MAE 91 Su13: HW 4 Solutions

MAE 91 Summer 2013 Problem Set 4: Solutions

Problem 1

A small turbine, shown in the �gure is operate at part load by throttling a 0.25kg/s steam supply

at 1.4MPa, 250◦C down to 1.1MPa before it enters the turbine and the exhaust is at 10kPa. If the

turbine produces 110kW, �nd the exhaust temperature (and quality if saturated).

I. Problem description

� Given: Water

m = 0.25kg/s

P1 = 1.4MPa, T1 = 250°C

P2 = 1.1MPa

P3 = 10kPa

Wt = 110kW

� Find

T3 =?, x3 =?

� Figure

II. Assumptions

Steady state process: d()dt = 0.

III. Analytical Solution

1. From conservation of mass: dmdt =

∑min −

∑mout

� C.V. around the valve:

0 = m1 − m2

� C.V. around the turbine:

0 = m2 − m3

The same mass �ow rate m is found at any station:

→ m1 = m2 = m3 = m

1


2. First law for an open system, or control volume, can be used to �nd missing properties:

dEc.v.dt

= Qc.v. − Wc.v. +∑

minein −∑

mouteout + Wflow in − Wflow out

where e = u+ ke+ pe , Wflow =∑mPv, and in general, Wc.v. = Wshaft

� For a C.V. around the valve: Neglecting ke+ pe and Wshaft = 0.

0 = mu1 − mu2 + mP1v1 − mP2v2

u1 + P1v1 = u2 + P2v2

h1 = h2

� For a C.V. around the turbine: Neglecting ke+ pe and Wshaft = Wt.

0 = mu2 − mu3 − Wt + mP2v2 − mP3v3

0 = m (u2 + P2v2 − u3 − P3v3)− Wt

Wt = m (h2 − h3)

h3 = h2 − Wt

m

� State 3: Found from h3 and P3.

IV. Numerical Solution

� State 1: P1 = 1.4MPa, T1 = 250°C with the use of Table B.1.3 → h1 = 2927.2kJ/kg

� State 2: h2 = h1 = 2927.2kJ/kg

h3 = h2 − Wt

m = 2927.2− 1100.25 = 2487.2kJ/kg

� State 3: from h3 = 2487.2kJ/kg and P3 = 10kPa, and Table B.1.2

→ T3 = 45.8°C (Saturated)

x3 =h3−hf@P

hfg@P= 2487.2−191.8

2392.8 = 0.959

2


Problem 2

The compressor of a large gas turbine receives air from the ambient at 95kPa, 20◦C, with a low velocity.

At the compressor discharge, air exits at 1.52MPa, 430◦C, with velocity of 90m/s. The power input

to the compressor is 5000kW. Determine the mass �ow rate of air through the unit.

I Problem description

� Known: Air

P1 = 95kPa, T1 = 20°C

ke1 ≈ 0

P2 = 1.52MPa, T2 = 430°C

V2 = 90m/s

Wc = 5000kW

� Find

m =?

� Using

Cpo = 1.004kJ/kgK

II. Assumptions

Steady state process, d()dt = 0, negligible potential energy, and kinetic energy is only important at the

exit of the compressor. First approximation using constant Cpo, considering that ∆T is not too large,

∆T < 500K.


1. From conservation of mass, the same mass �ow rate m is found at any station: dmdt = min− mout

� C.V. around the compressor:

0 = m1 − m2 → m1 = m2 = m

2. First law for an open system, or control volume, can be used to �nd the �ow rate:

dE

dt= Q− Wshaft + minein − mouteout + Wflow in − Wflow out

where e = u+ ke+ pe = u+ V ²

2 + gz and Wflow = mPv

� For a C.V. around the compressor: Neglecting ke1 + pe1 and pe2, and Wshaft = −Wc.

