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At stagnation point heat transfer coefficient is high. With angular co-ordinate it starts decreasing due to thickening of boundary layer . then it starts increasing as laminar flow converts into turbulent flow. Then is starts decreases upto a point where separation occurs. Then increases again.
Answer : option B
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In the thermal boundary layer over a cylinder in cross flow (Ts >T..) with Rep 10%,...
3) (1 point) The boundary layer flow separates from the surface if (a) du/dy = 0...
3) (1 point) The boundary layer flow separates from the surface if (a) du/dy = 0 and dp/dx = 0 (b) du/dy = 0 and dp/dx > 0 (c) du/dy = 0 and dp/dx < 0 (d) The boundary layer thickness is zero
3) (1 point) The boundary layer flow separates from the surface if (a) du/dy=0 and dp/dx...
3) (1 point) The boundary layer flow separates from the surface if (a) du/dy=0 and dp/dx = 0 (b) du/dy = 0 and dp/dx > 0 (c) du/dy = 0 and dp/dx < 0 (d) The boundary layer thickness is zero
Apply separation of variables and solve the following boundary value problem 0 < x < t>...
Apply separation of variables and solve the following boundary value problem 0 < x < t> 0 t>O Ytt(x, t) = 25 yxx(x, t) ya(0,t) = y2(7,t) = y(x,0) = f(x) yt(x,0) = g(x) 0 << 0 <r<a
4. [10] Find the solution to given initial-boundary value problem: 4uxx = ut 0<x<TI, t> 0...
4. [10] Find the solution to given initial-boundary value problem: 4uxx = ut 0<x<TI, t> 0 u(0,t) = 5, uit, t) = 10, t> 0 u(x,0) = sin 3x - sin 5x, 0<x<T
(1 point) Solve the heat problem with non-homogeneous boundary conditions v (2,t) = (2,t), 0<=<4, t>0...
(1 point) Solve the heat problem with non-homogeneous boundary conditions v (2,t) = (2,t), 0<=<4, t>0 u(0,t) =0, u(4,t) = 2, t>0, ulz,0) = , 0 <I<4. Recall that we find h(2), set u(2,t) = u(2,t) – h(2), solve a heat problem for v(, t) and write u(2,t) = v(2,t) +h(2) Find h() (2) = The solution (I, t) can be written as uz, t) =h(2) + (,t), where (2,t) = »=Ecseh (a) v2,t) = Finally, find limu,t) = t-o
3) (1 point) The boundary layer flow separates from the surface if (a) du/dy = 0 and dp/dx = 0 (b) du/dy = 0 and dp/dx > 0 (c) du/dy = 0 and dp/dx < 0 (d) The boundary layer thickness is zero
3) (1 point) The boundary layer flow separates from the surface if (a) du/dy=0 and dp/dx = 0 (b) du/dy = 0 and dp/dx > 0 (c) du/dy = 0 and dp/dx < 0 (d) The boundary layer thickness is zero
Apply separation of variables and solve the following boundary value problem 0 < x < t> 0 t>O Ytt(x, t) = 25 yxx(x, t) ya(0,t) = y2(7,t) = y(x,0) = f(x) yt(x,0) = g(x) 0 << 0 <r<a
4. [10] Find the solution to given initial-boundary value problem: 4uxx = ut 0<x<TI, t> 0 u(0,t) = 5, uit, t) = 10, t> 0 u(x,0) = sin 3x - sin 5x, 0<x<T
(1 point) Solve the heat problem with non-homogeneous boundary conditions v (2,t) = (2,t), 0<=<4, t>0 u(0,t) =0, u(4,t) = 2, t>0, ulz,0) = , 0 <I<4. Recall that we find h(2), set u(2,t) = u(2,t) – h(2), solve a heat problem for v(, t) and write u(2,t) = v(2,t) +h(2) Find h() (2) = The solution (I, t) can be written as uz, t) =h(2) + (,t), where (2,t) = »=Ecseh (a) v2,t) = Finally, find limu,t) = t-o
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