BIRLA INSTITUTE OF TECHNOLOGY AND SCIENCE, PILANI
SEMESTER-I, 2020-21
BITS F111 THERMODYNAMICS
MID-SEMESTER EXAMINATION (OPEN BOOK)
Conceptual Questions (6 Marks Each)
1.
Two hydraulic piston/cylinders are connected through a hydraulic line. If they have
diameters of D1 and D2 respectively, such that D2= 4D1.
a) What would be the ratio of forces i.e., F2/F1?
b) Write the major assumption you have made while deducing this relationship
between piston forces.
Solution:
For each cylinder we have the total force as: F = PAcyl = P π D2/4
F1 = PAcyl,1 = P π D21/4
F2 = PAcyl,2 = P π D22/4 = P π 16 D21/4 = 16 F1
Therefore, F2/F1 =16
-----[4 M]
The hydraulic fluid should be incompressible.
The pressure should be same at both the pistons. -----[2 M]
2.
A piston cylinder assembly containing water
(saturated liquid-vapor mixture) initially at 125oC,
as shown in the figure, is being held on the top of a
burner. Now, the burner is put into flames and
arrangements are made such that the flame burns
with a constant temperature of 500oC. Estimate the
work done during the process.
Sol. C.V: Control mass system.
Governing eqn.: Boundary work, 1W2=PΔV
For saturated mixture at 135oC,
P=Psat=232.1 kPa ---[1M]
V1=AX1 =0.5x0.1= 0.05 m3 ---[1M]
V2=AX2 =0.5x0.4 =0.20 m3 ---[1M]
1W2=232.1(0.20-0.05) =34.82 kJ ---[3M]
3.
10 kg of R-410a is contained in a system at 20oC and 200 kPa. The refrigerant goes
through a process (in a single step) in such a way that the final stage is saturated vapor
with a specific volume 0.12591 m3/kg. (Note: No interpolation required)
a) Check whether the process (most appropriate) is isobaric/isochoric/isothermal.
Justify.
b) Draw T-v diagram of the process.
1
Sol.
State 1: From Table B.4.2: At 20oC and 200 kPa, Sp.
Vol is 0.16322 m3/kg
State 2: After process: Saturated Vapor, v2=0.12591
m3/kg,
This corresponds to 200 kPa and -37.01oC.
Therefore, the process is isobaric. ---[3M]
T-v diagram ---[3M]
4.
In a rigid adiabatic container with hydrogen and oxygen gases, a scientist is attempting
an explosive reaction to produce water by combining the elemental gases. What would
be the internal energy change? Justify your answer.
Ans. As procss is adiabatic, and no work is done due to rigid container, the internal
energy change has to be zero. ---[6M]
5.
The relation between the internal energy, u (kJ/kg) and temperature, T (in Kelvin) of
a gas, assuming it to be ideal gas, is identified as follows:
𝑢 = 𝑎 + 𝑏𝑇
where, a and b are found by experimental analysis as, -0.46 kJ/kg and 0.716 kJ/kgK. Determine Cp for the gas, without using any property table.
[Data: Specific gas constant of the gas =0.268 kJ/kg-K]
Given: u= -0.47+0.716 T ---(1)
By ideal gas relation: pv=RT=0.268T ---(2)
=
𝐶 =
(
)
---[3M]
Substituting u and pv from eqn. (1) and (2)
𝐶 =
(
.
.
.
)
=0.984 kJ/kg-K ---[3M]
Alternative answer:
𝐶 =
=
(
)
= 𝑏=0.716 kJ/kg-K ---[3M]
As, Cp=Cv+R = 0.716 +0.268 = 0.984 kJ/kg-K ---[3M]
6.
Consider a process moving from state M to state O via path
M-N-O and path M-P-O. If 1200 kJ of heat is added during
the path M-N-O and work produced is 500 kJ. How much
heat will flow into the system along the path M-P-O if the
work produced along this path is 100 kJ?
Sol.
