\n\n\n\n
0.85 = 8.8(\\alpha + 0.035)
]<\/p>\n\n\n\n
\\alpha + 0.035 = \\frac{0.85}{8.8} \\approx 0.0966
]<\/p>\n\n\n\n
\\alpha = 0.0966 – 0.035 = 0.0616 \\, \\text{rad}
]<\/p>\n\n\n\n
\\alpha = 0.0616 \\times \\frac{180}{\\pi} \\approx 3.53^\\circ
]<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answer:<\/h3>\n\n\n\n
\n\n\n\nExplanation (Approx. 300 Words):<\/h3>\n\n\n\n
C_L = C_{L_\\alpha} (\\alpha – \\alpha_0)
]<\/p>\n\n\n\n