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Inbound 2447839727759457219

REVIEW MATERIAL
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BS Mechanical Engineering (BSME)

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Academic year: 2018/2019

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ALCORCON ENGINEERING REVIEW CENTER

Cebu Main: 4th floor Coast Pacific Downtown Center, Sanciangko St, Cebu City Tel #(032) 254-33- Manila: 3rd floor JPD Bldg 1955, C M Recto corner N. Reyes St, Sampaloc, Manila Tel # (02) 736-

POWER PLANT ENGINEERING – DAY 1

I. THERMODYNAMICS

Thermodynamics is the study of heat and work.

A pure substance is a working substance whose chemical composition remains the same even if there is a change in phase. (most common example is water)

An ideal gas is a working substance which remains in gaseous state during its operating cycle and whose equation of state is PV = mRT. (most common example is air)

PROPERTIES OF WORKING SUBSTANCE

1. PRESSURE

Pressure, in general, is the ratio of force per unit area.

Gage pressure is the pressure reading from the gage pressure instrument which is higher or lower than the atmospheric pressure.

Vacuum pressure is a pressure that is less than atmospheric pressure.

Pressure = , KN/m 2 , lb/in 2

Absolute pressure (Pabs) = Gauge pressure + Atmospheric Pressure

Pabs = Pg + Patm

A. Pressure Conversions

1 atm = 0 kPag = 101 kPa = 14 psi = 29 in Hg = 760 mm Hg

B. Hydrostatic Pressure, P P = w h

where: w = density of fluid = (SG x ww) h = height of fluid ww = density of water = 1000 kg/m 3 = 9 KN/m 3 = 62 lb/ft 3 = 1 kg/li

Area

Force

P Ih

Pabs

Pabs

Pgage

Pvac

Iatmospheric

Above atmospheric

Below atmospheric

P h

This file is only for viewing and printing.

2. TEMPERATURE

Temperature is the degree of hotness or coldness of a substance or body.

A thermometer is an instrument used to measure the temperature of a body or a substance.

A pyrometer is an instrument used to measure high temperature gases.

A. Relation between °C and °F scales

°C = 5/9 (F – 32) °F = 9/ 5 C + 32

B. Absolute temperatures:

°K = °C + 273 °R = °F + 460

C. Temperature Difference:

D°C = 5/9 D°F D°F = 9/5 D°C D°C = D°K D°F = D°R

D. Absolute-Zero Temperature = -273oC = -460oF

3. DENSITY, SPECIFIC VOLUME AND SPECIFIC GRAVITY:

Density is defined, in thermodynamic terms, as the mass per unit volume of a substance

Specific volume is the volume per unit of mass in a substance. It is also the reciprocal of density.

Specific gravity is the ratio of the density of a substance to the density of a standard (water for liquid and air for gas).

A. w = B. v = C. SG =

If two fluids are mixed together:

wm = vm = SGm =

SGm = specific gravity of mixture wm = density of mixture

4. INTERNAL ENERGY(U)

Internal energy is the heat energy due to the movement of molecules within the brought about its temperature.

v

1 V

m = w

1 m

V = Densityofwater

Densityofanyfluid

1 1 2 2

1 2 1 2

1 2 m/v m / v

m m V V

m m +

  • =

1 2

1 1 2 2 1 2

1 2 m m

m/v m /v m m

V V +

  • =

water

m w

w

This file is only for viewing and printing.

II. CONSERVATION OF MASS AND ENERGY

The law of conservation of mass states that the total mass is a constant. This means that the total mass entering to a system is equal to the total mass leaving.

FORMS OF ENERGY

####### 1. Potential Energy, P - is an energy produced due to the change in elevation.

