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Other Types of Work Transfer
1. Shaft Work :
Shaft work Let Ft = Tangential force on the shaft R = Radius of the shaft d = Angular displacement of the shaft in an interval of time ‘dt’ Shaft work in time interval ‘dt’, is dWs = Ft. AA 1 = Ft. R.d i.e., Ws =T.d Work done / unit time = dt dWs T. dt d = T.ω where ω = angular velocity, T = Torque But ω = 60
2 N
where N = rpm of the shaft Shaft work, Ws = 60
2 NT
watts
2. Stirring Work : Stirring work is nothing but shaft work is done on the system by using a
stirrer which is driven by a shaft. Figure3: Paddle-wheel work As the weight is lowered, and the paddle wheel turns, there is work transfer into the system which gets stirred. Since the volume of the system remains constant, pdv = 0. If m is the mass
of the weight lowered through a distance dz and T is the torque transmitted by the shaft in rotating through an angle d, the differential work transfer to the fluid is given by w = mgdz = Td and the total work transfer is w = 12 mgdz = 1 2 W^1 dz = 12 Td where W^1 is the weight lowered
3. Electrical Work : When a current flows through a resistor, taken as system, there is work
transfer into the system. This is because the current can drive a motor, the motor can drive a pulley and the pulley can raise a weight. The current I, flows is given by, I = d dC where C = charge in coulombs = time in seconds Thus dC is the charge crossing a system boundary during time d. If E is the voltage potential, the work is w = E.dC = EI d w = 2 1 EI d The electrical work is, w = lim d dw = EI This is the rate at which work is transferred.
4. Work done in stretching a wire : Consider a wire as the system. If the length of the
wire in which there is a tension Ŧ is changed from L to L + dL, the infinitesimal amount of work that is done is equal to, w = - Ŧ dL The - ve sign is used because a positive value of dL means an expansion of the wire, for which work must be done on the wire i.e., negative work. For a finite change of length, w = - 21 Ŧ dL With in the elastic limits, if E is the modulus of the elasticity, is the stress, ε is the strain, and A is the cross sectional area, then Ŧ = A = E.ε.A Therefore w = - E.ε.AdL But dε = dL/L or dL = L x dε w = - Ŧ dL = - E.ε.A. L dε i.e., w = - EAL 1 2 ε d ε δ 0
I
I
System boundary
12 q represents the total heat transfer that takes place when the system undergoes a change of state from state 1 to state 2. Sign Convention : Heat transfer is considered as positive if it takes place from the surroundings to the system and it is considered as negative if it takes place from the system to the surroundings. During an adiabatic process, Q = 0
Units : Since heat is a form of energy transfer it will have the same units as that of energy. In SI
units it is expressed in Joules (J) or Kilo Joules (kJ).
Comparison between work and heat :
Similarities :
Both are path functions and inexact differentials. Both are boundary phenomenon i.e., both are recognized at the boundaries of the system as they cross them. Both represent transient phenomenon; these energy interactions occur only when a system undergoes change of state i.e., both are associated with a process, not a state. Unlike properties, work or heat has no meaning at a state. A system possesses energy, but not work or heat. Concepts of heat and work are associated not with a ‘store’ but with a ‘process’.
Dissimilarities :
Heat is energy interaction due to temperature difference only; work is by reasons other than temperature difference. In a stable system, there cannot be work transfer; however there is no restriction for the transfer of heat. The sole effect external to the system could be reduced to rise of a weight but in the case of a heat transfer other effects are also observed. Heat is a low grade energy whereas work is a high grade energy.
Problems :
- Evaluate the work done in the following processes. The systems to be considered are underlined. a) An agent slowly raises a body of mass 2 kg a distance of 3 mts in a gravitational field of standard acceleration. Solution: By definition, considering agent as the system, it does positive work. The magnitude of work is measured by the product of the weight its lifts and the distance through which it is lifted. W = + mg. l = 2 x 9.81 x 3 = + 58.86 J System System +ve - ve
Work done by the agent = + 58.86 J Because, work is done by the agent, work is done on the body to the same amount. Work done on the body = 58.86 J or Work done by the body = - 58.86 J b) A mass of 1 tonne is suspended from a pulley block. An agent slowly raises the mass against the standard gravitational acceleration by 2m. Solution: By definition, considering agent as the system, it does positive work. W = + mg. l = 1000 x 9. 81 x 2 = 1962 0 J = 19.62 kJ Work done by the agent = 19.62 kJ Because, work is done by the agent, work is done on the mass to the same amount. Work done on the mass = 19.62 kJ or Work done by the mass = - 19.62 kJ c) After raising the mass as in (b), the mass falls freely through the same vertical distance of 2 m. The drag force of the atmosphere on the body is 50 N. Actual Surroundings Fictitious Surroundings Solution: Considering mass as a system (w r t. fictitious surroundings) W = F x l = 50 x 30 = 1500 J = 1.5 kJ Work done by the mass = 1.5 kJ Work done by the atmosphere = - 1.5 kJ Or Work done on the atmosphere = 1.5 kJ d) A body of mass 15 kg falls freely in a vacuum through a vertical distance of 30m. The gravitational acceleration is 6 m/s 2 . Solution: Considering body as a system, as it is falling freely there is no interaction with the system boundary and hence work done by the body is zero. (In other words, the work done by the body is zero as it can lift no weight. All that is happening as the body is falling freely is that its PE is decreasing and its KE is increasing accordingly). e) A rat weighing 5.0 N climbs a stair 0.2 m in height. Solution: Wrat = 0 (since there is no interaction between the system and its surroundings). 1 tonne Zero mass 50 N
1 tonne
50 N
No drag Force
