ME 300 -D

Midterm –I Review

Problem # 1.22

A spherical balloon holding 35 lbm of air has a diameter of 10 ft. For the air, determine the specific volume in ft3/lbm and ft3/lbmol, and b)the weight in lbf. Let g = 31.0 ft/s2.

m = 35lbm; volume of sphere = 4/3 R3 = 4/3 (3.1417)(5)3 = 523.6 ft3

v = specific volume = 523.6 / 35 = 14.96 ft3/lbmM air = 28.97 lbm / lbmol. Hence molar specific volume = M v = 28.97 x 14.96 = 433.4 ft3/lbmolWeight = m g = 35 x 31 lbm. ft/s2 = 1085 lbm ft/s2

= 1085 / 32.2 lbf = 33.696 lbf.

Problem 1.27

A closed system consisting of 5 kg. of gas undergoes a process during which the relationship between pressure and specific volume is p v 1.3 = constant. The process begins with p1 = 1 bar, and v1 = 0.2 m3 / kg. and ends with p2 = 0.25 bar. Determine the final volume in m3, and plot the process on a graph with pressure and volume as axes.m = 5 kg. p1 v1

1.3 = p2 v21.3 ; (1) (0.2)1.3 = (0.25) (v2) 1.3

(v2) 1.3 = (1/0.25)(0.2) 1.3; v2 = (4)1/1.3 (0.2) = 2.9039 (0.2)

= 0.5808 m3 /kgV = m v = (5)(0.5808) = 2.9039m3

Problem 1.31A gas contained within a piston-cylinder assembly undergoes three processes in series.Process 1-2: compression with pV = constant from p1 = 1 bar, V1 = 1.0m3 to V2 = 0.2m3

Process 2-3: Constant pressure expansion to V3 = 1.0m3

Process 3-1 Constant VolumeSketch the processes on a p-V diagram and label the pressures and volumes at each point.

p2 = p1 V1 / V2 = (1) (1) / 0.2 = 5 barsp3 = p2 Back to state 1.

Problem 1.57

What is the maximum increase and maximum decrease in body tmperature each in oC from a normal body temperature of 37 oC that humans can experience before serious medical complications result?Maximum body temperature =104 oF =(104-32)/1.8 oC = 40 oCMinimum temperature = 96oF = (96–32) /1.8=35.55 oC.Maximum increase from 37 oC = 3 oCMaximum decrease from 37 oC = 1.45 oC

Problem 2.8An object whose mass is 1 lb has a velocity of 100ft/s. Determine the final velocity in ft/s. if the kinetic energy of the object decreases by 100 ft. lbf. Determine the change in elevation in ft. associated with a 100 ft. lbf change in potential energy. Let g = 32.0 ft/s2.m = 1 lb. V1 = 100 ft/s. (KE)1 = ½ m V1

2= ½ (1)(100)(100) = 5000 lbm ft2/s2.1 ft. lbf = (1) (ft) (32.2) lbm ft/s2

KE 2 = 5000 – 32.2 x 100 = 1780 lbm. ft2/s2

V2 2 = 2 x (1780) / (1); V2 = 59.66 ft/sm g z = 32.2 x 100; z = 32.2 x 100 / (32.0 x 1) = 100.625 ft.

Problem 2.19

An object of mass 80 lb. initially at rest experiences a constant acceleration of 12 ft/s2 due to the action of a resultant force applied for 6.5 s. Determine the work of the resultant force in ft. lbf and in Btu.

m = 80 lb. Acceleration = 12.0 ft/s2. time of application = 6.5sTerminal velocity V2 = 12.0 x 6.5 = 78 ft/sHence, Work = change in KE = ½ (80) (78.0)2 lbm ft2/s2

= (40) (78)(78) / 32.2 ft. lbf = 7557.76 ft.lbf = 7557.76 / 778 Btu = 9.714 Btu.

Problem 2.33

Carbon monoxide contained within a piston-cylinder assembly undergoes three processes in series:Process 1-2: Expansion from p1 = 5 bar, V1 = 0.2 m3, to V2 = 1 m3, during which the pressure volume relationship is pV = constant.Process 2-3: constant-volume heating from state 2 to state 3, where p3 = 5 bar.

Process 3-1: constant pressure compression to the initial state. Sketch the process in series on p-V coordinates and evaluate work for each process in kJ.

Problem 2.33

Process 1-2: p V = constant. p2 = p1 V1 / V2

= (5 bars) (0.2m3)/(1m3) = 1 bar = 100 kPaSince pV = constant, Work = p1 V1 ln (V2/V1)

Work = (500kPa) (0.2m3) ln (1/0.2) = 100 ln 5 = 160.94 kJProcess 2-3: Constant volume heating. dV = 0. Hence work = 0.p3 = 5 bars

Process 3-1: Constant pressure compression. Work = p3 (V1-V3)

V3 = V2 = 1 m3; V1 = 0.2 m3; Work = -(500 kPa) (1 – 0.2) = -400 kJ

Problem 2.37

A 10-V battery supplies a constant current of 0.5 amp to a resistance for 30 min. a) determine the resistance, in ohms. b) for the battery, determine the amount of energy transfer by work in kJ.