P = m g h DP = P 2 - P 1 = mg(h 2 - h 1 ) where: m = mass of the body h = height or elevation

Unit Analysis: (SI Unit – English Unit)

,

kJ, Btu kW, Btu/min

m Not known (m = 1 kg/kg, 1lb/lb)

kg, lb kg/s, lb/min

h m, ft m, ft m, ft

####### 2. Kinetic Energy, KE - is an energy produced due to the mass and velocity.

where: m = mass W = weight v = velocity g = acceleration due to gravity

####### 3. Work, W - is the product of the displacement of the body and the component of the force in the

direction of the displacement.

Work = Force x Distance

Note: 1. Work done by the system is positive (out from the system) 2. Work done on the system is negative (entering to the system)

####### 5. Heat, Q - heat is formed due to the temperature difference.

Q = mcp(t 2 - t 1 )

where: m = mass cp = specific heat t = temperature Note: 1. Heat is positive when heat is added to the system. 2. Heat is negative when rejected from the system.

III. LAWS OF THERMODYNAMICS

1. Zeroth Law of Thermodynamics

The zeroth law of thermodynamics states that when the two bodies are in thermal equilibrium with the third body, they are in thermal equilibrium with each other and hence are at the same temperature.

####### 1000

####### mgh

####### PE =

####### 778

####### mh

####### PE =

####### kg

####### kJ

####### lb

####### Btu

mv 2 2

1 KE = m(v v ) 2

1 DKE = 12 - 22 m(v v ) 2

1 DKE = 12 - 22

This file is only for viewing and printing.

2. First Law of Thermodynamics

States that one form of energy may be converted into another form. All energy entering = All energy leaving

A. P 1 + K 1 + Wf1 + U 1 + W = P 2 + K 2 + Wf2 + U 2 + Q B. W = (P 2 - P 1 ) + (K 2 - K 1 ) + (U 2 - U 1 ) + (Wf2 -Wf1) + Q C. W = DP + DK + DU + DWf + Q But: Dh = DU + DWf D. W = DP + DK + Dh + Q E. Neglecting Potential energy, Kinetic energy & Heat: W = Dh = h 2 - h 1

3. Second Law of Thermodynamics

Kelvin-Planck Statement: “It is impossible to construct a heat engine which operates in a cycle and receives a given amount of heat from a high temperature body and does an equal amount of work”

4. Third Law of Thermodynamics

States that the absolute entropy of a pure crystalline substance in complete internal equilibrium is zero at zero degrees absolute.

IV. IDEAL GAS

Ideal gas is a substance that has the equation of state:

####### PV = mRT PV = 8 n T

where: P = absolute pressure m = mass of gas V = volume of gas R = gas constant T = absolute temperature n = number of mols

####### Boyle’s Law: (T = C ) P 1 V 1 = P 2 V 2

####### Charles’s Law: (P = C and V = C)

BASIC PROPERTIES OF IDEAL GAS
  1. Relationship between cp, cv, R and k

A. cp = cv + R B. cv = C. cp = D. k =

k = 1 for cold air k = 1 for hot air where: cp = constant pressure specific heat cv = constant volume specific heat k = specific heat ratio R = gas constant

For air: cp = 1 KJ/kg-K = 0 Btu/lb-R cv = 0 KJ/kg-K = 0 Btu/lb-R R = 0 KJ/kg-K = 53 ft-lb/lb-R

2

1 2

1 T

T V

V = 2

1 2

1 T

T P

P =

####### k 1

####### R

  • k 1

####### Rk

  • v

p

####### c

####### c

This file is only for viewing and printing.

  1. A pipe has a diameter of 4” at section AA, and a diameter of 2” at section BB. For an ideal fluid flow, the velocity is given as 1 ft/sec at section AA. What is the flow velocity at section BB? A. 4 ft/sec B. 0 ft/sec C. 1 ft/sec D. 2 ft/sec

####### CONSERVATION OF ENERGY (FIRST LAW)

  1. Steam turbine is receiving 1014 lbm/hr of steam, determine the horsepower output of the turbine if the work done by steam is 251 Btu/lbm A. 100 Hp B. 462 Hp C. 200 Hp D. 6002 Hp

