In deriving Bernoulli's equation , we equated the workdone on the fluid in the tube to its change in the potential and kinetic energy (a) How does the pressure change as the fluid moves along the tube if dissipative forces are present ? (b) Do the dissipative forces becomes more important as the fluid velocity increase? Discuss qualitatively.
In deriving Bernoulli's equation , we equated the workdone on the fluid in the tube to its change in the potential and kinetic energy (a) How does the pressure change as the fluid moves along the tube if dissipative forces are present ? (b) Do the dissipative forces becomes more important as the fluid velocity increase? Discuss qualitatively.
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(a). If dissipative forces are present, then some forces in liquid flow due to pressure difference is spent against dissipative forces, due to which the pressure drop becomes large
(b). The dissipative forces become more important with increasing flow velocity, because of turbulence.
(b). The dissipative forces become more important with increasing flow velocity, because of turbulence.
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The word fluid means a substance having particles which readily of its magnitude (a small shear stress, which may appear to be of negligible will cause deformation in the fluid). Fluids are charactrised by such properties as density. Specific weight, specific gravity, viscosity etc. Density of a substance is defined as mass per unit volume and it is denoted by. The specific gravity represents a numerical ratio of two densities, and water is commonly taken as a reference substance. Specific gravity of a substance in written as the ratio of density of substance to the density of water. Specific weight represents the force exerted by gravity on a unit volume of fluid. It is related to the density as the product of density of a fluid and acceleration due to gravity. Viscosity is the most important and is recognized as the only single property which influences the fluid motion to a great extent. The viscosity is the property by virtue of which a fluid offers resistance to deformation under the influenece if shear force. The force between the layers opposing relative motion between them are known as forces of viscosity. When a boat moves slowly on the river remains at rest. Velocities of different layers are different. Let v be the velocity of the level at a distance y from the bed and V+dv be the velocity at a distance y+dy . The velocity differs by dv in going through a distance by perpendicular to it. The quantity (dv)/(dy) is called velocity gradient. The force of viscosity between two layers of a fluid is proportional to velocity gradient and Area of the layer. F prop A & F prop (dv)/(dy) F= -etaA(dv)/(dy) ( -ve sign shown the force is frictional in nature and opposes relative motion. eta coefficient of dynamic viscosity Shear stress (F)/(A)= -eta(dv)/(dy) and simultaneously kinematic viscosity is defined as the dynamic viscosity divided by the density. If is denoted as v . The viscosity of a fluid depends upon its intermolecular structure. In gases, the molecules are widely spaced resulting in a negligible intermolecular cohesion, while in liquids the molecules being very close to each other, the cohesion is much larger with the increases of temperature, the cohesive force decreases rapidly resulting in the decreases of viscosity. In case of gases, the viscosity is mainly due to transfer of molecular momentum in the transerve direction brought about by the molecular agitation. Molecular agitation increases with rise in temperature. Thus we conclude that viscosity of a fluid may thus be considered to be composed of two parts, first due to intermolecuar cohesion and second due to transfer of molecular momentum. If the velocity profile is given by v=(2)/(3)y-y^(2)v is velocity in m//sec y is in meter above the bad. Determine shear stress at y=0.15m , & eta=0.863 Ns//m^(2)
The word fluid means a substance having particles which readily of its magnitude (a small shear stress, which may appear to be of negligible will cause deformation in the fluid). Fluids are charactrised by such properties as density. Specific weight, specific gravity, viscosity etc. Density of a substance is defined as mass per unit volume and it is denoted by. The specific gravity represents a numerical ratio of two densities, and water is commonly taken as a reference substance. Specific gravity of a substance in written as the ratio of density of substance to the density of water. Specific weight represents the force exerted by gravity on a unit volume of fluid. It is related to the density as the product of density of a fluid and acceleration due to gravity. Viscosity is the most important and is recognized as the only single property which influences the fluid motion to a great extent. The viscosity is the property by virtue of which a fluid offers resistance to deformation under the influenece if shear force. The force between the layers opposing relative motion between them are known as forces of viscosity. When a boat moves slowly on the river remains at rest. Velocities of different layers are different. Let v be the velocity of the level at a distance y from the bed and V+dv be the velocity at a distance y+dy . The velocity differs by dv in going through a distance by perpendicular to it. The quantity (dv)/(dy) is called velocity gradient. The force of viscosity between two layers of a fluid is proportional to velocity gradient and Area of the layer. F prop A & F prop (dv)/(dy) F= -etaA(dv)/(dy) ( -ve sign shown the force is frictional