dp/dz= - ρ g When the variation of density with elevation is known the pressure difference between points 1 and 2 can be determined by integration to be

Thermodynamics I

Spring 1432/1433H (2011/2012H)

Saturday, Wednesday 8:00am - 10:00am &

Monday 8:00am - 9:00am MEP 261 Class ZA

Dr. Walid A. Aissa Dr. Walid A. Aissa

Associate Professor, Mech. Engg. Dept.

Faculty of Engineering at Rabigh, KAU, KSA Chapter #1

April XX, 2012

1–2 ■ Units

Primary (Fundamental) (Main) (base) quantities: SI Units

S Quantity Designation Unit Symbol

1 mass m kilogram kg

2 length L meter m

3 time t second s

4 temperature T Kelvin K

Secondary

(Derived) quantities ^{: SI Units}

s Quantity Designation& Equation Unit

1 velocity V = L/t m/s

2 Acceleration a = L/t² m/s²

3 Volume V = L³ m³

* g (Gravitational acceleration) = 9.807 m/s^{2 ≈} 9.81 m/s²

SI Unit Prefixes

Factor Prefix Symbol Factor Prefix Symbol

10⁹ giga G 10^-1 deci d

10⁶ mega M 10^-2 centi C

10³ kilo k 10^-3 milli m

10³ kilo k 10^-3 milli m

10^-6 micro µµµµ

Force

1N = Kg m/s

²

, Newton

s Quantity Designation& Equation SI Symbol

4 Force F = m a ^{Kg m/s2}

Secondary quantities

: SI Units: (Continued)

1 kgf = 9.807 N

Weight is

W = m g (N) (1-2)

Work, 1 J = 1 (1-3)

(1 kJ = 10

³

J).

Density ; ρ is mass of a unit volume of a substance

ρ ⁼ m /V ^(Kg/m

³

^{) (1-4)}

1–5 ■ DENSITY, SPECIFIC VOLUME,

SPECIFIC GRAVITY & SPECIFIC WEIGHT

(m

³

/kg)

^(1-5)

Specific volume; v is volume per unit mass (reciprocal of density)

ρ ⁼ m /V ^(Kg/m

³

^{) (1-4)}

v = V /m

Specific gravity ; SG (relative density) defined as the ratio of the density of a substance to the density of water at

4 ° C, ( ρ

_H2O

= 1000 kg/m

³

)

(1-6)

)at 4°C

(N/m

³

)

^(1-#1)

Specific weight ; γγγγ is weight of a unit volume of a substance

γγγγ ⁼ W /V = m g/V = ρ ρ ρ ρg

)at 4°C

Temperature Scales

( )

^{K = T}

( )

^{°C + 273.15}

T

∆T (K) = ∆T (°C).

Multiples of Pa, Multiples of Pa,

kilopascal (1 kPa = 10³ Pa) and megapascal (1 Mpa = 10⁶ Pa)

1 bar = 10⁵ Pa = 0.1 MPa = 100 kPa

1 atm =101,325 Pa =101.325 kPa =1.01325 bars

Absolute, gage, and vacuum pressures

Movable datum (zero gage)

P_gage

P_vac

Fixed datum (zero absolute)

P_atm P_abs

P_abs

P_abs = P_atm + P_gage

= P_atm - P_vac

1 atm = 101,325 Pa = 101.325 kPa =1.01325 bars

=14.7 psi.

p_atm

p_atm γh p

h

p

p= p_atm + γh, Hence, p_gage =p- p_atm = γh

(1-19)

p

dp= -ρ g dz dz

p+dp

dp/dz= - ρ g

or

(1-20)

p

The -ve sign is due to our taking the +ve z direction to be upward so that dp is -ve

when dz is +ve (since pressure decreases in an upward direction).

dp/dz= - ρ g _(1-20)

When the variation of density with elevation is known the

pressure difference between points 1 and 2 can be

determined by integration to be.

z

z₁ z₂

From Eq. (1-20) dp/dz= - ρ g From Eq. (1-20) dp/dz= - ρ g

i.e.

Pascal’s law:

Lifting of a large weight by a small force by the

application of Pascal’s law.

Hence, Hence, Hence, But,

Hence,

1–10

■

THE MANOMETER

p_atm

Owing to Pascal’ law

p₁=p₂=p_atm+ρgh

Fluid of density; density; density; density; ρρρρ A

p_2)gage=p₂-p_atm=ρgh

The basic manometermanometermanometermanometer Fluid of density; density; density; density; ρρρρ

h

W = mg =(ρV) g =(ρAh) g

p_2)gage= W /A= ρAh) g /A= ρgh

Hence; ; ; ; Fluid of

density;

density;

density;

density; ρρρρ

×1

×2

3

×

In stacked-up fluid layers

3

p₁=p_atm+ρ₁gh₁ ×

p₂=p₁+ρ₂gh₂ =(p_atm+ρ₁gh₁)+ ρ₂gh₂= p_atm+ρ₁gh₁ +ρ₂gh₂

p₃=p₂+ρ₃gh₃=(p_atm+ρ₁gh₁)+ρ₂gh₂ )+ρ₃gh₃

= p_atm+ρ₁gh₁ +ρ₂gh₂ +ρ₃gh₃

Differential manometer

p_A=p_B

Used to measure pressure differential

P₁+ρ₁g(h+a) =p₂+ρ₁ga +ρ₂gh Hence,

p₁- p₂ =(ρ₁ga +ρ₂gh)- ρ₁g(h+a)

i.e., ∆p=p₁- p₂ =ρ₂gh- ρ₁gh=(ρ₂-ρ₁) gh

manometer

reading

1–11

■

THE BAROMETER AND ATMOSPHERIC PRESSURE

p_atm=p_D=p_B=0 +ρ _Mercury gh

W/A=mg/A=

×

×

×

×DDDD

i.e.,

W/A=mg/A=

(ρ_MercuryV) g/A = (ρ_MercuryAh) g/A=

ρ _Mercury gh

P_atm=ρ _Mercury gh