  2. Steam enters a turbine stage with an enthalpy of 3628 kJ/kg at 70 m/s and leaves the same stage with an enthalpy of 2846 kJ/kg and a velocity of 124 m/s. Calculate the power if there are 5 kg/s steam admitted at the turbine throttle? A. 4597 kW B. 3976 kW C. 3883 kW D. 1675 kW

####### IDEAL GAS

  1. If the initial volume of an ideal gas is compressed to one-half its original volume and to twice its temperature, the pressure: A. Doubles B. Quadruples C. Remains constant D. Halves

  2. Find the mass of carbon dioxide having a pressure of 20 psia at 200oF with 10 ft 3 volume. A. 1 lbs B. 1 lbs C. 1 lbs D. 1 lbs

  3. Find the work possessed for a Helium gas at 20oC. A. 609 KJ/kg B. 168 KJ/kg C. 229 KJ/kg D. 339 KJ/kg

This file is only for viewing and printing.

ALCORCON ENGINEERING REVIEW CENTER

Cebu Main: 4th floor Coast Pacific Downtown Center, Sanciangko St, Cebu City Tel #(032) 254-33- Manila: 3rd floor JPD Bldg 1955, C M Recto corner N. Reyes St, Sampaloc, Manila Tel # (02) 736-

POWER PLANT ENGINEERING – DAY 2

I. PROCESSES OF IDEAL GAS

  1. Reversible Process: No friction loss

  2. Adiabatic Process: No heat gain. No heat loss. The system is perfectly insulated

  3. For Any Process: Is a process where there is no change in mass (m 1 = m 2 )

A. B. Dh = m cp (T 2 - T 1 ) C. DU = m cv (T 2 - T 1 )

D.

  1. Constant Volume Process (V 1 = V 2 ): Another term: Isometric, Isochoric, Isovolumic. For example: rigid tank, air tank, rigid vessel.

A. Relation between P and T, B. Q = m cv (T 2 - T 1 ) C. W = 0

D. Dh = m cp (T 2 - T 1 ) E. DU = m cv (T 2 - T 1 ) F. Ds =

  1. Constant Pressure Process (P 1 = P 2 ) Another term: Isobaric process

A. Relation between V and T,

B. Q = mcp(T 2 - T 1 ) C. W = P(V 2 - V 1 ) D. Dh = mcp(T 2 - T 1 )

E. DU = mcv(T 2 - T 1 ) F. Ds = mcp ln

  1. Constant temperature process (T 2 =T 1 ) Another term: Isothermal process, Hyperbolic process

A. Relation between P and V: P 1 V 1 = P 2 V 2

B. W = P 1 V 1 ln = P 1 V 1 ln = mRT ln = mRT ln

C. Q = P 1 V 1 ln = P 1 V 1 ln = mRT ln = mRT ln

D. Dh = 0 E. DU = 0 F. Ds = mR ln = mR ln

2

2 2 1

11 T

PV T

P V =

v

p c

c U

h = D

D

2

1 2

1 T

T P

P =

1

2 v T

T mc ln

2

1 2

1 T

T V

V =

1

2 T

T

2

1 P

P

1

2 V

V

1

2 V

V

2

1 P

P

2

1 P

P

1

2 V

V

1

2 V

V

2

1 P

P

1

2 V

V

2

1 P

P

This file is only for viewing and printing.

E. Cycle Efficiency = = =

F. Mean Effective pressure, Pm Pm =

where: VD = V 2 – V 3

Note: 1. High temperature source is known as a temperature source. 2. Low temperature source is known as a temperature sink. 3. Heat is added during constant temperature process. 4. Heat is rejected during constant temperature process. 5. The process is two constant temperature and two isentropic process. 6. The efficiency of the cycle is a function of temperature. 7. If the temperature difference increases, its cycle efficiency will increase.

2. Otto Cycle

Is a spark-ignition type of engine.

