Sample Problem in Pressure
1. A force of 1.5 x 106 is acting on an area 0.75 m2. What is the pressure in Pa?
Solution:
P=F/A
P=1.5 x106 N/m2 / 0.75 m2
P= 2.0 x106 Pa
2. What is the absolute pressure if the gauge pressure is reading 6.8 bar and the atmospheric pressure is 1000 millibars?
Solution:
Pabs=Pgage + Patm
Pabs=6.8 bar + 1000 x 10-3 bar
Pabs= 7.8 bar x 100 kPa/1 bar
Pabs= 780 kPa
3. Temperature is kept constant during an expansion process. The pressure starts 0.7 bar and volume is 0.2 m3. The volume at the end of the process is 0.8 m3. What is the pressure in Pascals?
Solution:
P1V1/T1=P2V2/T2
T1=T2
P2= 70,000 N/m2 (0.2 m3) / 0.8 m3
P2= 17,500 Pa
4. A mass of 75 g is acting on an area 75 cm2. What is the pressure in N/m2?
Solution:
P=[75g(9.81)/1000(75)]N/cm2 x (100)2cm2/m2
P= 98.1 N/m2
Polytropic Process Sample Problem
One kg/sec of air initially at 101 kPa and 300K is compressed polytropically according to the process PV1.3 = C. a.) Calculate the power necessary to compress the air to 13080 kPa. An after cooler removes 100 kW from the air before it enters the storage tank. b.) Determine the change in internal energy of the air across the after cooler.
a.) compressor power
change in kinetic energy = 0
change in potential energy = 0
WSF = IP = nWNF (polytropic process)
IP = indicated power
IP = [ n (P2V2 – P1V1) / 1-n]
IP = nmR (T2 – T1) / 1 –n
PVT relationship, Polytropic
T2/T1 = (P2/P1) n-1/n
T2 = 300 (1380/101) 1.3-1/1.3
= 548.5 K
IP = (1.3) (1) (0.287) (548.5 – 300) / 1-1.3
IP = - 309.05 kW ; 309.05 kW (supplied)
b.) Q = mCp(T2 – T1)
Change in internal energy = mCv (T2-T1)
- 100 = m(1.0) (T2 – T1)
Cv air = 0.718 kJ/kg K
Change in internal energy = -100 (0.718)
Change in internal energy = -71.8 kW
Change in internal energy = 71.8 kW (decrease)
Isentropic Process Sample Problem
An ideal gas having a mass of 2 kg at 465K and 415 kPa expands in a reversible adiabatic process to 138 kPa. The ideal gas constant is 242 J/kg. K and k = 1.4. Determine
a. Final temperature
b. Change in internal energy
c. The work
d. Cp
e. Cv
Non-flow isometric
a. T2 : PVT
T2 / T1 = (P2 / P1) k-1/k
= 465 (138/415) 0.4/1.4
= 339.5 K
b. change in internal energy = mCv (T2 – T1)
Cp / Cv = k = 1.4
Cp = Cv (1.4)
Cp – Cv = R
1.4 Cv – Cv = 0.242 kJ / kg K
Cv = 0.605 kJ / kg K
change in internal energy = 2(0.605) (339.5 – 465)
change in U = -151.85 kJ = 1.51.85 kJ (decrease)
c. Wnf = P2V2 – P1V1 / 1 – k = mR (T2 – T1) / 1 – k
Q = Wnf + change in U
Wnf – Au = (-151.85)
Wnf = 151.85 kJ
d. Cp = 1.4 (0.605) = 0.847 kJ / kg K
e. Cv = 0.605 kJ / kg K
Isothermal Process Sample Problem
Four pounds of air gain 0.491 Btu/°R of entropy during a non-flow isothermal process. If P1 = 120 psia and V2 = 42.5 ft³, find a. V1 and T1 b. Wnf c. Q and d. Change in U.