0 = Wc + m (u1)− m (u2 + ke2) + mP1v1 − mP2v2

0 = m (u1 + P1v1 − u2 − P2v2 − ke2) + Wc

Since ke = V 2

2

0 = m(h1 − h2 − V 2

2

2

)+ Wc

3


m = Wc

h2−h1+V 222

� For air, assume ideal gas, constant Cpo, then h2 − h1 = Cpo (T2 − T )

IV. Numerical solution

The mass �ow rate is given by,

m = Wc

Cpo(T2−T )+V 222

= 5000

1.004(430−20)+ 902

2 × 11000

= 12.0kg/s

Note the division 1/1000:

The units of [kJ/kg] = [kN−m/kg] = 1000 [N−m/kg] = 1000[kg m2

s2 /kg]

= 1000[m2

s2

]

4


Problem 3

The main waterline into a tall building has a pressure of 600kPa at 5m below ground level. A pump

brings the pressure up so the water can be delivered at 200kPa at the top �oor 150m above ground

level. Assume a �ow rate of 10kg/s liquid water at 10°C and neglect any di�erence in kinetic energy

and internal energy u. Find the power of the pump.


� Given:

Water m = 10kg/s

P1 = 600kPa

z1 = −5m

P2 = 200kPa

z2 = 150m

T = 10°C

∆ke = 0, ∆u = 0

� Find:

W =?

� Figure

II. Assumptions

Steady state process, negligible change of the kinetic energy and the internal energy. Inlet and outlet

temperature is the same.



C.V. around the pump and the pipe:

0 = m1 − m2 → m1 = m2 = m

2. First law for an open system, or control volume, can be used to �nd the power of the pump:

5


dEC.V.dt

= Q− Wshaft + minein − mouteout + Wflow in − Wflow out

where e = u+ ke+ pe and Wflow = mPv

� For a CV around the whole pipe, including the pump: Neglecting ke1 and ke2, Q = 0

and Wshaft = −Wp. Note that the sign convention for the pump was already chosen and

negative means it goes into the system.

0 = Wp + m(u1 +

V 21

2 + gz1

)− m

(u2 +

V 22

2 + gz2

)+ mP1v1 − mP2v2

0 = m(u1 +

V 21

2 + P1v1 + gz1 − u2 − V 22

2 − P2v2 − gz2)

+ Wp

Since ∆ke =V 22

2 −V 21

2 = 0, and ∆u = u2 − u1 = 0

0 = m (P1v1 + gz1 − P2v2 − gz2) + Wp

Since the temperature is assumed to be the same, v1 = v2 = v

0 = m [v (P1 − P2) + g (z1 − z2)] + Wp

Then,

Wp = m [v (P2 − P1) + g (z2 − z1)]


Wp = m [v (P2 − P1) + g (z2 − z1)] = 10{

0.001 (200− 600) + 9.8 [100− (−5)] 11000

}= 6.29kW

6


Problem 4

A modern jet engine has a temperature after combustion of about 1500K at 3200kPa as it enters the

turbine section, see state 3 Figure shown. The compressor inlet is 80kPa, 260K state 1 and outlet state

2 is 3300kPa, 780K; the turbine outlet state 4 into the nozzle is 400kPa, 900K and nozzle exit (state 5)

at 80kPa, 640K. Neglect any heat transfer and neglect kinetic energy except out of the nozzle. Find

the compressor and turbine speci�c work terms and the nozzle exit velocity.


� Given: Air

P1 = 80kPa, T1 = 260K

P2 = 3.3MPa, T2 = 780K

P3 = 3.2MPa, T3 = 1500K

P4 = 400kPa, T4 = 900K

P5 = 80kPa, T5 = 640K

� Find:

wt =?, wc =?

V5 =?

II. Assumption

No heat transfer on the compressor or turbine. Kinetic energy is only important at the exit of the

nozzle. Potential energy is negligible. ∆T is large, constant Cp is not recommendable.




0 = m1 − m2

� C.V. around the combustor:

0 = m2 − m3

7


� C.V. around the turbine:

0 = m3 − m4

� C.V. around the nozzle:

0 = m4 − m5 → m1 = m2 = m3 = m4 = m5 = m

2. First law for an open system, or control volume, can be used to �nd the power of the devices:

dE



dE

dt= Q− Wshaft + min (h+ ke+ pe)in − mout (h+ ke+ pe)out

� C.V. around the compressor: Neglecting ke1 and ke2, Q = 0 and Wshaft = −Wc.

0 = Wc + m(h1 +

V 21

2

)− m

(h2 +

V 22

2

)0 = Wc + m (h1 − h2)

wc = Wc

m = h2 − h1

� C.V. around the combustor: Wshaft = 0, and neglecting ke2 and ke3.

