thermo exams and answers

8/19/2019 Thermo Exams and Answers

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Mansoura University Engineering Faculty Mechanical Eng. Department

1st year Thermodynamics 2013 Time 3 hours

Thermodynamic tables are allowed

1.

A rigid tank of volume 0.2 m3 initially contains 0.652 kg of H20 at 2 bar. A heat

reservoir at 500oC is used to heat the tank until the pressure inside it reaches 10 bar.

Find initial and final temperatures inside the tank, heat and work exchanged by H 2O,

entropy change of the universe as well as availability change due to this process

(ambient temperature 27oC). Is the process reversible?

2.

In a jet engine (نفاث

محرك

) enters 0.3 kg/s of fuel (وقود

) as well as 3.6 kg/s of air. Aftercombustion (قارتحالا) exhaust gases (مداعلا تازاغ) are at 6 bar and 517oC. Their properties

may be considered as those of air, semi-ideal. Gases expand in a nozzle in a reversible

adiabatic process to exit from it at 1 bar and very high speed to produce a thrust (عفد ةوق).

Find exit temperature and exit velocity. Find exit density as well as exit cross-sectional

area. What is the entropy change during expansion?

3. Ocean (طیحملا) temperature at its surface is 27oC. Temperature decreases as we go deep

inside the ocean (المحیط

في عماق

), because solar rays are absorbed (الشمس

امتصاص شعة

) to

reach 17oC. It is possible to use this temperature difference to produce electricity, by

placing a thermocouple between those temperatures. Two models exist in the market;

each of them absorbs 30 W of heat at the hot temperature. Catalog of model 1 states it

delivers 0.72 W of electricity, while catalog of model 2 states 1 W of electricity. Which

one would you recommend?

4. In an insulated mixer, 0.2 kg/s of CO2 enter at 1.1 bar and 50oC as well as 0.3 kg/s of N2

at 1.05 bar at 40oC. Mixture exits at 1 bar. Find exit temperature, partial pressure of each

gas at exit, heat and work exchanged, and entropy production due to this process. If

ambient temperature was 27oC, what is the availability loss due to this process?

5.

In order to measure the dryness fraction of steam flowing in a pipe, part of this steam is

extracted to flow through a throttle valve (called throttling calorimeter) before exiting to

atmosphere. If steam pressure inside the pipe was 20 bar, while conditions at exit of the

throttle valve were 1 bar and 120oC, what is the dryness fraction of steam in the pipe?

What is the entropy production due to throttling? If outside air temperature was 27oC,

what is the availability loss due to throttling?

Figure problem 2 Figure problem 5


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Mansoura University Engineering Faculty Mechanical Eng. Department

1st year Thermodynamics 2014 Time 3 hours

Thermodynamic tables are allowed

Question No. (1) (25 Marks)

A rigid tank of volume 17.15 liter contains 0.5 kg of steam initially at 200oC. Heat from a

reservoir at 500oC is added to steam until its pressure reaches 80 bar. Find final steam

temperature, heat and work exchanged by steam, steam entropy change, entropy

production of

Universe

and

availability

loss.


In an insulated air compressor, air is admitted at ambient conditions (1 bar, 27oC). If the mass

flow rate was 2kg/s, final pressure was 75.9 bar and isentropic efficiency was 75%, find

exit temperature as well as heat and work exchanged by air. Find also total entropy

production as well as Second law efficiency. Assume air is an ideal gas.


It is required to heat a room by adding 2kW at 29oC, while outside temperature is 5oC. Heat

will be provided for 1000 hours per year over 2 years. Two options are available. The first

option is to buy an electric heater of price 150 EGP. The second option is to buy an air

conditioning system of price 2 000 EGP. The system can be considered as a heat pump

having a COP that is 40% that of a Carnot engine working at the same temperatures.

