# chap 5 - CHAPTER CHAPTER First Law of Thermodynamics Ch 5...

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HAPTER CHAPTER First Law of Thermodynamics

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Ch. 5 First Law of Thermodynamics (Energy balance) In one process form E  nit = Joule or the rate form: system out in EE     Unit Joule or Btu in the rate form:  nit = Watt or btu/h in system out   Unit = Watt or btu/h The dot means per unit time
First Law of Thermodynamics (Energy balance) Energy transfer (cross the system boundary) by heat (Q), work (W) and mass (m) oundary    boundary in out system EE     System hange in total energy E surroundings Change in total energy E =Change in( internal (U)+ kinetic (KE)+ potential (PE))

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First Law of Thermodynamics (Energy balance) Total System Energy system EU K E P E   Or E system = m(u 2 -u 1 ) + m (ke 2 -ke 1 ) + m(pe 2 -pe 1 ) r Q = m(u m(ke e + m(pe e Or Q - W = m(u 2 -u 1 )+m(ke 2 -ke 1 ) + m(pe 2 -pe 1 )
Example 5-1 (KE to PE) A steel ball weighing 5 kg rolls horizontally with a velocity of 10 m/s as it approaches an incline. How high up the incline will it be when it comes to rest assuming standard gravitation. Control Volume (C.V.) Steel ball. nergy Eq : E 0 =0 Energy Eq.: E 2 – E 1 = 1 Q 2 1 W 2 = 0 – 0 = 0 Assumes no Heat transfer between the ball and surroundings and no work interaction (e.g. shaft work) therefore: E 1 = mu 1 + mgZ 1 + 0.5 mV 1 2 E = mu + mgZ + 0 2 2 2 We assume the steel ball does not change temperature (u 2 = u 1 ) therefore u u mgZ gZ 5mV 2 0 or mu 2 – mu 1 + mgZ 2 – mgZ 1 – 0.5 mV 1 = 0 or mg (Z 2 –Z 1 ) = 0.5 mV 1 2 Z 2 1 = 0.5 V 1 2 /g = 0.5 × 10 2 (m 2 /s 2 ) / (9.81 m/s 2 ) = 5.1 m

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Example 5-2 A hydraulic hoist raises a 1750 kg car 1.8 m in an auto repair shop. The hydraulic pump has a constant ressure of 800 kPa on its piston What is the increase pressure of 800 kPa on its piston. What is the increase in potential energy of the car and how much volume should the pump displace to deliver that amount of work?
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chap 5 - CHAPTER CHAPTER First Law of Thermodynamics Ch 5...

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