Thermodynamics Advanced HomeWork 1

7/21/2019 Thermodynamics Advanced HomeWork 1

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Question No. 3: To control an isentropic steam turbine, a throttle valve is placed in steam lineleading to the turbine inlet. Steam at 6 MPa and 700oC is supplied to the throttle inlet, and theturbine exhaust pressure is set at 70 kPa. What is the effect of stream exergy at the turbine inletwhen the throttle valve is partially closed such that the pressure at the turbine inlet is 3 MPa?Compare the second law efficiency of this system when the valve is partially open to when it is fullyopen. Take To= 25oC



Question No. 4: In a heat ‐powered refrigerator, a turbine is used to drive the compressor using the same working fluid. Consider the combination shown in Figure 4 where the turbine produces just enough power to drive the compressor and the two exit flows are mixed

together. List any assumptions made and find the ratio of mass flowrates ⁄ and T5 (x5 if in two ‐phase region) if:

a) The turbine and the compressor are reversible and adiabatic.

b) The turbine and the compressor both have an isentropic efficiency of 70%.

3

Compressor

1

24

5

Turbine

T3=100oC

Saturated Vapor T1= ‐20

oC

Saturated Vapor

To condenser

P2 = P4 = P5 = 1.0MPa



10-27

Question No. 6

Refrigerant-22 is flowing in a pipeline at 10 °C, 600 kPa, with a velocity of 200m/s, at a steady flowrate of 0.1 kg/s. It is desired to decelerate the fluid and increase its pressure by installing a diffuser in the line (a diffuser is basically the

opposite of a nozzle in this respect). The R-22 exits the diffuser at 30 °C, with avelocity of 100 m/s. It may be assumed that the diffuser process is SSSF,

polytropic, and internally reversible. Determine the diffuser exit pressure and therate of irreversibility for the process.

C.V. Diffuser out to T 0, Int. Rev. flow ⇒ sgen R-22 = 0/

m• 1 = m•2 , h1 + (1/2) V 1

2 + q = h 2 + (1/2) V 22

s1 + ∫ dq/T + s gen = s 2 = s 1 + q/T 0 + s gen

Rev. Polytropic Process: w 12 = 0/ = - ∫ v dP + ( V 12 - V 2

2)/2

LHS = ∫ v dP = nn-1

(P 2v2 - P1v1) = 12(V 2

1 - V 22) = 1

2(200 2 - 100 2)/1000 = 15 kJ/kg

P1v1n = P 2v2

n⇒ n = ln(P 2/P1)/ln(v 1/v2)

Inlet state: v 1 = 0.04018, s 1 = 0.9295, h 1 = 255.1

Exit state: T 2,V2, ? ⇒ only P 2 unknown. ⇒ v2 = v(P 2), n = n(P 2)

Trial and error on P 2 to give LHS = 15:

P2 = 1 MPa ⇒ v2 = 0.0246, n = ln

1000

600 / ln

0.04018

0.0246 = 1.0412

LHS = (1.0412/0.0412)(1000 ×0.0246 - 600 ×0.04018) = 12.43

LHS too small so P 2 > 1.0 MPa

Assume sat. vapor @ 30 °C, P 2 = 1.1919 MPa

v2 = 0.01974, n = ln

1.1919

0.6 /ln

0.04018

0.01974 = 0.9657

LHS =0.9657-0.0343 (1191.9 ×0.01974 - 600 ×0.04018) = 16.33 > 15

P2 = 1.1 MPa ⇒ v2 = 0.021867, n = 0.9963, LHS = 14.596

Interpolate to get the final state:P2 = 1.121 MPa, v 2 = 0.021353, n = 0.9887, LHS = 15.01 OK

s2 = 0.8967 h 2 = 260.56

Find q from energy equation (Assume T 0 = T1 = 10°C)

q = h 2 - h1 + 12(V 22 - V1

2) = 260.56 - 255.1 - 15 = -9.54 kJ/kg

I• = m.

[T0(s2 - s1) - q ]= 0.1[283.15(0.8967 - 0.9295) + 9.54] = 0.0253 kW



Question No. 7: c onsider two rigid containers each of volume 1 m 3 containing air at 100 kPa, 400 K.An internally reversible Carnot heat pump is then thermally connected between themso it heats one up and cools the other down. In order to transfer heat at a reasonablerate, the temperature difference between the working substance inside the heat pumpand the air in the containers is set to 20 °C. The process stops when the air in thecoldest tank reaches 300 K. Find the final temperature of the air that is heated up, thework input to the heat pump, and the overall second-law efficiency.

A BH.P.

WH.P.

Q A

QB

⇑

⇐⇐The high and the low temperatures in the heat

pump are T A+20 and T B-20, respectively.