Given:
m = 4 lbm
Change in S = 0.491 Btu/°R
P1 = 120 psia
V2 = 42.5 ft³
R = 53.432 ft – lbf/lbm - °R
Solution:
Change in S/ mR = ln V2/V1
(0.491 Btu/°R x 778 ft – lbf /1 Btu)/(4 lbmx53.342 ft-lbf/ lbm - °R)= ln V2/V1
e raised to 1.79 = V2/V1
2.718 raised 1.79 = V2/V1; V2=42.5ft³
V1 = 7.093 ft³
Q = T ( Change in S )
T = Q/Change in S
= (PV ln V2/V1)/Change in S
= 120 lbf/in² ( 7.093 ft³)( 144 in² ) ln (42.5/7.93)/0.491 Btu/ °R ( 778 ft – lbf/1Btu )
= 574.5 °R
Q = T ( Change in S )
= ( 574.5 °R ) ( 0.491 Btu/°R )
= 282.1 Btu
Change in U = 0
Isobaric Sample Problem
While the pressure remains constant at 689.5 kPa, the volume of a system of air changes from 0.567 m³ to 0.283 m³, what are a. Change in U b. Change in H c. Q d. Change in S e. if the process is non-flow and internally reversible, what is the work?
Given:
P = 689.5 kPa
V1 = 0.567 m³
V2 = 0.283 m³
m = 1 kg
Rair = 0.28707 kJ/kg . K
Cv = 0.7181 kJ/kg . K
Cp = 1.0062 kJ/kg. K
Solution:
T1 = P1V1/mR
= ( 689.5 kN/m2) ( 0.567 m³ )/( 1 kg ) ( 0.28708 kJ/kg . K )
= 1361.80 K
T2 = ( 689.5kN/m2) ) ( 0.283m³ )/( 1 kg ) ( 0.28708 kJ/kg . K )
= 679.70 K
Change in U = mCv ( T2 – T1 )
= 1 kg ( 0.7816 kJ/kg . K ) ( 679.70 – 1361.80 ) K
= - 490.2 kJ
Q=Change in H = mCp ( T2 – T1 )
= 1 ( 1.0062 ) ( 679.70 – 1361.80)
= - 686.33 kJ
Change in S = mCn ln T2/T1
= 1 ( 1.0062 ) ln 679.70/1361.80
= - 0.6992 kJ/kg . K
Wnf = P ( V2 – V1 )
= 689.5 ( 0.283 – 0.567 )
= - 195.8 kJ
Isometric Process Sample Problem
A perfect gas has a value of R = 58.8 ft – lbf/lbm - °R and k = 1.26. If 20 Btu are added to 5 lbm of this gas at constant volume when the initial temperature is 90 °F, find (a) T2, Change in H, Change in S, Change in U and (b) Work for a nonflow process.
Given:
R = 58.8 ft – lbf/lbm - °R
k = 1.26
m = 5 lbm
Q = 20 Btu
T1 = 90 + 460 = 550 °R
Solution:
a.) Cv = R/k-1
= 58.8 ft – lbf/lbm - °R/(1.26-1)
= 226.15 ft – lbf/ lbm - °R
T2 = (Q / mCv)+ T1
= 20/ 5 ( 226.1 /778)+500
T2= 563.8 °R
Change in H = mCp ( T2 – T1 )
= 5 ( 284.95/778)x(563.8 – 550 )
=25.27 Btu
Change in S = mCv ln T2/T1
= 5 ( 226.15/778 ) ln ( 563.8 )/( 550 )
= 0.036 Btu / °R
Change in U = mCv ( T2 – T1 )
= 5 {( 226.15 )/( 778 )} ( 563.8 -550 )
= 20.01 Btu
b.)
