Day 2
REVIEWER
1. Determine the percentage overshoot of a step change with magnitude 4 introduced into a
system having a transfer function of
𝑌(𝑠)
𝑋(𝑠)
=
9
.
𝑠2 +2.4𝑠+3
A. 4.21%
B. 5.81%
C. 9.48%
D. 10.11%
Solution:
𝑌(𝑠)
8
1/4
= 2
𝑋(𝑠) 𝑠 + 2.4𝑠 + 4 1/4
𝑌(𝑠)
2
=
2
𝑋(𝑠) 0.25𝑠 + 0.6𝑠 + 1
4
𝑋(𝑠) =
𝑠
4
2
𝑌(𝑠) =
2
𝑠 0.25𝑠 + 0.6𝑠 + 1
𝜏 2 = 0.25
𝜏 = 0.5
2𝜁𝜏 = 0.6
𝜁 = 0.6
−𝜋𝜁
%𝑂𝑆 = exp (
) × 100%
√1 − 𝜁 2
%𝑂𝑆 = 9.48%
2. Determine the volume of liquid that can be placed in a tank with hemispherical bottom. The
dimensions of the tank are 𝐷 = 2.5 , 𝐻 = 7𝑚, 𝑎 = 1.5𝑚. Assume a 20% freeboard
A. 36 m3
B. 49 m3
C. 51 m3
D. 29 m3
Solution:
𝑉𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 = 𝜋𝑟 2 𝐿
2.5𝑚 2
𝑉𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 = 𝜋 (
) (7𝑚)
2
𝑉𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 = 34.36 𝑚2
𝜋𝑎
𝑉ℎ𝑒𝑚𝑖𝑠𝑝ℎ𝑒𝑟𝑒 =
(3𝑟 2 + 𝑎2 )
3
𝜋(1.5𝑚)
2.5
𝑉ℎ𝑒𝑚𝑖𝑠𝑝ℎ𝑒𝑟𝑒 =
(3( 𝑚)2 + (1.5𝑚)2 )
3
2
𝑉ℎ𝑒𝑚𝑖𝑠𝑝ℎ𝑒𝑟𝑒 = 10.8974 𝑚3
𝑉𝑓𝑢𝑙𝑙 = 34.36𝑚3 + 10.90𝑚3 = 45.26𝑚3
𝑉𝑓𝑟𝑒𝑒𝑏𝑜𝑎𝑟𝑑 = 45.26𝑚3 (1 − 0.20) = 36.208 𝑚3
3. Calculate the number of moles of theoretical air and theoretical oxygen required for the
complete combustion of 1 kg of fuel supplied to an engine, which contains 68.3% C, 19.9% H,
7.5% S, 1.67% N, and the remaining percentage in O.
A. 38.23 kmol
B. 34.58 kmol
C. 35.21 kmol
D. 24.31 kmol
Solution:
𝐵𝑎𝑠𝑖𝑠: 100 𝑘𝑔 𝑓𝑢𝑒𝑙
𝐻
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑂2 = 𝐶 + + 𝑆 − 𝑂
4
68.3 19.9 7.5 2.63
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑂2 =
+
+
−
= 7.2616 𝑘𝑚𝑜𝑙 𝑂2
32
4
32
32
100 𝑘𝑚𝑜𝑙 𝐴𝑖𝑟
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑎𝑖𝑟 = 7.2616 𝑘𝑚𝑜𝑙 𝑂2 ×
= 34.58 𝑘𝑚𝑜𝑙
21 𝑘𝑚𝑜𝑙 𝑂2
4. Checal problems
5. A thermometer of mass 0.065 kg and heat capacity 46.1 J/kg°C reads 17.0°C. It is then
completely immersed in 0.310 kg of water and it comes to the same final temperature as the
water. If the thermometer reads 45°C, what was the temperature of the water before insertion
of the thermometer, neglecting other heat losses?
