Question

2. A tape is located between two parallel plates moving in opposite directions (see figure). A tape separates two fluids with different viscosity (u, = 1 mPas, uz = 5 mPa s). The distance between the plates is H = 5 cm and the tape is not subjected to any external force. (a) Determine the value of h if the tape is not moving. (b) Determine the tape velocity if h = 2.5 cm, its surface area is A = 1 m², and it is pulled in the x direction with an external force Fx = 1 N. v=1 cm/s - Tape H2 V2-2 cm/s

Answer 1

Solution criven in the concertion Mi = 1 mPa.sec.a.1x103 Pa. See illy = 5m Parsee = 5xios Parsee H=5cm = 5x1oºm Since I mitt-to? -3 I millimeter a 10m Imparsesna to? Parsee 1 cm = 102m Vai misec = misec - tape h V = 2 cm) see = 3x10 ml see Scanned with CamScanner

According to Moveston's Creed of viscosity Shear stress T oll all av (Mlm2). 0 = 1 ค!.44 ----57 T- - - --- by -- ocks Oferpe via -- = = - = - = MIIII111111911444446 na eee - 2xta mec 0 - terpe palate ☺ - 130 Home folobe let's assume that area of tape is (A) fi= force acted on teepe deee to tope plate Fa = force acted on tape due to 130 Hom plate a = shear stress on tape faute of tope falak 9,2-4, Co-v} = M N O Na Scanned in Shear Stress on tape cause ay Bottome CamScanner S plase.

To = - Mg x lo-3 = Mga al 9 C . Ho Hah . we know that a Force = are ax shear stress his Rorce on the tape because cause of tope plate FIAT, Ally - 3 fg = force on the teipe cause of Bottome plack- fat Ai Cg = Albino - given in the cleation - h = fg because tape is at rent. Hah from equation a s equation - Miro Hrhole, y = h lg ₂ 1 Scanned with CamScanner htelV+ Hly vs] = U, V,.H.

M = 4, y H ( *,*43 % put the values of M,,N,,V2, sta 41 - 10 x 110 x xi 115.6 x หอ 4 เอว 28 ] 5 X 16 1" } x (ox 10 - 5 x16 11 x เอ้ s x16 ๆ 11 5 0, 4S4” xt อิ | 4 . 49 cm /= 1x10 mise งงงงงงงงงงงง h=25cm - 1 22 ty - (*, . - - - - -+V, A - ] H-h - 1 --5.0 | - - 7-fฯ 4 - assume V constant Greint ve x-Direction Scanned with CamScanner

Cniven in Hees part- 1 career ong tape seer tercer 21 m2 he 2.5cm = 9.5x10²m 666 $b bb bb bbbbbbbbbbbb Ho h= 5-2.5=2,5cm = 2.5x152 m = 103 Parsee (Don. viscosity of upper liquid on tapos Mg = 5x103 porsee. (riscosity of lower lequ of cover lequid from type) Assume velocity of teipe is (v) in al-Direction - fes I ne FRorce ueting on teipe (external)] f = force acting on tape dele to tape plank = AT = shear stress on tape due to topo plak fa = force acting on tape duo to Bottome plan - A Cg to shear stress on tape due to Bottomalar According to rowlton's Colee of viscosity C = 4 CVV) - y = Scanned with Canal canner Mg C-vi (Hang

from equation © FEAT, A LV,V) -☺ 6 fa = AEG e Allo (V. rv) - (Haug Block Diagram of tape. 19999999911111111111111111111 - Assumption Force Balance on tape - 1 Ret. P a Ro lif va constanse 7 I moving in the l pure fe, h s far - Direction t am, (1,-1) = All (V₂ -V) CR-h) loute all the value of N, N, kh, Hah, A, M, M2 xl 9xjö? vy 1 + 1x103 x 9 x 1?-V) - 1x 5x103 x 64 - 2.50102 2.5x152 Scanned what its* (102-v) = (2x05?_~] CamScanner

the one = 10 xið?-102, Come to ser xro2-/ 4w = 9x10²-95 4V5-94.91 v=-6.2275 m/sec so we 1953u me that teppae is moving in t vel--Direction & Ror external Rorco apply on tape in the her birection But velocity sign is freg which is - coming cut so plate moving in Eveg se-Direction e force is Applped interesa - Direction so Repeat N, 210² mln. UU _=159 toscan van focal ne drop la a - = 2x15² m/sec Scanned with CamScanner

Bloere Dicgrom af teper Apply Rorce Balance for constande velocity a fet fig = R pure value of fx, fa s fir 1 te A (Vgr va M, A (N-V) H- pouth value of l, lly, A, Hoh, u, v, 18. - l+ spo**, *(?**?v) = 10%, (1832-07 1 + (2x16?_v7 = xf (162_V3 2.5x102 1+ long top 10 Scanned with CamScanner qu= 25.09

NE 5.09 N= 35.00 v = 6.2995) misec Scanned with CamScanner