Wnf = 0
Sample Problems: Ideal Gas, Enthalpy and Internal energy of Ideal Gas
1. A gas initially at 15 psia and 2 ft³ undergoes a process to 90 psia and 0.60 ft³, during which the enthalpy increases by 15.5 Btu; Cv = 2.44 Btu/ lb . °R. Determine (a) Change in U; (b) Cp and (c) R
Given:
P1 = 15 psia
V1 = 2 ft³
Cv = 2.44 Btu/ lb . °R
P2 = 90 psia
V2 = 0.60 ft³
Change in H = 15.5 Btu
Solution:
a. Change in H = Change in U + P2V2 – P1V1
15.5 Btu =Change in U +[ { 90 lb/in² x 0.60 ft³ - 15 lb/in² x 2 ft³}{ 12² in²/1² ft² x 1 Btu/778 lbf – ft}]
Change in U = 11.05 Btu
b. k = Change in H/Change in U
= 15.5/11.05
= 1.40
Cp = kCv
= 1.40 ( 2.44 Btu/lb . °R)
= 3.42 Btu/lb . °R
c. Cp = Cv + R
R = Cp – Cv
= 3.42 – 2.44
= 0.98 Btu/lb . °R x (778 lbf – ft/1 Btu)
= 762.4 lbf – ft/lbm - °R
2. An automobile tire is inflated to 32 psig pressure at 50° F. After being driven, the temperature rise to 75 ° F. Determine the final gage pressure assuming the volume remains constant.
Given:
P1 = 32 psig + 14.696 = 46.696 psia
T1 = 50 + 460 = 510 °R
T2 = 75 + 460 = 535 °R
P2 = ? in psig
Solution:
P2 = P1T2/T1
= 46.696 psia ( 535 °R )/510 °R
= 48.99 psia
P2 = 48.99 psia – 14.696
= 32.49 psig
Ideal Gas Sample Problems
Problem Solving:
1. A drum 6 in. in diameter and 40 in. long contained acetylene at 250 psia and 90°F. After some of the acetylene was used, the pressure was 200 psia and the temperature was 85°F, (a) what proportion of the acetylene was used? (b) what volume would the used acetylene occupy at 14.7 psia and 80°F? R for acetylene is 59.35 ft – lbf/lbm - °R
Given:
D = 6 in
h = 40 in
P1 = 250 psia
T1 = 90°F + 460 = 550 °R
P2 = 200 psia
T2 = 85°F + 460 = 545 °R
m1 = mass of the acetylene initially in the drum
m2 = mass of the acetylene left in the drum
m3 = mass of the acetylene used
Solution:
a. Vol. of tank = ? D²h/4
= 3.1416 ( 6 in) ² ( 40 in) (1ft³)/4 (12³) in³
= 0.6545 ft³
m1 = [250 lbf/in² ( 0.6545 ft³) ( 12² in² )/1 ft²]/{59.35 ft – lbf /lbm - °R( 550 °R )}
= 0.7218 lbm
m2 = 200 ( 0.6545 ) (12 ² )/59.35 ( 545 )
= 0.5828 lbm
m3 = m1 – m2
= 0.7128 – 0.5828
= 0.139 lbm
Acetylene used = (m3/m1)x100
= 19.26 %
b. P3 = 14.7psi
T3 = 80 + 460 = 540 °R
V3 = m3RT3/P3
= 0.139 lbm ( 59.35 lbf – ft/lbm - °R)( 540 °R)/14.7 lbf/in²x 12² in²/1² ft²
= 2.10 ft³
2. Two vessels A and B of different sizes are connected by a pipe with a valve. Vessel A contains 142 litres of air at 2, 767.92 kPa, and 93.33 °C. Vessel B, of unknown volume, contains air at 68.95 kPa, 4.44 °C. The valve is open and when the properties have been determined, it is found that Pm = 1378.96 kPa, tm = 43.33 °C. What is the volume of vessel B?
Given:
For vessel A
VA = 142 L
PA = 2,767.92 kPa
TA = 93.33 °C + 273 = 366.33 K
For vessel B
VB = ?