Formulas:

In terms of compression ratio pressure ratio:

rk = compression ratio =

rp = compression pressure ratio =

A. QA = heat added = m cv (T 3 - T 2 )

B. QR = heat rejected = m cv (T 4 - T 1 )

C. W = work = QA - QR

D. e = cycle efficiency

e = = = =

E. rk = compression ratio = = where: c = clearance volume

F. Pm = mean effective pressure = where: V 1 = mRT 1 / P 1

G. T 3 = maximum temperature, P 3 = maximum pressure

Note: 1. Maximum temperature occur after combustion or before expansion. 2. Heat is added during constant volume process. 3. Heat is rejected during constant volume process. 4. The process is two constant volume and two isentropic process.

Q A

W A

A R

####### Q

####### Q - Q

W Q R

W

  • 1

1 2 T

T - T

V D

W

k 2 2

k

####### P 1 V 1 =PV

k 1

2

1 1

2 V

V T

T

÷ ÷ ø

ö ç ç è

æ

k

k 1

1

2 1

2 P

P T

T

÷ ÷ ø

ö ç ç è

æ

( k)k

1

2 r P

P

= ( k)k 1

1

2 r T

T - = rp =(rk)k

####### V 1 /V 2 =V 4 /V 3

####### P 2 /P 1

Q A

W

A

A R

####### Q

####### Q - Q

W Q R

W + k

k 1 p

k 1 k r

1 1 r

1 1 - - = - -

2

1 V

V

c

1 +c

D V 1 V 2

W V

W -

=

Pressure, P

Volume, V

TDC BDC

QA

QR

Isentropic

Isentropic V = C

V = C

####### Process 1 - 2: S = c

####### Process 2 - 3: T = c

####### Process 3 - 4: S = c

####### Process 4 - 1: T = c

This file is only for viewing and printing.

  1. If compression ratio increases, its cycle efficiency will increase.
  2. The cycle efficiency depends on compression ratio and its specific heat ration.
PROBLEMS:

####### PROCESSES OF IDEAL GAS

  1. Air is compressed adiabatically from 30oC to 100oC. If mass of air being compressed is 5 kg. Find the change of entropy. A. 1 KJ/oK B. 0 KJ/oK C. 0 D. 1 KJ/kg

  2. Two kilogram of air in a rigid tank changes its temperature from 32oC to 150oC. Find the work done during the process. A. 236 B. 170 C. 195 D. 0

  3. A perfect gas has a value of R = 58 ft-lb/lb-°R and k = 1. If 20 Btu are added to 10 lbs of this gas at constant volume when initial temperature is 90°F, find the final temperature. A. 97°F B. 104°F C. 154°F D. 185°F

  4. Helium gas is compressed in an adiabatic compressor from an initial state of 14 psia and 50oF to a final temperature of 320oF in a reversible manner. Determine the exit pressure of Helium. A. 38 psia B. 40 psia C. 42 psia D. 44 psia

  5. Find the enthalpy of Helium if its internal energy is 200 kJ/kg. A. 144 kJ/kg B. 223 kJ/kg C. 333 kJ/kg D. 168 kJ/kg

####### CARNOT CYCLE

  1. A Carnot cycle operates between the temperature limits of 300oK and 1500oK, and produces 600 KW of net power. The rate of entropy change of the working fluid during the heat addition process is: A. 0 B. 0 KW/K C. 0 KW/K D. 2 KW/K

  2. A Carnot cycle has a maximum temperature of 550 oF and minimum temperature of 100 oF. If the heat added is 4200 Btu/min, find the horsepower output of the engine. A. 34 B. 40 C. 44 D. 65.

OTTO CYCLE

  1. A spark-ignition engine operates on an Otto cycle with a compression ratio of 9 and a temperature limits of 30oC and 1000oC. If the power input is 500 KW, calculate the mass flow rate of air. A. 1 kg/s B. 2 kg/s C. 2 kg/s D. 2 kg/s

  2. An ideal gasoline engine operates with an initial cycle temperature of 48oC and exhaust temperature of 150oC. The change in temperature during combustion is 150oK. Calculate the ideal thermal efficiency. A. 32% B. 33% C. 34% D. 35%

This file is only for viewing and printing.