in nature and opposes relative motion. eta coefficient of dynamic viscosity Shear stress (F)/(A)= -eta(dv)/(dy) and simultaneously kinematic viscosity is defined as the dynamic viscosity divided by the density. If is denoted as v . The viscosity of a fluid depends upon its intermolecular structure. In gases, the molecules are widely spaced resulting in a negligible intermolecular cohesion, while in liquids the molecules being very close to each other, the cohesion is much larger with the increases of temperature, the cohesive force decreases rapidly resulting in the decreases of viscosity. In case of gases, the viscosity is mainly due to transfer of molecular momentum in the transerve direction brought about by the molecular agitation. Molecular agitation increases with rise in temperature. Thus we conclude that viscosity of a fluid may thus be considered to be composed of two parts, first due to intermolecuar cohesion and second due to transfer of molecular momentum. Ten litres of a liquid of specific gravity 1.3 is mixed with 6 litres of a liquid of specific gravity 0.8 the specific gravity of mixture is
The word fluid means a substance having particles which readily of its magnitude (a small shear stress, which may appear to be of negligible will cause deformation in the fluid). Fluids are charactrised by such properties as density. Specific weight, specific gravity, viscosity etc. Density of a substance is defined as mass per unit volume and it is denoted by. The specific gravity represents a numerical ratio of two densities, and water is commonly taken as a reference substance. Specific gravity of a substance in written as the ratio of density of substance to the density of water. Specific weight represents the force exerted by gravity on a unit volume of fluid. It is related to the density as the product of density of a fluid and acceleration due to gravity. Viscosity is the most important and is recognized as the only single property which influences the fluid motion to a great extent. The viscosity is the property by virtue of which a fluid offers resistance to deformation under the influenece if shear force. The force between the layers opposing relative motion between them are known as forces of viscosity. When a boat moves slowly on the river remains at rest. Velocities of different layers are different. Let v be the velocity of the level at a distance y from the bed and V+dv be the velocity at a distance y+dy . The velocity differs by dv in going through a distance by perpendicular to it. The quantity (dv)/(dy) is called velocity gradient. The force of viscosity between two layers of a fluid is proportional to velocity gradient and Area of the layer. F prop A & F prop (dv)/(dy) F= -etaA(dv)/(dy) ( -ve sign shown the force is frictional in nature and opposes relative motion. eta coefficient of dynamic viscosity Shear stress (F)/(A)= -eta(dv)/(dy) and simultaneously kinematic viscosity is defined as the dynamic viscosity divided by the density. If is denoted as v . The viscosity of a fluid depends upon its intermolecular structure. In gases, the molecules are widely spaced resulting in a negligible intermolecular cohesion, while in liquids the molecules being very close to each other, the cohesion is much larger with the increases of temperature, the cohesive force decreases rapidly resulting in the decreases of viscosity. In case of gases, the viscosity is mainly due to transfer of molecular momentum in the transerve direction brought about by the molecular agitation. Molecular agitation increases with rise in temperature. Thus we conclude that viscosity of a fluid may thus be considered to be composed of two parts, first due to intermolecuar cohesion and second due to transfer of molecular momentum. Viscosity of liquids
The potential energy at a point, relative to the reference point is always defined as the negative of work done by the force as the object moves from the reference point to the point considered. The value of potential energy at the reference point itself can be set equal to zero because we are always concerned only with differences of potential energy between two points and the associated change of kinetic energy. A particles A is fixed at origin of a fixed coordinate system. A particle B which is free to move experiences an force vec(F) = (-(2a)/(r^(3)) + (beta)/(r^(2))) hat(r) due to particle A where hat(r) is the position vector of particle B relative to A. It is given that the force is conservative in nature and _potential energy at infinity is zero. If B has to be removed from the influence of A, energy has to be supplied for such a process. The ionization energy E 0 is work that has to be done by an external agent to move the particle from a distance r_(0) to infinity slowly. Here r_(0) is the equilibrium position of the particle If particle B is transferred slowly from point P_(1) (sqrt2 r_(0), sqrt2 r_(0)) to point P_(2) ((r_(0))/(sqrt2), (r_(0))/(sqrt2)) in the xy-plane by an external agent, calculate work required to be done by it in the process:
