Jamuna Oil Comany Ltd recruitment question & ans (5th july 2018)

This solution is done by
Engr. Md. Al-amin, ME, BUET.
(Obviously with the help of some talented people)
If any mistake/wrong solution is found please feel free to comment/discuss at alaminbuet2008@gmail.com
Jamuna Oil Company Ltd
Exam date: 5th
July 2018
Exam Time:
3.30 PM ~ 5.00 PM ( 1Hr 30 min)
Exam venue: BUET
Question pattern:
MCQ = 50*0.5 = 25 Marks
Dept. Written (12 question) = 4*10+5*2 = 50
Marks
-------------------------------------------------------
Total = 75 Marks
Questions
MCQ:
1. Where cash memory is placed?
2. Who is the writer of “love and prejudice”?
3. if 𝑒6𝑙𝑛2
= x , then find the value of
𝑥
32
+ 9 ?
4. বাাংলাদেশ ভারদের মদযে পানি চুনি কদব সাক্ষনরে হয় ?
5. নিউনিয়ার পাওয়ার প্লান্টে Coolant নিসান্টে নি েযেিার িরা িয় ?
6. মুনিযুন্টের সময় ঢািা িত িম্বর সসক্টন্টর নিল ?
7. জন্টয়ে স ান্টসের সসক্টর িমান্ডার সি নিন্টলি ?
8. বঙ্গবন্ধু কদব জেল জেদক মুনি পাি ?
Departmental Question:
1. For the following hydraulic block diagram draw energy gradient line (EGD) & Hydraulic
gradient line (HGD).
2. For pipe flow having diameter 8 cm, flow rate 15L/s, length of pipe 20 m find the flow
whether it is laminar or turbulent ? f= 0.032
3. If area of a square pipe and a circular pipe is same & parameters like density of flow, velocity
& viscosity if flow is same then in which pipe Reynolds number will be larger?

4. Write down the three dimensional heat transfer equation in Cartesian co-ordinates & state
its boundary conditions.
5. A pressure vessel having dia 400 mm, internal pressure 50 MPa, Max. value of tangential
and longitudinal stress is 140 MPa and 190 MPa, find out the thickness of the pressure vessel.
6. The bob of a 2m simple pendulum describes an arc of a circle in a vertical plane. If the
tension in the cord is 2.5 times the weight of the bob for the position shown, find the velocity
of the bob at this position.
7. Derive the equation for elongation of bar due to its own weight hanging vertically having
length L, mass m, area A, modulus of elasticity E.
8. A parallel flow pipe heat exchanger is to heat water by ethanol of 10000 Kg/hr from 64.5 ֯C
to 38.2֯C having specific heat 1905 J /Kg.K. Water at a rate of 7000 Kg/hr is heated. If water
enter at 8֯C and have specific heat 4180 J/Kg.K then find out the exit temperature of water &
LMTD.
9. Draw the T-s diagram of a vapor compression refrigeration cycle. Calculate COP of the
bellow p-h diagram.
10. A steel pipes is under load of 2000N, length of the bar is 2m & is under maximum stress
of 150MPa. For those condition maximum deflection is 30 mm, then find the required
diameter.E=200GPa
11. For a belt system initial pull is 2500N, angle of smaller pulley is 150 degree, coefficient
0.25, velocity 6.5 m/s. then find the power transmitted.
12. Draw top view and front view of the following drawing.

Solutions
Come.
Let’s solve it together…

Departmental Question:
1. For the following hydraulic block diagram draw energy gradient line (EGD) & Hydraulic
gradient line (HGD).
Ans:
2. For pipe flow having diameter 8 cm, flow rate 15L/s, length of pipe 20 m, specific weight
0.8 then find loss in KPa & the flow whether it is laminar or turbulent ? f= 0.0032
Ans:
We know,
Q=AV⇒ V= Q/A = Q/(πD2
/4)=4Q/ πD2
= 4 x 0.015/π x 0.082
= 2.984 m/s
Head loss, Hf =
4𝑓𝑙𝑣2
2𝑔𝑑
=
4 𝑥 0.0032 𝑥20𝑥2.9842
2𝑥9.81𝑥0.08
= 9.558 m of H2O
head loss in KPa,
P=
𝐻⍴𝑔
1000
=
9.558𝑥800𝑥9.81
1000
= 75.24 KPa
flow type:
need to determine reynolds no, not enough is
data was provided.
Given Data:
dia, d = 8 cm = 0.08 m
flow rate, Q = 15 L/s = 0.015 m3
/s
length, l = 20 m
sp. wt = 0.8
density , ⍴ = sp. wt x 1000 Kg/ m3
= 800 Kg/ m3
f= 0.0032
loss , Hf = ?
flow type= ? ?