II. BRAYTON CYCLE
Open Cycle Gas Turbine

A. Compression Process from 1 to 2 using Compressor

In terms of compression ratio pressure ratio:

WC = compressor work = m cp (T 2 - T 1 )

where: cp = 1 KJ/kg-K

B. Heat Added in the Combustor: 2 to 3 (Constant Pressure process: P 2 = P 3 ) QA = m cp (T 3 - T 2 )

C. Turbine Expansion: 3 to 4 (Isentropic Process: s 3 = s 4 ) WT = turbine work = m cp (T 3 - T 4 )

D. Net Turbine Work, WN WN = WT - WC

E. Cycle Efficiency, e

e = =

In terms of temperature and enthalpy:

e =

e =

rp = pressure ratio = rk = compression ratio =

F. Backwork Ratio, BW

k 2 2

k P 1 V 1 =PV

k 1

2

1 1

2 V

V T

T

÷ ÷ ø

ö ç ç è

æ

k

k 1

1

2 1

2 P

P T

T

÷ ÷ ø

ö ç ç è

æ

( k)k

1

2 r P

P

= ( k)k 1

1

2 r T

T - = rp =(rk)k

k 4 4

k P 3 V 3 = PV

k

k 1

4

3 4

3 P

P T

T

÷

÷ ø

ö ç

ç è

æ = k 1 3

4 4

3 ) V

V ( T

T - =

A

T c A

N Q

W W Q

W - = k 1 k k

k 1 p

r

1 1 r

1 1 - - = - -

3 2

4 1 3 2

3 4 2 1 T T

T T 1 (T T)

(T T) (T T) -

  • = -

3 2

4 1 3 2

3 4 2 1 h h

h h 1 (h h)

(h h) (h h) -

  • = -

4

3 1

2 P

P P

P = 2

1 V

V

T

c W

W TurbineWork

CompressorWork BW = =

Temperature, T

WT

Entropy, s

QA

Wc

P = c

P = c

This file is only for viewing and printing.

III. GAS MIXTURE
Partial Pressure

P = Total pressure = P 1 + P 2 + P 3 V = V 1 + V 2 + V 3

Partial Pressure of each Gas:

####### A. Partial pressure of gas 1:

####### B. Partial pressure of gas 2:

####### C. Partial pressure of gas 3:

Percent Volume & Mass

A. PERCENT VOLUME Vm = total volume of gas mixture = V 1 + V 2 + V 3 Percent volume of gas 1: %V 1 = V 1 / Vm = P 1 / Pm Percent volume of gas 2: %V 2 = V 2 / Vm = P 2 / Pm Percent volume of gas 3: %V 3 = V 2 / Vm = P 3 / Pm

B. PERCENT MASS: mm = total mass of mixture = m 1 + m 2 + m 3 Percent mass of gas 1: %m 1 = m 1 / mm Percent mass of gas 2: %m 2 = m 2 / mm Percent mass of gas 3: %m 3 = m 3 / mm

Where: P 1 = partial pressure of gas 1 V 1 = volume of gas 1 P 2 and V 2 are pressure and volume of gas 2.

Conversion of Volume to Mass or Vice Versa

A. Conversion of volume to mass basis: multiply the molecular weight, V (M) B. Conversion of mass to volume basis: divide the molecular weight, m / M

Molecular weight. M

A. If percent volume given: Mm = V 1 M 1 + V 2 M 2 + V 3 M 3

B. If percent mass given: Mm

Where: Mm = molecular weight of the mixture

Gas constant, R

Rm = gas constant of mixture Rm = 8/Mm

m m

1

####### 1 P

####### V

####### V

####### P =

m m

1

####### 2 P

####### V

####### V

####### P =

m m

1

####### 3 P

####### V

####### V

####### P =

3

3 2

2 1

1

####### M

####### m

####### M

####### m

####### M

####### m

####### 1

####### + +

####### =

This file is only for viewing and printing.