PB = 68.95 kPa
TB = 4.44 °C + 273 = 277.4 K
For the mixture
PM = 1,378.96 kPa
TM = 43.33 °C + 273 = 310.33 K
mm = ma + mb
PV = mRT
mm = PmVm/RTm
PmVm/RTm = PAVA/RTA= PBVB/ RTB
1,378.96 Vm/310.33 = 2767.92 ( 142 )/ 366.33 = 68.95 Vb/277.44
4.36 Vm = 1072.93 + 0.25 Vb1
Vm = VA+ VB 2
Vm = 142 + VB2’
4.36 ( 142 + VB ) = 1072.93 + 0.25 VB
619.12 + 4.31 VB = 1072.93 + 0.25 VB
619.12 + 1, 072.93 = 4.36 VB – 0.25 VB
453.81 = 4.11 VB
VB = 110.42 L
Problem Solving - Conservation of Energy
Problem Solving:
1. 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 2486 kJ/kg and a velocity of 124 m/s. Calculate the work done by the steam.
Given:
h1 = 3628 kJ/kg
v1= 70 m/s
h2 = 2846 kJ/kg
V2 = 124 m/s
Solution:
KE1 + h1 = KE2 + h2 + W
KE1 = mv²/2k
= ( 70m/s ) ²/2 ( 1 kg . m )/N . s²
= 2.45 kJ/kgm
KE2 = mV²/2k
= ( 124 m/s ) ²/2 ( 1 kg . m )/N . s²
= 7.68 kJ/kgm
W = KE1 – KE2 + h1 – h2
= 2.45 kJ/kgm – 7.68 kJ/kgm + 3628 kJ/kgm – 2846 kJ/kgm
= 776.8 kJ/kgm
2. A thermodynamic steady flow system receives 4.56 kg/min of a fluid where P1 = 137.90 kPa, V1 = 0.0388 m³/kg, v1 = 122 m/s and u1 = 17.16 kJ/kg. The fluid leaves the system at a boundary where P2 = 551.6 kPa, V2 = 0.193 m³/kg, v2 = 183 m/s and u2 = 52.80 kJ/kg. During passage through the system, the fluid receives 180 kJ/min of heat. Determine the work.
Given:
P1 = 137.90 kPa
V1 = 0.0388 m³/kg
v1 = 122 m/s
u1 = 17.16 kJ/kg
m = 4.56 kg/min
P2 = 551.6 kPa
V2 = 0.193 m³/kg
v2 = 183 m/s
u2 = 52.80 kJ/kg
Q = 180 kJ/min
Solution:
KE1 = mv²/2k
= 4.56 kgm/min ( 122 m/s) ² x 1 kJ/min
= 33.94 kJ/min
KE2 = mv²/2k
= 4.56 ( 183 ) ²/2000
= 76.35 kJ/min
Wf1 = PmV
= 137.90 kN/min x 4.56 kgm/min x 0.0388 m³/kgm
= 24.39 kJ/min
Wf2 = 551.6 ( 4.56) ( 0.193)
= 485.45 kJ/min
Change in total internal energy = m(u1 – u2)
= 4.56 kgm/min ( 17.16 – 52.80 ) kJ/kgm
= - 162.52 kJ/min
W = KE1 – KE2 + Wf1 – Wf2 -162.52 kJ/min + 180 kJ/min
= 33.94 – 76.35 + 24.39 – 485.45 – 162.52 + 180 kJ/min
= - 485.89 kJ/min
Problem Solving - Pressure
Problem Solving - Pressure
A manometer is used to measure the pressure in a tank. The fluid used has a specific gravity of 0.85 and the manometer column height is 55 cm. if the local atmospheric pressure is 96 kPa, determine the absolute pressure within the tank.
Pabs = Patm – Pgage
= 96 kN/m² + 0.85 ( 1000 kg/m³) ( 9.81 m/s²) ( 0.55 m)
1 kgm. m
N. s²
= 100. 59 kPa
* Note:
In thermodynamic relation and tables, absolute pressure is most always used.
e.g.
P will denote absolute pressure unless specified otherwise. Letter a for absolute pressure and g for gage pressure are added to pressure units.
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