3. If area of a square pipe and a circular pipe is same & parameters like density of flow, velocity
& viscosity if flow is same then in which pipe Reynolds number will be larger?
Ans:
we know, Reynolds no , Re=
⍴𝑉𝐷′
𝜇
if area, density of flow, velocity&
viscosity of a square & circular
pipe is same then reynolds will
vary only for hydraulic diameter
= D’
hydraulic diameter for square
pipe,
D’ =
4 𝑥 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
𝑤𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟
=
4 𝑥 𝐴
𝑃
=
4 𝑥 𝑎2
4𝑎
= a = √πr2
= 1.77248r
hydraulic diameter for circular
pipe,
D’ =
4 𝑥 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
𝑤𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟
=
4 𝑥 𝐴
𝑃
=
4 𝑥 𝜋𝑟2
2𝜋𝑟
= 2r
⇒ As the hydraulic diameter for circular pipe (=2r) is
greater then the hydraulic diamtere for square pipe (=
1.77248r) So Reynolds no will be greater for circular
pipe.
4. Write down the three dimensional heat transfer equation in Cartesian co-ordinates & state
its different boundary conditions.
Ans:
Three dimensional heat conduction equation in Cartesian co-ordinates is
δ2
T
+
δ2
T
+
δ2
T
+
egen
=
1
.
δT
δx2
δy2
δz2
k α δt
Where,
k= thermal conductivity
α = thermal diffusivity =
𝐾
⍴𝐶
Different boundary conditions are given bellow:
Conditions Equation
1. Steady state
(Poission equation)
δ2
T
+
δ2
T
+
δ2
T
+
egen
= 0
δx2
δy2
δz2
k
2. Transient, no heat generation
(diffusion equation)
δ2
T
+
δ2
T
+
δ2
T
=
1
.
δT
δx2
δy2
δz2
α δt
3. Steady state, no heat generation
(Laplace equation)
δ2
T
+
δ2
T
+
δ2
T
= 0
δx2
δy2
δz2

5. A pressure vessel having dia 400 mm, internal pressure 50 MPa, Max. value of tangential
and longitudinal stress is 140 MPa and 190 MPa, find out the thickness of the pressure vessel.
Ans:
we know tangential stress, σt =
𝑃𝐷
2𝑡
⇒ t =
𝑃𝐷
2σt
t =
50 𝑥 400
2 x 140
= 71.42 mm
we know longitudinal stress, σL =
𝑃𝐷
4𝑡
⇒ t =
𝑃𝐷
4σL
t =
50 𝑥 400
4 x 190
= 52.63 mm
safe thickness for the pressure vessel = 71.42 mm
Given Data:
D = 400 mm
P = 50 MPa
σt = 140 MPa
σL = 190 MPa
thickness, t = ?
6. The bob of a 2m simple pendulum describes an arc of a circle in a vertical plane. If the
tension in the cord is 2.5 times the weight of the bob for the position shown, find the velocity
of the bob at this position.
Ans:
We know,
∑Ft = mat
⇒ mg sinꝊ = mat
⇒ at = g sinꝊ = 9.81 x sin30֯
⇒ at = 4.9 m/s2
Again,
∑Fn = man
⇒ T- mg cosꝊ = man
⇒ 2.5 mg- mg cosꝊ = man
⇒ 2.5 g- g cosꝊ = an
⇒ an =2.5g-gcos30֯= 2.5x9.81 -
9.81xcos30֯ = 16.03 m/s2
⇒ an = 16.03 m/s2
Given Data:
r = 2 m
T = 2.5W = 2.5 mg
v=?
Again,
an = v2
/r
⇒ v= √( an . r)
⇒ v= √( 16.03 x 2)
⇒ v= 5.66 m/s
7. Derive the equation for elongation of bar due to its own weight hanging vertically having
length L, mass m, area A, modulus of elasticity E.
Ans:
As the Self weight works on centroid of the bar then for
elongation active length will be half if the total length.
length for elongation L’= L/2
Load = W = mg
We know,
elongations , δ =
𝑃𝐿′
𝐴𝐸
=
𝑚𝑔𝐿/2
𝐴𝐸
=
𝑚𝑔𝐿
2𝐴𝐸
δ =
𝒎𝒈𝑳
𝟐𝑨𝑬