ALCORCON ENGINEERING REVIEW CENTER

Cebu Main: 4th floor Coast Pacific Downtown Center, Sanciangko St, Cebu City Tel #(032) 254-33- Manila: 3rd floor JPD Bldg 1955, C M Recto corner N. Reyes St, Sampaloc, Manila Tel # (02) 736-

POWER PLANT ENGINEERING – DAY 4

I. PURE SUBSTANCE

####### DEFINITIONS:

Sub-cooled liquid or compressed liquid – if the liquid is not about to vaporized. Saturated liquid - if the liquid is about to vaporized. Saturated vapor - if the vapor is about to condense. Superheated vapor - if the vapor is not about to condense. Vapor - is the name given to a gaseous phase that is in contact with the liquid phase. Degrees superheat - is the difference between the actual temperature of superheated vapor. Degrees Sub-cooled - is the difference between the saturation temperature for the given pressure and the actual sub- cooled liquid temperature. Wet vapor or mixture - is a combination of saturated vapor and saturated liquid. Critical point - is the point at which the saturated liquid and saturated vapor are identical.

####### T-s DIAGRAM OF PURE SUBSTANCE

  1. Degree sub-cooled = Saturation temperature - Actual liquid temperature = tsat – ta

  2. Degree Superheat = Actual superheated temperature - Saturation temperature = ta - tsat

  3. Properties of wet steam:

    1. v = specific volume

v = vf + x vfg = vf + x (vg – vf) or

  1. s = entropy

s = sf + x sfg = sf + x (sg – sf) or

  1. h = enthalpy h = hf + x hfg = hf + x (hg – hf) or

  2. U = internal energy U = Uf + x Ufg = Uf + x (Ug – Uf) or

f = represents properties of saturated liquid

g = represents properties of saturated vapor

fg = represents the difference by evaporation

U = internal energy s = entropy

h = enthalpy v = specific volume

x = quality of steam

fg

f

####### v

####### v v

####### x

####### -

####### =

fg

f

####### s

####### s s

####### x

####### -

####### =

fg

f

####### h

####### h h

####### x

####### -

####### =

fg

f

####### U

####### U U

####### x

####### -

####### =

This file is only for viewing and printing.

II. QUALITY AND MOISTURE

Quality - it is the ratio of the mass of vapor to the total mass of liquid and vapor.

A. x = quality =

Moisture - it is the ratio of mass of liquid to the total mass of vapor and liquid.

B. y = moisture = 1 - x = C. VL = volume of liquid = mL vL = mL vf

D. Vv = volume of vapor = mv vv = mv vg E. V = total volume = VL + Vv

where: mL = mass of liquid mv = mass of vapor vf = saturated liquid specific volume vg = saturated vapor specific volume

Note: For saturated liquid: ( y = 100%; x = 0%) For saturated vapor: ( y = 0%; x = 100%)

III. APPROXIMATE ENTHALPY, hf

A. Given temperature in oC (SI)

hf = cp t, KJ/kg where: cp = 4 KJ/kg-K

B. Given temperature in oF (English Unit)

hf = cp (t – 32) , Btu/lb where: cp = 1 Btu/lb-R

IV. PROCESSES OF STEAM

  1. Constant Pressure Process:

    1. W = P(v 2 - v 1 ) 2. DU = U 2 - U 1
    2. Dh = h 2 - h 1 4. Q = h 2 - h 1
    3. Ds = s 2 - s 1 6. Dv = v 2 - v 1

  2. Constant Volume Process:

    1. W = 0 2. Q = U 2 - U 1
    2. Dh = h 2 - h 1 4. DU = U 2 - U 1
    3. Ds = s 2 - s 1 6. Dv = v 2 - v 1

v L

v

####### m m

####### m

####### +

L v

L

####### m m

####### m

####### +

####### T

####### S

####### P = c

####### V = c

####### 1

####### 2

####### T

####### S

####### 2

####### 1

This file is only for viewing and printing.

