thermodynamics prepared by: engr. darwin mañaga

11Maxim Confidential

THERMODYNAMICS

Prepared by: Engr. Darwin Mañaga

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Thermodynamics – transformation of heat energy into other forms of energy and vice versa

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Forms of Energy

1. Heat ; Q

2. Potential ; PE

3. Kinetic ; KE

4. Internal ; U

5. Flow ; Wf

6. Enthalpy ; H = U + Wf

7. Mechanical Work ; W

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Power ; P = energy / unit time

Energy ; E = the ability or capacity to do work

Work = Force*distance ( always parallel to the axis of motion )

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Unit of Power

English MkS-cgs SI

1 hp = 1 eng hp = 1 arabian horse

1 french horse = 1 Chueval Vapuer

Watt, J/s, Nm/s

1 mule = 1 mola 1 Pfer Starkey

1 hp = 550 lbf – 1 ft / sec

= 33000 lbf – 1 ft / min

1 met hp = 75kgs – m/1 sec

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Energy Units Conversions

English MKS SI

BTU Cal J

Ft-lbf Dyne-cm N-m

Therm erg

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Conversion of Energy Units

1 BTU = 778.16 ft-lbf 1 kgf = 9.8066 N1 BTU = 1.055 KJ 1 erg = 1 dyne -

cm1 J = 1 N-m1 J = 107 dyne-cm1 J = 107 erg1 cal = 4.18 J

1 lbf = 4.4482 N

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Heat Energy ; Q

Qs = sensible heat = MCp(t2-t1)

Cp - specific heat at constant pressure

QL = latent heat ( only phase change )

- solid to liquid:

QL = mhfi ; hfi – latent heat of fusion of ice

hfi = 144 BTU/lbm = 8000Kcal/kgm = 335 KJ / kgm

- liquid to gas:

QL = mhfg ; hfg – latent heat of fusion of air

hfg = 970.3 BTU/lbm = 540Kcal/kgm = 2257 KJ / kgm

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Kinetic Energy – due to velocity

KE = 0.5 mv2/k ; k – proportional constant

eng mks SI

1 slug-ft/lbf-s2 = 1 gm – cm/dyne - s2 = 1 kgm – m/N - s2 =

32.174 lbm – ft/lbf-s2 = 9.8066 kgm – m/kgf - s2 = 9.8066 kgm – m/kgf - s2

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1. If the mass of a typical neutron is 1.66x10-24 gram and the velocity is approximately 2200m/s. What is the kinetic energy of this neutron in electron volt. Note: 1 ev = 1.6x10-

12 erg.

a. 1.025b. 0.025c. 0.889d. 1.228

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2. The ratio of the density of a substance to the density of some standard substance is called

a. relative densityb. specific gravityc. specific densityd. relative density

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3. The S.I unit of force

a. poundb. Newtonc. kilogramd. Dyne

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Remember the conversions:

1 pound = 1 lb

1 pond = 1 gram

1 poundal = 32.174-1 pound

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4. Heat normally flowing from a high temperature body to a low temperature body wherein it is impossible to convert heat without other effects is called

a. second law of thermodynamicsb. first law of thermodynamicsc. third low of thermodynamicsd. zeroth law of thermodynamics

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Laws of Thermodynamics:

1st Law: Law of conservation of energy

- Energy can neither be created nor destroyed but it can be transformed from on form to another

Ein = Eout

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2nd Law: Law of degradation of energy

- Heat is moving from heat body to cold body

- No engine w/ 100% thermal efficiencyn = (W/QA)x100% = ((QA – QR)/QA)*100%

= ((TMAX – TMIN)/ TMAX)*100%

Note:

Use absolute temp for T

R = deg F + 460 ; K = deg C + 273

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3rd Law: Law of absolute zero temperature

- @ absolute zero temp. line, all molecules of a crystalline substance are dead & the degree of disorder of molecules is equal to 0 & heat is eventually equal to 0.