Since T A and T B change during the process,

the coefficient of performance changes, and so it must be integrated.

dU A = m AduA = m ACvdTA = dQ A dUB = m BduB = m BcvdTB = -dQ B

Carnot heat pump: dQ AdQ B

= TA + 20TB - 20 = -mACvdTA

mBCvdTB = -dT A

dTB

⇒ (dT A)/(T A + 20) = - (dT B)/(T B - 20)

⇒⌡⌠ dTA

TA + 20 = lnTA2 + 20

TA1 + 20 = -⌡⌠ dTB

TB - 20 = -lnTB2 - 20

TB1 - 20

TA1 = T B1 = 400 K, T B2 = 300 K ⇒

lnTA2 + 20

420 = -ln300 - 20

380 ⇒ T A2 = 550 K

mA = m B = P 1V1/RT 1 = (100 ×1)/(0.287 ×400) = 0.871 kg

WH.P. = Q A - QB = mC v(TA2 - T A1) + mC v(TB2 - T B1)

= 0.871 × 0.717 (550 - 400 + 300 - 400) = 31.2 kJ

Second law efficiency is total increase in air availability over the work input.

∆φ = mC v(T2 - T 1) + P 0(V2 - V 1) - T 0(s2 - s1)

∆φA = Q A - T 0m(s2 - s1)A ∆φB = -Q B - T 0m(s2 - s1)B

∆φA + ∆φB = W H.P. - mT 0[ ](s2 - s1)A + (s 2 - s1)B

To find entropies we need the pressures (P 2/P1) = T 2/T1 for both A and B

s2 - s1 = C pln(T 2/T1) - R ln(T 2/T1) = C vln(T 2/T1)

∆φtot = W H.P. - mT 0Cv[ ]ln(T 2/T1)A + ln(T 2/T1)B

= 31.2 - 0.871 ×298.15 ×0.7175 ×

ln550400 + ln

300400 = 25.47

ηII = ∆φtot/W H.P. = 25.47/31.2 = 0.816



Question No. 8: Chickens with an average mass of 1.6 kg and average specific heat of 3.54 kJ/kg ⋅oC are to be cooled by chilled water that enters a continuous ‐flow ‐type immersion chiller at 0.5 oC and leaves at 2.5 oC. Chickens are dropped into the chiller at a uniform temperature of 15 oC at a rate of 700 chickens per hour and are cooled to an average temperature of 3oC before they are taken out. The chiller gains heat from the surroundings at a rate of 400 kJ/h. Determine (a) The amount of heat transfer and (b) The exergy destroyed during this process



8-84 A rigid tank initially contains saturated R-134a vapor at a specified pressure. The tank is connected toa supply line, and R-134a is allowed to enter the tank. The amount of heat transfer with the surroundingsand the exergy destruction are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 Thedirection of heat transfer is from the tank (will be verified).

R-134a

R-134a0.2 m 3 1 MPa

Sat. vapor

1.4 MPa60 C

Properties The properties of refrigerant are (Tables A-11 through A-13)

kg/m0.020313=

K kJ/kg0.91558=

kJ/kg250.68=

sat.vapor

MPa1

3MPa1@1

MPa1@1

MPa1@11

g

g

g

s s

uu P

v v

K kJ/kg93889.0

kJ/kg47.285

C60

MPa4.1

i

i

i

i

s

h

T

P Q

Analysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed as

Mass balance : 12systemoutin mmmmmm i

Energy balance :

)0 peke (since 1122out

energiesetc. potential, kinetic,internal,inChange

system

massandwork,heat, bynsfer energy tra Net

outin

W umumQhm

E E E

ii

The initial and the final masses in the tank are

kg846.9kg/m020313.0

m2.03

3

11

v

V m

kg91.117983.593.111kg/m016715.0

m1.0

kg/m0008934.0

m1.03

3

3

3

2

g

g

f

f

g f mmm

v

V

v

V

kJ/K 967.5291303.0983.542441.093.111kJ581,1481.253983.570.11693.111

222

222

g g f f

g g f f

sm sm smS umumumU

Then from the mass and energy balances,kg06108.846.991.11712 mmmi

The heat transfer during this process is determined from the energy balance to bekJ18,73768.250.846914,58147.28506.1081122out umumhmQ ii

(b) The exergy destroyed during a process can be determined from an exergy balance or directly from itsdefinition . The entropy generation S gen0destroyed S T X gen in this case is determined from an entropy

balance on an extended system that includes the cylinder and its immediate surroundings so that the boundary temperature of the extended system is the surroundings temperature T surr at all times. It gives

0

out1122gen

tank 1122tank genout b,

out

entropyinChange

system

generationEntropy

gen

massandheat byansfer entropy tr Net

outin )(=

T Q

sm sm smS

sm smS S smT QS S S S

ii

ii

Substituting, the exergy destruction is determined to be

kJ1599]298/737,180.9388906.10891558.0846.9K)[52.967298(0

out11220gen0destroyed T

Q sm sm smT S T X ii

Question No. 10: A 0.2 m3 rigid tank initially contains saturated refrigerant-134a vapor at 1 MPa. The tank is connectedby a valve to a supply line that carries refrigerant-134a at 1.4 MPa and 60oC. The valve is now opened, and therefrigerant is allowed to enter the tank. The valve is closed when one-half of the volume of the tank is filled with liquidand the rest with vapor at 1.2 MPa. The refrigerant exchanges heat during this process with the surroundings at 25oC.Determine (a) the amount of heat transfer and (b) the exergy destruction associated with this process.

Thermodynamics Advanced HomeWork 1

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