The velocity distribution for laminar flow between parallel plates is given by: u u max = 1 − ( 2 y h ) 2 where h is the distance separating the plates and the origin is placed midway between the plates. Consider a flow of water at 15 ∘ C , with u max = 0.30 m s e c and h = 0.50 m m . Calculate the shear stress on the upper plate and give its direction.

Question

The velocity distribution for laminar flow between parallel plates is given by:      u          u              max              =  1  −            (                        2          y                h            )              2      where h is the distance separating the plates and the origin is placed midway between the plates. Consider a flow of water at       15          ∘        C , with       u          max        =  0.30      m          s      e      c       and   h  =  0.50     m  m . Calculate the shear stress on the upper plate and give its direction.

dobradisamgn · Accepted Answer

Step 1Given data:The velocity distribution for laminar flow is,      u          u              max              =  1  −            (                        2          y                h            )              2      The temperature at which the flow is to be considered is,  T  =      15          ∘        CThe value of maximum velocity is,      u          max        =  0.30     m      /    sThe value of the distance between the plates is,  h  =  0.50     m  mOrigin is midway between the plates, so the distance of upper plate from the axis is,  y  =      h    2    =  0.25     m  mStep 2The shear stress in terms of velocity for a Newtonian fluid like water is given by formula,  τ  =  μ            d      u              d      y      Here,   τ represents the shear stress and   μ is the dynamic viscosity of the flow of water which is known to be   1.14  ×      10          −      3           k  g  ×      m          −      1        ×      s          −      1       at given temperature.Substitute the expression for u to find the expression for shear stress.Therefore,  τ  =  μ      d          d      y            (          u              max                    (      1      −                        4                      y                          2                                                h                      2                              )        )      =  −            8      μ              u                  max                    y              h              2            Substitute the known values,  τ  =  −            8      ×      1.14      ×              10                  −          3                    k      g                     m                  −          1                            s                  −          1                    0.30      m              /            s      ×      0.25             m      m              (                                            10                              −                3                                      m                                m            m                          )                            (        0.50                 m        m        ×                  (                                                    10                                  −                  3                                            m                                      m              m                                )                )                    2                =  −  2.74  k  g  ×      m          −      1        ×      s          −      2            (          N              k        g        ×        m        ×                  s                      −            2                                )      =  −  2.74  N  ×      m          −      2          =  −  2.74  P  aThe negative sign shows that the shear stress on the upper plate is opposite to the direction of the flow of water.The shear stress on the upper plate is,   τ  =  −  2.74  P  a acting opposite to the direction of laminar flow of water.