Direction : Resistive force proportional to object velocity At low speeds, the resistive force acting on an object that is moving a viscous medium is effectively modeleld as being proportional to the object velocity. The mathematical representation of the resistive force can be expressed as R = -bv Where v is the velocity of the object and b is a positive constant that depends onthe properties of the medium and on the shape and dimensions of the object. The negative sign represents the fact that the resistance froce is opposite to the velocity. Consider a sphere of mass m released frm rest in a liquid. Assuming that the only forces acting on the spheres are the resistive froce R and the weight mg, we can describe its motion using Newton's second law. though the buoyant force is also acting on the submerged object the force is constant and effect of this force be modeled by changing the apparent weight of the sphere by a constant froce, so we can ignore it here. Thus mg - bv = m (dv)/(dt) rArr (dv)/(dt) = g - (b)/(m) v Solving the equation v = (mg)/(b) (1- e^(-bt//m)) where e=2.71 is the base of the natural logarithm The acceleration becomes zero when the increasing resistive force eventually the weight. At this point, the object reaches its terminals speed v_(1) and then on it continues to move with zero acceleration mg - b_(T) =0 rArr m_(T) = (mg)/(b) Hence v = v_(T) (1-e^((vt)/(m))) In an experimental set-up four objects I,II,III,IV were released in same liquid. Using the data collected for the subsequent motions value of constant b were calculated. Respective data are shown in table. {:("Object",I,II,II,IV),("Mass (in kg.)",1,2,3,4),(underset("in (N-s)/m")("Constant b"),3.7,1.4,1.4,2.8):} If buoyant force were also taken into account then value of terminal speed would have
Direction : Resistive force proportional to object velocity At low speeds, the resistive force acting on an object that is moving a viscous medium is effectively modeleld as being proportional to the object velocity. The mathematical representation of the resistive force can be expressed as R = -bv Where v is the velocity of the object and b is a positive constant that depends onthe properties of the medium and on the shape and dimensions of the object. The negative sign represents the fact that the resistance froce is opposite to the velocity. Consider a sphere of mass m released frm rest in a liquid. Assuming that the only forces acting on the spheres are the resistive froce R and the weight mg, we can describe its motion using Newton's second law. though the buoyant force is also acting on the submerged object the force is constant and effect of this force be modeled by changing the apparent weight of the sphere by a constant froce, so we can ignore it here. Thus mg - bv = m (dv)/(dt) rArr (dv)/(dt) = g - (b)/(m) v Solving the equation v = (mg)/(b) (1- e^(-bt//m)) where e=2.71 is the base of the natural logarithm The acceleration becomes zero when the increasing resistive force eventually the weight. At this point, the object reaches its terminals speed v_(1) and then on it continues to move with zero acceleration mg - b_(T) =0 rArr m_(T) = (mg)/(b) Hence v = v_(T) (1-e^((vt)/(m))) In an experimental set-up four objects I,II,III,IV were released in same liquid. Using the data collected for the subsequent motions value of constant b were calculated. Respective data are shown in table. {:("Object",I,II,II,IV),("Mass (in kg.)",1,2,3,4),(underset("in (N-s)/m")("Constant b"),3.7,1.4,1.4,2.8):} At the start of motion when object is released in the liquid, its acceleration is :
A 100 kg block is started with a speed of 2.0 m s^(-1) on a long, rough belt kept fixed in a horizontal position. The coefficient of kinetic friction between the block and the belt is 0.20. (a) calculate the change in the internal energy of the block=-belt system as the block comes to a stop on the belt. (b) consider the sutuation from a frame of reference moving at 2.0 m s^(-1) along the initial velocity of the block. as seen from this frame, the block is gently put on a moving belt and in due time the block starts moving with the belt at 2.0 m s^(-1) , calculate the increase in the kinetic energy of the block as it stops slipping past the belt. (c ) find the work done in this frame by the external force holding the belt.
Direction : Resistive force proportional to object velocity At low speeds, the resistive force acting on an object that is moving a viscous medium is effectively modeleld as being proportional to the object velocity. The mathematical representation of the resistive force can be expressed as R = -bv Where v is the velocity of the object and b is a positive constant that depends onthe properties of the medium and on the shape and dimensions of the object. The negative sign represents the fact that the resistance froce is opposite to the velocity. Consider a sphere of mass m released frm rest in a liquid. Assuming that the only forces acting on the spheres are the resistive froce R and the weight mg, we can describe its motion using Newton's second law. though the buoyant force is also acting on the submerged object the force is constant and effect of this force be modeled by changing the apparent weight of the sphere by a constant froce, so we can ignore it here. Thus mg - bv = m (dv)/(dt) rArr (dv)/(dt) = g - (b)/(m) v Solving the equation v = (mg)/(b) (1- e^(-bt//m)) where e=2.71 is the base of the natural logarithm The acceleration becomes zero when the increasing resistive force eventually the weight. At this point, the object reaches its terminals speed v_(1) and then on it continues to move with zero acceleration mg - b_(T) =0 rArr m_(T) = (mg)/(b) Hence v = v_(T) (1-e^((vt)/(m))) In an experimental set-up four objects I,II,III,IV were released in same liquid. Using the data collected for the subsequent motions value of constant b were calculated. Respective data are shown in table. {:("Object",I,II,II,IV),("Mass (in kg.)",1,2,3,4),(underset("in (N-s)/m")("Constant b"),3.7,1.4,1.4,2.8):} If an object of mass 2 kg and constant b = 4 (N-s)/(m) has terminal speed v_(T) in a liquid then time required to reach 0.63 v_(T) from start of the motion is :
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