0 = Qcomb + m(h2 +

V 22

2

)− m

(h3 +

V 23

2

)Qcomb = m (h3 − h2)

� C.V. around the turbine: Neglecting ke3 and ke4, Q = 0 and Wshaft = Wt.

0 = −Wt + m(h3 +

V 23

2

)− m

(h4 +

V 24

2

)0 = −Wt + m (h3 − h4)

wt = Wt

m = h3 − h4

� C.V. around the nozzle: Neglecting ke4, Q = 0 and Wshaft = 0.

0 = m(h4 +

V 24

2

)− m

(h5 +

V 25

2

)0 = m

(h4 − h5 − V 2

5

2

)V 25

2 = h4 − h5V5 =

√2 (h4 − h5)


States are obtained from from Table A.7.1, h = h (T )

At T1 = 260K, h1 = 260.32kJ/kg

8


At T2 = 780K, h2 = 800.28kJ/kg

At T3 = 1500K, h3 = 1635.80kJ/kg

At T4 = 900K, h4 = 933.15kJ/kg

At T5 = 640K, h5 = 649.53kJ/kg

wc = Wc

m = h2 − h1 = 800.28− 260.32 = 539.36kJ/kg

wt = Wt

m = h3 − h4 = 1635.8− 933.15 = 702.65kJ/kg

V5 =√

2 (h5 − h4) =√

2 (933.15− 649.53)× 1000 = 753m/s

9


Problem 5

A heat exchanger, shown in Figure, is used to cool an air �ow from 800K to 360K, both states at

1MPa. The coolant is a water �ow at 15°C, 0.1MPa. If the water leaves as saturated vapor, �nd the

ratio of the �ow rates mH2O/mair.


� Given:

-Air:

P1 = 1MPa, T1 = 800K

P2 = 1MPa, T2 = 360K

-Water:

P3 = 100kPa, T3 = 15°C

P4 = 100kPa, x4 = 1

� Find:

mH2O/mair

1. From conservation of mass dmdt = min − mout

� C.V. for air:

0 = m1 − m2→ m1 = m2 = mair

� C.V. for water:

0 = m3 − m4 → m3 = m4 = mwater

1. First law for an open system, or control volume, can be used to �nd the power of the devices:

dEc.v.dt

= Qc.v. − Wshaft + minein − mouteout + Wflow in − Wflow out


dEc.v.dt

= Qc.v. − Wshaft + min (h+ ke+ pe)in − mout (h+ ke+ pe)out

10


� C.V. around the heat exchanger (Water+Air): Neglecting ke and pe, the whole device is

rigid Wshaft = 0 and does not exchange heat with the exterior Qc.v. = 0, only between

water and air.

0 = m1 (h1)− m2 (h2) + m3 (h3)− m4 (h4)

0 = mair (h1 − h2) + mwater (h3 − h4)

mwater (h4 − h3) = mair (h1 − h2)

mwater

mair= h1−h2

h4−h3


� States:

� Air: obtained from from Table A.7.1 for air, h = h (T )

At T1 = 800K, h1 = 822.29kJ/kg

At T2 = 360K, h2 = 360.86kJ/kg

� Water: obtained from from Table B.1

Compressed liquid at T3 = 15°C, Table B.1.1 h3 = hf@T3= 62.98kJ/kg

Saturated vapor at P4 = 100kPa, Table B.1.2 h4 = hg@P4= 2675.5kJ/kg

� Then,

mwater

mair=h1 − h2h4 − h3

=822.29− 360.86

2675.5− 62.98= 0.1766

11


Problem 6

A R-410a heat pump cycle shown in Figure has a R − 410a �ow rate of 0.05kg/s with 5kW into the

compressor. The following data are given,

Calculate the heat transfer from the compressor, the heat transfer from theR−410a in the condenser

and the heat transfer to the R− 410a in the evaporator.