E = 10 V; i = 0.5 amp. R = E / i = 10/0.5 = 20 ohms.Work = Rate of work done x time = E i x time = 10 x 0.5 x 30 x 60 = 9000 Watts .s = 9kJ

Fig08_02

Fig08_E8

Fig10_03

Thermodynamic Cycle

The energy balance for a thermodynamic cycle states:

Ecycl = Qcycl – W cycl

Since the energy change for a cycle is zero, we can write this as:

Qcycl = W cycl

Fig02_17

Problem 2.78

For a power cycle, the heat transfers are Q in = 50 kJ and Q out = 35 kJ. Determine the net work and thermal efficiency.

W cycl = Q in – Q out = 50 – 35 = 15 kJ.

Efficiency = W cycl / Q in = 15/50 = 30 %

Problem 2.81

For a power cycle W cycl = 800 Btu, and Q out = 1800 Btu. What is the thermal efficiency?Work = 800 Btu. Heat rejected = 1800 Btu.Wcycl = Q in – Q out

Q in = Q out + W cycl = 1800 + 800 = 2600 Btu;Thermal efficiency = Wcycl / Q in = 800 / 2600 = 0.3077 or 30.77%

Problem 2.88

A refrigeration cycle operates with a coefficient of performance of 1.5. For the cycle, Qout = 500kJ. Determine Qin and Wcycl, each in kJ

= 1.5 = Qin / Wcycl = (Qout – Wcycl)/Wcycl

= 1.5 = Qout/Wcycl – 1

Qout/Wcycl= 2.5; Wcycl = 500/2.5 = 200 kJ

Qin = Qout – Wcycl = 500 – 200 = 300kJ

Problem 2.92A heat pump delivers energy by heat transfer to a dwelling at a rate of 60,000 Btu/h. The power input to the cycle is 7.8 hp. Determine the COP of the cycle; At $0.08 per kW.h determine the cost of electricity in a month operating for 200 hours.Q out = 60,000 Btu/h; W = 7.8 hp = 2545 x 7.8 Btu/h

COP = Q out /W = 60000/(2545 x7.8) = 3.022

1 hp = 0.7457 kW; kWh/month = 7.8 x 0.7457 x 200 = 1163.292 kW; Cost = 1163.292 x 0.08 = $93.06

Problem 2.64

Two kilograms of air is contained in a rigid well-insulated tank with a volume of 0.6 m3. The tank is fitted with a paddle wheel that transfers energy to the air at a constant rate of 10 W for 1 hour. If no changes in kinetic and potential energy occur, determinea)Specific volume of the final state in m3 /kg;b)The energy transfer by work, in kJc)The changes in internal energy of the air, in kJ/kg

m = 2 kg. V = 0.6 m3. The volume and mass remain constant throughout the process; v = 0.6 / 2 = 0.3 m3 /kg

Problem 2.64 contd.

Work done = - 10 J/s x 3600 s = - 36 kJ

b) Energy transfer by work = E = Q – WQ = 0 because it is well-insulated. E = 36 kJ.

c) Since no changes in KE or PE occur, all of the change in energy is in internal energy. Increase in specific internal energy is: 36kJ/2 = 18.0 kJ/kg

Ideal Gas Model

The definitions of internal energy and enthalpy simplify for an ideal gas model. For ideal gases, it is observed empirically that the internal energy depends only on temperature.p v = RTh = u + pv = u + RT = u(T) + RT = h(T) onlyHence enthalpy is also a function of temperature only. To verify if a gas can be considered as an ideal gas, compute Z and ensure that it is near unity.

Problem 3-105A tank contains 10 lbm of air at 70oF with a pressure of 30 lbf/in2. Determine the volume of air in ft3. Verify that ideal gas behavior can be assumed for these conditions.First let us compute Z, the compressibility factor. For air, pcrit = 37.2 atm; Tcrit = 239 R;p/pcrit = (30 lbf/in2)/(37.2 x 14.696 lbf/in2) = 0.0548T/Tcrit = (460 + 70) / 239 = 2.2176 Z from the figure is: 0.99; Hence ideal gas behavior can be assumed.V = m R T / p = 10 x (1545/28.97) x (530)/(30x144) = 65.42 ft3

lbm x (ft. lbf/lbmol oR) (lbmol/lbm) (oR) (ft2/lbf) = ft3

Fig03_12

Problem 3.116A piston-cylinder assembly contains air at a pressure of 30 lbf/in2. and a volume of 0.75 ft3. The air is heated at constant pressure until its volume is doubled. Assuming the ideal gas behavior with constant specific heat ratio = k = 1.4, determine the work and heat transfer.cp = cv + R; cp/ cv = k; cv k = cv + R; hence cv = R/(k-1) cv/R = 1/(k-1) = 1/0.4 = 2.5pV = mRT; Work = p(V2-V1) = 30x144 x (1.5 – 0.75) = 3240 ft. lbf Change in internal energy = cv (T2 – T1) = cv (pV2/R – pV1/R) = (cv)(p)(V2-V1)/R = (30 x 144) (0.75) (1/0.4) = 8100ft.lbfQ = change in internal energy + W = 8100 + 3240 = 11,340 ft. lbf.