A. 32.18 °C
B. 39.21 °C
C. 49.21 °C
D. 44.35 °C
Solution:
𝑄𝑡ℎ𝑒𝑟𝑚𝑜𝑚𝑒𝑡𝑒𝑟 = 𝑄𝑤𝑎𝑡𝑒𝑟
𝑚𝑡 𝐶𝑡 ∆𝑇𝑡 = 𝑚𝑤 𝐶𝑤 ∆𝑇𝑤
𝐽
𝐽
) (45℃ − 17℃) = (0.31𝑘𝑔)(4184
)( 45℃ − 𝑇𝑖 )
𝑘𝑔 − ℃
𝑘𝑔 − ℃
𝑇𝑖 = 44.35℃
6. Given that the helium liquefaction apparatus in a laboratory is at a temperature of 298 K and the
helium within the apparatus is at 4.3 K, if 156 mJ of heat is transferred from the helium,
determine the minimum amount of heat delivered to the laboratory.
A. 12.87 J
B. 10.81 J
C. 12.87 mJ
D. 10.81 mJ
Solution:
𝑇𝐻
𝑄𝐻 = 𝑄𝐿 ( )
𝑇𝐿
298𝐾
𝑄𝐻 = (0.156 𝐽) (
)
4.3𝐾
𝑄𝐻 = 10.81 𝐽
(0.65𝑘𝑔) (46.1
7. The first-order decomposition of acetonedicarboxylic acid are 𝑘 = 5.75 × 10−4 𝑠 −1 at 292K and
𝑘 = 1.93 × 10−3 at 304 K. determine the activation energy (𝐸𝑎 ) for this reaction
CO(CH2COOH)2(aq) → CO(CH3)2(aq) + 2 CO2(g)
A. 76.21 kJ/mol
B. 90.26 kJ/mol
C. 91.57 kJ/mol
D. 74.47 kJ/mol
Solution:
𝑘2 𝐸𝑎 1
1
ln =
[ − ]
𝑘1
𝑅 𝑇1 𝑇2
−3
1.93 × 10
𝐸𝑎
1
1
ln
=
−
[
]
−4
5.75 × 10
8.314 292 304
𝐽
𝑘𝐽
𝐸𝑎 = 74472.40
= 74.47
𝑚𝑜𝑙
𝑚𝑜𝑙
8. The first-order decomposition of dinitrogen pentoxide at 335K is as follows.
N2O5(g) → NO2(g) + O2(g)
What mass on 𝑁2 𝑂5 will remain after 5 mins if the reaction started with a 2.53-gram sample of
𝑁2 𝑂5 and have 1.45 g remaining after 207 s.
A. 2.65 g
B. 1.13 g
C. 3.98 g
D. 4.14 g
Solution:
1
N2O5(g) → 2NO2(g) + O2(g)
2
1.45
ln
= −𝑘(207)
2.53
𝑘 = 2.69 × 10−3 𝑠 −1
[𝑁2 𝑂5 ]5
60𝑠
𝑙𝑛
= −(2.69 × 10−3 𝑠) (5𝑚𝑖𝑛 ×
)
2.53
𝑚𝑖𝑛
[𝑁2 𝑂5 ]5 = 1.13 𝑔
9. A double-pipe counterflow heat exchanger has water flowing through it at a rate of 68 kg/min.
The oil flows through the tube and heats the water from 57°C to 79°C. The specific heat of the oil
is 1.770 kJ/kg·K. The oil enters the heat exchanger at 115°C and leaves at 70°C. Given an overall
heat transfer coefficient of 340 W/m²·K, calculate the heat exchanger area in 𝑚2 .