Knowing that the price of a kWh is 0.6 EGP, calculate the sum of initial and running costs

التشغیل)

تكلفة

زا د

الشرا

?) of both options over the two years. Which is cheaperتكلفة


A rigid insulated tank is internally separated into two rooms by a membrane. Room A has a

volume of 0.2 m3, containing 0.3 kg of N2 initially at 1.5 bar. Room B has a volume of 0.6

m3, containing 0.1 kg of H2 initially at 2.5 bar. Membrane ruptures, gases mix. Find

mixture pressure and temperature, partial pressures of each gas in the mixture as well as

entropy production due to this process.


In a vertical frictionless piston and cylinder arrangement, 0.5 kg of H2O initially at

120oC occupies 0.3m3. It loses heat to atmosphere at 27oC until it becomes

saturated liquid. Find the work and heat exchanged as well as entropy changes of

H2O. Find also the entropy production for the universe during this process.

Best Wishes – Prof. Mohamed‐Nabil Sabry


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Test

1)

Given : H2O, m = 0.652 kg, V=0.2 m3, Trecevoir = 500 ͦ

C, Tambient = 27 ͦ C

State 1: P1 = 2 bar, V1 = 0.2 m3

State 2: P2 = 10 bar

Process: Isochoric (constant volume)

Required: T1, T2, W, Q, ΔSUniverse, availability change

Solution

State 1:

= =

= .

. v1 = 0.30675 m3 /kg

Enter table (2) by P1 = 2 bar:

vf = 0.00106049 m3 /kg, vg = 0.8858 m3 /kg

∵ < < Point 1 is in the wet zonea)

Then,T1 = Tsat = 120.211546 ͦ C

Then, from table: uf = 504.47 kJ/kg, ug = 2529.21

kJ/kg, sf = 1.53 kJ/kg.K, sg = 7.127 kJ/kg.K

= + ( − ) 0.30675=0.00106049 + x (0.8858 – 0.00106049)

X = 0.3455

= + ( − ) u1=504.47 + 0.3455 * (2529.21 - 504.47)

u1 = 1204 kJ/kg

= + ( − )


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vf = 0.001127 m3 /kg, vg = 0.1944 m3 /kg

∵ > Point 2 is in the super-heated zoneEnter table 3 by P2 and v2

b) T2 = 400 ͦ C

u2 = 2957.79 kJ/kg, s2 = 7.46597 kJ/kg. K

c)

W = Zero Constant volume process.

d)

Q = m (u2 – u1) = 0.652 * (2957.79 – 1204)

Q = 1143.47 kJ

e) ΔSSystem = m * (s2 – s1)

= 0.652 (7.46597 – 3.4638)

ΔSSystem = 2.6094 kJ/K

ΔSheat recevoir = -Q / Trecevoir = - 1143.47 / 773

ΔS heat recevoir = -1.479 kJ/K

ΔSuniverse = ΔSsystem + ΔSheat recevoir

= 2.6094 + (-1.479) ΔSuniverse = 1.1304 kJ/K


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f) Exergy change for closed system (ΔX)

ΔX = m*((u2 – u1) + P0 (v2 – v1) – T0 (s2 – s1))

= 0.652 * ((2957.79 – 1204) + 100(0.30675 -

0.30675) – 300(7.46597 – 3.4638))

ΔX = 114.19 kJ

2)

Given: Air (semi- ideal), mf ͦ = 0.3 kg/s, mair ͦ = 3.6 kg/s.

State 1: T1 = 517 ͦ C = 790 K, P1 = 6 bar

State 2: P2 = 1 bar

Process: reversible adiabatic (isentropic)

Required: T2, C2, ρ2, A2, ΔS

Solution

- For semi-ideal gas:

Enter table (page 6) by T1 = 790 K


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=

= 7.592

Enter table (page 5) by = 7.592

a)

Get by interpolation:

− 480490−480 = 7.592 − 7.2687.824−7.268 => = 485.83

Get ℎ by interpolation: ℎ = 488.463 / Get ∘ by interpolation: ∘ = 2.18993 .

b)

Applying first law for open system:

Q – P = m ͦ (Δh + ΔK.E + ΔP.E)

Q = zero (adiabatic process), P = Zero, ΔP.E = ZeroThen, Δh = - ΔK.E

h1 – h2 = 0.5 * (C22 – C12)/1000

Let C1 = zero (very law compared to exit velocity)

= 2 ∗ ( ℎ − ℎ) = 2000∗(810.99−488.463)

C2 = 803.15 m/s

c) Applying the equation of state on the exit:

P2 = ρ2*R2*T2 = ∗.∗.ρ2=0.717 kg/m3

d)

m ͦ = ρ2 * A2 * C2 = ...∗. A2 = 0.00677 m2

e) ΔS = Zero (because it is an isentropic process).