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Zeroth or 4th Law:

- made by Filipino

- keyword: 3rd body

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Potential Energy – due to elevation

= Weight*distance ; N*m or J

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5. The weight of a 100kg mass body at a given elevation is 0.9804kN. If the variation from the standard gravitational acceleration is –0.004mps2 per 1000m. Determine the elevation at this point.

a. 550mb. 50mc. 660md. 650m

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Pressure

Pabs = Pbaro + Pg = Pbaro – Pvac

Note:

If Pbaro is not mentioned, use Pbaro = 1 atm

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Pressure Conversions:Eng Mks SI

14.7 lbf/in2 = 1.0332 kgf/cm2 = 101.325 kPa or KN/m2 =

29.92 inHg = 760 mmHg = 0.76 mHg =

34 ftH20 = 10330 mmH20 = 10.33 mH20

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Remember:

1 TORR = 1mmHg

1 bar = 100kPa

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6. The pressure of a boiler is 9.5 kg/cm2 . The barometric pressure of the atmosphere is 768mm of mercury. Find the absolute pressure in the boiler in MPa.

a. 1.03b. 2.03c. 3.03d. 4.03

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7. How many metric horse power are there in 10 kW?

a. 13.4b. 13.59c. 15.16d. 16.33

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8. A pressure of 1 millibar is equivalent to

a. 1000 dynes/cm2

b. 1000 cm Hgc. 1000psid. 1000kg/sq.cm

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8. What is the minimum pressure required to force blood from the heart to top of the head if the vertical distance is 27 in? Assume the density of the blood to be 1.04gm/cm3 and neglect friction. Express the answer in mmHg.

a. 54.42b. 52.44c. 42.25d. 25.54

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Remember:

Pressure in mmHg = (Specific Gravity of substance)(h)

13.6

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9. A water temperature rise of 18 deg F in the water cooled condenser is equivalent in deg C to

a. -7.78b. 10c. 265.56d. -9.44

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10. If the temperature is held constant and increased beyond the saturation pressure, we have a

a. saturated vaporb. compressed liquidc. saturated liquidd. sub-cooled liquid

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Flow Energy ; Wf

- aka resistance energy

Wf = (Pressure)(Volume)

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11. A piston was moved at a distance of 75cm by a gas pressure of a 450kPa. If the work done is 15kJ, the piston diameter in cm is

a. 23.79b. 27.93c. 29.37d. 32.79

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12. The change of water from the solid to vapor phase.

a. always must occur in two steps-solid to b. liquid to liquid then liquid to vaporb. can occur directly at elevated pressurec. can occur directly at extremely low pressured. can occur directly at room temperature

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13. Which of the following is an example of path function?

a. temperatureb. volumec. workd. entropy

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Point Functions

• Temp

• Pressure

• Volume

• Entropy

• Internal energy

• Enthalpy

Path Functions

• Heat

• Work

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14. What is the process if n = negative to positive infinity?

a. isobaricb. isothermalc. polytropicd. isentropic

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General Equation:

PVn = C

0 1 K -∞ to + ∞ + ∞

Isobaric/Isopiestic

Isothermal Isentropik/reversible adiabatic

Polytropic Isometric/Isochoric/

Isovolumeric

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15. In a sealed and rigid system, the constant is

a. volumeb. pressurec. temperatured. entropy

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17. In a sealed and rigid system, determine the output atmospheric pressure if the input atmospheric pressure is 2 atm. The temperature at the input is 27 deg C w/c increases to 100 deg C at the output.

a. 0.725 atmb. 0.625 atmc. 0.825 atmd. 0.925 atm

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18. Non flow work is applicable to

a. closed systemb. open systemc. impact systemd. solar system

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Work

Non Flow Work:

Wn = ∫ Pdv

Steady Flow Work:

Ws = - ∫ Vdp

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NFW (Closed System)

• Piston

- mass will not cross the boundary

SFW (Open System)

• Turbine

• Pump

- mass and energy cross the boundary

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19. Pressure of octane and air were mixed and produced 86.1 kPa. Determine the pressure of air considering the following parameters: mass of octane = 0.0624 kg. R of octane = 8.314/114; mass of air = 0.91 kg. R of octane = 8.314/29. At a temperature of 290 K.

a. 84.6 kPab. 94.6 kPac. 64.6 kPad. 44.6 kPa

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Daltons Law

- mixture of gases is equal to the summation of the pressures

PTOTAL = ∑ P = P1 + P2 + P3 . . . + PN

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20. A 20x10-3 kg bullet running at 250m/sec hit a target and travelled 12 cm after the impact. Determine the force.

a. 4280Nb. 5280Nc. 6280Nd. 7280N

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Internal Energy; U- due to the movement of molecules

U = mCV(t2 – t1)

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Enthalpy; U- sum of internal energy and flow energy

H = U + Wf = mCpt + PV = mCp(t2 – t1)

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Mechanical work or energy ; W- due to shaft rotation or piston movement

W = (force)(distance)

ω = (Frequency of rotation)(radius) = Torque = rad/sec

Power = (Torque)(ω) = watt

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