2
By first law of thermodynamics:
Along the path MNO:
Heat in – Work Out = ΔU  1200 – 500 = ΔU i.e., ΔU=700 kJ ---[3M]
Therefore, the internal energy change between the state M and state O = 700 kJ
Along the path MPO:
Heat in – Work Out = ΔU  Heat in = Work Out + ΔU = 100 + 700 = 800 kJ ---[3M]
Section-2 (Subjective Questions)
1. Consider you are an engineer at the cloud computing hardware firm. You have been given a task
of designing a server room. Imagine a server room you are in charge of has a volume 20 m3 and
it dissipates energy to the outside air at a rate of 18 kW. The room has 1 m3 aluminum and 1 m3
granite with all material at 25oC and 1 atm. Assume control mass system, constant specific heat
with all the mass heating up uniformly.
Estimate:
a) mass of air in the room in kg
b) change in internal energy if the server room temperature increases by 5°C
c) the time required (in minutes) to increase the server room temperature by 5°C?
[5+5+5 =15M]
Solution: C.V. air, aluminum and granite. m2 = m1
a)
Vair = 20 - 1 - 1 =18 m3
---[2M]
Mair = PV/RT
= 101.325 × 18/(0.287 × 298.15)
= 21.3144 kg
---[3M]
b)
malumnium = 1 m3 x 2700 kg/m3 = 2700 kg
mgranite = 1 m3 x 2750 kg/m3 = 2750 kg
We do not have a u table for aluminum or granite so use heat capacity from A.3.
ΔU = [mair Cv + malumnium Cv + mgrantite Cv ] ΔT ---[2M]
= (21.3144 × 0.717 + 2700 × 0.90 + 2750 × 0.89) 5
= 24,463.9121 kJ
---[3M]
c)
Energy Eq.5.11: U2 - U1 = 1Q2 = Q. Δt ---[2M]
= Q. × Δt = 18 kW × Δt
=> Δt = 24463.9121 /18
= 1359.10 sec
= 22.65 min or 0.378 hr
3
---[3M]
2. A piston and cylinder arrangement as shown in the schematic
representation below, contains R-410a which is initially saturated liquid at
-45oC, with volume 0.03 m3 (State-1). The electric heater shown, is
switched ON and supplies 5000 kJ of electric work done on the system,
until the piston just touches the linear spring. Then it is switched OFF
(State-2). During process 1-2, there is no heat interaction. Then an external
heating device adds heat such that a final state of 200 kPa and 40oC is
reached (State-3).
a) Draw P-V diagram for the process
Find:
b) Volume (m3) when piston just touches the spring
c) Heat added during process 2-3 (kJ)
d) Total boundary work (kJ)
e) Time (in minutes) that the electric heater is ON with a supply of 20 Ampere current and 200
Volts
f) The slope (kPa/m3) for process 2-3 on the P-V diagram
Solution:
Note: no marks for simply getting any properties, but only as shown in solution. No marks for
expressions in OB, but only correct substitution followed by correct answer. PARTIAL
MARKS ARE GIVEN ONLY AS SHOWN; AND NOT FOR EVERY PROPERTY VALUE.
CM: R-410a process: first constant pressure expansion, then expansion against spring
3
State 1 : given :sat. liq T1=-45°C, V1=0.03 m , P1 =138.8 kPa (B.4.1)
3
v1=vf=-45°C=0.000752 m /kg; m=V1/v1=0.03/0.000752=39.894 kg
h1= hf=-6.92 kJ/kg;
Process 1-2 :5000 kJ electric work added, boundary work done, no heat interaction(1Q2=0).