8. A parallel flow pipe heat exchanger is to heat water by ethanol of 10000 Kg/hr from 64.5 ֯C
to 38.2֯C having specific heat 1905 J /Kg.K. Water at a rate of 7000 Kg/hr is heated. If water
enter at 8֯C and have specific heat 4180 J/Kg.K then find out the exit temperature of water &
LMTD.
Heat released by ethanol = heat absorbed by water
me Se∆ Te = mw Sw∆ Tw
⟹ 10000 x 1905 x 26.3 = 7000 x 4180 x (t-281)
⟹t=281 +
10000 x 1905 x 26.3
7000 x 4180
= 281 + 17.1 = 298.1K
t= 298.1 K = 25.1 ֯C
Now ,
for LMTD ,
∆T1 = 64.5-8 = 56.5
∆T2 = 38.2-25.1 = 13.1
we know,
LMTD =
∆T1−∆T2
ln
∆T1
∆T1
=
56.5−13.1
ln
56.5
13.1
=
43.4
ln
56.5
13.1
=
43.4
1.46
= 29.7
Given Data:
me =10000 Kg/hr
Te,in = 64.5 ֯C = 64.5+273 = 337.5 K
Te,out = 38.2 ֯C = 38.2+273 = 311.2 K
∆ Te = Te,in- Te,out = 337.5 – 311.2 = 26.3
Se =1905 J /Kg.K
mw =7000 Kg/hr
Tw,in = 8 ֯C = 8+273 = 281 K
Tw,out = t (let) = ?
∆ Tw = Tw,out- Te,in = t- 281
Sw =4180 J /Kg.K
LMTD = ?
9. Draw the T-s diagram of a vapor compression refrigeration cycle. Calculate COP of the
bellow p-h diagram.
According to the P-h figures,
h1= enthalpy at suction of compressor = 234 KJ/Kg
h2= enthalpy at exit part of compressor = 290 KJ/Kg
h4= enthalpy at the entry port of evaporator = 65.5
KJ/Kg
We have,
COP =
𝐶𝑜𝑜𝑙𝑖𝑛𝑔 𝑒𝑓𝑓𝑒𝑐𝑡
𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 𝑤𝑜𝑟𝑘
=
ℎ1−ℎ4
ℎ2−ℎ1
=
234−65.5
290−234
=
168.5
56
= 3.0089

10. A steel pipe is under load of 2000N, length of the bar is 2m & is under maximum stress of
150MPa. For those condition maximum deflection is 30 mm, then find the required diameter.
E=200GPa
We know,
σmax =
𝑃
𝐴
=
𝑃
𝜋𝐷2/4
⟹ D = √
4𝑃
𝜋 𝑥 σmax
= √
4𝑃
𝜋 𝑥 σmax
= √
4𝑥2000
𝜋 𝑥 150x1000000
= 4.12 mm
Again,
δmax =
𝑃𝐿
𝐴𝐸
=
𝑃𝐿
𝜋𝐷2
4𝐸
⟹ D = √
4𝑃𝐿
𝜋 𝐸δmax
= √
4𝑥2000𝑥2
𝜋𝑥
200𝑥1000000000𝑥30
1000
= 0.92 mm
Required dia = safe dia = 4.12 mm
Given Data:
load, P = 2000 N
L= 2 m
σmax = 150 MPa
δmax = 30 mm
E = 200 GPa
dia , d = ?
11. For a belt system initial pull is 2500N, angle of smaller pulley is 150 degree, coefficient
0.25, velocity 6.5 m/s. then find the power transmitted.
We know,
𝑇1
𝑇2
= 𝑒 𝜇Ꝋ
⇒ T2 =
𝑇1
𝑒 𝜇Ꝋ =
2500
𝑒
0.25𝑥150 x
𝜋
180
=
2500
1.92418
= 1299.25 N
again, power ,
p = (T1-T2) v
= (2500-1299.25) x 6.5
= 7.804 KW
Given Data:
T1 = 2500 N
Ꝋ = 150֯ = 150 x
𝜋
180
μ= 0.25
v= 6.5 m/s
Power, p = ?
12. Draw top view and front view of the following drawing.

Md. Iqbal Hussain
B. Sc in Mechanical Engineer, RUET
Senior Cooling Engineer, R&D Refrigerator
Walton Hi-Tech Industries Ltd.
Email: iqbal092067@gmail.com
Tafhimul Hossain Siddiqi
B. Sc in Mechanical Engineer, CUET
Senior Cooling Engineer, R&D Refrigerator
Email: tafhim.hossain@gmail.com
Abdul Aziz
B. Sc in Mechanical Engineer, IUT
Product Engineer
Samsung
Email: azizfuad111439@gmail.com
Arnab Das
B. Sc in Mechanical Engineer, KUET
Senior Design Engineer, R&D Refrigerator
Email: arnabs2009@gmail.com
Wasek Ahmed Atul
B. Sc in Mechanical Engineer, RUET
Creative Design Engineer, R&D Refrigerator
Email: wasekahmed07@gmail.com

Jamuna Oil Comany Ltd recruitment question & ans (5th july 2018)

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