A. 14.68
B. 17.98
C. 13.59
D. 12.79
Solution:
𝑄 = 𝑚𝑐 𝐶𝑝𝑐 (𝑡2 − 𝑡1 )
𝑘𝑔 𝑚𝑖𝑛
𝐽
𝑄 = (68
×
) (4184
) (79℃ − 57℃)
𝑚𝑖𝑛 60𝑠
𝑘𝑔 ℃
𝑄 = 104321.07 𝑊
(𝑇1 − 𝑡2 ) − (𝑇2 − 𝑡1 )
∆𝑇𝑙𝑚 =
𝑇 − 𝑡2
ln [ 1
𝑇2 − 𝑡1 ]
(115 − 79) − (70 − 57)
∆𝑇𝑙𝑚 =
115 − 79
ln [
]
70 − 57
∆𝑇𝑙𝑚 = 22.58
𝑄 = 𝑈𝐴(∆𝑇𝑙𝑚 )
𝑊
104321.07 𝑊 = (340 2 )(𝐴)(22.58𝐾)
𝑚 𝐾
𝐴 = 13.59 𝑚2
10. Assuming that a 74 W light bulb filament behaves as a black body radiating into a black
enclosure at 74°C, and given that the filament has a diameter of 0.12 mm and a length of 4.75
cm, calculate the temperature of the filament taking radiation into consideration.
A. 893 C
B. 302 C
C. 369 C
D. 656 C
Solution:
𝜖 = 1 𝑓𝑜𝑟 𝑏𝑙𝑎𝑐𝑘 𝑏𝑜𝑑𝑦
𝑄 = 𝜎𝜖𝐴(𝑇14 − 𝑇24 )
(74𝑊) = (5.67 × 10−8 )(1)(𝜋 × 0.012 × 0.0475)[𝑇14 − (74 + 273𝐾)4 ]
𝑇1 = 928.53𝐾 = 655.53℃
11. A 100-gram ball was dropped from the height of 20cm. The ball hit the floor and was reflected
upwards with a velocity of 1 m/s and acceleration due to gravity of 10 𝑚𝑠 −2 . What is the change
in the momentum of the ball.
A. 2.6 kg m/s
B. 1.4 kg m/s
C. 0.3 kg m/s
D. 2.1 kg m/s
Solution:
𝑣 2 = 2𝑔ℎ = 2(10𝑚𝑠 −2 )(0.20)
𝑣 = −2 𝑚/𝑠
∆𝑃 = 𝑚(𝑣𝑡 − 𝑣𝑜 )
∆𝑃 = (0.1)(1 − (−2))
∆𝑃 = 0.3 𝑁𝑠
12. A 7.1 kg ball moving at 0.6 m/s collides with a 4.8 kg ball at rest. If the heavier ball moves at 0.27
m/s after the collision, how fast does the lighter ball move?
A. 0.32 m/s
B. 0.68 m/s
C. 0.49 m/s
D. 0.52 m/s
Solution:
𝑚2 𝑣2′ + 𝑚1 𝑣1′ = 𝑚1 𝑣1 + 𝑚2 𝑣2
𝑚
𝑚
𝑚
(4.8𝑘𝑔)𝑣2′ + (7.1𝑘𝑔)(0.27 ) = (7.1𝑘𝑔)(0.6 ) + (4.8𝑘𝑔)(0 )
𝑠
𝑠
𝑠
𝑣2′ = 0.488 𝑚/𝑠
13. Sepro
14. Sepro
15. A material initially having an average size of 25 mm is crushed to an average size of 6 mm,
requiring an energy input of 34 kJ/kg. Considering Kick’s Law, calculate the energy required if the
desired reduction is to an average size of 3.2 mm instead.
A. 43.66 kJ/kg
B. 34.66 kJ/kg
C. 56.66 kJ/kg
D. 65.66 kJ/kg
Solution:
𝑝
𝐷𝑠𝑎
= 𝐾𝑘 [ln (
)]
𝑚
𝐷𝑠𝑏
𝑘𝐽
25
34
= 𝐾𝑘 [ln ( )]
𝑘𝑔
6
𝑘𝐽
𝐾𝑘 = 23.82
𝑘𝑔
𝑝
𝑘𝐽
20
= (23.82 ) [ln ( )]
𝑚
𝑘𝑔
3.2
𝑝
= 43.66 𝑘𝐽/𝑘𝑔
𝑚