� Given:

R-410a

m = 0.05kg/s

Table

Wcomp = 5kW

� Find:

Qcomp, Qcond, Qevap

� Sketch

II. Analytical Solution



0 = m6 − m1

12


� C.V. around the condenser:

0 = m2 − m3

� C.V. around the valve:

0 = m3 − m4

� C.V. around the evaporator:

0 = m4 − m5 → m1 = m2 = m3 = m4 = m5 = m

2. First law for an open system, or control volume, can be used to �nd the rate of heat transfer:

dE



dE

dt= Q− Wshaft + min (h+ ke+ pe)in − mout (h+ ke+ pe)out

� C.V. around the compressor: Neglecting ke and pe, and Wshaft = −Wcomp.

0 = Qcomp + Wcomp + m (h6)− m (h1)

Qcomp = m (h1 − h6)− Wcomp

� C.V. around the condenser: Wshaft = 0, and neglecting ke, pe.

0 = Qcond + m (h2)− m (h3)

Qcond = m (h3 − h2)

� C.V. around the valve: Neglecting ke and pe, Qvalve = 0 and Wshaft = 0.

0 = m (h3)− m (h4)

h4 = h3

� C.V. around the evaporator: Neglecting ke and pe, and Wshaft = 0.

0 = Qevap + m (h4)− m (h5)

Qevap = m (h5 − h4)

III. Numerical Solution

Qcomp = m (h1 − h6)− Wcomp = 0.05 (377− 367)− 5.0 = −0.35kW

Qcond = m (h3 − h2) = 0.05 (134− 367) = −11.65kW

Qevap = m (h5 − h4) = 0.05 (280− 134) = 7.3kW

13


Problem 7

A 25-L tank, shown in Figure bellow, that is initially evacuated is connected by a valve to an air

supply line �owing air at 20◦C, 800kPa. The valve is opened, and air �ows into the tank until the

pressure reaches 600kPa. Determine the �nal temperature and mass inside the tank, assuming the

process is adiabatic. Develop an expression for the relation between the line temperature and the �nal

temperature using constant speci�c heats.


� Given: Air

� VTank = 0.025m3

� TLine = 20°C, Pline = 800kPa

� Pfinal = 600kPa

� Find:

� Tfinal =?

� Sketch

II. Assumption

Constant speci�c heat. Tank is adiabatic. Tank is initially empty. Air can be treated as an ideal gas.


1. From conservation of mass, the amount of mass that �lled up the tank can be found: dmdt =

min − mout

� C.V. around the Tank:

dmdt = min

� Integrating, from the initial time, to, to the �nal time, tf ,

14


´ tfto

dmdt dt =

´ tftomindt → mf Tank −mo Tank =

´ tfto

dmin

dt dt → mf −mo =´mf in

mo indmin

mf −mo = min

� Since the tank is initially empty, mo = 0, so

mf = min

1. First law for an open system, or control volume, can be used to �nd the �nal state:

dE

dt= Q− Wc.v. + min (h+ ke+ pe)in − mout (h+ ke+ pe)out

� C.V. around the tank: Neglecting ke and pe, Wc.v. = 0 (rigid boundary) and Q = 0

(adiabatic wall). Consider that E = U +KE + PE, so in this case, dEdt = dUdt .

dUdt = minhin − mouthout

� Considering mout = 0, the energy balance reduces to,

dUdt = minhin

� Integrating from the initial time, to, to the �nal time, tf ,´ tfto

dUdt dt =

´ tftominhindt

� Since the line is always at the same properties, hin is constant, and we already know

that´ tftomindt = min

Uf − Uo = hin´ tftomindt = hinmin → Uo = mouo = 0

mfuf = hinmin, since mf = min, then

uf = hline

2. Mass is obtained from ideal gas law PV = mRT ,

mfinal =PfVTank

RairTf

3. Using constant speci�c heats ∆u = Cvo∆T and ideal gas law Pv = RT :

uf = hline → uf = uline + Plinevline = uline +RairTline

uf − uline = RairTline → Cvo (Tf − Tline) = RairTline

CvoTf = CvoTline +RairTline → Tf = (Cvo+Rair)Cvo

Tline

Since Cpo − Cvo = R and Cpo/Cvo = k

Tf =Cpo

CvoTline → Tf = kTline

15



Since, uf = hline the �nal state can be obtained by table A.7. So, uf = hline = 293.64kJ/kg.