V. STEAM CYCLES
1. RANKINE CYCLE

The Rankine cycle is the most common steam cycle. The working fluid is water.

A. Heat Added in the Boiler, QA

QA = h 1 – h4, KJ/kg = m(h 1 – h 4 ), KW

B. Turbine Work, Wt

Wt = ideal turbine work

Wt = h 1 - h 2 , KJ/kg = m(h 1 - h 2 ), KW

Quality after Turbine Expansion s 2 = s 1 = sf + xsfg

or x = h 2 = hf + x hfg g f

f fg

f s s

s s s

s s -

  • =

This file is only for viewing and printing.

Wo = turbine work output = Wt x e Where: e = turbine efficiency

C. Heat Rejected in the Condenser, QR

QR = h 2 - h 3 , KJ/kg = m(h 2 - h 3 ), KW where: h 3 = hf at given condenser temperature or pressure

Qw = heat carried by cooling water = mw cp (t 2 – t 1 )

mw = mass of cooling water requirement in Condenser

mw =

D. Pump Work, Wp

Wpo = pump work output = h 4 - h 3 , KJ/kg

Wpo = v 3 (P 4 - P 3 ), KJ/kg = m(h 4 - h 3 ), KW

ep = pump efficiency =

Wpi = pump work input

E. Enthalpy, h 4 h 4 = v 3 (P 4 - P 3 ) + h 3 where: P 4 = boiler pressure P 3 = condenser pressure

F. Cycle Efficiency, e

e = =

G. Cogeneration efficiency, ec ec =

H. Overall thermal efficiency, eo eo =

I. Back work ratio = Pump work / Turbine work = Wp / W

####### c(t t )

####### m(h h)

p 2 1

2 3

####### -

####### -

pi

po

####### W

####### W

Heat Added

Turbine Work-Pump Work A

A R A

T p

####### Q

####### Q Q

####### Q

####### ( W W) -

####### =

####### -

A

R T

####### Q

####### (Q +W)

####### Q g

####### G

####### HeatInput

####### Net PlantOutput

####### =

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Inbound 2447839727759457219

Course: BS Mechanical Engineering (BSME)

526 Documents
Students shared 526 documents in this course

University: Batangas State University

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Page 1
ALCORCON ENGINEERING REVIEW CENTER
Cebu Main: 4th floor Coast Pacific Downtown Center, Sanciangko St, Cebu City Tel #(032) 254-33-84
Manila: 3rd floor JPD Bldg 1955, C M Recto corner N. Reyes St, Sampaloc, Manila Tel # (02) 736-4438
POWER PLANT ENGINEERING – DAY 1
I. THERMODYNAMICS
Thermodynamics is the study of heat and work.
A pure substance is a working substance whose chemical composition remains the same even if there is a change
in phase. (most common example is water)
An ideal gas is a working substance which remains in gaseous state during its operating cycle and whose equation
of state is PV = mRT. (most common example is air)
PROPERTIES OF WORKING SUBSTANCE
1. PRESSURE
Pressure, in general, is the ratio of force per unit area.
Gage pressure is the pressure reading from the gage pressure instrument
which is higher or lower than the atmospheric pressure.
Vacuum pressure is a pressure that is less than atmospheric pressure.
Pressure = , KN/m2, lb/in2
Absolute pressure (Pabs) = Gauge pressure + Atmospheric Pressure
Pabs = Pg + Patm
A. Pressure Conversions
1 atm = 0 kPag = 101.325 kPa = 14.7 psi = 29.92 in Hg = 760 mm Hg
B. Hydrostatic Pressure, P P = w h
where: w = density of fluid = (SG x ww)
h = height of fluid ww = density of water
= 1000 kg/m3 = 9.81 KN/m3 = 62.4 lb/ft3 = 1 kg/li
Area
Force
PIh
Pabs
Pabs
Pgage
Pvac
Iatmospheric
Above atmospheric
Below atmospheric
h
P

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