Properties of Liquid Vapor Mixture

In the vapor dome, the pressure and temperature stay constant during the evaporation. The quantity that changes is the quality of the mixture. Going from left to right, the quality increases; going from right to left, the quality decreases. The pressure and temperature are uniquely related at the saturation values. The other properties can be obtained via the relations:

)()(

)()(

)()(

fgffg

fgffg

fgffg

hhxhhxhxh

uuxuuxuxu

vvxvvxvxv

1

1

1

Properties of Liquids and Solids

Liquids and solids are essentially incompressible, i.e. the effect of pressure is very small. Hence we can write the properties specific volume and internal energy purely as a function of fluid temperature. However, the enthalpy has a pressure dependence, as follows:

)()()(

)()(),(

)(),(

)(),(

TvppTh

TpvTupTh

TupTu

TvpTv

lsatl

ll

l

l

Incompressible Substance Model

For an incompressible substance, the specific heats are functions of only temperature. Following relations can be derived for incompressible fluids.

vp

vp

vv

ccsolidsandliquidsfor

TcdT

duc

pvTupTh

TcdT

duc

);(

)(),(

)(

Internal energy and enthalpy

The changes in internal energy and enthalpy for an incompressible substance can be written as:

For constant specific heats,

)()(

)(

)(

12

121212

12

2

1

2

1

ppvdTTc

ppvuuhh

dTTcuu

T

T

T

T

)()(

)(

121212

1212

ppvTTchh

TTcuu

Example of Interpolation

Given following table of values

p = 1.0MPa p = 1.5 Mpa

ToC v (m3/kg) T(oC) v (m3/kg)

200 0.2060 200 0.1325240 0.2275 240 0.1483280 0.2480 280 0.1627

Interpolation

v at 240 oC, 1.25 MPa;Take the middle row: v at 240 oC = 0.2275 + (0.1483 – 0.2275)/(1.5-1.0) x

(1.25 – 1.0) = 0.18915 m3/kg T for p = 1.5 MPa, v = 0.1555 m3/kgTake right columns, bracket the specific volume

between 240 and 280 oC;T = 240 + (0.1555 – 0.1483) x (280-240)/(0.1627 –

0.1483) = 240 + 20 = 260 oC

Problem 3.13

Determine the mass in kg of 0.1 m3 of Refrigerant 134 a, at 4 bar, 100oC.

p = 4 bar, Tsat = 8.93 oC;

at 100 oC, v = 0.07327 m3 /kg; m = V / v = 0.1 / 0.07327 = 1.3648 kg.

Problem 3.58

A closed, rigid tank contains 2kg of water initially at 80oC and a quality of 0.6. Heat transfer occurs until the tank contains only saturated vapor at a higher pressure. KE and PE effects are negligible. For the water as a system, determine the amount of energy transfer by heat, in kJ.u1 = uf + x(ug – uf) = 334.86 + (0.60) (2482.2 – 334.86) = 1623.26 kJ/kg v1 = vf + x(vg – vf) = 1.029 x 10-3 + (0.6) (3.407 – 1.0291 x 10-3) = 2.0446 m3/kg.v2 = v1, interpolation gives p2 = 0.82 bars, u2 = 2499.54 kJ/kg.

Q = m (u2 – u1) = 2 (2499.54 – 1623.26) = 1752.56 kJ

Fig03_03

1

2

Problem 3.63Saturated water vapor is contained in a closed rigid tank. The water is cooled to a temperature of 110oC. At this temperature the masses of saturated vapor and liquid are each 1000kg. Determine the heat transfer for the process, in kJ.At state 2, the quality is 0.5; at 110oC, vf=1.0516 x10-3m3/kg, vg = 1.210m3/kg; v = 1.0516x10-3 + 0.5 (1.210-1.0516x10-3) = 0.6055 m3/kg.uf = 461.14 kJ/kg; ug = 2518.1 kJ/kg; u = 461.14 + 0.5 x (2518.1-461.14) = 1489.62 kJ/kg.At state 1, v = 0.6055 m3/kg, saturated vapor; p = 3.00 bars, T = 133.6 oC. u = 2543.6 kJ/kg.Heat transfer = 2000 x (2543.6 – 1489.62) = 2107.96 MJ