Given that uf = 293.64kJ/kg, then Tf = 410.0K

Then, the �nal mass in the tank is,

mfinal =PfVTank

RairTf= 600×0.025

0.287×410 = 0.1275kg

From constant speci�c heat, kair = 1.40, then

Tf = kTline = 1.40× 293.2 = 410.5K

16


Problem 8

A mass-loaded piston/cylinder, shown in Figure, containing air is at 300kPa, 17◦C with a volume of

0.25m3, while at the stops V = 1m3. An air line, 500kPa, 600K, is connected by a valve that is then

opened until a �nal inside pressure of 400kPa is reached, at which point T = 350K. Find the air mass

that enters, the work, and heat transfer.


� Given: Air

P1 = 300kPa, T1 = 17°C, V1 = 0.25m3

P2 = 400kPa, T2 = 350K

Pline = 500kPa, Tline = 600K

Vstops = 1m3

� Find:

min =? , 1W2 =?, 1Q2 =?

� Sketch

II. Assumption

Air can be treated as an ideal gas. Enthalpy is a function of temperature only. Potential and kinetic

energy of air is neglected.


1. Conservation of mass is used to balance the initial mass with the exchanged mass: dmdt = min −

mout

� C.V. around the air contained in the piston:

dmdt = min

� Integrating, from the initial time, to, to the �nal time, tf ,

17


´ tfto

dmdt dt =

´ tftomindt → mf cylinder −mo cylinder =

´ tfto

dmin

dt dt → mf −mo =´mf in

mo indmin

mf −mo = min

� The initial mass is found from ideal gas: PV = mRairT

mo = P1V1

RairT1

� The volume inside the cylinder expands at constant pressure, P1, until the piston reaches

the stops. Then is when the pressure increases. Since P2 > P1, the stop is reached and

V2 = Vstops. The �nal mass is,

mf = P2V2

RairT2

� Then,

min = P2V2

RairT2− P1V1

RairT1

2. Work for a constant pressure process 1→ 1a and constant volume 1a→ 2 is,

1W2 =´ 21δW =

´ 1a1δW +

´ 21aδW =

´ 1a1δW =

´ VStops

V1PdV = P1 (V2 − V1)

3. First law for an open system, or control volume, can be used to �nd the heat transfer:

dE

dt= Q− Wc.v. + min (h+ ke+ pe)in − mout (h+ ke+ pe)out

� C.V. around the air: Neglecting ke and pe. Consider that E = U + KE + PE, so in this

case, dEdt = dUdt .

dUdt = Q− Wc.v. + minhin − mouthout

� Considering mout = 0, the energy balance reduces to,dUdt = Q− Wc.v. + mlinehin

� Integrating from the initial time, to, to the �nal time, tf ,´ tfto

dUdt dt =

´ tfto

δQδt dt−

´ tfto

δWδt dt+

´ tftominhindt

� Since the line is always at the same properties, hin is constant and´ tftominhindt = hinmin.

Also,´ tfto

δQδt dt =

´ 21δQ =1 Q2, and

´ tfto

δWδt dt =

´ 21δW =1 W2.

Uf − Uo =1 Q2 − 1W2 + hinmin

� Solving for 1Q2:

1Q2 = Uf − Uo + 1W2 − hinmin

� Since, Uf = mfuf = m2u2, Uo = m1u1, and hin = hline then

1Q2 = m2u2 −m1u1 + 1W2 −minhline

� Enthalpy and internal energy is obtained from table A.7.1 at the given temperature and

pressure.

18



Computing the initial and �nal mass,

m1 = P1V1

RairT1= 300×0.25

0.287×290.2 = 0.90kg

m2 = P2V2

RairT2= 400×1

0.287×350 = 3.982kg

The mass that enters is,

min = m2 −m1 = 3.0982− 0.9 = 3.082kg

The work done by the air is,

1W2 = P1 (V2 − V1) = 300 (1− 0.25) = 225kJ

States are:

T1 = 290.2K → u1 = 207.2kJ/kg

T2 = 350K→ u2 = 250.3kJ/kg

Tline = 600K→ hline = 607.3kJ/kg

The heat transfer going into the system is,

1Q2 = m2u2 −m1u1 + 1W2 −minhline

1Q2 = 3.982× 250.3− 0.9× 207.2 + 225− 3.082× 607.3 = −836.5kJ/kg

Note: the heat goes out of the system.

19

MAE 91 Su13: HW 4 Solutions

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