Meccanica(2013)48:2149-2158 2151 on 0<x<oo,0<y<oo subject to the boundary f”+9+产9s)ds)+f-2 conditions aT =v=0.kay=-hf()(-T)ony=0 (12) 4→Ux).T→Too asy→o on applying the boundary conditions as.We can then put P(n)=(s)ds to obtain where x and y are respectively the cartesian coordi nates measured along the surface and normal to it,u and v are the velocity components in the x and y di- f"+三0+P)+2ff"+(M2-f)=0(13) rections.T is the fluid temperature.p is the pre ssure c p'=-9 with then P→0asn→o (14) is the acceleration due to gravity.a is the thermal dif- fusivity.8 is the coefficient of thermal expansion.v is It is the problem given by Eqs.(1).(13).(14)subject the kinematic viscosity,k is the thermal conductivity to boundary conditions(11)that we now consider. and x)is outer flow.We can eliminate the Before discussing the solution this problem in de sure p from Fas ().(3)by differentiating ()with tail we can make some general observations about the nature of the solution.If we put g(n)=of(s)ds, 03 we can formally express the solution to Eq.(10)as 2u 6 gm=-Ae-9(15 To reduce (1).(4).(5).(6)to similarity form we need to specify specific functional forms for both h for some constant A.Expression(15)shows that 'is and U as of one sign on 0sn<oo.Boundary conditions (11) hf(x)=kCox-2/5. Uo(x)=/5 give We then put A=1+yloo where loo= 6 e-9mdm>0(16 =(u2g△T)5x35fm) Since y>0,we have A>0 and hence,when a solu- tion to (10).(13).(14).(11)exists,it must have> T-T=△T 2 and0'<0for0≤n<o. where AT=T-To.This gives fm+2+f"+5f”=0 3 Results (9) +0-0 (10) We start by considering the free convection,M=0. limit before considering the general problem where primes denote differentiation with re on 3.1 Free convection limit,M=0 ditions (5)beco f=f'=0. 0'=-y(1-0)0nn=0 Equations (10).(13).(14)with M=0 subject to olved n f→M, →0as0→0 nditions(11) ing a standard bo the sults,shown by plots of f(and0) =10 re given in Fig.1.We nd e(0)ir non M U y=Co 21s appe (w(g△T)P)5 for large y. to zero asy de This lead o consider We can integrate Eq.(9)to get SpringerMeccanica (2013) 48:2149–2158 2151 on 0 ≤ x < ∞, 0 ≤ y < ∞ subject to the boundary conditions u = v = 0, k ∂T ∂y = −hf (x)(Tf − T ) on y = 0 u → U∞(x), T → T∞ as y → ∞ (5) where x and y are respectively the Cartesian coordi￾nates measured along the surface and normal to it, u and v are the velocity components in the x and y di￾rections, T is the fluid temperature, p is the pressure, g is the acceleration due to gravity, α is the thermal dif￾fusivity, β is the coefficient of thermal expansion, ν is the kinematic viscosity, k is the thermal conductivity and U∞(x) is outer flow. We can eliminate the pres￾sure p from Eqs. (2), (3) by differentiating (2) with respect to y and using (1) to get u ∂2u ∂x∂y + v ∂2u ∂y2 = −gβ ∂T ∂x + ν ∂3u ∂y3 (6) To reduce (1), (4), (5), (6) to similarity form we need to specify specific functional forms for both hf and U∞ as hf (x) = kC0x−2/5, U∞(x) = U0x1/5 (7) We then put ψ = ν3gβ T 1/5 x3/5f (η) T − T∞ = T θ, η = gβ T ν2 1/5 y x2/5 (8) where T = Tf − T∞. This gives f  + 2 5 ηθ + 3 5 ff  + 1 5 f  f  = 0 (9) 1 σ θ + 3 5 f θ = 0 (10) where primes denote differentiation with respect to η and where σ is the Prandtl number. The boundary con￾ditions (5) become f = f  = 0, θ = −γ(1 − θ) on η = 0 f  → M, θ → 0 as η → ∞ (11) where M is the constant positive (assisting flow) mixed convection parameter and γ is the Biot num￾ber, which are defined as M = U0 (ν(gβ T )2)1/5 , γ = C0  ν2 gβ T 1/5 We can integrate Eq. (9) to get f  + 2 5  ηθ +  ∞ η θ(s)ds + 3 5 ff  − 1 5 f  2 = constant = −1 5 M2 (12) on applying the boundary conditions as η → ∞. We can then put P(η) =  ∞ η θ(s)ds to obtain f  + 2 5 (ηθ + P) + 3 5 ff  + 1 5 M2 − f  2 = 0 (13) P = −θ with then P → 0 as η → ∞ (14) It is the problem given by Eqs. (10), (13), (14) subject to boundary conditions (11) that we now consider. Before discussing the solution this problem in de￾tail we can make some general observations about the nature of the solution. If we put q(η) = 3 5σ  η 0 f (s)ds, we can formally express the solution to Eq. (10) as θ(η) = A  ∞ η e−q(s) ds, θ (η) = −Ae−q(η) (15) for some constant A. Expression (15) shows that θ is of one sign on 0 ≤ η < ∞. Boundary conditions (11) give A = γ 1 + γI∞ where I∞ =  ∞ 0 e−q(η) dη> 0 (16) Since γ > 0, we have A > 0 and hence, when a solu￾tion to (10), (13), (14), (11) exists, it must have θ > 0 and θ < 0 for 0 ≤ η < ∞. 3 Results We start by considering the free convection, M = 0, limit before considering the general problem. 3.1 Free convection limit, M = 0 Equations (10), (13), (14) with M = 0 subject to boundary conditions (11) were solved numerically us￾ing a standard boundary-value problem solver [29] and the results, shown by plots of f (0) and θ(0) against γ for σ = 1.0, are given in Fig. 1. We see that both f (0) and θ(0) increase monotonically as γ is increased with both appearing to approach a finite asymptotic limit for large γ . Also both f (0) and θ(0) appear to reduce to zero as γ decreases to zero. This leads us to consider the asymptotic limits of both large and small γ .