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3) Given: TH=300 K, TC = 290 K,

QH = 30 W

Model 1: W = 0.72 W

Model 2: W = 1 W

Required: recommend a model

Solution

a) For ideal heat engine (Carnot cycle):

Ƞ, = 1 − = 1 − ƞc,th = 3.333 %

b) For first model:

Ƞ, = = .

ƞth,I = 2.4 % ƞth,I < ƞc,th

c)

For second model:

Ƞ, = = ƞth,II = 3.333 % ƞth,II = ƞc,th

- Although the second model efficiency is greater than

the first, it is practically impossible because the

efficiency cannot be equal to or greater than a Carnot

cycle heat engine that works under the same

conditions. Therefore, I recommend the first model.

4) Given: - mixture contents:

State 1:

CO2 m ͦ = 0.2 kg/s, P = 1.1 bar, T = 50 ͦ C

N2 m ͦ = 0.3 kg/s, P = 1.05 bar, T = 40 ͦ C

State 2: Pmix = 1 bar

Tamb = 300 K


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Solution

For CO2: = = .∗.

. (γ=1.33 for polyatomic gases)

cp) co2 = 0.7615 kJ/kg. K

For N2: = = .∗.. (γ=1.4 for diatomic gases)cp) N2 = 1.03925 kJ/kg. K

a) Applying first law for open system:

Q – P = m ͦ (Δh + ΔK.E + ΔP.E)

Q = zero, P = Zero, ΔK.E = Zero, ΔP.E = Zero

Then, ΔH = Zero

Or: ΔHCO2 = ΔHN2

(m ͦ

* cp * (T1 – Tmix)) CO2 = (m ͦ

* cp * (Tmix – T1)) N2 0.2 * 0.7615 *(50 – T) = 0.3 * 1.03925 * (T – 40)

Solving for T Tmix = 43.28 ͦ C

b)

=

=

.

. XCO2 = 0.4

= =

.. XN2 = 0.6

content X % µ x/µ y = / /% = ∗ (bar)

CO2 40 44 0.909 29.78 0.2978

N2 60 28 2.143 70.22 0.7022

=3.052


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Δ = ∗

−

ΔSCO2 = 0.04618 kW/K

ΔSCO2 = 0.03909 kW/K

ΔSsystem = ΔSCO2 + ΔSN2 ΔSsystem = 0.08527 kW/K

d)

Exergy change for closed system (ΔX)

ΔX = T0 * ΔSsystem ΔX = 25.581 kW

5)

-

Given: H2O, Tambient = 27 C

State 1: P1 = 20 bar

State 2: P2 = 1 bar, T2 = 120 C

Process: constant enthalpy

Required: X1, ΔS, availability change

Solution


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Applying first law for open system:

Q – P = m (Δh + ΔK.E + ΔP.E)

Q = zero, P = Zero, ΔK.E = Zero, ΔP.E = Zero

Then, ΔH = Zero h1 = h2 = 2716.37 kJ/kg


hf = 908.720 kJ/kg, hg = 2798.75 kJ/kg

∵ ℎ ℎ ℎ Point 1 is in the wet zoneThen, from table: sf =2.44719 kJ/kg. K, sg = 6.33962

kJ/kg. K.

ℎ ℎ ℎ ℎ) 2716.37 =908.720 + x (2798.75 – 908.720)

X = 0.9564

) s1 = 6.1699 kJ/kg. K

Δs = s2 – s1 Δs = 1.29663 kJ/kg. K

Exergy change for closed system (ΔX)

ΔX = T0 * ΔSsystem ΔX = 388.989 kJ/kg


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