m(u2-u1) = 1Q2 - 1W2 - We; m(u2-u1) +1W2=-We
m(u2-u1) + Pm (v2-v1) = m(h2-h1) =-We [We=-5000kJ; electric work done on system]
h2=(+5000/39.894)-6.92=118.412 kJ/kg ---[2M]
State 2: h2=118.412 kJ/kg (between hf and hg), hence two phase:
h2=h2f+x2h2fg =(h2-h2f)/hfg = (118.412+6.92)/267.27; x2= 0.469 ---[2M]
v2=v2,f +x2v2,fg =0.000752+0.469(0.17729)=0.0839 m3/kg
u2=u2f+x2u2,fg= -7.02+0.469(242.67)=106.792 kJ/kg
4
3
V2=mv2=3.347 m =volume at stops ---[1M]
Ans(b)
_______________________
Process 2-3: heat 2Q3 added and boundary work against spring
m(u3-u2) = 2Q3-2W3;
State 3: given : T3=40°C , P3 =200KPa (If sat@T3= 40°C , Psat>200kpa , so superheated)
v3=0.17545 m3/kg; V3=mv3 =6.999 m3;
u3=293.59 kJ/kg ---[2M] u2=106.792 kJ/kg ---[2M]
2Q3-2W3=U3-U2= m(u3-u2); 2Q3 = m(u3-u2) + [(P2+P3)/2 x (V3-V2)] = [39.894(293.59-106.792)]
+ [(138.8+200)/2 x(6.999-3.347)]; 2Q3=7452.119+618.649=8070.768 kJ ---[1M]
Ans(c)
_____________________
WB =1W2+2W3 =[P1(V2-V1)]+[(P2+P3)/2 x (V3-V2)] Total boundary work
=[138.8(3.347-0.03)+(138.8+200)/2 x(6.999-3.347)]
(correct v’s substituted; (2) marks, correct P’s substituted (2) marks)
=460.4 + 618.649 =1079.049 kJ ---[1M] Ans(d)
___________________________
We=5000x103 J= VIt x 60 ---[2M]
=200x20xtx60 ---[2M]
t=20.83 min ---[1M] Ans(e)
__________________________
((4) marks for correct substitution of correct P’s and V’s)
Slope = (P3 – P2) / (V3-V2) = (200-138.8)/(6.999-3.347)= 16.758 kPa/m3 ---[1M] Ans(f)
------------------------------------------------------------------------------------------------------------(6 x 5 =30 marks) Answers: (a) P-V diagram for the process (5 marks, 1 each for: saturation
dome, st1, st2, st3, connecting straight lines, all correct)
(b) Volume when piston touches the stops = 3.347 m3 (5 marks)
(c) Heat added during process 2-3 = 8071 kJ (5 marks)
(d) Total boundary work = 1079 kJ (5 marks)
(e) t=21 min (5 marks)
(f) Slope for process 2-3 = 16.76 kPa/m3 (5 marks)
====================================================================
5
3. Two kg of water contained in an insulated piston-cylinder arrangement (with a massless piston)
at 50oC with volume of 0.02 m3 when the piston is initially resting on the stops. Take Pfloat=100
kPa. The paddle (shaft) work is supplied such that water reaches to 200oC in its final state.
a) Find the boundary work done (kJ) by the water
b) Find the paddle work (kJ) supplied to the water
c) Show the process on the P-v diagram
d) Show the process on the T-v diagram.
e) Draw P-T diagram (with reference to critical and triple point) for the process.
[25 marks]
3
3
o
Solution: State 1: v1 = 0.02 m / 2 kg = 0.01 m /kg, Since vf<v1<vg @ 50 C
x1 = 0.000747, u1 = uf + x × ufg = 210.9689 kJ/kg
Intermediate State 2: the piston leaves the stop at P = 100 kPa
@ 100 KPa, vf = 0.001043 m3/kg, vg = 1.69296 m3/kg; therefore vf < v2 = v1 <vg at 100 kPa
x2 = v2 - vf/vfg = 0.00529
State 3: Now we heat isobarically until T3 = 200 oC, P3 = P2 = 100 kPa
v3 = 2.17226 m3/kg, V3 = m × v3 = 2.0 kg × 2.17226 = 4.34452 m3, u3 =2658.05 kJ/kg
(a)
Work done by the water 1W3 = 1W2 + 2W3,
since 1W2 = 0 (no boundary work) ---[2M]
1W3 = P2 (V3-V2) = m.P2 (v3-v2)  1W3 = 2.0 x 100 (2.17226-0.01) = 432.452 kJ ---[3M]
(b)
Paddle work on the water, m(u3-u1) = 1Q2  (1W31W3, paddle) ---[2M]
2 x (2658.05 – 210.9689)+432.452 = 1W3 (paddle),
1W3 (paddle) = 5326.6142 kJ ---[3M]
(c) ---[5M]
(d) ---[5M]
*****End of Question Paper*****
6
(e) ---[5M]