SAIF ALDIN ALI MADIN سيف الدين علي ماضي S96aif@gmail.com Experiment Name: - Water / Water Cross Flow Shell and Tube Heat Exchanger 1. Abstract Studying the performance of this type of heat exchanger 2. Introduction Types of heat exchangers: Onetype of heat exchanger is that of a double pipe arrangement with either counter or parallel flow and with either the hot or cold fluid occupying the annular space and the other fluid occupying the inside of the inner pipe. A type of heat exchanger widely used in the chemical process inches is that of the shell and tube arrangement

1. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 1 | P a g e
[Heat-Transfer Laboratory]
University of Baghdad
Name: - Saif Al-din Ali -B-

2. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 2 | P a g e
TABLE OF CONTENTS
ABSTRACT.......................................................................I
INTRODUCTION...........................................................II
THEORY........................................................................III
APPARATUS...................................................................V
Calculations and results................................................VI
DISCUSSION ...............................................................VII

3. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 3 | P a g e
Experiment Name: - Water / Water Cross Flow Shell and Tube Heat
Exchanger
1. Abstract
Studying the performance of this type of heat exchanger
2. Introduction
Types of heat exchangers:
Onetype of heat exchanger is that of a double pipe arrangement with either
counter or parallel flow and with either the hot or cold fluid occupying the annular
space and the other fluid occupying the inside of the inner pipe. A type of heat
exchanger widely used in the chemical process inches is that of the shell and tube
arrangement (Fig. (1))
One fluid flows in the inside of the tubes, while the other fluid is forced through the shell
and over the outside of the tubes. To insure that the shell side fluid flow will across the
tubes and thus induct higher heat transfer, baffles are placed in the shell as shown in Fig.
(2). Cross flow heat exchangers are commonly used in air or gas heating and cooling
applications, where a gas may be forced across a tube handle, while another fluid is used
inside the tubes for heating or cooling purposes.
Shell Type Cross Flow Heat Exchanger
Heat transfer surfaces very higher could be obtained with this type of heat
exchanger which is built with a compact tube bundle fastened at its ends to two
circular plates. This tube bundle is placed inside the cylinder shall. The flows of the
Two fluids are showed in the picture above. Usually orthogonal diaphragms are
placedtoincrease the turbulence of the External flow with the purpose to increase
the convection flow

4. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 4 | P a g e
By means of deflecting plates you can obtain that the internal flow courses the
length of the exchanger more than one time: in this way you can obtain good values
of the velocity of the flow, higher values of the correction
The schematic of a shell-and-tube heat exchanger (one-shell pass and one-tube pass)
Different flow regimes and associated temperature profiles in a double-pipe heat exchanger.
3. Theory

5. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 5 | P a g e
Determination of the heat convection coefficient between the water flows and tube
surfaces: Heat transfer coefficient for both inner surface and surface of the tube bundles
could be calculated using the following procedure, this procedure could be connected for
the inner surface and then repeated surface using the appropriate formula. However, it is
necessary to make distinction between laminar and turbulent flow
1. Laminar Flow
As far as the laminar flow received there two zones where the flow is quite different with
respect to the other one
a. Entrance length which length is given by the following: 𝑳𝒊 =
𝑹 𝒆 𝒑 𝒓 𝒅
𝟐𝟎
Where you can use; 𝑵𝒖 = 𝟏. 𝟖𝟔 (𝑹 𝒆 𝒑 𝒓
𝒅
𝑳
)
𝟏
𝟑
(
𝒖
𝒖 𝒘
)
𝟎⋅𝟏𝟒
For 𝑹 𝒆 𝒑 𝒓(d/l) >= 10, l/d>2, 100<Re<2100, 0.48<𝑝𝑟<16700
𝑢 = The dynamic viscosity at average temperature
𝑢 𝑤= the dynamic viscosity at wall temperature
b. The Zone where the steady state is generated completely. Here as the convective flow is
constant, you have: Nu = 3.66 for constant wall temperature
Nu = 4.53 for constant heat flux
2. Turbulent Flow –
a. Entrance length whose length is given always by the previous formula where you can
apply
𝐍 𝐮 = 𝟎. 𝟎𝟑𝟔 𝐑𝐞 𝟎.𝟖
𝐩 𝐫
𝟏/𝟑
(𝐝 ∕ 𝐋) 𝟏/𝟖
With; 10<(l/d) <400,Re>104
,0.7<Pr<16700
b. The zone where the steady state is generated completely. Here you can obtain the convective
coefficient by 𝐍 𝐮 = 𝟎. 𝟎𝟐𝟑𝐑 𝐞
𝟎.𝟖
𝐩 𝐫
𝟎.𝟒
[𝟏 + (
𝐝
𝟐
)
𝟎.𝟕
]
𝐡−
=
𝐡 𝐌 𝐋 𝐌+𝐡 𝐆 𝐋 𝐆
𝐋 𝐌+𝐋 𝐆
. For 10000<Re<120000, 0.7<Pr<120, (l/d)>60 Where
𝐡 𝐌 = Mouth zone coefficient convection (W / m^2*c)
𝐡 𝐆 =Steady state completely generated zone coefficient convection
𝐋 𝐌 =Mouth zone length
𝐋 𝐆 = Steady state completely generated zone length
The Properties of the now as the Viscosity. The density. Thermal capacity and the thermal
Conductivity will be taken from relative tables at average temperature TM

6. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 6 | P a g e
𝑇 𝑀 =
𝑇𝐼𝑁 + 𝑇𝑂𝑈𝑇
2
Determination of the overall heat transfer coefficient Where:
𝑸 = 𝑼𝑨∆𝑻 𝑳𝑴𝑻𝑫 = 𝑸 𝟏 = 𝑸 𝟐
𝑸 𝟏 = 𝒎∗
𝒉𝒐𝒕−𝒘𝒂𝒕𝒆𝒓 ∗ 𝑪 𝒉𝒐𝒕−𝒘𝒂𝒕𝒆𝒓 ∗ ∆𝑻 𝒉𝒐𝒕−𝒘𝒂𝒕𝒆𝒓
𝑸 𝟐 = 𝒎∗
𝑪𝑶𝑶𝑳−𝒘𝒂𝒕𝒆𝒓 ∗ 𝑪 𝑪𝑶𝑶𝑳−𝒘𝒂𝒕𝒆𝒓 ∗ ∆𝑻 𝑪𝑶𝑶𝑳−𝒘𝒂𝒕𝒆𝒓
𝑸 = Thermal flow
𝑸 𝟏 = Thermal flow through the primary circuit (hot water circuit)
𝑸 𝟐 = Thermal flow through the secondary circuit (cool water circuit)
∆𝑻 𝑳𝑴𝑻𝑫 = The log mean temperature difference ∆𝑻 𝑳𝑴𝑻𝑫 =
∆2−∆1
𝑙𝑛(
∆2
∆1
)
U = the overall heat transfer coefficient (𝑤/𝑚2
𝑐`
)
𝑸 = 𝑼𝒊 𝑨𝒊∆𝑻 𝑳𝑴𝑻𝑫 = 𝑼 𝒐 𝑨 𝒐∆𝑻 𝑳𝑴𝑻𝑫
𝑼 𝒐 =
1
1
ℎ𝑖
𝑟𝑜
𝑟𝑖
+
𝑟𝑜(ln(
𝑟𝑜
𝑟𝑖
))
𝑘
+
1
ℎ 𝑜
𝑸 = 𝑼 𝒐 𝑨 𝒐∆𝑻 𝑳𝑴𝑻𝑫 ∗ 𝐹
K= 349 (𝑤/𝑚𝑐`
) for the copper, 𝑟𝑜 = 8mm, 𝑟𝑖= 7 mm, A = 67380 m𝑚2
F = correction factor
4. APPARATUS

7. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 7 | P a g e
The Procedure
A. Test rig (Apr Apparatus) preparation for measurement
Mixing Device To avoid that the temperature inside the secondary circuit increases
until unwanted values which would not allow to obtain a good cooling because the
difference in temperature between the primary and the secondary circuits would
decrease more and more during the working of the unit, this group is provided with an
adjustable valves system, Connect (58) with the water supply and feed with water the
unit (secondary circuit) Close the valve (114), You can control the cool water flow rate
by the thermostatic valve (18) or by the hand operated one (19). If you want to use the
first one (18), deviate the Water flow by means of the three - Way valve (117) and
adjust the interference temperature by regulating the thermostatic valve (18) itself.
Close completely the valve (19). On the other hand, if you want to use the hand-
operated Valve (19), deviate the Water flow rate by means of the three - way valve
(117) and adjust it mainly using the valve (19) itself. In any case you can read the
refreshing Water flow rate by means of the flow meter (8). Connect (63) with a
discharging pipe. The choking valves (20) and (56) must be adjusted in order to obtain
water flow rates (both the circuits) just over the flow meter scale maximum value (300 l
/ h) when the valves (23) and (24) are completely open.
Note: In order to avoid any formation of air bubbles inside the Water circuits, the unit is
provided with air discharge valves (31) and (32) for both the circuits. You can eliminate
the air bubbles by means of these valves as soon as they appear through the
transparent flow meters. Their presence in fact makes it impossible for the flow meters
to give the correct readings of the Water flow rates.
Water Supply:
• Primary Circuit: Connect (61) with the Water supply by means of a pipe. Open the
Adjust the valve (56) as described for the mixing. Fill with the water (37) Jam primary
circuit and tank (26) as well, until the water begins to exit from the valve (34). Close the
valve (37) and close the water supply, Connect (62) with the discharge pipe.
secondary Circuit: Connect (58) with the Water supply and keep the valve (114) They
are the valve (21) close in order to fill all the secondary circuit until the valve (35) begins
to discharge the Water from the circuit. You must take care that (58) is always
connected to water supply and the valve (114) is always open, Adjust the valve (20) as
described for the mixing. Connect (63) with the discharge pipe.
water Discharge:

8. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 8 | P a g e
primary Circuit; Turn off the pump (7). Connect (61) with the discharge by means of a suitable
pipe, Open the valve (37). .
Secondary Circuit: Tum off the pump (6), disconnect (58) from the water supply Close the valve
(114). Connect (60) with the discharge by means of a suitable PIPS Open the valve (21).
Safety Thermostat: Adjust the safety thermostat.
Adjustable U - differential manometer: Fill with manometric liquid having ROM density such as
mercury.
B. Test Rig operation:
 Connect the group with the supply mains by connecting the plug of the plug of the electric
supply cable (730 to the socket
 Tum on the main switch (74)
 Turn on the primary and secondary circuit water pumps by means of switches (9) ) and (81)
placed on the control board,
 Turn on the electrical resistances by means of their switches and adjust the value of the
adjustable one from 0 up to 800 W by means of the control knob (71),
 Adjust the water flow rates through the two circuits using the control valves (23) and (24) and
read their values on their graduated scales,
 Read the values of the temperature of the water relating to the most important point of the
circuits by the temperature displays (123) and select them by the probe selector (93) and (94)
as illustrated on the control board itself.
 Use the U-differential manometer and the valves system for measuring the pressure losses of
the water flow through the two heat exchangers
 Parallel Flow and Counter Flow Through The Water - Water Exchanger:
By means of the valves (39) and (114) you can easily choose how to make the water exchanger
work as parallel and counter flow as shown in the figure below,

9. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 9 | P a g e
5. Calculations and results
Parallel flow ; hot
𝑸 𝐡𝐨𝐭= 300 l/h  𝑸 𝐡𝐨𝐭 = 𝟖. 𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟓 + 𝑻 𝟕
𝟐
=
𝟒𝟔 + 𝟒𝟐. 𝟗
𝟐
= 𝟒𝟒. 𝟒𝟓 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝑨 = 𝑨 𝒔𝒉𝒆𝒆𝒍 − 𝑨𝒕𝒖𝒃𝒆
=
𝝅
𝟒
𝒅 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
𝒅 𝒐
𝟐
)
=
𝝅
𝟒
(𝟓𝟎 ∗ 𝟏𝟎−𝟑
) 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
(𝟏𝟔 ∗ 𝟏𝟎−𝟑
) 𝒐
𝟐
)
=1.15866∗ 𝟏𝟎−𝟑
𝒗 =
𝟖. 𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟓
𝟏. 𝟏𝟓𝟖𝟔𝟔 ∗ 𝟏𝟎−𝟑
= 𝟎. 𝟎𝟕𝟏𝟗𝟏 𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝒅 𝟎
𝒖
=
𝟗𝟗𝟎∗𝟎.𝟎𝟕𝟏𝟗𝟏∗𝟎.𝟎𝟏𝟔
𝟓.𝟖𝟒𝟗𝟔∗𝟏𝟎−𝟒
→ 𝑹 𝒆 = 𝟏𝟗𝟒𝟖. 𝟏𝟑𝟑< 2100 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
𝑳𝒊 =
𝟏𝟗𝟒𝟖. 𝟏𝟑𝟑 ∗ 𝟑. 𝟕𝟐𝟗𝟓 ∗ 𝟎. 𝟎𝟏𝟔
𝟐𝟎
= 𝟓. 𝟖𝟏𝟑𝟎𝟑 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
Let 𝑻 𝒘 =𝑻 𝒇 − 𝟓  𝑻 𝒘 = 𝟑𝟗. 𝟒𝟓 𝒄  Dynamic Viscosity= 6.4211∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟏𝟗𝟒𝟖. 𝟏𝟑𝟑 ∗ 𝟑. 𝟕𝟐𝟗𝟓 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟓.𝟖𝟒𝟗𝟔∗𝟏𝟎−𝟒
𝟔.𝟒𝟐𝟏𝟏∗𝟏𝟎−𝟒 )
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟐. 𝟗𝟎𝟏𝟏
𝒉 𝟎 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 =
𝟏𝟐.𝟗𝟎𝟏𝟏∗𝟎.𝟔𝟑𝟕𝟖𝟒
𝟎.𝟎𝟏𝟔
= 𝟓𝟏𝟒. 𝟑𝟎𝟐𝟒𝐰/𝐦 𝟐
∗ 𝐜
Q=m`*Cp*(To-Ti)
Q=𝟗𝟗𝟎 ∗ 𝟎. 𝟎𝟖𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟑
∗ 𝟒. 𝟎𝟔𝟔 ∗ 𝟏𝟎 𝟑
∗ (𝟒𝟔 − 𝟒𝟐. 𝟗) =1374.3 w
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
1374.3 =(3.14∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 (𝟒𝟒. 𝟒𝟓 − 𝑻 𝒘
𝟏
)𝟓𝟏𝟒. 𝟑𝟎𝟐𝟒) ∗ 𝟒
𝑻 𝒘
𝟏
= 𝟒. 𝟕 Dynamic Viscosity=𝟏. 𝟒𝟕𝟐𝟐 ∗ 𝟏𝟎−𝟑
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟏𝟗𝟒𝟖. 𝟏𝟑𝟑 ∗ 𝟑. 𝟕𝟐𝟗𝟓 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟓.𝟖𝟒𝟗𝟔∗𝟏𝟎−𝟒
𝟏.𝟒𝟕𝟐𝟐∗𝟏𝟎−𝟑
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟏. 𝟒𝟖𝟔𝟐
𝒉 =
𝟏𝟏. 𝟒𝟖𝟔𝟐 ∗ 𝟎. 𝟔𝟑𝟕𝟖𝟒
𝟎. 𝟎𝟏𝟔
= 𝟒𝟓𝟕. 𝟖𝟗𝟕𝟒 𝐰/𝐦 𝟐
∗ 𝐜

10. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 10 | P a g e
Parallel flow; cold
𝑸 𝐜𝐨𝐥𝐝 = 300 l/h  𝑸 𝐜𝐨𝐥𝐝 = 𝟖. 𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟏 + 𝑻 𝟑
𝟐
=
𝟐𝟖. 𝟔 + 𝟐𝟑. 𝟖
𝟐
= 𝟐𝟔. 𝟐 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝒗 =
𝟖. 𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟓
𝟑. 𝟏𝟒 ∗ (𝟎. 𝟎𝟏𝟒) 𝟐
= 𝟎. 𝟏𝟑𝟓𝟑𝟑 𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝒅 𝟎
𝒖
=
𝟗𝟗𝟔. 𝟐𝟕𝟎𝟎 ∗ 𝟎. 𝟏𝟑𝟓𝟑𝟑 ∗ 𝟎. 𝟎𝟏𝟒
𝟖. 𝟒𝟓𝟎𝟕 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟐𝟐𝟑𝟑. 𝟔 < 2100
𝑳𝒊 =
𝟐𝟐𝟑𝟑. 𝟔 ∗ 𝟓. 𝟔𝟎𝟒𝟏 ∗ 𝟎. 𝟎𝟏𝟒
𝟐𝟎
= 𝟖. 𝟕𝟔𝟐𝟏
𝟖. 𝟕𝟔𝟐𝟏 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & Turbulent flow
𝐍 𝐮 = 𝟎. 𝟎𝟑𝟔 ∗ 𝟐𝟐𝟑𝟑. 𝟔 𝟎.𝟖
∗ 𝟓. 𝟔𝟎𝟒𝟏 𝟏/𝟑
∗ (𝟎. 𝟎𝟏𝟒 𝟎⁄ . 𝟑𝟑𝟓) 𝟏/𝟖
 𝐍 𝐮 = 𝟐𝟎. 𝟓𝟒𝟐𝟏
𝒉𝒊 = (𝟐𝟎. 𝟓𝟒𝟐𝟏 ∗ 𝟎. 𝟏𝟒𝟎𝟑)/𝟎. 𝟎𝟏𝟒 = 𝟐𝟎𝟓. 𝟖𝟔𝟏𝟐 𝐰/𝐦 𝟐
∗ 𝐜
∆𝟏 = 𝐓𝟓 − 𝐓𝟑  ∆𝟏 = 𝟒𝟔 − 𝟐𝟑. 𝟖 = 𝟐𝟐. 𝟐
∆𝟐 = 𝐓𝟕 − 𝐓𝟏 ∆𝟐 = 𝟒𝟐. 𝟗 − 𝟐𝟖. 𝟔 = 𝟏𝟒. 𝟑
∆𝑻 𝑳𝑴𝑻𝑫 =
∆ 𝟐−∆ 𝟏
𝒍𝒏(
∆ 𝟐
∆ 𝟏
)
∆𝑻 𝑳𝑴𝑻𝑫 =
𝟏𝟒.𝟑−𝟐𝟐.𝟐
𝒍𝒏(
𝟏𝟒.𝟑
𝟐𝟐.𝟐
)
=17.96
𝑼 𝒐 =
𝟏
𝟏
𝒉 𝒊
𝒓 𝒐
𝒓 𝒊
+
𝒓 𝒐(𝐥𝐧(
𝒓 𝒐
𝒓 𝒊
))
𝒌
+ 𝟏
𝒉 𝒐
 𝑼 𝒐 =
𝟏
𝟏
𝟒𝟓𝟕.𝟖𝟗𝟕𝟒
∗ 𝟎.𝟎𝟖
𝟎.𝟎𝟕
+
(𝟎.𝟎𝟖𝐥𝐧( 𝟎.𝟎𝟖
𝟎.𝟎𝟕
))
𝟑𝟒𝟗
+ 𝟏
𝟐𝟎𝟓.𝟖𝟔𝟏𝟐
= 𝟏𝟑𝟓. 𝟒𝟐𝟓
𝑸 = 𝑼 𝒐 𝑨 𝒐∆𝑻 𝑳𝑴𝑻𝑫 ∗ 𝑭
𝑸 = 𝟏𝟑𝟓. 𝟒𝟐𝟓 ∗ 𝝅 ∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 ∗ 𝟏𝟕. 𝟗𝟔 ∗ 𝑭
𝑸 = 𝟒𝟎. 𝟗𝑭 from fig at R=0.645 &p=0.216 F=1
𝑸 = 𝟒𝟎. 𝟗 𝒘

11. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 11 | P a g e
Parallel flow ; hot 𝑸 𝐡𝐨𝐭 = 300 l/h
at 𝑻 𝟓 = 𝟒𝟔 𝑻 𝟕 = 𝟒𝟒. 𝟐
𝑸 𝐡𝐨𝐭 = 300 l/h  𝑸 𝐡𝐨𝐭 = 𝟖. 𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟓 + 𝑻 𝟕
𝟐
=
𝟒𝟔 + 𝟒𝟒. 𝟐
𝟐
= 𝟒𝟓. 𝟏 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝑨 = 𝑨 𝒔𝒉𝒆𝒆𝒍 − 𝑨𝒕𝒖𝒃𝒆
=
𝝅
𝟒
𝒅 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
𝒅 𝒐
𝟐
)
=
𝝅
𝟒
(𝟓𝟎 ∗ 𝟏𝟎−𝟑
) 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
(𝟏𝟔 ∗ 𝟏𝟎−𝟑
) 𝒐
𝟐
)
=1.15866∗ 𝟏𝟎−𝟑
𝒗 =
𝟖. 𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟓
𝟏. 𝟏𝟓𝟖𝟔𝟔 ∗ 𝟏𝟎−𝟑
= 𝟎. 𝟎𝟕𝟏𝟗𝟏 𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝒅 𝟎
𝒖
=
𝟗𝟖𝟗.𝟕∗𝟎.𝟎𝟕𝟏𝟗𝟏∗𝟎.𝟎𝟏𝟔
𝟓.𝟕𝟖𝟕𝟒∗𝟏𝟎−𝟒
→ 𝑹 𝒆 = 𝟏𝟗𝟔𝟕. 𝟔 < 2100 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
𝑳𝒊 =
𝟏𝟗𝟔𝟕. 𝟔 ∗ 𝟑. 𝟔𝟖𝟓𝟕 ∗ 𝟎. 𝟎𝟏𝟔
𝟐𝟎
= 𝟓. 𝟖𝟎𝟏𝟔 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
Let 𝑻 𝒘 =𝑻 𝒇 − 𝟓  𝑻 𝒘 = 𝟑𝟗. 𝟒𝟓 𝒄  Dynamic Viscosity= 6.3424∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟏𝟗𝟔𝟕. 𝟔 ∗ 𝟑. 𝟔𝟖𝟓𝟕 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟓.𝟕𝟖𝟕∗𝟏𝟎−𝟒
𝟔.𝟑𝟒𝟐𝟒∗𝟏𝟎−𝟒)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟐. 𝟗𝟏𝟓𝟑
𝒉 𝟎 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 =
𝟏𝟐.𝟗𝟏𝟓𝟑∗𝟎.𝟔𝟑𝟖𝟓𝟑
𝟎.𝟎𝟏𝟔
= 𝟓𝟏𝟓. 𝟒𝟐𝟓𝟒 𝐰/𝐦 𝟐
∗ 𝐜
Q=m`*Cp*(To-Ti)
Q=𝟗𝟖𝟗. 𝟕 ∗ 𝟎. 𝟎𝟖𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟑
∗ 𝟒. 𝟎𝟔𝟔 ∗ 𝟏𝟎 𝟑
∗ (𝟒𝟔 − 𝟒𝟒. 𝟐) =603.5939w
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
603.5939=(3.14∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 (𝟒𝟒. 𝟒𝟓 − 𝑻 𝒘
𝟏
)𝟓𝟏𝟓. 𝟒𝟐𝟓𝟒 ) ∗ 𝟒
𝑻 𝒘
𝟏
= 𝟐𝟕. 𝟎𝟏𝟓  Dynamic Viscosity=𝟖. 𝟐𝟗𝟖𝟕 ∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟏𝟗𝟔𝟕. 𝟔 ∗ 𝟑. 𝟔𝟖𝟓𝟕 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟓.𝟖𝟕𝟖𝟕∗𝟏𝟎−𝟒
𝟖.𝟐𝟗𝟖𝟕∗𝟏𝟎−𝟒 )
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟐. 𝟒𝟑𝟖𝟑
𝒉 𝒐
𝟏
=
𝟏𝟐. 𝟒𝟑𝟖𝟑 ∗ 𝟎. 𝟔𝟑𝟕𝟖𝟒
𝟎. 𝟎𝟏𝟔
= 𝟒𝟗𝟔. 𝟑𝟖𝟗𝟐 𝐰/𝐦 𝟐
∗ 𝐜
𝒉 𝒐
𝒊𝒗
= 𝟒𝟕𝟖. 𝟖𝟗𝟕𝟓

12. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 12 | P a g e
Parallel flow; cold
𝑸 𝐜𝐨𝐥𝐝 = 150 l/h  𝑸 𝐜𝐨𝐥𝐝 = 4.1667 ∗ 10−5
m3/s
𝑇𝑓 =
𝑇1 + 𝑇3
2
=
29.8 + 24.4
2
= 27.1 𝑐0
𝒗 =
𝑸
𝑨
𝑣 =
4.1667 ∗ 10−5
3.14 ∗ (0.014)2
= 0.0677 𝑚/𝑠
𝑅 𝑒 =
𝜌𝑣 ⅆ0
𝑢
=
996.027 ∗ 0.0677 ∗ 0.014
8.2820 ∗ 10−4
𝑅 𝑒 = 1117.1 < 2100
𝐿𝑖 =
1117.1 ∗ 5.48 ∗ 0.014
20
= 4.2852
8.7621 > 𝐿(0.33𝑚)
.`.Developed & 𝑙𝑎𝑚𝑖𝑛𝑎𝑟 flow
Let 𝑇 𝑤 =𝑇𝑓 + 5  𝑇 𝑤 = 32.1 𝑐  Dynamic Viscosity= 7.4352∗ 10−4
Nu = 1.86 (1117.1 ∗ 5.48 ∗
0.014
0.335
)
1/3
(
8.282∗10−4
7.4352∗10−4)
0⋅14
 Nu = 11.9873
ℎ𝑖 =
𝑁𝑢∗𝑘
𝑑0
 ℎ =
11.9873∗0.61533
0.014
= 526.8675 w/m2
∗ c
Q=m`*Cp*(To-Ti)
Q=996.027 ∗ 4.1667 ∗ 10−5
∗ 4179.5 ∗ (29.8 − 24.4) =912.4552 w
Q=(3.14∗ ⅆ0 ∗ 𝑙 (𝑇𝑓 − 𝑇 𝑤)ℎ0) ∗ 4
912.4552/4 =3.14∗ 0.014 ∗ 0.335 (𝑇 𝑤
1
− 27.1)526.8675
𝑇 𝑤
1
= 56.48  Dynamic Viscosity=8.2987 ∗ 10−4
Nu = 1.86 (1117.1 ∗ 5.48 ∗
0.014
0.335
)
1/3
(
8.282∗10−4
8.2987∗10−4)
0⋅14
 Nu = 12.7482
ℎ𝑖
1
=
12.7482 ∗ 0.61533
0.014
= 560.3107 w/m2
∗ c
∆1 = T5 − T3  ∆1 = 46 − 24.4 = 21.6
∆2 = T7 − T1 ∆2 = 44.2 − 29.8 = 14.4

13. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 13 | P a g e
∆𝑇𝐿𝑀𝑇𝐷 = ∆2−∆1
𝑙𝑛(
∆2
∆1
)
∆𝑇𝐿𝑀𝑇𝐷 = 14.4−21.6
𝑙𝑛(14.4
21.6
)
= 16.523
𝑼 𝒐 =
𝟏
𝟏
𝒉 𝒊
𝒓 𝒐
𝒓 𝒊
+
𝒓 𝒐(𝐥𝐧(
𝒓 𝒐
𝒓 𝒊
))
𝒌
+
𝟏
𝒉 𝒐

𝑼 𝒐 =
𝟏
𝟏
𝟓𝟔𝟎.𝟑𝟏𝟎𝟕
∗
𝟎.𝟎𝟖
𝟎.𝟎𝟕
+
(𝟎.𝟎𝟖∗𝐥 𝐧(
𝟎.𝟎𝟖
𝟎.𝟎𝟕))
𝟑𝟒𝟗
+
𝟏
𝟒𝟕𝟖.𝟖𝟗𝟕𝟓
=𝟐𝟒𝟎
𝑄 = 𝑈 𝑜 𝐴 𝑜∆𝑇𝐿𝑀𝑇𝐷 ∗ 𝐹
𝑄 = 𝟐𝟒𝟎 ∗ 𝜋 ∗ 0.016 ∗ 0.335 ∗16.523 ∗ 𝐹
𝑄 = 66.775 𝐹 from fig at R=0.333 &p=0.25 F=1
𝑄 = 66.775 𝑤
Parallel flow; cold
𝑸 𝐜𝐨𝐥𝐝 = 50 l/h  𝑸 𝐜𝐨𝐥𝐝 = 𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟏 + 𝑻 𝟑
𝟐
=
𝟑𝟐. 𝟐 + 𝟐𝟒. 𝟔
𝟐
= 𝟐𝟖. 𝟒 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝒗 =
𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
𝟑. 𝟏𝟒 ∗ (𝟎. 𝟎𝟏𝟒) 𝟐
= 𝟎. 𝟎𝟐𝟐𝟔𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝒅 𝟎
𝒖
=
𝟗𝟗𝟓. 𝟎𝟐𝟕 ∗ 𝟎. 𝟎𝟐𝟐𝟔 ∗ 𝟎. 𝟎𝟏𝟒
𝟖. 𝟎𝟒𝟕𝟔 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟑𝟗𝟏. 𝟐𝟏𝟓𝟐< 2100
𝑳𝒊 =
𝟑𝟗𝟏. 𝟐𝟏𝟓𝟐 ∗ 𝟓. 𝟑𝟎𝟖𝟐 ∗ 𝟎. 𝟎𝟏𝟒
𝟐𝟎
= 𝟏. 𝟒𝟓𝟑𝟕
𝟖. 𝟕𝟔𝟐𝟏 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓 flow
Let 𝑻 𝒘 =𝑻 𝒇 + 𝟓  𝑻 𝒘 = 𝟐𝟖. 𝟒 + 𝟓 = 𝟑𝟑. 𝟒 𝒄  Dynamic Viscosity= 7.2375∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟑𝟗𝟏. 𝟐𝟏𝟓𝟐 ∗ 𝟓. 𝟑𝟎𝟖𝟐 ∗
𝟎.𝟎𝟏𝟒
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟖.𝟎𝟖𝟐𝟖∗𝟏𝟎−𝟒
𝟕.𝟐𝟑𝟕𝟓∗𝟏𝟎−𝟒)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟖. 𝟑𝟔𝟑𝟑
𝒉 𝟎 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 =
𝟖.𝟑𝟔𝟑𝟑∗𝟎.𝟔𝟏𝟕𝟏𝟖
𝟎.𝟎𝟏𝟒
= 𝟑𝟔𝟖. 𝟔𝟗𝟎𝟏 𝐰/𝐦 𝟐
∗ 𝐜

14. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 14 | P a g e
Q=m`*Cp*(To-Ti)
Q=𝟗𝟗𝟓. 𝟎𝟐𝟕 ∗ 𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
∗ 𝟒. 𝟎𝟕𝟎𝟗 ∗ 𝟏𝟎 𝟑
∗ (𝟑𝟐. 𝟐 − 𝟐𝟒. 𝟔) =427.5726w
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
427.5726/4 =3.14∗ 𝟎. 𝟎𝟏𝟒 ∗ 𝟎. 𝟑𝟑𝟓 (𝑻 𝒘
𝟏
− 𝟐𝟖. 𝟒)𝟑𝟔𝟖. 𝟔𝟗𝟎𝟏
𝑻 𝒘
𝟏
= 𝟒𝟖. 𝟎𝟖𝟓𝟏  Dynamic Viscosity=𝟓. 𝟒𝟗𝟑𝟖 ∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟏𝟏𝟏𝟕. 𝟏 ∗ 𝟓. 𝟒𝟖 ∗
𝟎.𝟎𝟏𝟒
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟖.𝟐𝟖𝟐∗𝟏𝟎−𝟒
𝟓.𝟒𝟗𝟑𝟖∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟖. 𝟔𝟗𝟐𝟒
𝒉𝒊
𝟏
=
𝟖. 𝟔𝟗𝟐𝟒 ∗ 𝟎. 𝟔𝟏𝟕𝟏𝟖
𝟎. 𝟎𝟏𝟒
= 𝟑𝟖𝟑. 𝟏𝟗𝟖𝟐 𝐰/𝐦 𝟐
∗ 𝐜
∆𝟏 = 𝐓𝟓 − 𝐓𝟑  ∆𝟏 = 𝟒𝟔. 𝟏 − 𝟐𝟒. 𝟔 = 𝟐𝟏. 𝟓
∆𝟐 = 𝐓𝟕 − 𝐓𝟏 ∆𝟐 = 𝟒𝟓. 𝟐 − 𝟑𝟐. 𝟐 = 𝟏𝟑
∆𝑻 𝑳𝑴𝑻𝑫 = ∆ 𝟐−∆ 𝟏
𝒍𝒏(
∆ 𝟐
∆ 𝟏
)
∆𝑻 𝑳𝑴𝑻𝑫 =
𝟏𝟑−𝟐𝟏.𝟓
𝒍𝒏( 𝟏𝟑
𝟐𝟏.𝟓
)
= 16.8951
𝑼 𝒐 =
𝟏
𝟏
𝒉 𝒊
𝒓 𝒐
𝒓 𝒊
+
𝒓 𝒐(𝐥𝐧(
𝒓 𝒐
𝒓 𝒊
))
𝒌
+ 𝟏
𝒉 𝒐
𝑼 𝒐 =
𝟏
𝟏
𝟑𝟖𝟑.𝟏𝟗𝟖𝟐
∗
𝟎.𝟎𝟖
𝟎.𝟎𝟕
+
(𝟎.𝟎𝟖∗𝐥 𝐧(
𝟎.𝟎𝟖
𝟎.𝟎𝟕))
𝟑𝟒𝟗
+
𝟏
𝟒𝟕𝟖.𝟖𝟗𝟕𝟓
=𝑼 𝒐 = 𝟏𝟗𝟔. 𝟎𝟎𝟑
𝑸 = 𝑼 𝒐 𝑨 𝒐∆𝑻 𝑳𝑴𝑻𝑫 ∗ 𝑭
𝑸 = 𝟏𝟗𝟔. 𝟎𝟎𝟑 ∗ 𝝅 ∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 ∗ 𝟏𝟔. 𝟖𝟗𝟓𝟏 ∗ 𝑭
𝑸 = 𝟓𝟓. 𝟕𝟔 𝑭 From fig at R=0.10526 &p=0.353 F=1
𝑸 = 𝟓𝟓. 𝟕𝟔 𝒘

15. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 15 | P a g e
Parallel flow ; hot
𝑸 𝐡𝐨𝐭 = 150 l/h  𝑸 𝐡𝐨𝐭 = 4.1667∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟓 + 𝑻 𝟕
𝟐
=
𝟒𝟔 + 𝟒𝟎. 𝟖
𝟐
= 𝟒𝟑. 𝟒 𝒄 𝟎
𝐯 =
𝐐
𝐀
𝑨 = 𝑨 𝒔𝒉𝒆𝒆𝒍 − 𝑨𝒕𝒖𝒃𝒆
=
𝝅
𝟒
𝒅 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
𝒅 𝒐
𝟐
)
=
𝝅
𝟒
(𝟓𝟎 ∗ 𝟏𝟎−𝟑
) 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
(𝟏𝟔 ∗ 𝟏𝟎−𝟑
) 𝒐
𝟐
)
=1.15866∗ 𝟏𝟎−𝟑
𝒗 =
𝟒. 𝟏𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟓
𝟏. 𝟏𝟓𝟖𝟔𝟔 ∗ 𝟏𝟎−𝟑
= 𝟎. 𝟎𝟑𝟔𝟎 𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝐝 𝟎
𝒖
=
𝟗𝟗𝟎. 𝟒 ∗ 𝟎. 𝟎𝟑𝟔𝟎 ∗ 𝟎. 𝟎𝟏𝟔
𝟓. 𝟗𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟗𝟓𝟔. 𝟎𝟗𝟎𝟑< 2100 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
𝑳𝒊 =
𝟗𝟓𝟔. 𝟎𝟗𝟎𝟑 ∗ 𝟑. 𝟖𝟏𝟏𝟗 ∗ 𝟎. 𝟎𝟏𝟔
𝟐𝟎
= 𝟐. 𝟗𝟏𝟓𝟔 > 𝑳( 𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
Let 𝑻 𝒘 =𝑻 𝒇 − 𝟓  𝑻 𝒘 = 𝟒𝟑. 𝟒 − 𝟓 = 𝟑𝟖. 𝟒 𝒄  Dynamic Viscosity=6.5516∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 ( 𝟗𝟓𝟔. 𝟎𝟗𝟎𝟑 ∗ 𝟑. 𝟖𝟏𝟏𝟗 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟓.𝟖𝟒𝟗𝟔∗𝟏𝟎−𝟒
𝟔.𝟓𝟓𝟏𝟔∗𝟏𝟎−𝟒 )
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟎. 𝟐𝟓𝟎𝟐
𝒉 𝟎 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 =
𝟏𝟎.𝟐𝟓𝟎𝟐∗𝟎.𝟔𝟑𝟔𝟓𝟓
𝟎.𝟎𝟏𝟔
= 𝟒𝟎𝟕. 𝟕𝟗𝟕𝟖 𝐰/𝐦 𝟐
∗ 𝐜
Q=m`*Cp*(To-Ti)
Q= 𝟗𝟗𝟎. 𝟒 ∗ 𝟒. 𝟏𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟓
∗ 𝟒. 𝟎𝟔𝟔𝟕 ∗ 𝟏𝟎 𝟑
∗ ( 𝟒𝟔 − 𝟒𝟎. 𝟖) = 872.6666w
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
872.6666=(3.14∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 ( 𝟒𝟑. 𝟒 − 𝑻 𝒘
𝟏
) 𝟓𝟏𝟒. 𝟑𝟎𝟐𝟒) ∗ 𝟒
𝑻 𝒘
𝟏
= 𝟕𝟓. 𝟏𝟖𝟕𝟎  Dynamic Viscosity=𝟑. 𝟔𝟕𝟕𝟐 ∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 ( 𝟗𝟓𝟔. 𝟎𝟗𝟎𝟑 ∗ 𝟑. 𝟖𝟏𝟏𝟗 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟓.𝟖𝟒𝟗𝟔∗𝟏𝟎−𝟒
𝟑.𝟔𝟕𝟕𝟐∗𝟏𝟎−𝟒 )
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟏. 𝟏𝟏𝟑𝟒
𝒉 𝒐 =
𝟏𝟏. 𝟏𝟏𝟑𝟒 ∗ 𝟎. 𝟔𝟑𝟔𝟓𝟓
𝟎. 𝟎𝟏𝟔
= 𝟒𝟒𝟐. 𝟏𝟑𝟗𝟕 𝐰/𝐦 𝟐
∗ 𝐜

16. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 16 | P a g e
Parallel flow; cold
𝑸 𝐜𝐨𝐥𝐝 = 300 l/h  𝑸 𝐜𝐨𝐥𝐝 = 𝟖. 𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟏 + 𝑻 𝟑
𝟐
=
𝟐𝟖. 𝟏 + 𝟐𝟒. 𝟐
𝟐
= 𝟐𝟔. 𝟏𝟓 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝒗 =
𝟖. 𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟓
𝟑. 𝟏𝟒 ∗ (𝟎. 𝟎𝟏𝟒) 𝟐
= 𝟎. 𝟏𝟑𝟓𝟑𝟑 𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝒅 𝟎
𝒖
=
𝟗𝟗𝟔. 𝟐𝟖 ∗ 𝟎. 𝟏𝟑𝟓𝟑𝟑 ∗ 𝟎. 𝟎𝟏𝟒
𝟖. 𝟒𝟔𝟎𝟐 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟐𝟐𝟑𝟏. 𝟏 < 2100
𝑳𝒊 =
𝟐𝟐𝟑𝟏. 𝟏 ∗ 𝟓. 𝟔𝟏𝟏𝟏 ∗ 𝟎. 𝟎𝟏𝟒
𝟐𝟎
= 𝟖. 𝟕𝟔𝟐𝟏
𝟖. 𝟕𝟔𝟑𝟐 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & Turbulent flow
𝐍 𝐮 = 𝟎. 𝟎𝟑𝟔 𝐑𝐞 𝟎.𝟖
𝐩 𝐫
𝟏/𝟑
(𝐝 ∕ 𝐋) 𝟏/𝟖
𝐍 𝐮 = 𝟎. 𝟎𝟑𝟔 ∗ 𝟐𝟐𝟑𝟏. 𝟏 𝟎.𝟖
∗ 𝟓. 𝟔𝟏𝟏𝟏 𝟏/𝟑
∗ (𝟎. 𝟎𝟏𝟒 𝟎⁄ . 𝟑𝟑𝟓) 𝟏/𝟖
 𝐍 𝐮 = 𝟐𝟎. 𝟓𝟑𝟐𝟐
𝒉𝒊 = (𝟐𝟎. 𝟓𝟑𝟐𝟐 ∗ 𝟎. 𝟔𝟏𝟑𝟗𝟔)/𝟎. 𝟎𝟏𝟒 = 𝟗𝟎𝟎. 𝟒𝟐𝟓𝐰/𝐦 𝟐
∗ 𝐜
∆𝟏 = 𝐓𝟓 − 𝐓𝟑  ∆𝟏 = 𝟒𝟔 − 𝟐𝟒. 𝟐 = 𝟐𝟏. 𝟖
∆𝟐 = 𝐓𝟕 − 𝐓𝟏 ∆𝟐 = 𝟒𝟎. 𝟖 − 𝟐𝟖. 𝟏 = 𝟏𝟐. 𝟕
∆𝑻 𝑳𝑴𝑻𝑫 =
∆ 𝟐−∆ 𝟏
𝒍𝒏(
∆ 𝟐
∆ 𝟏
)
∆𝑻 𝑳𝑴𝑻𝑫 =
𝟏𝟐.𝟕−𝟐𝟏.𝟖
𝒍𝒏(
𝟏𝟐.𝟕
𝟐𝟏.𝟖
)
=16.8
𝑼 𝒐 =
𝟏
𝟏
𝒉 𝒊
𝒓 𝒐
𝒓 𝒊
+
𝒓 𝒐(𝐥𝐧(
𝒓 𝒐
𝒓 𝒊
))
𝒌
+ 𝟏
𝒉 𝒐
 𝑼 𝒐 =
𝟏
𝟏
𝟗𝟎𝟎.𝟒𝟐𝟓
∗
𝟎.𝟎𝟖
𝟎.𝟎𝟕
+
(𝟎.𝟎𝟖𝐥𝐧(
𝟎.𝟎𝟖
𝟎.𝟎𝟕))
𝟑𝟒𝟗
+
𝟏
𝟒𝟒𝟐.𝟏𝟑𝟗𝟕
= 𝟐𝟖𝟎. 𝟕
𝑸 = 𝑼 𝒐 𝑨 𝒐∆𝑻 𝑳𝑴𝑻𝑫 ∗ 𝑭
𝑸 = 𝟐𝟖𝟎. 𝟕 ∗ 𝝅 ∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 ∗ 𝟏𝟔. 𝟖 ∗ 𝑭
𝑸 = 𝟕𝟎. 𝟒 ∗ 𝟎. 𝟗𝟖 from fig at R=1.333 &p=0.17888 F=0.98
𝑸 = 𝟔𝟖. 𝟗𝟗 𝒘

17. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 17 | P a g e
Parallel flow; cold
𝑸 𝐜𝐨𝐥𝐝 = 150 l/h  𝑸 𝐜𝐨𝐥𝐝 = 𝟒. 𝟏𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟏 + 𝑻 𝟑
𝟐
=
𝟐𝟗. 𝟓 + 𝟐𝟒. 𝟐
𝟐
= 𝟐𝟔. 𝟖𝟓 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝒗 =
𝟒. 𝟏𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟓
𝟑. 𝟏𝟒 ∗ (𝟎. 𝟎𝟏𝟒) 𝟐
= 𝟎. 𝟎𝟔𝟕𝟕 𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝒅 𝟎
𝒖
=
𝟗𝟗𝟔. 𝟗 ∗ 𝟎. 𝟎𝟔𝟕𝟕 ∗ 𝟎. 𝟎𝟏𝟒
𝟖. 𝟑𝟐𝟖𝟑 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟏𝟏𝟑𝟒. 𝟓 < 2100
𝑳𝒊 =
𝟏𝟏𝟑𝟒. 𝟓 ∗ 𝟓. 𝟓𝟏𝟒𝟏 ∗ 𝟎. 𝟎𝟏𝟒
𝟐𝟎
= 𝟒. 𝟑𝟕𝟗𝟎
𝟒. 𝟑𝟕𝟗𝟎 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓 flow
Let 𝑻 𝒘 =𝑻 𝒇 + 𝟓  𝑻 𝒘 = 𝟐𝟔. 𝟖𝟓 + 𝟓 = 𝟑𝟏. 𝟖𝟓  Dynamic Viscosity= 7.4742∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟏𝟏𝟑𝟒. 𝟓 ∗ 𝟓. 𝟓𝟏𝟒𝟏 ∗
𝟎.𝟎𝟏𝟒
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟖.𝟑𝟐𝟖𝟑∗𝟏𝟎−𝟒
𝟕.𝟒𝟕𝟒𝟐∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟐. 𝟎𝟕𝟒𝟖
𝒉𝒊 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 =
𝟏𝟐.𝟎𝟕𝟒𝟖∗𝟎.𝟔𝟏𝟒𝟗
𝟎.𝟎𝟏𝟒
= 𝟓𝟑𝟎. 𝟑𝟒𝟐𝟓 𝐰/𝐦 𝟐
∗ 𝐜
Q=m`*Cp*(To-Ti)
Q=𝟗𝟗𝟔. 𝟗 ∗ 𝟒. 𝟏𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟓
∗ 𝟒𝟎𝟕𝟏. 𝟕 ∗ (𝟐𝟗. 𝟓 − 𝟐𝟒. 𝟐) =896.3868w
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
896.3868/4 =3.14∗ 𝟎. 𝟎𝟏𝟒 ∗ 𝟎. 𝟑𝟑𝟓 (𝑻 𝒘
𝟏
− 𝟐𝟔. 𝟖𝟓)𝟓𝟑𝟎. 𝟑𝟒𝟐𝟓
𝑻 𝒘
𝟏
= 𝟓𝟓. 𝟓𝟒𝟑𝟎  Dynamic Viscosity=𝟒. 𝟖𝟔𝟏𝟑 ∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟏𝟏𝟑𝟒. 𝟓 ∗ 𝟓. 𝟓𝟏𝟒𝟏 ∗
𝟎.𝟎𝟏𝟒
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟖.𝟑𝟐𝟖𝟑∗𝟏𝟎−𝟒
𝟒.𝟖𝟔𝟏𝟑∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟐. 𝟖𝟐𝟒𝟑
𝒉𝒊
𝟏
=
𝟏𝟐. 𝟖𝟐𝟒𝟑 ∗ 𝟎. 𝟔𝟏𝟒𝟗
𝟎. 𝟎𝟏𝟒
= 𝟓𝟔𝟑. 𝟐𝟔𝟏𝟔 𝐰/𝐦 𝟐
∗ 𝐜
∆𝟏 = 𝐓𝟓 − 𝐓𝟑  ∆𝟏 = 𝟒𝟔 − 𝟐𝟒. 𝟐 = 𝟐𝟏. 𝟖
∆𝟐 = 𝐓𝟕 − 𝐓𝟏 ∆𝟐 = 𝟒𝟐. 𝟕 − 𝟐𝟗. 𝟓 = 𝟏𝟑. 𝟐

18. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 18 | P a g e
∆𝑻 𝑳𝑴𝑻𝑫 =
∆ 𝟐−∆ 𝟏
𝒍𝒏(
∆ 𝟐
∆ 𝟏
)
∆𝑻 𝑳𝑴𝑻𝑫 =
𝟏𝟑.𝟐−𝟐𝟏.𝟖
𝒍𝒏(
𝟏𝟑.𝟐
𝟐𝟏.𝟖
)
= 17.14
𝑼 𝒐 =
𝟏
𝟏
𝒉 𝒊
𝒓 𝒐
𝒓 𝒊
+
𝒓 𝒐(𝐥𝐧(
𝒓 𝒐
𝒓 𝒊
))
𝒌
+ 𝟏
𝒉 𝒐
𝑼 𝒐 =
𝟏
𝟏
𝟗𝟎𝟎. 𝟒𝟐𝟓
∗
𝟎. 𝟎𝟖
𝟎. 𝟎𝟕
+
( 𝟎. 𝟎𝟖 ∗ 𝐥 𝐧 (
𝟎. 𝟎𝟖
𝟎. 𝟎𝟕
))
𝟑𝟒𝟗
+
𝟏
𝟓𝟔𝟑. 𝟐𝟔𝟏𝟔
𝑼 𝒐 = 𝟑𝟐𝟓. 𝟏𝟗𝟒
𝑸 = 𝑼 𝒐 𝑨 𝒐∆𝑻 𝑳𝑴𝑻𝑫 ∗ 𝑭
𝑸 = 𝟑𝟐𝟓. 𝟏𝟗𝟒 ∗ 𝝅 ∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 ∗ 𝟏𝟕. 𝟏𝟒 ∗ 𝑭
𝑸 = 𝟗𝟑. 𝟖𝟓𝟕 𝑭 from fig at R=0.622 &p=0.2431 F=1
𝑸 = 𝟗𝟑. 𝟖𝟓𝟕 𝒘
Parallel flow ; hot
𝑸 𝐡𝐨𝐭 = 150 l/h  𝑸 𝐡𝐨𝐭 = 𝟒. 𝟏𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟓 + 𝑻 𝟕
𝟐
=
𝟒𝟔. 𝟏 + 𝟒𝟒. 𝟓
𝟐
= 𝟒𝟓. 𝟑 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝑨 = 𝑨 𝒔𝒉𝒆𝒆𝒍 − 𝑨𝒕𝒖𝒃𝒆
=
𝝅
𝟒
𝒅 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
𝒅 𝒐
𝟐
)
=
𝝅
𝟒
(𝟓𝟎 ∗ 𝟏𝟎−𝟑
) 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
(𝟏𝟔 ∗ 𝟏𝟎−𝟑
) 𝒐
𝟐
)
=1.15866∗ 𝟏𝟎−𝟑
𝒗 =
𝟒. 𝟏𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟓
𝟏. 𝟏𝟓𝟖𝟔𝟔 ∗ 𝟏𝟎−𝟑
= 𝟎. 𝟎𝟑𝟔 𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝐝 𝟎
𝒖
=
𝟗𝟖𝟗.𝟕∗𝟎.𝟎𝟑𝟔∗𝟎.𝟎𝟏𝟔
𝟓.𝟕𝟔𝟔𝟗∗𝟏𝟎−𝟒 → 𝑹 𝒆 = 𝟗𝟖𝟖. 𝟓𝟏𝟓𝟖 < 2100 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
𝑳𝒊 =
𝟗𝟖𝟖. 𝟓𝟏𝟓𝟖 ∗ 𝟑. 𝟔𝟕𝟏𝟑 ∗ 𝟎. 𝟎𝟏𝟔
𝟐𝟎
= 𝟐. 𝟗𝟎𝟑𝟑 > 𝑳(𝟎. 𝟑𝟑𝒎)

19. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 19 | P a g e
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
Let 𝑻 𝒘 =𝑻 𝒇 − 𝟓  𝑻 𝒘 = 𝟒𝟓. 𝟑 − 𝟓 = 𝟒𝟎. 𝟑 𝒄  Dynamic Viscosity=6.3185∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟗𝟖𝟖. 𝟓𝟏𝟓𝟖 ∗ 𝟑. 𝟔𝟕𝟏𝟑 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟓.𝟕𝟔𝟔𝟗∗𝟏𝟎−𝟒
𝟔.𝟑𝟏𝟖𝟓∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟎. 𝟐𝟑𝟖𝟗
𝒉 𝟎 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 =
𝟏𝟎.𝟐𝟑𝟖𝟗∗𝟎.𝟔𝟑𝟖𝟕𝟔
𝟎.𝟎𝟏𝟔
= 𝟒𝟎𝟖. 𝟕𝟔𝟐𝟓 𝐰/𝐦 𝟐
∗ 𝐜
Q=m`*Cp*(To-Ti)
Q=𝟗𝟖𝟗. 𝟕 ∗ 𝟒. 𝟏𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟓
∗ 𝟒. 𝟎𝟔𝟔 ∗ 𝟏𝟎 𝟑
∗ (𝟒𝟔. 𝟏 − 𝟒𝟒. 𝟓) = 268.2768 w
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
268.2768=(3.14∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 (𝟒𝟓. 𝟑 − 𝑻 𝒘
𝟏
)𝟒𝟎𝟖. 𝟕𝟔𝟐𝟓) ∗ 𝟒
𝑻 𝒘
𝟏
= 𝟑𝟔  Dynamic Viscosity=𝟔. 𝟖𝟔𝟔𝟗 ∗ 𝟏𝟎−𝟒
𝐮 = 𝟏. 𝟖𝟔 (𝟗𝟖𝟖. 𝟓𝟏𝟓𝟖 ∗ 𝟑. 𝟔𝟕𝟏𝟑 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟓.𝟕𝟔𝟔𝟗∗𝟏𝟎−𝟒
𝟔.𝟖𝟔𝟔𝟗∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟎. 𝟏𝟐𝟎𝟑
𝒉 𝒐
𝟏
=
𝟏𝟎. 𝟏𝟐𝟎𝟑 ∗ 𝟎. 𝟔𝟑𝟖𝟕𝟔
𝟎. 𝟎𝟏𝟔
= 𝟒𝟎𝟒. 𝟎𝟐𝟕𝟕 𝐰/𝐦 𝟐
∗ 𝐜
Parallel flow; cold
𝑸 𝐜𝐨𝐥𝐝 = 50 l/h  𝑸 𝐜𝐨𝐥𝐝 = 𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟏 + 𝑻 𝟑
𝟐
=
𝟑𝟏. 𝟕 + 𝟐𝟒. 𝟔
𝟐
= 𝟐𝟖. 𝟏𝟓 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝒗 =
𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
𝟑. 𝟏𝟒 ∗ (𝟎. 𝟎𝟏𝟒) 𝟐
= 𝟎. 𝟎𝟐𝟐𝟔𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝒅 𝟎
𝒖
=
𝟗𝟗𝟓. 𝟕𝟐 ∗ 𝟎. 𝟎𝟐𝟐𝟔 ∗ 𝟎. 𝟎𝟏𝟒
𝟖. 𝟎𝟗𝟏𝟗 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟑𝟖𝟗. 𝟑𝟑𝟒𝟖< 2100
𝑳𝒊 =
𝟑𝟖𝟗. 𝟑𝟑𝟒𝟖 ∗ 𝟓. 𝟑𝟒𝟎𝟔 ∗ 𝟎. 𝟎𝟏𝟒
𝟐𝟎
= 𝟏. 𝟒𝟓𝟓𝟓
𝟏. 𝟒𝟓𝟓𝟓 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓 flow
Let 𝑻 𝒘 =𝑻 𝒇 + 𝟓  𝑻 𝒘 = 𝟐𝟖. 𝟏𝟓 + 𝟓 = 𝟑𝟑. 𝟏𝟓 𝒄  Dynamic Viscosity= 7.2749∗ 𝟏𝟎−𝟒

20. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 20 | P a g e
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟑𝟖𝟗. 𝟑𝟑𝟒𝟖 ∗ 𝟓. 𝟑𝟒𝟎𝟔 ∗
𝟎.𝟎𝟏𝟒
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟖.𝟎𝟗𝟏𝟗∗𝟏𝟎−𝟒
𝟕.𝟐𝟕𝟒𝟗∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟖. 𝟕𝟒𝟐𝟕
𝒉 𝟎 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 =
𝟖.𝟕𝟒𝟐𝟕∗𝟎.𝟔𝟏𝟔𝟖𝟐
𝟎.𝟎𝟏𝟒
= 𝟑𝟖𝟓. 𝟏𝟗𝟎𝟗 𝐰/𝐦 𝟐
∗ 𝐜
Q=m`*Cp*(To-Ti)
Q=𝟗𝟗𝟓. 𝟕𝟐 ∗ 𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
∗ 𝟒. 𝟎𝟕𝟎𝟗 ∗ 𝟏𝟎 𝟑
∗ (𝟑𝟏. 𝟕 − 𝟐𝟒. 𝟔) =399.7210 w
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
399.7210/4 =3.14∗ 𝟎. 𝟎𝟏𝟒 ∗ 𝟎. 𝟑𝟑𝟓 (𝑻 𝒘
𝟏
− 𝟐𝟖. 𝟏𝟓)𝟑𝟗𝟗. 𝟕𝟐𝟏𝟎
𝑻 𝒘
𝟏
= 𝟒𝟓. 𝟏𝟐𝟔𝟏  Dynamic Viscosity=𝟓. 𝟕𝟖𝟒 ∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟑𝟖𝟗. 𝟑𝟑𝟒𝟖 ∗ 𝟓. 𝟑𝟒𝟎𝟔 ∗
𝟎.𝟎𝟏𝟒
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟖.𝟎𝟗𝟏𝟗∗𝟏𝟎−𝟒
𝟓.𝟕𝟖𝟒𝟕∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟗. 𝟎𝟐𝟕𝟖
𝒉 𝟎
𝟏
=
𝟗. 𝟎𝟐𝟕𝟖 ∗ 𝟎. 𝟔𝟏𝟔𝟖𝟐
𝟎. 𝟎𝟏𝟒
= 𝟑𝟗𝟕. 𝟕𝟓𝟐𝟐 𝐰/𝐦 𝟐
∗ 𝐜
∆𝟏 = 𝐓𝟓 − 𝐓𝟑  ∆𝟏 = 𝟒𝟔. 𝟐 − 𝟐𝟒. 𝟔 = 𝟐𝟏. 𝟔𝟎
∆𝟐 = 𝐓𝟕 − 𝐓𝟏 ∆𝟐 = 𝟒𝟒. 𝟓 − 𝟑𝟏. 𝟕 = 𝟏𝟐. 𝟖
∆𝑻 𝑳𝑴𝑻𝑫 =
∆ 𝟐−∆ 𝟏
𝒍𝒏(
∆ 𝟐
∆ 𝟏
)
∆𝑻 𝑳𝑴𝑻𝑫 =
𝟏𝟐.𝟖−𝟐𝟏.𝟔𝟎
𝒍𝒏(
𝟏𝟐.𝟖
𝟐𝟏.𝟔𝟎
)
= 16.818
𝑼 𝒐 = 𝟏
𝟏
𝒉 𝒊
𝒓 𝒐
𝒓 𝒊
+
𝒓 𝒐(𝐥𝐧(
𝒓 𝒐
𝒓 𝒊
))
𝒌
+ 𝟏
𝒉 𝒐
𝑼 𝒐 =
𝟏
𝟏
𝟒𝟎𝟒. 𝟎𝟐𝟕𝟕
∗
𝟎. 𝟎𝟖
𝟎. 𝟎𝟕
+
(𝟎. 𝟎𝟖 ∗ 𝐥 𝐧 (
𝟎. 𝟎𝟖
𝟎. 𝟎𝟕))
𝟑𝟒𝟗
+
𝟏
𝟑𝟗𝟕. 𝟕𝟓𝟐𝟐
𝑼 𝒐 = 𝟏𝟖𝟔. 𝟏
𝑸 = 𝑼 𝒐 𝑨 𝒐∆𝑻 𝑳𝑴𝑻𝑫 ∗ 𝑭
𝑸 = 𝟏𝟖𝟔. 𝟏 ∗ 𝝅 ∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 ∗ 𝟏𝟔. 𝟖𝟏𝟖 ∗ 𝑭
𝑸 = 𝟓𝟐. 𝟕 𝑭 From fig at R=0.2394 &p=0.328 F=1
𝑸 = 𝟓𝟐. 𝟕 𝒘

21. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 21 | P a g e
Parallel flow ; hot
𝑸 𝐡𝐨𝐭 = 50 l/h  𝑸 𝐡𝐨𝐭 = 𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟓 + 𝑻 𝟕
𝟐
=
𝟒𝟓. 𝟖 + 𝟑𝟓. 𝟗
𝟐
= 𝟒𝟎. 𝟖𝟓 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝑨 = 𝑨 𝒔𝒉𝒆𝒆𝒍 − 𝑨𝒕𝒖𝒃𝒆
=
𝝅
𝟒
𝒅 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
𝒅 𝒐
𝟐
)
=
𝝅
𝟒
(𝟓𝟎 ∗ 𝟏𝟎−𝟑
) 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
(𝟏𝟔 ∗ 𝟏𝟎−𝟑
) 𝒐
𝟐
)
=1.15866∗ 𝟏𝟎−𝟑
𝒗 =
𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
𝟏. 𝟏𝟓𝟖𝟔𝟔 ∗ 𝟏𝟎−𝟑
= 𝟎. 𝟎𝟏𝟐𝟎 𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝐝 𝟎
𝒖
=
𝟗𝟗𝟏. 𝟒 ∗ 𝟎. 𝟎𝟏𝟐𝟎 ∗ 𝟎. 𝟎𝟏𝟔
𝟔. 𝟐𝟓𝟑 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟑𝟎𝟒. 𝟒𝟏𝟐𝟎< 2100 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
𝑳𝒊 =
𝟑𝟎𝟒. 𝟒𝟏𝟐𝟎 ∗ 𝟒. 𝟎𝟏𝟒𝟖 ∗ 𝟎. 𝟎𝟏𝟔
𝟐𝟎
= 𝟎. 𝟗𝟕𝟕𝟕 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
Let 𝑻 𝒘 =𝑻 𝒇 − 𝟓  𝑻 𝒘 = 𝟒𝟎. 𝟖𝟓 − 𝟓 = 𝟑𝟓. 𝟖𝟓 𝒄  Dynamic Viscosity=6.8875∗
𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟑𝟎𝟒. 𝟒𝟏𝟐𝟎 ∗ 𝟒. 𝟎𝟏𝟒𝟖 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟔.𝟐𝟓𝟑∗𝟏𝟎−𝟒
𝟔.𝟖𝟖𝟕𝟓∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟕. 𝟏𝟏𝟖𝟐
𝒉 𝟎 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 =
𝟕.𝟏𝟏𝟖𝟐∗𝟎.𝟔𝟑𝟑𝟓𝟏
𝟎.𝟎𝟏𝟔
= 𝟐𝟖𝟏. 𝟖𝟒𝟎𝟕 𝐰/𝐦 𝟐
∗ 𝐜
Q=m`*Cp*(To-Ti)
Q=𝟗𝟗𝟏. 𝟒 ∗ 𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
∗ 𝟒. 𝟎𝟔𝟕𝟕 ∗ 𝟏𝟎 𝟑
∗ (𝟒𝟓. 𝟖 − 𝟑𝟓. 𝟗) = 554.5031 w
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
554.5031=(3.14∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 (𝟒𝟎. 𝟖𝟓 − 𝑻 𝒘
𝟏
)𝟓𝟏𝟒. 𝟑𝟎𝟐𝟒) ∗ 𝟒
𝑻 𝒘
𝟏
= 𝟕𝟒. 𝟐𝟒𝟗𝟑  Dynamic Viscosity=𝟑. 𝟕𝟐𝟐𝟐 ∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟑𝟎𝟒. 𝟒𝟏𝟐𝟎 ∗ 𝟒. 𝟎𝟏𝟒𝟖 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏
𝟑
(
𝟔.𝟐𝟓𝟑∗𝟏𝟎−𝟒
𝟑.𝟕𝟐𝟐𝟐∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟕. 𝟕𝟓𝟖𝟕
𝒉 𝒐 =
𝟕. 𝟕𝟓𝟖𝟕 ∗ 𝟎. 𝟔𝟑𝟑𝟓𝟏
𝟎. 𝟎𝟏𝟔
= 𝟑𝟎𝟕. 𝟐𝟎𝟎𝟗 𝐰/𝐦 𝟐
∗ 𝐜

22. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 22 | P a g e
Parallel flow; cold
𝑸 𝐜𝐨𝐥𝐝 = 300 l/h  𝑸 𝐜𝐨𝐥𝐝 = 𝟖. 𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟏 + 𝑻 𝟑
𝟐
=
𝟐𝟕 + 𝟐𝟑. 𝟔
𝟐
= 𝟐𝟓. 𝟑 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝒗 =
𝟖. 𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟓
𝟑. 𝟏𝟒 ∗ (𝟎. 𝟎𝟏𝟒) 𝟐
= 𝟎. 𝟏𝟑𝟓𝟑𝟑 𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝒅 𝟎
𝒖
=
𝟗𝟗𝟔. 𝟓𝟏𝟖 ∗ 𝟎. 𝟏𝟑𝟓𝟑𝟑 ∗ 𝟎. 𝟎𝟏𝟒
𝟖. 𝟔𝟐𝟒 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟐𝟏𝟖𝟗. 𝟑 < 2100
𝑳𝒊 =
𝟐𝟏𝟖𝟗. 𝟑 ∗ 𝟓. 𝟕𝟑𝟐𝟓 ∗ 𝟎. 𝟎𝟏𝟒
𝟐𝟎
= 𝟖. 𝟕𝟖𝟓𝟏
𝟖. 𝟕𝟖𝟓𝟏 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & Turbulent flow
𝐍 𝐮 = 𝟎. 𝟎𝟑𝟔 𝐑𝐞 𝟎.𝟖
𝐩 𝐫
𝟏/𝟑
(𝐝 ∕ 𝐋) 𝟏/𝟖
𝐍 𝐮 = 𝟎. 𝟎𝟑𝟔 ∗ 𝟐𝟏𝟖𝟗. 𝟑 𝟎.𝟖
∗ 𝟓. 𝟕𝟑𝟐𝟓 𝟏/𝟑
∗ (𝟎. 𝟎𝟏𝟒 𝟎⁄ . 𝟑𝟑𝟓) 𝟏/𝟖
 𝐍 𝐮 = 𝟐𝟎. 𝟑𝟔𝟖𝟕
𝒉𝒊 =
𝟐𝟎. 𝟑𝟔𝟖𝟕 ∗ 𝟎. 𝟔𝟏𝟐𝟕𝟑
𝟎. 𝟎𝟏𝟒
= 𝟖𝟗𝟏. 𝟒𝟔𝟓𝟑 𝐰/𝐦 𝟐
∗ 𝐜
∆𝟏 = 𝐓𝟓 − 𝐓𝟑  ∆𝟏 = 𝟒𝟓. 𝟖 − 𝟐𝟑. 𝟔 = 𝟐𝟐. 𝟐
∆𝟐 = 𝐓𝟕 − 𝐓𝟏 ∆𝟐 = 𝟑𝟓. 𝟗 − 𝟐𝟕 = 𝟖. 𝟗
∆𝑻 𝑳𝑴𝑻𝑫 =
∆ 𝟐−∆ 𝟏
𝒍𝒏(
∆ 𝟐
∆ 𝟏
)
∆𝑻 𝑳𝑴𝑻𝑫 =
𝟖.𝟗−𝟐𝟐.𝟐
𝒍𝒏(
𝟖.𝟗
𝟐𝟐.𝟐
)
=14.55
𝑼 𝒐 = 𝟏
𝟏
𝒉 𝒊
𝒓 𝒐
𝒓 𝒊
+
𝒓 𝒐(𝐥𝐧(
𝒓 𝒐
𝒓 𝒊
))
𝒌
+ 𝟏
𝒉 𝒐
 𝑼 𝒐 =
𝟏
𝟏
𝟖𝟗𝟏.𝟒𝟔𝟓𝟑
∗
𝟎.𝟎𝟖
𝟎.𝟎𝟕
+
(𝟎.𝟎𝟖𝐥𝐧(
𝟎.𝟎𝟖
𝟎.𝟎𝟕))
𝟑𝟒𝟗
+
𝟏
𝟑𝟎𝟕.𝟐𝟎𝟎𝟗
= 𝟐𝟏𝟖. 𝟗
𝑸 = 𝑼 𝒐 𝑨 𝒐∆𝑻 𝑳𝑴𝑻𝑫 ∗ 𝑭
𝑸 = 𝟐𝟏𝟖. 𝟗 ∗ 𝝅 ∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 ∗ 𝟏𝟒. 𝟓𝟓 ∗ 𝑭
𝑸 = 𝟓𝟑. 𝟔𝟑 ∗ 𝟏 from fig at R=2.911 &p=0.151 F=0.98
𝑸 = 𝟓𝟑. 𝟔𝟑 𝒘

23. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 23 | P a g e
Parallel flow; cold
𝑸 𝐜𝐨𝐥𝐝 = 150 l/h  𝑸 𝐜𝐨𝐥𝐝 = 𝟒. 𝟏𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟏 + 𝑻 𝟑
𝟐
=
𝟐𝟖 + 𝟐𝟑. 𝟖
𝟐
= 𝟐𝟓. 𝟗 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝒗 =
𝟒. 𝟏𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟓
𝟑. 𝟏𝟒 ∗ (𝟎. 𝟎𝟏𝟒) 𝟐
= 𝟎. 𝟎𝟔𝟕𝟕 𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝒅 𝟎
𝒖
=
𝟗𝟗𝟔. 𝟑 ∗ 𝟎. 𝟎𝟔𝟕𝟕 ∗ 𝟎. 𝟎𝟏𝟒
𝟖. 𝟓𝟎𝟖𝟏 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟏𝟏𝟎𝟗. 𝟗 < 2100
𝑳𝒊 =
𝟏𝟏𝟎𝟗. 𝟗 ∗ 𝟓. 𝟔𝟒𝟔 ∗ 𝟎. 𝟎𝟏𝟒
𝟐𝟎
= 𝟒. 𝟑𝟖𝟔𝟓
𝟒. 𝟑𝟖𝟔𝟓 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓 flow
Let 𝑻 𝒘 =𝑻 𝒇 + 𝟓  𝑻 𝒘 = 𝟐𝟓. 𝟗 + 𝟓 = 𝟑𝟎. 𝟗  Dynamic Viscosity= 7.6254∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟏𝟏𝟎𝟗. 𝟗 ∗ 𝟓. 𝟔𝟒𝟔 ∗
𝟎.𝟎𝟏𝟒
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟖.𝟓𝟎𝟖𝟏∗𝟏𝟎−𝟒
𝟕.𝟔𝟐𝟓𝟒∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟐. 𝟎𝟖𝟑𝟗
𝒉𝒊 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉𝒊 =
𝟏𝟐.𝟎𝟖𝟑𝟗∗𝟎.𝟔𝟏𝟑𝟔
𝟎.𝟎𝟏𝟒
= 𝟓𝟐𝟗. 𝟔𝟐𝟎𝟏 𝐰/𝐦 𝟐
∗ 𝐜
Q=m`*Cp*(To-Ti)
Q=𝟗𝟗𝟔. 𝟑 ∗ 𝟒. 𝟏𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟓
∗ 𝟒𝟎𝟕𝟐. 𝟐 ∗ (𝟐𝟖 − 𝟐𝟑. 𝟖) =710.0039 w
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
710.0039/4 =3.14∗ 𝟎. 𝟎𝟏𝟒 ∗ 𝟎. 𝟑𝟑𝟓 (𝑻 𝒘
𝟏
− 𝟐𝟓. 𝟗)𝟓𝟐𝟗. 𝟔𝟐𝟎𝟏
𝑻 𝒘
𝟏
= 𝟒𝟖. 𝟔𝟓𝟖  Dynamic Viscosity=𝟓. 𝟒𝟒𝟎𝟒 ∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟏𝟏𝟎𝟗. 𝟗 ∗ 𝟓. 𝟔𝟒𝟔 ∗
𝟎.𝟎𝟏𝟒
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟖.𝟓𝟎𝟖𝟏∗𝟏𝟎−𝟒
𝟓.𝟒𝟒𝟎𝟒∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟐. 𝟔𝟔𝟖𝟖
𝒉𝒊
𝟏
=
𝟏𝟐. 𝟔𝟔𝟖𝟖 ∗ 𝟎. 𝟔𝟏𝟑𝟔
𝟎. 𝟎𝟏𝟒
= 𝟓𝟓𝟓. 𝟐𝟓𝟓𝟒 𝐰/𝐦 𝟐
∗ 𝐜

24. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 24 | P a g e
∆𝟏 = 𝐓𝟓 − 𝐓𝟑  ∆𝟏 = 𝟒𝟔. 𝟐 − 𝟐𝟑. 𝟖 = 𝟐𝟐. 𝟒𝟎
∆𝟐 = 𝐓𝟕 − 𝐓𝟏 ∆𝟐 = 𝟑𝟖. 𝟖 − 𝟐𝟖 = 𝟏𝟎. 𝟖
∆𝑻 𝑳𝑴𝑻𝑫 = ∆ 𝟐−∆ 𝟏
𝒍𝒏(
∆ 𝟐
∆ 𝟏
)
∆𝑻 𝑳𝑴𝑻𝑫 =
𝟏𝟎.𝟖−𝟐𝟐.𝟒
𝒍𝒏(
𝟏𝟎.𝟖
𝟐𝟐.𝟒
)
=15.9
𝑼 𝒐 =
𝟏
𝟏
𝒉 𝒊
𝒓 𝒐
𝒓 𝒊
+
𝒓 𝒐(𝐥𝐧(
𝒓 𝒐
𝒓 𝒊
))
𝒌
+ 𝟏
𝒉 𝒐
 𝑼 𝒐 =
𝟏
𝟏
𝟓𝟓𝟓.𝟐𝟓𝟓𝟒
∗
𝟎.𝟎𝟖
𝟎.𝟎𝟕
+
(𝟎.𝟎𝟖𝐥𝐧(
𝟎.𝟎𝟖
𝟎.𝟎𝟕))
𝟑𝟒𝟗
+
𝟏
𝟑𝟎𝟕.𝟐𝟎𝟎𝟗
= 𝟏𝟖𝟕. 𝟏𝟐
𝑸 = 𝑼 𝒐 𝑨 𝒐∆𝑻 𝑳𝑴𝑻𝑫 ∗ 𝑭
𝑸 = 𝟏𝟖𝟕. 𝟏𝟐 ∗ 𝝅 ∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 ∗ 𝟏𝟓. 𝟗 ∗ 𝑭
𝑸 = 𝟓𝟎 ∗ 𝟏 from fig at R=1.76 &p=0.1875F=0.98
𝑸 = 𝟓𝟎 𝒘
Parallel flow ; hot
𝑸 𝐡𝐨𝐭 = 50 l/h  𝑸 𝐡𝐨𝐭 = 1.3889∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟓 + 𝑻 𝟕
𝟐
=
𝟒𝟔 + 𝟒𝟏. 𝟗
𝟐
= 𝟒𝟑. 𝟗𝟓𝒄 𝟎
𝒗 =
𝑸
𝑨
𝑨 = 𝑨 𝒔𝒉𝒆𝒆𝒍 − 𝑨𝒕𝒖𝒃𝒆
=
𝝅
𝟒
𝒅 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
𝒅 𝒐
𝟐
)
=
𝝅
𝟒
(𝟓𝟎 ∗ 𝟏𝟎−𝟑
) 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
(𝟏𝟔 ∗ 𝟏𝟎−𝟑
) 𝒐
𝟐
)
=1.15866∗ 𝟏𝟎−𝟑
𝒗 =
𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
𝟏. 𝟏𝟓𝟖𝟔𝟔 ∗ 𝟏𝟎−𝟑
= 𝟎. 𝟎𝟏𝟐 𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝐝 𝟎
𝒖
=
𝟗𝟗𝟎. 𝟐𝟔 ∗ 𝟎. 𝟎𝟏𝟐 ∗ 𝟎. 𝟎𝟏𝟔
𝟓. 𝟗𝟎𝟕 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟑𝟐𝟏. 𝟖𝟕𝟐𝟐< 2100 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
𝑳𝒊 =
𝟑𝟐𝟏. 𝟖𝟕𝟐𝟐 ∗ 𝟑. 𝟕𝟕𝟎𝟑 ∗ 𝟎. 𝟎𝟏𝟔
𝟐𝟎
= 𝟎. 𝟗𝟕𝟎𝟖 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
Let 𝑻 𝒘 =𝑻 𝒇 − 𝟓  𝑻 𝒘 = 𝟒𝟑. 𝟗𝟓 − 𝟓 = 𝟑𝟖. 𝟗𝟓 𝒄  Dynamic Viscosity=6.4827∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟑𝟐𝟏. 𝟖𝟕𝟐𝟐 ∗ 𝟑. 𝟕𝟕𝟎𝟑 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟓.𝟗𝟎𝟕∗𝟏𝟎−𝟒
𝟔.𝟒𝟖𝟐𝟕∗𝟏𝟎−𝟒)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟔. 𝟕𝟗𝟓𝟖

25. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 25 | P a g e
𝒉 𝟎 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 =
𝟔.𝟕𝟗𝟓𝟖∗𝟎.𝟔𝟑𝟕𝟐
𝟎.𝟎𝟏𝟔
= 𝟐𝟕𝟎. 𝟔𝟒𝟐𝟕 𝐰/𝐦 𝟐
∗ 𝐜
Q=m`*Cp*(To-Ti)
Q=𝟗𝟗𝟎. 𝟐𝟔 ∗ 𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
∗ 𝟒. 𝟎𝟔𝟔𝟕 ∗ 𝟏𝟎 𝟑
∗ (𝟒𝟔 − 𝟒𝟏. 𝟗) = 229.3223 w
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
229.3223 =(3.14∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 (𝟒𝟑. 𝟗𝟓 − 𝑻 𝒘
𝟏
)𝟐𝟕𝟎. 𝟔𝟒𝟐𝟕 ) ∗ 𝟒
𝑻 𝒘
𝟏
= 𝟐𝟗. 𝟓𝟔𝟓𝟕 Dynamic Viscosity=𝟕. 𝟖𝟒𝟔𝟑 ∗ 𝟏𝟎−𝟒
𝐮 = 𝟏. 𝟖𝟔 (𝟑𝟐𝟏. 𝟖𝟕𝟐𝟐 ∗ 𝟑. 𝟕𝟕𝟎𝟑 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏
𝟑
(
𝟓.𝟗𝟎𝟕∗𝟏𝟎−𝟒
𝟕.𝟖𝟒𝟔𝟑∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟔. 𝟔𝟏𝟔𝟔
𝒉 𝒐 =
𝟔. 𝟔𝟏𝟔𝟔 ∗ 𝟎. 𝟔𝟑𝟕𝟐
𝟎. 𝟎𝟏𝟔
= 𝟐𝟔𝟑. 𝟓𝟎𝟔𝟏 𝐰/𝐦 𝟐
∗ 𝐜
Parallel flow; cold
𝑸 𝐜𝐨𝐥𝐝 = 50 l/h  𝑸 𝐜𝐨𝐥𝐝 = 𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟏 + 𝑻 𝟑
𝟐
=
𝟑𝟎. 𝟑 + 𝟐𝟒. 𝟏
𝟐
= 𝟐𝟕. 𝟐 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝒗 =
𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
𝟑. 𝟏𝟒 ∗ (𝟎. 𝟎𝟏𝟒) 𝟐
= 𝟎. 𝟎𝟐𝟐𝟔𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝒅 𝟎
𝒖
=
𝟗𝟗𝟓. 𝟗𝟗 ∗ 𝟎. 𝟎𝟐𝟐𝟔 ∗ 𝟎. 𝟎𝟏𝟒
𝟖. 𝟐𝟔𝟑𝟔 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟑𝟖𝟏. 𝟑𝟒𝟖𝟔< 2100
𝑳𝒊 =
𝟑𝟖𝟏. 𝟑𝟒𝟖𝟔 ∗ 𝟓. 𝟒𝟔𝟔 ∗ 𝟎. 𝟎𝟏𝟒
𝟐𝟎
= 𝟏. 𝟒𝟓𝟗𝟏
𝟏. 𝟒𝟓𝟗𝟏 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓 flow
Let 𝑻 𝒘 =𝑻 𝒇 + 𝟓  𝑻 𝒘 = 𝟐𝟕. 𝟐 + 𝟓 = 𝟑𝟐. 𝟐 𝒄  Dynamic Viscosity= 7.4197∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟑𝟖𝟏. 𝟑𝟒𝟖𝟔 ∗ 𝟓. 𝟒𝟔𝟔 ∗
𝟎.𝟎𝟏𝟒
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟖.𝟐𝟔𝟑𝟔∗𝟏𝟎−𝟒
𝟕.𝟒𝟏𝟗𝟕∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟖. 𝟑𝟕𝟎𝟔
𝒉 𝟎 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 =
𝟖.𝟑𝟕𝟎𝟔∗𝟎.𝟔𝟏𝟓𝟒𝟕
𝟎.𝟎𝟏𝟒
= 𝟑𝟔𝟕. 𝟗𝟖𝟗𝟓 𝐰/𝐦 𝟐
∗ 𝐜
Q=m`*Cp*(To-Ti)

26. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 26 | P a g e
Q=𝟗𝟗𝟓. 𝟗𝟗 ∗ 𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
∗ 𝟒. 𝟎𝟕𝟎𝟗 ∗ 𝟏𝟎 𝟑
∗ (𝟑𝟎. 𝟑 − 𝟐𝟒. 𝟏) =349.1468 w
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
349.1468/4 =3.14∗ 𝟎. 𝟎𝟏𝟒 ∗ 𝟎. 𝟑𝟑𝟓 (𝑻 𝒘
𝟏
− 𝟐𝟖. 𝟏𝟓)𝟑𝟔𝟕. 𝟗𝟖𝟗𝟓
𝑻 𝒘
𝟏
= 𝟐𝟖. 𝟏𝟔  Dynamic Viscosity=𝟖. 𝟎𝟗𝟎𝟏 ∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟑𝟖𝟏. 𝟑𝟒𝟖𝟔 ∗ 𝟓. 𝟒𝟔𝟔 ∗
𝟎.𝟎𝟏𝟒
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟖.𝟐𝟔𝟑𝟔∗𝟏𝟎−𝟒
𝟖.𝟎𝟗𝟎𝟏∗𝟏𝟎−𝟒)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟖. 𝟐𝟔𝟗𝟖
𝒉𝒊
𝟏
=
𝟖. 𝟐𝟔𝟗𝟖 ∗ 𝟎. 𝟔𝟏𝟓𝟒𝟕
𝟎. 𝟎𝟏𝟒
= 𝟑𝟔𝟑. 𝟓𝟓𝟖𝟏 𝐰/𝐦 𝟐
∗ 𝐜
∆𝟏 = 𝐓𝟓 − 𝐓𝟑  ∆𝟏 = 𝟒𝟔 − 𝟐𝟒. 𝟏 = 𝟐𝟏. 𝟗
∆𝟐 = 𝐓𝟕 − 𝐓𝟏 ∆𝟐 = 𝟒𝟏. 𝟗 − 𝟑𝟎. 𝟑 = 𝟏𝟏. 𝟔
∆𝑻 𝑳𝑴𝑻𝑫 = ∆ 𝟐−∆ 𝟏
𝒍𝒏(
∆ 𝟐
∆ 𝟏
)
∆𝑻 𝑳𝑴𝑻𝑫 = 𝟏𝟏.𝟔−𝟐𝟏.𝟗
𝒍𝒏( 𝟏𝟏.𝟔
𝟐𝟏.𝟗
)
= 16.208
𝑼 𝒐 = 𝟏
𝟏
𝒉 𝒊
𝒓 𝒐
𝒓 𝒊
+
𝒓 𝒐(𝐥𝐧(
𝒓 𝒐
𝒓 𝒊
))
𝒌
+ 𝟏
𝒉 𝒐
𝑼 𝒐 =
𝟏
𝟏
𝟑𝟔𝟑.𝟓𝟓𝟖𝟏
∗
𝟎.𝟎𝟖
𝟎.𝟎𝟕
+
(𝟎.𝟎𝟖∗𝐥 𝐧(
𝟎.𝟎𝟖
𝟎.𝟎𝟕))
𝟑𝟒𝟗
+
𝟏
𝟐𝟔𝟑.𝟓𝟎𝟔𝟏
−→ 𝑼 𝒐 = 𝟏𝟒𝟑. 𝟒𝟗
𝑸 = 𝑼 𝒐 𝑨 𝒐∆𝑻 𝑳𝑴𝑻𝑫 ∗ 𝑭
𝑸 = 𝟏𝟒𝟑. 𝟒𝟗 ∗ 𝝅 ∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 ∗ 𝟏𝟔. 𝟐𝟎𝟖 ∗ 𝑭
𝑸 = 𝟑𝟗. 𝟏𝟔 𝑭 From fig at R=0.661 &p=0.283F=1
𝑸 = 𝟑𝟗. 𝟏𝟔 𝒘

27. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 27 | P a g e
Counter flow; hot
𝑸 𝐡𝐨𝐭 = 300 l/h  𝑸 𝐡𝐨𝐭 = 𝟖. 𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟓 + 𝑻 𝟕
𝟐
=
𝟒𝟔 + 𝟒𝟐. 𝟒
𝟐
= 𝟒𝟒. 𝟐 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝑨 = 𝑨 𝒔𝒉𝒆𝒆𝒍 − 𝑨𝒕𝒖𝒃𝒆
=
𝝅
𝟒
𝒅 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
𝒅 𝒐
𝟐
)
=
𝝅
𝟒
(𝟓𝟎 ∗ 𝟏𝟎−𝟑
) 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
(𝟏𝟔 ∗ 𝟏𝟎−𝟑
) 𝒐
𝟐
)
=1.15866∗ 𝟏𝟎−𝟑
𝒗 =
𝟖. 𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟓
𝟏. 𝟏𝟓𝟖𝟔𝟔 ∗ 𝟏𝟎−𝟑
= 𝟎. 𝟎𝟕𝟏𝟗𝟏 𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝐝 𝟎
𝒖
=
𝟗𝟗𝟎. 𝟏 ∗ 𝟎. 𝟎𝟕𝟏𝟗𝟏 ∗ 𝟎. 𝟎𝟏𝟔
𝟓. 𝟖𝟖𝟏𝟐 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟏𝟗𝟑𝟕 < 2100 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
𝑳𝒊 =
𝟏𝟗𝟑𝟕 ∗ 𝟑. 𝟕𝟓𝟏𝟕 ∗ 𝟎. 𝟎𝟏𝟔
𝟐𝟎
= 𝟓. 𝟖𝟏𝟑𝟔 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
Let 𝑻 𝒘 =𝑻 𝒇 − 𝟓  𝑻 𝒘 = 𝟒𝟒. 𝟐 − 𝟓 = 𝟑𝟗. 𝟐 𝒄  Dynamic Viscosity= 6.4518∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟏𝟗𝟑𝟕 ∗ 𝟑. 𝟕𝟓𝟏𝟕 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟓.𝟖𝟖𝟏𝟐∗𝟏𝟎−𝟒
𝟔.𝟒𝟓𝟏𝟖∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟐. 𝟑𝟒𝟏𝟒
𝒉 𝟎 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 𝟎 =
𝟏𝟐.𝟑𝟒𝟏𝟒∗𝟎.𝟔𝟑𝟕𝟒𝟗
𝟎.𝟎𝟏𝟔
= 𝟒𝟗𝟏. 𝟕𝟏𝟗𝟗𝐰/𝐦 𝟐
∗ 𝐜
Q=m`*Cp*(To-Ti)
Q=𝟗𝟗𝟎. 𝟏 ∗ 𝟖. 𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟓
∗ 𝟒. 𝟎𝟔𝟔 ∗ 𝟏𝟎 𝟑
∗ (𝟒𝟔 − 𝟒𝟐. 𝟒) =1207.7 w
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
1207.7 =(3.14∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 (𝟒𝟒. 𝟐 − 𝑻 𝒘
𝟏
) ∗ 𝟒𝟗𝟏. 𝟕𝟏𝟗𝟗) ∗ 𝟒
𝑻 𝒘
𝟏
= 𝟕. 𝟗𝟏  Dynamic Viscosity=𝟏. 𝟑𝟒𝟐𝟒 ∗ 𝟏𝟎−𝟑
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟏𝟗𝟑𝟕 ∗ 𝟑. 𝟕𝟓𝟏𝟕 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟓.𝟖𝟖𝟏𝟐∗𝟏𝟎−𝟒
𝟏.𝟑𝟒𝟐𝟒∗𝟏𝟎−𝟑
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟏. 𝟏𝟑𝟖𝟐
𝒉 𝟎 =
𝟏𝟏. 𝟏𝟑𝟖𝟐 ∗ 𝟎. 𝟔𝟑𝟕𝟒𝟗
𝟎. 𝟎𝟏𝟔
= 𝟒𝟒𝟑. 𝟕𝟖𝟎𝟕 𝐰/𝐦 𝟐
∗ 𝐜

28. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 28 | P a g e
Counter flow; cold
𝑸 𝐜𝐨𝐥𝐝 = 300 l/h  𝑸 𝐜𝐨𝐥𝐝 = 𝟖. 𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟏 + 𝑻 𝟑
𝟐
=
𝟐𝟓. 𝟒 + 𝟐𝟕. 𝟓
𝟐
= 𝟐𝟔. 𝟒𝟓 𝒄 𝟎
𝒗 =
𝑸
𝑨
 𝐯 =
𝟖.𝟑𝟑𝟑 ∗𝟏𝟎−𝟓
𝟑.𝟏𝟒∗(𝟎.𝟎𝟏𝟒) 𝟐
= 𝟎. 𝟏𝟑𝟓𝟑𝟑 𝐦/𝐬
𝑹 𝒆 =
𝝆𝒗 𝒅 𝟎
𝒖
=
𝟗𝟗𝟔𝟐 ∗ 𝟎. 𝟏𝟑𝟓𝟑𝟑 ∗ 𝟎. 𝟎𝟏𝟒
𝟖. 𝟒𝟎𝟑 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟐𝟐𝟒𝟔. 𝟏 < 2100
𝑳𝒊 =
𝟐𝟐𝟒𝟔. 𝟏 ∗ 𝟓. 𝟓𝟔𝟗𝟐 ∗ 𝟎. 𝟎𝟏𝟒
𝟐𝟎
= 𝟖. 𝟕𝟓𝟔𝟑
𝟖. 𝟕𝟓𝟔𝟑 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & Turbulent flow
𝐍 𝐮 = 𝟎. 𝟎𝟑𝟔 𝐑𝐞 𝟎.𝟖
𝐩 𝐫
𝟏/𝟑
(𝐝 ∕ 𝐋) 𝟏/𝟖
𝐍 𝐮 = 𝟎. 𝟎𝟑𝟔 ∗ 𝟐𝟐𝟒𝟔. 𝟏 𝟎.𝟖
∗ 𝟓. 𝟓𝟔𝟗𝟐 𝟏/𝟑
∗ (𝟎. 𝟎𝟏𝟒 𝟎⁄ . 𝟑𝟑𝟓) 𝟏/𝟖
 𝐍 𝐮 = 𝟐𝟎. 𝟓𝟗𝟏𝟏
𝒉𝒊 = (𝟐𝟎. 𝟓𝟗𝟏𝟏 ∗ 𝟎. 𝟔𝟏𝟒𝟒)/𝟎. 𝟎𝟏𝟒 = 𝟗𝟎𝟑. 𝟔𝟓𝟓𝟏 𝐰/𝐦 𝟐
∗ 𝐜
∆𝟏 = 𝐓𝟓 − 𝐓𝟑  ∆𝟏 = 𝟒𝟔 − 𝟐𝟕. 𝟓 = 𝟏𝟖. 𝟓
∆𝟐 = 𝐓𝟕 − 𝐓𝟏 ∆𝟐 = 𝟒𝟐. 𝟒 − 𝟐𝟓. 𝟒 = 𝟏𝟕
∆𝑻 𝑳𝑴𝑻𝑫 =
∆ 𝟐−∆ 𝟏
𝒍𝒏(
∆ 𝟐
∆ 𝟏
)
∆𝑻 𝑳𝑴𝑻𝑫 =
𝟏𝟕−𝟏𝟖.𝟓
𝒍𝒏(
𝟏𝟕
𝟏𝟖.𝟓
)
=17.7
𝑼 𝒐 = 𝟏
𝟏
𝒉 𝒊
𝒓 𝒐
𝒓 𝒊
+
𝒓 𝒐(𝐥𝐧(
𝒓 𝒐
𝒓 𝒊
))
𝒌
+ 𝟏
𝒉 𝒐
 𝑼 𝒐 =
𝟏
𝟏
𝟗𝟎𝟑.𝟔𝟓𝟓𝟏
∗
𝟎.𝟎𝟖
𝟎.𝟎𝟕
+
(𝟎.𝟎𝟖𝐥𝐧(
𝟎.𝟎𝟖
𝟎.𝟎𝟕))
𝟑𝟒𝟗
+
𝟏
𝟒𝟒𝟑.𝟕𝟖𝟎𝟕
= 𝟐𝟖𝟏. 𝟕
𝑸 = 𝑼 𝒐 𝑨 𝒐∆𝑻 𝑳𝑴𝑻𝑫 ∗ 𝑭
𝑸 = 𝟐𝟖𝟏. 𝟕 ∗ 𝝅 ∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 ∗ 𝟏𝟕. 𝟕 ∗ 𝑭
𝑸 = 𝟖𝟑. 𝟗𝟔𝑭 from fig at R=0.5833 &p=0.174 F=1
𝑸 = 𝟖𝟑. 𝟗𝟔 𝒘

29. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 29 | P a g e
Counter flow; cold
𝑸 𝐜𝐨𝐥𝐝 = 150 l/h  𝑸 𝐜𝐨𝐥𝐝 = 𝟒. 𝟏𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟏 + 𝑻 𝟑
𝟐
=
𝟐𝟓. 𝟓 + 𝟐𝟖. 𝟖
𝟐
= 𝟐𝟕. 𝟏𝟓 𝒄 𝟎
𝒗 =
𝑸
𝑨
 𝐯 =
𝟒.𝟏𝟔𝟔𝟕∗𝟏𝟎−𝟓
𝟑.𝟏𝟒∗(𝟎.𝟎𝟏𝟒) 𝟐 = 𝟎. 𝟎𝟔𝟕𝟕 𝐦/𝐬
𝑹 𝒆 =
𝝆𝒗 𝒅 𝟎
𝒖
=
𝟗𝟗𝟔. 𝟎𝟏 ∗ 𝟎. 𝟎𝟔𝟕𝟕 ∗ 𝟎. 𝟎𝟏𝟒
𝟖. 𝟐𝟕𝟐𝟖 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟏𝟏𝟒𝟏. 𝟏 < 2100
𝑳𝒊 =
𝟏𝟏𝟒𝟏. 𝟏 ∗ 𝟓. 𝟒𝟕𝟑𝟑 ∗ 𝟎. 𝟎𝟏𝟒
𝟐𝟎
= 𝟒. 𝟑𝟕𝟏𝟗
𝟒. 𝟑𝟕𝟏𝟗 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓 flow
Let 𝑻 𝒘 =𝑻 𝒇 + 𝟓  𝑻 𝒘 = 𝟐𝟕. 𝟏𝟓 + 𝟓 = 𝟑𝟐. 𝟏𝟓  Dynamic Viscosity= 7.4274∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟏𝟏𝟒𝟏. 𝟏 ∗ 𝟓. 𝟒𝟕𝟑𝟑 ∗
𝟎.𝟎𝟏𝟒
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟖.𝟐𝟕𝟐𝟖∗𝟏𝟎−𝟒
𝟕.𝟒𝟐𝟕𝟒∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟐. 𝟎𝟔𝟕𝟔
𝒉𝒊 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 =
𝟏𝟐.𝟎𝟔𝟕𝟔∗𝟎.𝟔𝟏𝟒𝟓
𝟎.𝟎𝟏𝟒
= 𝟓𝟐𝟗. 𝟔𝟖𝟏𝟒 𝐰/𝐦 𝟐
∗ 𝐜
Q=m`*Cp*(To-Ti)
Q=𝟗𝟗𝟔. 𝟎𝟏 ∗ 𝟒. 𝟏𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟓
∗ 𝟒𝟎𝟕𝟏. 𝟓 ∗ (𝟐𝟖. 𝟖 − 𝟐𝟓. 𝟓) = 557.6020 w
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
557.6020 /4 =3.14∗ 𝟎. 𝟎𝟏𝟒 ∗ 𝟎. 𝟑𝟑𝟓 (𝑻 𝒘
𝟏
− 𝟐𝟕. 𝟏𝟓)𝟓𝟓𝟕. 𝟔𝟎𝟐𝟎
𝑻 𝒘
𝟏
= 𝟒𝟒. 𝟏𝟐  Dynamic Viscosity=𝟓. 𝟖𝟖𝟗𝟔 ∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟏𝟏𝟒𝟏. 𝟏 ∗ 𝟓. 𝟒𝟕𝟑𝟑 ∗
𝟎.𝟎𝟏𝟒
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟖.𝟐𝟕𝟐𝟖∗𝟏𝟎−𝟒
𝟓.𝟖𝟖𝟗𝟔∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟐. 𝟒𝟔𝟓𝟗
𝒉𝒊
𝟏
=
𝟏𝟐. 𝟒𝟔𝟓𝟗 ∗ 𝟎. 𝟔𝟏𝟒𝟓
𝟎. 𝟎𝟏𝟒
= 𝟓𝟒𝟕. 𝟏𝟔𝟒 𝐰/𝐦 𝟐
∗ 𝐜
∆𝟏 = 𝐓𝟓 − 𝐓𝟑  ∆𝟏 = 𝟒𝟔. 𝟏 − 𝟐𝟖. 𝟖 = 𝟏𝟕. 𝟑
∆𝟐 = 𝐓𝟕 − 𝐓𝟏 ∆𝟐 = 𝟒𝟒. 𝟏 − 𝟐𝟓. 𝟓 = 𝟏𝟖. 𝟔
∆𝑻 𝑳𝑴𝑻𝑫 = ∆ 𝟐−∆ 𝟏
𝒍𝒏(
∆ 𝟐
∆ 𝟏
)
 ∆𝑻 𝑳𝑴𝑻𝑫 =
𝟏𝟖.𝟔−𝟏𝟕.𝟑
𝒍𝒏(
𝟏𝟖.𝟔
𝟏𝟕.𝟑
)
= 17.942

30. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 30 | P a g e
𝑼 𝒐 = 𝟏
𝟏
𝒉 𝒊
𝒓 𝒐
𝒓 𝒊
+
𝒓 𝒐(𝐥𝐧(
𝒓 𝒐
𝒓 𝒊
))
𝒌
+ 𝟏
𝒉 𝒐
𝑼 𝒐 =
𝟏
𝟏
𝟓𝟒𝟕. 𝟏𝟔𝟒
∗
𝟎. 𝟎𝟖
𝟎. 𝟎𝟕
+
(𝟎. 𝟎𝟖 ∗ 𝐥 𝐧 (
𝟎. 𝟎𝟖
𝟎. 𝟎𝟕
))
𝟑𝟒𝟗
+
𝟏
𝟒𝟎𝟒. 𝟎𝟐𝟕𝟕
= 𝟐𝟏𝟕. 𝟔𝟓𝟕
𝑸 = 𝑼 𝒐 𝑨 𝒐∆𝑻 𝑳𝑴𝑻𝑫 ∗ 𝑭
𝑸 = 𝟐𝟏𝟕. 𝟔𝟓𝟕 ∗ 𝝅 ∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 ∗ 𝟏𝟕. 𝟗𝟒𝟐 ∗ 𝑭
𝑸 = 𝟔𝟓. 𝟕𝟓𝟗𝟒 𝑭 From fig at R=0.215 &p=0.097 F=1
𝑸 = 𝟔𝟓. 𝟕𝟓𝟗𝟒 𝒘
Counter flow; hot
𝑸 𝐡𝐨𝐭 = 300 l/h  𝑸 𝐡𝐨𝐭 = 𝟖. 𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟓 + 𝑻 𝟕
𝟐
=
𝟒𝟔. 𝟐 + 𝟒𝟓. 𝟐
𝟐
= 𝟒𝟓. 𝟕 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝑨 = 𝑨 𝒔𝒉𝒆𝒆𝒍 − 𝑨𝒕𝒖𝒃𝒆
=
𝝅
𝟒
𝒅 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
𝒅 𝒐
𝟐
)
=
𝝅
𝟒
(𝟓𝟎 ∗ 𝟏𝟎−𝟑
) 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
(𝟏𝟔 ∗ 𝟏𝟎−𝟑
) 𝒐
𝟐
)
=1.15866∗ 𝟏𝟎−𝟑
𝒗 =
𝟖. 𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟓
𝟏. 𝟏𝟓𝟖𝟔𝟔 ∗ 𝟏𝟎−𝟑
= 𝟎. 𝟎𝟕𝟏𝟗𝟏 𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝐝 𝟎
𝒖
=
𝟗𝟖𝟗. 𝟓𝟒 ∗ 𝟎. 𝟎𝟕𝟏𝟗𝟏 ∗ 𝟎. 𝟎𝟏𝟔
𝟓. 𝟕𝟐𝟔𝟑 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟏𝟗𝟖𝟖. 𝟐 < 2100 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
𝑳𝒊 =
𝟏𝟗𝟖𝟖. 𝟐 ∗ 𝟑. 𝟔𝟒𝟐𝟖 ∗ 𝟎. 𝟎𝟏𝟔
𝟐𝟎
= 𝟓. 𝟕𝟗𝟒𝟏 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
Let 𝑻 𝒘 =𝑻 𝒇 − 𝟓  𝑻 𝒘 = 𝟒𝟓. 𝟕 − 𝟓 = 𝟒𝟎. 𝟕 𝒄  Dynamic Viscosity= 6.2712∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟏𝟗𝟖𝟖. 𝟐 ∗ 𝟑. 𝟔𝟒𝟐𝟖 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟓.𝟕𝟐𝟔𝟑∗𝟏𝟎−𝟒
𝟔.𝟐𝟕𝟏𝟐∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟐. 𝟑𝟑𝟎𝟓
𝒉 𝟎 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 𝟎 =
𝟏𝟐.𝟑𝟑𝟎𝟓∗𝟎.𝟔𝟑𝟗𝟐𝟐
𝟎.𝟎𝟏𝟔
= 𝟒𝟗𝟐. 𝟔𝟏𝟖𝟗 𝐰/𝐦 𝟐
∗ 𝐜

31. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 31 | P a g e
Q=m`*Cp*(To-Ti)
Q=𝟗𝟖𝟗. 𝟓𝟒 ∗ 𝟖. 𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟓
∗ 𝟒. 𝟎𝟔𝟔 ∗ 𝟏𝟎 𝟑
∗ (𝟒𝟔. 𝟐 − 𝟒𝟓. 𝟐) = 335.275 w
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
335.275=(3.14∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 (𝟒𝟓. 𝟕 − 𝑻 𝒘
𝟏
) ∗ 𝟒𝟗𝟐. 𝟔𝟏𝟖𝟗) ∗ 𝟒
𝑻 𝒘
𝟏
= 𝟔𝟐. 𝟔𝟕𝟔𝟏 Dynamic Viscosity=𝟒. 𝟑𝟔𝟒𝟒 ∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟏𝟗𝟖𝟖. 𝟐 ∗ 𝟑. 𝟔𝟒𝟐𝟖 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟓.𝟕𝟐𝟔𝟑∗𝟏𝟎−𝟒
𝟒.𝟑𝟔𝟒𝟒∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟐. 𝟗𝟕𝟐𝟒
𝒉 𝟎 =
𝟏𝟐. 𝟗𝟕𝟐𝟒 ∗ 𝟎. 𝟔𝟑𝟗𝟐𝟐
𝟎. 𝟎𝟏𝟔
= 𝟓𝟏𝟖. 𝟐𝟔𝟑𝟔 𝐰/𝐦 𝟐
∗ 𝐜
Counter flow;; cold
𝑸 𝐜𝐨𝐥𝐝 = 50 l/h  𝑸 𝐜𝐨𝐥𝐝 = 𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟏 + 𝑻 𝟑
𝟐
=
𝟐𝟔. 𝟐 + 𝟑𝟏. 𝟐
𝟐
= 𝟐𝟖. 𝟕 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝒗 =
𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
𝟑. 𝟏𝟒 ∗ (𝟎. 𝟎𝟏𝟒) 𝟐
= 𝟎. 𝟎𝟐𝟐𝟔𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝒅 𝟎
𝒖
=
𝟗𝟗𝟓. 𝟓𝟓𝟕 ∗ 𝟎. 𝟎𝟐𝟐𝟔 ∗ 𝟎. 𝟎𝟏𝟒
𝟕. 𝟗𝟗𝟓 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟑𝟗𝟑. 𝟗𝟖𝟗< 2100
𝑳𝒊 =
𝟑𝟗𝟑. 𝟗𝟖𝟗 ∗ 𝟓. 𝟐𝟔𝟗 ∗ 𝟎. 𝟎𝟏𝟒
𝟐𝟎
= 𝟏. 𝟒𝟓𝟑𝟏
𝟏. 𝟒𝟓𝟑𝟏 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓 flow
Let 𝑻 𝒘 =𝑻 𝒇 + 𝟓  𝑻 𝒘 = 𝟐𝟖. 𝟕 + 𝟓 = 𝟑𝟑. 𝟕 𝒄  Dynamic Viscosity=7.1931∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟑𝟗𝟑. 𝟗𝟖𝟗 ∗ 𝟓. 𝟐𝟔𝟗 ∗
𝟎.𝟎𝟏𝟒
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟕.𝟗𝟗𝟓∗𝟏𝟎−𝟒
𝟕.𝟏𝟗𝟑𝟏∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟖. 𝟑𝟓𝟔𝟖
𝒉 𝟎 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 =
𝟖.𝟑𝟓𝟔𝟖∗𝟎.𝟔𝟏𝟕𝟔
𝟎.𝟎𝟏𝟒
= 𝟑𝟔𝟖. 𝟔𝟓𝟒𝟑 𝐰/𝐦 𝟐
∗ 𝐜
Q=m`*Cp*(To-Ti)
Q=𝟗𝟗𝟓. 𝟓𝟓𝟕𝟗 ∗ 𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
∗ 𝟒. 𝟎𝟕𝟎𝟗 ∗ 𝟏𝟎 𝟑
∗ (𝟑𝟏. 𝟐 − 𝟐𝟔. 𝟐) = 𝟐𝟖𝟏. 𝟒𝟒𝟕𝟗 𝒘

32. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 32 | P a g e
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
281.4479/4 =3.14∗ 𝟎. 𝟎𝟏𝟒 ∗ 𝟎. 𝟑𝟑𝟓 (𝑻 𝒘
𝟏
− 𝟐𝟖. 𝟕) ∗ 𝟑𝟔𝟖. 𝟔𝟓𝟒𝟑
𝑻 𝒘
𝟏
= 𝟒𝟏. 𝟔𝟔  Dynamic Viscosity=𝟔. 𝟏𝟔 ∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟑𝟗𝟑. 𝟗𝟖𝟗 ∗ 𝟓. 𝟐𝟔𝟗 ∗
𝟎.𝟎𝟏𝟒
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟕.𝟗𝟗𝟓∗𝟏𝟎−𝟒
𝟔.𝟏𝟔∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟖. 𝟓𝟒𝟎𝟏
𝒉𝒊
𝟏
=
𝟖. 𝟓𝟒𝟎𝟏 ∗ 𝟎. 𝟔𝟏𝟕𝟔
𝟎. 𝟎𝟏𝟒
= 𝟑𝟕𝟔. 𝟕𝟒𝟎𝟒 𝐰/𝐦 𝟐
∗ 𝐜
∆𝟏 = 𝐓𝟓 − 𝐓𝟑  ∆𝟏 = 𝟒𝟔. 𝟐 − 𝟑𝟏. 𝟐 = 𝟏𝟓
∆𝟐 = 𝐓𝟕 − 𝐓𝟏 ∆𝟐 = 𝟒𝟓. 𝟐 − 𝟐𝟔. 𝟐 = 𝟏𝟗
∆𝑻 𝑳𝑴𝑻𝑫 =
∆ 𝟐 − ∆ 𝟏
𝒍𝒏(
∆ 𝟐
∆ 𝟏
)
∆𝑻 𝑳𝑴𝑻𝑫 =
𝟏𝟗−𝟏𝟓
𝒍𝒏(
𝟏𝟗
𝟏𝟓
)
= 16.92
𝑼 𝒐 =
𝟏
𝟏
𝒉 𝒊
𝒓 𝒐
𝒓 𝒊
+
𝒓 𝒐(𝐥𝐧(
𝒓 𝒐
𝒓 𝒊
))
𝒌
+
𝟏
𝒉 𝒐

𝑼 𝒐 =
𝟏
𝟏
𝟑𝟕𝟔.𝟕𝟒𝟎𝟒
∗
𝟎.𝟎𝟖
𝟎.𝟎𝟕
+
(𝟎.𝟎𝟖∗𝐥 𝐧(
𝟎.𝟎𝟖
𝟎.𝟎𝟕))
𝟑𝟒𝟗
+
𝟏
𝟓𝟏𝟖.𝟐𝟔𝟑𝟔
−→ 𝑼 𝒐 = 𝟐𝟎𝟎. 𝟐𝟓
𝑸 = 𝑼 𝒐 𝑨 𝒐∆𝑻 𝑳𝑴𝑻𝑫 ∗ 𝑭
𝑸 = 𝟐𝟎𝟎. 𝟐𝟓 ∗ 𝝅 ∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 ∗ 𝟏𝟔. 𝟗𝟐 ∗ 𝑭
𝑸 = 𝟓𝟕. 𝟎𝟓𝟒𝟐 𝑭 From fig at R=0.357 &p=0.166F=1
𝑸 = 𝟓𝟕. 𝟎𝟓𝟒𝟐 𝒘

33. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 33 | P a g e
Counter flow;; hot
𝑸 𝐡𝐨𝐭 = 150 l/h  𝑸 𝐡𝐨𝐭 = 4.1667∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟓 + 𝑻 𝟕
𝟐
=
𝟒𝟔 + 𝟒𝟎. 𝟑
𝟐
= 𝟒𝟑. 𝟏𝟓 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝑨 = 𝑨 𝒔𝒉𝒆𝒆𝒍 − 𝑨𝒕𝒖𝒃𝒆
=
𝝅
𝟒
𝒅 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
𝒅 𝒐
𝟐
)
=
𝝅
𝟒
(𝟓𝟎 ∗ 𝟏𝟎−𝟑
) 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
(𝟏𝟔 ∗ 𝟏𝟎−𝟑
) 𝒐
𝟐
)
=1.15866∗ 𝟏𝟎−𝟑
𝒗 =
𝟒. 𝟏𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟓
𝟏. 𝟏𝟓𝟖𝟔𝟔 ∗ 𝟏𝟎−𝟑
= 𝟎. 𝟎𝟑𝟔𝟎 𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝐝 𝟎
𝒖
=
𝟗𝟗𝟎. 𝟓𝟖 ∗ 𝟎. 𝟎𝟑𝟔𝟎 ∗ 𝟎. 𝟎𝟏𝟔
𝟓. 𝟗𝟗𝟑𝟖 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟗𝟓𝟏. 𝟗𝟒𝟎𝟓< 2100 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
𝑳𝒊 =
𝟗𝟓𝟏. 𝟗𝟒𝟎𝟓 ∗ 𝟑. 𝟖𝟑𝟏𝟏 ∗ 𝟎. 𝟎𝟏𝟔
𝟐𝟎
= 𝟐. 𝟗𝟏𝟕𝟔 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
Let 𝑻 𝒘 =𝑻 𝒇 − 𝟓  𝑻 𝒘 = 𝟒𝟑. 𝟏𝟓 − 𝟓 = 𝟑𝟖. 𝟏𝟓𝒄  Dynamic Viscosity=6.5833∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟗𝟓𝟏. 𝟗𝟒𝟎𝟓 ∗ 𝟑. 𝟖𝟑𝟏𝟏 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟓.𝟗𝟗𝟑𝟖∗𝟏𝟎−𝟒
𝟔.𝟓𝟖𝟑𝟑∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟗. 𝟖𝟎𝟓𝟖
𝒉 𝟎 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 =
𝟗.𝟖𝟎𝟓𝟖∗𝟎.𝟔𝟑𝟔𝟐
𝟎.𝟎𝟏𝟔
= 𝟑𝟖𝟗. 𝟗𝟎𝟑𝟏 𝐰/𝐦 𝟐
∗ 𝐜
Q=m`*Cp*(To-Ti)
Q=𝟗𝟗𝟎. 𝟓𝟖 ∗ 𝟒. 𝟏𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟓
∗ 𝟒. 𝟎𝟔𝟔𝟕 ∗ 𝟏𝟎 𝟑
∗ (𝟒𝟔 − 𝟒𝟎. 𝟑) = 956.7507 w
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
956.7507 =(3.14∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 (𝟒𝟑. 𝟏𝟓 − 𝑻 𝒘
𝟏
) 𝟑𝟖𝟗. 𝟗𝟎𝟑𝟏) ∗ 𝟒
𝑻 𝒘
𝟏
= 𝟔. 𝟓𝟖  Dynamic Viscosity=𝟏. 𝟑𝟗𝟒𝟐 ∗ 𝟏𝟎−𝟑
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟗𝟓𝟏. 𝟗𝟒𝟎𝟓 ∗ 𝟑. 𝟖𝟑𝟏𝟏 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟓.𝟗𝟗𝟑𝟖∗𝟏𝟎−𝟒
𝟏.𝟑𝟗𝟒𝟐∗𝟏𝟎−𝟑
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟗. 𝟐𝟐𝟗𝟕
𝒉 𝒐 =
𝟗. 𝟐𝟐𝟗𝟕 ∗ 𝟎. 𝟔𝟑𝟔𝟐
𝟎. 𝟎𝟏𝟔
= 𝟑𝟔𝟔. 𝟗𝟗𝟓𝟗𝐰/𝐦 𝟐
∗ 𝐜

34. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 34 | P a g e
Counter flow; cold
𝑸 𝐜𝐨𝐥𝐝 = 300 l/h  𝑸 𝐜𝐨𝐥𝐝 = 𝟖. 𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟏 + 𝑻 𝟑
𝟐
=
𝟐𝟕. 𝟑 + 𝟐𝟓. 𝟒
𝟐
= 𝟐𝟔. 𝟒𝟓 𝒄 𝟎
𝒗 =
𝑸
𝑨
 𝐯 =
𝟖.𝟑𝟑𝟑 ∗𝟏𝟎−𝟓
𝟑.𝟏𝟒∗(𝟎.𝟎𝟏𝟒) 𝟐
= 𝟎. 𝟏𝟑𝟓𝟑𝟑 𝐦/𝐬
𝑹 𝒆 =
𝝆𝒗 𝒅 𝟎
𝒖
=
𝟗𝟗𝟔. 𝟐𝟐 ∗ 𝟎. 𝟏𝟑𝟓𝟑𝟑 ∗ 𝟎. 𝟎𝟏𝟒
𝟖. 𝟒𝟏𝟗𝟑 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟐𝟐𝟒𝟏. 𝟖 < 2100
𝑳𝒊 =
𝟐𝟐𝟒𝟏. 𝟖 ∗ 𝟓. 𝟓𝟖𝟏𝟏 ∗ 𝟎. 𝟎𝟏𝟒
𝟐𝟎
= 𝟖. 𝟕𝟓𝟖𝟐
𝟖. 𝟕𝟓𝟖𝟐 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & Turbulent flow
𝐍 𝐮 = 𝟎. 𝟎𝟑𝟔 𝐑𝐞 𝟎.𝟖
𝐩 𝐫
𝟏/𝟑
(𝐝 ∕ 𝐋) 𝟏/𝟖
𝐍 𝐮 = 𝟎. 𝟎𝟑𝟔 ∗ 𝟐𝟐𝟒𝟏. 𝟖 𝟎.𝟖
∗ 𝟓. 𝟓𝟖𝟏𝟏 𝟏/𝟑
∗ (𝟎. 𝟎𝟏𝟒 𝟎⁄ . 𝟑𝟑𝟓) 𝟏/𝟖
 𝐍 𝐮 = 𝟐𝟎. 𝟓𝟕𝟒𝟐
𝒉𝒊 = (𝟐𝟎. 𝟓𝟕𝟒𝟐 ∗ 𝟎. 𝟔𝟏𝟒𝟒)/𝟎. 𝟎𝟏𝟒 = 𝟗𝟎𝟐. 𝟗𝟏𝟑𝟓 𝐰/𝐦 𝟐
∗ 𝐜
∆𝟏 = 𝐓𝟓 − 𝐓𝟑  ∆𝟏 = 𝟒𝟔 − 𝟐𝟕. 𝟑 = 𝟏𝟖. 𝟕
∆𝟐 = 𝐓𝟕 − 𝐓𝟏 ∆𝟐 = 𝟒𝟎. 𝟑 − 𝟐𝟓. 𝟒 = 𝟏𝟒. 𝟗
∆𝑻 𝑳𝑴𝑻𝑫 =
∆ 𝟐−∆ 𝟏
𝒍𝒏(
∆ 𝟐
∆ 𝟏
)
∆𝑻 𝑳𝑴𝑻𝑫 =
𝟏𝟒.𝟗−𝟏𝟖.𝟕
𝒍𝒏(
𝟏𝟒.𝟗
𝟏𝟖.𝟕
)
=16.728
𝑼 𝒐 =
𝟏
𝟏
𝒉 𝒊
𝒓 𝒐
𝒓 𝒊
+
𝒓 𝒐(𝐥𝐧(
𝒓 𝒐
𝒓 𝒊
))
𝒌
+
𝟏
𝒉 𝒐
𝑼 𝒐 =
𝟏
𝟏
𝟗𝟎𝟐.𝟗𝟏𝟑𝟓
∗
𝟎.𝟎𝟖
𝟎.𝟎𝟕
+
(𝟎.𝟎𝟖𝐥𝐧(
𝟎.𝟎𝟖
𝟎.𝟎𝟕))
𝟑𝟒𝟗
+
𝟏
𝟑𝟔𝟔.𝟗𝟗𝟓𝟗
= 𝟐𝟒𝟖. 𝟔𝟖𝟑
𝑸 = 𝑼 𝒐 𝑨 𝒐∆𝑻 𝑳𝑴𝑻𝑫 ∗ 𝑭
𝑸 = 𝟐𝟒𝟖. 𝟔𝟖𝟑 ∗ 𝝅 ∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 ∗ 𝟏𝟔. 𝟕𝟐𝟖 ∗ 𝑭
𝑸 = 𝟕𝟎. 𝟎𝟒𝟗𝑭 from fig at R=0.33 &p=0.2766 F=1
𝑸 = 𝟕𝟎. 𝟎𝟒𝟗 𝒘

35. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 35 | P a g e
Counter flow; cold
𝑸 𝐜𝐨𝐥𝐝 = 150 l/h  𝑸 𝐜𝐨𝐥𝐝 = 𝟒. 𝟏𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟏 + 𝑻 𝟑
𝟐
=
𝟐𝟗. 𝟓 + 𝟐𝟒. 𝟐
𝟐
= 𝟐𝟔. 𝟖𝟓 𝒄 𝟎
𝒗 =
𝑸
𝑨
 𝐯 =
𝟒.𝟏𝟔𝟔𝟕∗𝟏𝟎−𝟓
𝟑.𝟏𝟒∗(𝟎.𝟎𝟏𝟒) 𝟐
= 𝟎. 𝟎𝟔𝟕𝟕 𝐦/𝐬
𝑹 𝒆 =
𝝆𝒗 𝒅 𝟎
𝒖
=
𝟗𝟗𝟔. 𝟎𝟗 ∗ 𝟎. 𝟎𝟔𝟕𝟕 ∗ 𝟎. 𝟎𝟏𝟒
𝟖. 𝟑𝟐𝟖𝟑 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟏𝟏𝟑𝟑. 𝟔 < 2100
𝑳𝒊 =
𝟏𝟏𝟑𝟑. 𝟔 ∗ 𝟓. 𝟓𝟏𝟒𝟏 ∗ 𝟎. 𝟎𝟏𝟒
𝟐𝟎
= 𝟒. 𝟑𝟕𝟓𝟓
𝟒. 𝟑𝟕𝟓𝟓 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓 flow
Let 𝑻 𝒘 =𝑻 𝒇 + 𝟓  𝑻 𝒘 = 𝟐𝟔. 𝟖𝟓 + 𝟓 = 𝟑𝟏. 𝟖𝟓  Dynamic Viscosity=7.4742∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟏𝟏𝟑𝟑. 𝟔 ∗ 𝟓. 𝟓𝟏𝟒𝟏 ∗
𝟎.𝟎𝟏𝟒
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟖.𝟑𝟐𝟖𝟑∗𝟏𝟎−𝟒
𝟕.𝟒𝟕𝟒𝟐∗𝟏𝟎−𝟒)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟐. 𝟔𝟐𝟏
𝒉𝒊 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 =
𝟏𝟐.𝟔𝟐𝟏∗𝟎.𝟔𝟏𝟒𝟗
𝟎.𝟎𝟏𝟒
= 𝟓𝟓𝟒. 𝟑𝟑𝟐𝟒 𝐰/𝐦 𝟐
∗ 𝐜
Q=m`*Cp*(To-Ti)
Q=𝟗𝟗𝟔. 𝟎𝟗 ∗ 𝟒. 𝟏𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟓
∗ 𝟒𝟎𝟕𝟏. 𝟕 ∗ (𝟐𝟗. 𝟓 − 𝟐𝟒. 𝟐) = 895.6585 w
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
895.6585/4 =3.14∗ 𝟎. 𝟎𝟏𝟒 ∗ 𝟎. 𝟑𝟑𝟓 (𝑻 𝒘
𝟏
− 𝟐𝟕. 𝟏𝟓) ∗ 𝟓𝟓𝟒. 𝟑𝟑𝟐𝟒
𝑻 𝒘
𝟏
= 𝟓𝟑. 𝟑  Dynamic Viscosity=𝟓. 𝟎𝟑𝟕𝟗 ∗ 𝟏𝟎−𝟒
𝐍𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟏𝟏𝟑𝟑. 𝟔 ∗ 𝟓. 𝟓𝟏𝟒𝟏 ∗
𝟎.𝟎𝟏𝟒
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟖.𝟑𝟐𝟖𝟑∗𝟏𝟎−𝟒
𝟓.𝟎𝟑𝟕𝟗∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟐. 𝟕𝟓𝟕𝟎
𝒉𝒊
𝟏
=
𝟏𝟐. 𝟕𝟓𝟕𝟎 ∗ 𝟎. 𝟔𝟏𝟒𝟗
𝟎. 𝟎𝟏𝟒
= 𝟓𝟔𝟎. 𝟑𝟎𝟓𝟕 𝐰/𝐦 𝟐
∗ 𝐜
∆𝟏 = 𝐓𝟓 − 𝐓𝟑  ∆𝟏 = 𝟒𝟔. 𝟏 − 𝟐𝟒. 𝟐 = 𝟐𝟏. 𝟗
∆𝟐 = 𝐓𝟕 − 𝐓𝟏 ∆𝟐 = 𝟒𝟐. 𝟕 − 𝟐𝟗. 𝟓 = 𝟏𝟑. 𝟐
∆𝑻 𝑳𝑴𝑻𝑫 =
∆ 𝟐−∆ 𝟏
𝒍𝒏(
∆ 𝟐
∆ 𝟏
)
 ∆𝑻 𝑳𝑴𝑻𝑫 =
𝟏𝟑.𝟐−𝟐𝟏.𝟗
𝒍𝒏(
𝟏𝟑.𝟐
𝟐𝟏.𝟗
)
= 17.1845

36. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 36 | P a g e
𝑼 𝒐 =
𝟏
𝟏
𝒉 𝒊
𝒓 𝒐
𝒓 𝒊
+
𝒓 𝒐(𝐥𝐧(
𝒓 𝒐
𝒓 𝒊
))
𝒌
+
𝟏
𝒉 𝒐
𝑼 𝒐 =
𝟏
𝟏
𝟓𝟔𝟎. 𝟑𝟎𝟓𝟕
∗
𝟎. 𝟎𝟖
𝟎. 𝟎𝟕
+
(𝟎. 𝟎𝟖 ∗ 𝐥 𝐧 (
𝟎. 𝟎𝟖
𝟎. 𝟎𝟕
))
𝟑𝟒𝟗
+
𝟏
𝟑𝟔𝟔. 𝟗𝟗𝟓𝟗
= 𝟐𝟎𝟖. 𝟓𝟒𝟒
𝑸 = 𝑼 𝒐 𝑨 𝒐∆𝑻 𝑳𝑴𝑻𝑫 ∗ 𝑭
𝑸 = 𝟐𝟎𝟖. 𝟓𝟒𝟒 ∗ 𝝅 ∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 ∗ 𝟏𝟕. 𝟏𝟖𝟒𝟓 ∗ 𝑭
𝑸 = 𝟔𝟎. 𝟑𝟒 𝑭 From fig at R=0.34 &p=0.22 F=1
𝑸 = 𝟔𝟎. 𝟑𝟒 𝒘
Counter flow;; hot
𝑸 𝐡𝐨𝐭 = 150 l/h  𝑸 𝐡𝐨𝐭 = 4.1667∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟓 + 𝑻 𝟕
𝟐
=
𝟒𝟔 + 𝟒𝟑. 𝟖
𝟐
= 𝟒𝟒. 𝟗 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝑨 = 𝑨 𝒔𝒉𝒆𝒆𝒍 − 𝑨𝒕𝒖𝒃𝒆
=
𝝅
𝟒
𝒅 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
𝒅 𝒐
𝟐
)
=
𝝅
𝟒
(𝟓𝟎 ∗ 𝟏𝟎−𝟑
) 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
(𝟏𝟔 ∗ 𝟏𝟎−𝟑
) 𝒐
𝟐
)
=1.15866∗ 𝟏𝟎−𝟑
𝒗 =
𝟒. 𝟏𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟓
𝟏. 𝟏𝟓𝟖𝟔𝟔 ∗ 𝟏𝟎−𝟑
= 𝟎. 𝟎𝟑𝟔𝟎 𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝐝 𝟎
𝒖
=
𝟗𝟖𝟗. 𝟖𝟖 ∗ 𝟎. 𝟎𝟑𝟔𝟎 ∗ 𝟎. 𝟎𝟏𝟔
𝟓. 𝟖𝟎𝟖 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟗𝟖𝟏. 𝟔𝟗𝟗𝟐< 2100 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
𝑳𝒊 =
𝟗𝟖𝟏. 𝟔𝟗𝟗𝟐 ∗ 𝟑. 𝟕𝟎𝟎𝟐 ∗ 𝟎. 𝟎𝟏𝟔
𝟐𝟎
= 𝟐. 𝟗𝟎𝟔𝟎 > 𝑳( 𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
Let 𝑻 𝒘 =𝑻 𝒇 − 𝟓  𝑻 𝒘 = 𝟒𝟒. 𝟗 − 𝟓 = 𝟑𝟗. 𝟗𝒄  Dynamic Viscosity=6.3664∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 ( 𝟗𝟖𝟏. 𝟔𝟗𝟗𝟐 ∗ 𝟑. 𝟕𝟎𝟎𝟐 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟓.𝟖𝟎𝟖∗𝟏𝟎−𝟒
𝟔.𝟑𝟔𝟔𝟒∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟐. 𝟗𝟒𝟑𝟑
𝒉 𝟎 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 =
𝟏𝟐.𝟗𝟒𝟑𝟑∗𝟎.𝟔𝟑𝟖𝟑
𝟎.𝟎𝟏𝟔
= 𝟓𝟏𝟔. 𝟑𝟓𝟔𝟖 𝐰/𝐦 𝟐
∗ 𝐜

37. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 37 | P a g e
Q=m`*Cp*(To-Ti)
Q= 𝟗𝟖𝟗. 𝟖𝟖 ∗ 𝟒. 𝟏𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟓
∗ 𝟒. 𝟎𝟔𝟔𝟕 ∗ 𝟏𝟎 𝟑
∗ ( 𝟒𝟔 − 𝟒𝟑. 𝟖) = 369.0112w
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
369.0112=(3.14∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 ( 𝟒𝟒. 𝟗 − 𝑻 𝒘
𝟏
) 𝟓𝟏𝟔. 𝟑𝟓𝟔𝟖 ) ∗ 𝟒
𝑻 𝒘
𝟏
= 𝟒𝟑. 𝟗  Dynamic Viscosity=𝟓. 𝟗𝟏𝟑𝟎 ∗ 𝟏𝟎−𝟑
𝐍𝐮 = 𝟏. 𝟖𝟔 ( 𝟗𝟖𝟏. 𝟔𝟗𝟗𝟐 ∗ 𝟑. 𝟕𝟎𝟎𝟐 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟓.𝟖𝟎𝟖∗𝟏𝟎−𝟒
𝟓.𝟗𝟏𝟑𝟎∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟏𝟑. 𝟎𝟕𝟕𝟖
𝒉 𝒐 =
𝟏𝟑. 𝟎𝟕𝟕𝟖 ∗ 𝟎. 𝟔𝟑𝟖𝟑
𝟎. 𝟎𝟏𝟔
= 𝟓𝟐𝟏. 𝟕𝟐𝟐𝟓𝐰/𝐦 𝟐
∗ 𝐜
Counter flow;; cold
𝑸 𝐜𝐨𝐥𝐝 = 50 l/h  𝑸 𝐜𝐨𝐥𝐝 = 𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟏 + 𝑻 𝟑
𝟐
=
𝟑𝟎. 𝟖 + 𝟐𝟔. 𝟏
𝟐
= 𝟐𝟖. 𝟒𝟓 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝒗 =
𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
𝟑. 𝟏𝟒 ∗ (𝟎. 𝟎𝟏𝟒) 𝟐
= 𝟎. 𝟎𝟐𝟐𝟔𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝒅 𝟎
𝒖
=
𝟗𝟗𝟓. 𝟔𝟒 ∗ 𝟎. 𝟎𝟐𝟐𝟔 ∗ 𝟎. 𝟎𝟏𝟒
𝟖. 𝟎𝟑𝟖𝟖 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟑𝟗𝟏. 𝟖𝟕𝟓𝟎< 2100
𝑳𝒊 =
𝟑𝟗𝟏. 𝟖𝟕𝟓𝟎 ∗ 𝟓. 𝟑𝟎𝟏𝟖 ∗ 𝟎. 𝟎𝟏𝟒
𝟐𝟎
= 𝟏. 𝟒𝟓𝟒𝟒
𝟏. 𝟒𝟓𝟒𝟒 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓 flow
Let 𝑻 𝒘 =𝑻 𝒇 + 𝟓  𝑻 𝒘 = 𝟐𝟖. 𝟒𝟓 + 𝟓 = 𝟑𝟑. 𝟒𝟓𝒄  Dynamic Viscosity=7.2301∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟑𝟗𝟏. 𝟖𝟕𝟓𝟎 ∗ 𝟓. 𝟑𝟎𝟏𝟖 ∗
𝟎.𝟎𝟏𝟒
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟖.𝟎𝟑𝟖𝟖∗𝟏𝟎−𝟒
𝟕.𝟐𝟑𝟎𝟏∗𝟏𝟎−𝟒)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟖. 𝟑𝟓𝟗𝟒
𝒉 𝟎 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 =
𝟖.𝟑𝟓𝟗𝟒∗𝟎.𝟔𝟏𝟕𝟐𝟓
𝟎.𝟎𝟏𝟒
= 𝟑𝟔𝟖. 𝟓𝟔 𝐰/𝐦 𝟐
∗ 𝐜
Q=m`*Cp*(To-Ti)

38. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 38 | P a g e
Q=𝟗𝟗𝟓. 𝟔𝟒 ∗ 𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
∗ 𝟒. 𝟎𝟕𝟎𝟗 ∗ 𝟏𝟎 𝟑
∗ (𝟑𝟎. 𝟖 − 𝟐𝟔. 𝟏) = 𝟐𝟔𝟒. 𝟓𝟖𝟐𝟖𝒘
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
264.5828/4 =3.14∗ 𝟎. 𝟎𝟏𝟒 ∗ 𝟎. 𝟑𝟑𝟓 (𝑻 𝒘
𝟏
− 𝟐𝟖. 𝟒𝟓) ∗ 𝟑𝟔𝟖. 𝟓𝟔
𝑻 𝒘
𝟏
= 𝟑𝟗. 𝟏𝟏  Dynamic Viscosity=𝟔. 𝟒𝟔𝟐𝟗 ∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟑𝟗𝟏. 𝟖𝟕𝟓𝟎 ∗ 𝟓. 𝟑𝟎𝟏𝟖 ∗
𝟎.𝟎𝟏𝟒
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟖.𝟎𝟑𝟖𝟖∗𝟏𝟎−𝟒
𝟔.𝟒𝟔𝟐𝟗∗𝟏𝟎−𝟒)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟖. 𝟒𝟗𝟏𝟖
𝒉𝒊
𝟏
=
𝟖. 𝟒𝟗𝟏𝟖 ∗ 𝟎. 𝟔𝟏𝟕𝟐𝟓
𝟎. 𝟎𝟏𝟒
= 𝟑𝟕𝟒. 𝟑𝟗𝟕𝟒 𝐰/𝐦 𝟐
∗ 𝐜
∆𝟏 = 𝐓𝟓 − 𝐓𝟑  ∆𝟏 = 𝟒𝟔 − 𝟑𝟎. 𝟖 = 𝟏𝟓. 𝟐𝟎
∆𝟐 = 𝐓𝟕 − 𝐓𝟏 ∆𝟐 = 𝟒𝟑. 𝟖 − 𝟐𝟔. 𝟏 = 𝟏𝟕. 𝟕
∆𝑻 𝑳𝑴𝑻𝑫 =
∆ 𝟐−∆ 𝟏
𝒍𝒏(
∆ 𝟐
∆ 𝟏
)
∆𝑻 𝑳𝑴𝑻𝑫 =
𝟏𝟕.𝟕−𝟏𝟓.𝟐𝟎
𝒍𝒏(
𝟏𝟕.𝟕
𝟏𝟓.𝟐𝟎
)
= 16.418
𝑼 𝒐 =
𝟏
𝟏
𝒉 𝒊
𝒓 𝒐
𝒓 𝒊
+
𝒓 𝒐(𝐥𝐧(
𝒓 𝒐
𝒓 𝒊
))
𝒌
+
𝟏
𝒉 𝒐

𝑼 𝒐 =
𝟏
𝟏
𝟑𝟕𝟒.𝟑𝟗𝟕𝟒
∗
𝟎.𝟎𝟖
𝟎.𝟎𝟕
+
(𝟎.𝟎𝟖∗𝐥 𝐧(
𝟎.𝟎𝟖
𝟎.𝟎𝟕))
𝟑𝟒𝟗
+
𝟏
𝟓𝟐𝟏.𝟕𝟐𝟐
−→ 𝑼 𝒐 = 𝟐𝟗𝟔. 𝟗𝟗
𝑸 = 𝑼 𝒐 𝑨 𝒐∆𝑻 𝑳𝑴𝑻𝑫 ∗ 𝑭
𝑸 = 𝟐𝟗𝟔. 𝟗𝟗 ∗ 𝝅 ∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 ∗ 𝟏𝟔. 𝟒𝟏𝟖 ∗ 𝑭
𝑸 = 𝟖𝟐. 𝟏 𝑭 From fig at R=0.45 &p=0.35F=1
𝑸 = 𝟖𝟐. 𝟏 𝒘

39. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 39 | P a g e
Counter flow;; hot
𝑸 𝐡𝐨𝐭 = 50 l/h  𝑸 𝐡𝐨𝐭 = 𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟓 + 𝑻 𝟕
𝟐
=
𝟒𝟔 + 𝟑𝟓. 𝟑
𝟐
= 𝟒𝟎. 𝟔𝟓 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝑨 = 𝑨 𝒔𝒉𝒆𝒆𝒍 − 𝑨𝒕𝒖𝒃𝒆
=
𝝅
𝟒
𝒅 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
𝒅 𝒐
𝟐
)
=
𝝅
𝟒
(𝟓𝟎 ∗ 𝟏𝟎−𝟑
) 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
(𝟏𝟔 ∗ 𝟏𝟎−𝟑
) 𝒐
𝟐
)
=1.15866∗ 𝟏𝟎−𝟑
𝒗 =
𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
𝟏. 𝟏𝟓𝟖𝟔𝟔 ∗ 𝟏𝟎−𝟑
= 𝟎. 𝟎𝟏𝟐𝟎 𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝐝 𝟎
𝒖
=
𝟗𝟗𝟏. 𝟓𝟔 ∗ 𝟎. 𝟎𝟏𝟐𝟎 ∗ 𝟎. 𝟎𝟏𝟔
𝟔. 𝟐𝟕𝟕 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟑𝟎𝟑. 𝟐𝟗𝟕𝟎< 2100 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
𝑳𝒊 =
𝟑𝟎𝟑. 𝟐𝟗𝟕𝟎 ∗ 𝟒. 𝟎𝟑𝟏𝟓 ∗ 𝟎. 𝟎𝟏𝟔
𝟐𝟎
= 𝟎. 𝟗𝟕𝟖𝟐 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
Let 𝑻 𝒘 =𝑻 𝒇 − 𝟓  𝑻 𝒘 = 𝟒𝟎. 𝟔𝟓 − 𝟓 = 𝟑𝟓. 𝟔𝟓 𝒄  Dynamic Viscosity=6.9150∗
𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟑𝟎𝟑. 𝟐𝟗𝟕𝟎 ∗ 𝟒. 𝟎𝟑𝟏𝟓 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟔.𝟐𝟕𝟕∗𝟏𝟎−𝟒
𝟔.𝟗𝟏𝟓𝟎∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟕. 𝟏𝟏𝟗𝟐
𝒉 𝟎 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 =
𝟕.𝟏𝟏𝟗𝟐∗𝟎.𝟔𝟑𝟑𝟐𝟔
𝟎.𝟎𝟏𝟔
= 𝟐𝟖𝟏. 𝟕𝟔𝟗𝟎 𝐰/𝐦 𝟐
∗ 𝐜
Q=m`*Cp*(To-Ti)
Q=𝟗𝟗𝟏. 𝟓𝟔 ∗ 𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
∗ 𝟒. 𝟎𝟔𝟕𝟕 ∗ 𝟏𝟎 𝟑
∗ (𝟒𝟔 − 𝟑𝟓. 𝟑) = 599.408 w
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
599.408=(3.14∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 (𝟒𝟎. 𝟔𝟓 − 𝑻 𝒘
𝟏
)𝟐𝟖𝟏. 𝟕𝟔𝟗𝟎 ) ∗ 𝟒
𝑻 𝒘
𝟏
= 𝟓𝟐. 𝟗𝟖  Dynamic Viscosity=𝟓. 𝟎𝟔𝟒𝟎 ∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟑𝟎𝟑. 𝟐𝟗𝟕𝟎 ∗ 𝟒. 𝟎𝟑𝟏𝟓 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟔.𝟐𝟕𝟕∗𝟏𝟎−𝟒
𝟓.𝟎𝟔𝟒𝟎∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟕. 𝟒𝟑𝟔𝟔
𝒉 𝒐 =
𝟕. 𝟒𝟑𝟔𝟔 ∗ 𝟎. 𝟔𝟑𝟑𝟐𝟔
𝟎. 𝟎𝟏𝟔
= 𝟐𝟗𝟒. 𝟑𝟑𝟏𝟑 𝐰/𝐦 𝟐
∗ 𝐜

40. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 40 | P a g e
Counter flow; cold
𝑸 𝐜𝐨𝐥𝐝 = 300 l/h  𝑸 𝐜𝐨𝐥𝐝 = 𝟖. 𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟏 + 𝑻 𝟑
𝟐
=
𝟐𝟔 + 𝟐𝟓. 𝟏
𝟐
= 𝟐𝟓. 𝟓𝟓 𝒄 𝟎
𝒗 =
𝑸
𝑨
 𝐯 =
𝟖.𝟑𝟑𝟑 ∗𝟏𝟎−𝟓
𝟑.𝟏𝟒∗(𝟎.𝟎𝟏𝟒) 𝟐
= 𝟎. 𝟏𝟑𝟓𝟑𝟑 𝐦/𝐬
𝑹 𝒆 =
𝝆𝒗 𝒅 𝟎
𝒖
=
𝟗𝟗𝟔. 𝟒𝟒 ∗ 𝟎. 𝟏𝟑𝟓𝟑𝟑 ∗ 𝟎. 𝟎𝟏𝟒
𝟖. 𝟓𝟕𝟓𝟖 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟐𝟐𝟎𝟏. 𝟒 < 2100
𝑳𝒊 =
𝟐𝟐𝟎𝟏. 𝟒 ∗ 𝟓. 𝟔𝟗𝟔𝟒 ∗ 𝟎. 𝟎𝟏𝟒
𝟐𝟎
= 𝟖. 𝟕𝟕𝟖𝟎
𝟖. 𝟕𝟕𝟖𝟎 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & Turbulent flow
𝐍 𝐮 = 𝟎. 𝟎𝟑𝟔 𝐑𝐞 𝟎.𝟖
𝐩 𝐫
𝟏/𝟑
(𝐝 ∕ 𝐋) 𝟏/𝟖
𝐍 𝐮 = 𝟎. 𝟎𝟑𝟔 ∗ 𝟐𝟐𝟎𝟏. 𝟒 𝟎.𝟖
∗ 𝟓. 𝟔𝟗𝟔𝟒 𝟏/𝟑
∗ (𝟎. 𝟎𝟏𝟒 𝟎⁄ . 𝟑𝟑𝟓) 𝟏/𝟖
 𝐍 𝐮 = 𝟐𝟎. 𝟒𝟏𝟓𝟕
𝒉𝒊 = (𝟐𝟎. 𝟒𝟏𝟓𝟕 ∗ 𝟎. 𝟔𝟏𝟑𝟎𝟗)/𝟎. 𝟎𝟏𝟒 = 𝟖𝟗𝟒. 𝟎𝟒𝟕𝟑 𝐰/𝐦 𝟐
∗ 𝐜
∆𝟏 = 𝐓𝟓 − 𝐓𝟑  ∆𝟏 = 𝟒𝟔 − 𝟐𝟔 = 𝟐𝟎
∆𝟐 = 𝐓𝟕 − 𝐓𝟏 ∆𝟐 = 𝟑𝟓. 𝟑 − 𝟐𝟓. 𝟏 = 𝟏𝟎. 𝟐
∆𝑻 𝑳𝑴𝑻𝑫 =
∆ 𝟐−∆ 𝟏
𝒍𝒏(
∆ 𝟐
∆ 𝟏
)
∆𝑻 𝑳𝑴𝑻𝑫 =
𝟏𝟎.𝟐−𝟐𝟎
𝒍𝒏(
𝟏𝟎.𝟐
𝟐𝟎
)
=14.55
𝑼 𝒐 =
𝟏
𝟏
𝒉 𝒊
𝒓 𝒐
𝒓 𝒊
+
𝒓 𝒐(𝐥𝐧(
𝒓 𝒐
𝒓 𝒊
))
𝒌
+
𝟏
𝒉 𝒐
 𝑼 𝒐 =
𝟏
𝟏
𝟖𝟗𝟒.𝟎𝟒𝟕𝟑
∗
𝟎.𝟎𝟖
𝟎.𝟎𝟕
+
(𝟎.𝟎𝟖𝐥𝐧(
𝟎.𝟎𝟖
𝟎.𝟎𝟕))
𝟑𝟒𝟗
+
𝟏
𝟑𝟔𝟔.𝟗𝟗𝟓𝟗
= 𝟐𝟏𝟐. 𝟒𝟕
𝑸 = 𝑼 𝒐 𝑨 𝒐∆𝑻 𝑳𝑴𝑻𝑫 ∗ 𝑭
𝑸 = 𝟐𝟏𝟐. 𝟒𝟕 ∗ 𝝅 ∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 ∗ 𝟏𝟒. 𝟓𝟓 ∗ 𝑭
𝑸 = 𝟓𝟎. 𝟎𝟓𝑭 From fig at R=1.2 &p=0.22 F=1
𝑸 = 𝟓𝟎. 𝟎𝟓 𝒘

41. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 41 | P a g e
Counter flow; cold
𝑸 𝐜𝐨𝐥𝐝 = 150 l/h  𝑸 𝐜𝐨𝐥𝐝 = 𝟒. 𝟏𝟔𝟔𝟕 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟏 + 𝑻 𝟑
𝟐
=
𝟐𝟕. 𝟐 + 𝟐𝟓. 𝟏
𝟐
= 𝟐𝟔. 𝟏𝟓 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝒗 =
𝟖. 𝟑𝟑𝟑 ∗ 𝟏𝟎−𝟓
𝟑. 𝟏𝟒 ∗ (𝟎. 𝟎𝟏𝟒) 𝟐
= 𝟎. 𝟏𝟑𝟓𝟑𝟑 𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝒅 𝟎
𝒖
=
𝟗𝟗𝟔. 𝟐𝟖 ∗ 𝟎. 𝟏𝟑𝟓𝟑𝟑 ∗ 𝟎. 𝟎𝟏𝟒
𝟖. 𝟒𝟔𝟎𝟐 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟐𝟐𝟑𝟏. 𝟏 < 2100
𝑳𝒊 =
𝟐𝟐𝟑𝟏. 𝟏 ∗ 𝟓. 𝟔𝟏𝟏𝟏 ∗ 𝟎. 𝟎𝟏𝟒
𝟐𝟎
= 𝟖. 𝟕𝟔𝟐𝟏
𝟖. 𝟕𝟔𝟑𝟐 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & Turbulent flow
𝐍 𝐮 = 𝟎. 𝟎𝟑𝟔 𝐑𝐞 𝟎.𝟖
𝐩 𝐫
𝟏/𝟑
(𝐝 ∕ 𝐋) 𝟏/𝟖
𝐍 𝐮 = 𝟎. 𝟎𝟑𝟔 ∗ 𝟐𝟐𝟑𝟏. 𝟏 𝟎.𝟖
∗ 𝟓. 𝟔𝟏𝟏𝟏 𝟏/𝟑
∗ (𝟎. 𝟎𝟏𝟒 𝟎⁄ . 𝟑𝟑𝟓) 𝟏/𝟖
 𝐍 𝐮 = 𝟐𝟎. 𝟓𝟑𝟐𝟐
𝒉𝒊 = (𝟐𝟎. 𝟓𝟑𝟐𝟐 ∗ 𝟎. 𝟔𝟏𝟑𝟗𝟔)/𝟎. 𝟎𝟏𝟒 = 𝟗𝟎𝟎. 𝟒𝟐𝟓𝐰/𝐦 𝟐
∗ 𝐜
∆𝟏 = 𝐓𝟓 − 𝐓𝟑  ∆𝟏 = 𝟒𝟓. 𝟖 − 𝟐𝟕. 𝟐 = 𝟏𝟖. 𝟔
∆𝟐 = 𝐓𝟕 − 𝐓𝟏 ∆𝟐 = 𝟑𝟕. 𝟖 − 𝟐𝟓. 𝟏 = 𝟏𝟐. 𝟕
∆𝑻 𝑳𝑴𝑻𝑫 =
∆ 𝟐−∆ 𝟏
𝒍𝒏(
∆ 𝟐
∆ 𝟏
)
∆𝑻 𝑳𝑴𝑻𝑫 =
𝟏𝟐.𝟕−𝟏𝟖.𝟔
𝒍𝒏(
𝟏𝟐.𝟕
𝟏𝟖.𝟔
)
=15.462
𝑼 𝒐 =
𝟏
𝟏
𝒉 𝒊
𝒓 𝒐
𝒓 𝒊
+
𝒓 𝒐(𝐥𝐧(
𝒓 𝒐
𝒓 𝒊
))
𝒌
+
𝟏
𝒉 𝒐
 𝑼 𝒐 =
𝟏
𝟏
𝟗𝟎𝟎.𝟒𝟐𝟓
∗
𝟎.𝟎𝟖
𝟎.𝟎𝟕
+
(𝟎.𝟎𝟖𝐥𝐧(
𝟎.𝟎𝟖
𝟎.𝟎𝟕))
𝟑𝟒𝟗
+
𝟏
𝟑𝟔𝟔.𝟗𝟗𝟓𝟗
= 𝟐𝟒𝟖. 𝟒
𝑸 = 𝑼 𝒐 𝑨 𝒐∆𝑻 𝑳𝑴𝑻𝑫 ∗ 𝑭
𝑸 = 𝟐𝟒𝟖. 𝟒 ∗ 𝝅 ∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 ∗ 𝟏𝟓. 𝟒𝟔𝟐 ∗ 𝑭
𝑸 = 𝟔𝟒. 𝟔𝟕 ∗ 𝑭 from fig at R=1.23 &p=1.744 F=1
𝑸 = 𝟔𝟒. 𝟔𝟕 𝒘

42. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 42 | P a g e
Counter flow;; hot
𝑸 𝐡𝐨𝐭 = 50 l/h  𝑸 𝐡𝐨𝐭 = 𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟓 + 𝑻 𝟕
𝟐
=
𝟒𝟓. 𝟖 + 𝟒𝟎. 𝟖
𝟐
= 𝟒𝟑. 𝟑 𝒄 𝟎
𝒗 =
𝑸
𝑨
𝑨 = 𝑨 𝒔𝒉𝒆𝒆𝒍 − 𝑨𝒕𝒖𝒃𝒆
=
𝝅
𝟒
𝒅 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
𝒅 𝒐
𝟐
)
=
𝝅
𝟒
(𝟓𝟎 ∗ 𝟏𝟎−𝟑
) 𝒔𝒉𝒆𝒆𝒍
𝟐
− (𝟒
𝝅
𝟒
(𝟏𝟔 ∗ 𝟏𝟎−𝟑
) 𝒐
𝟐
)
=1.15866∗ 𝟏𝟎−𝟑
𝒗 =
𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
𝟏. 𝟏𝟓𝟖𝟔𝟔 ∗ 𝟏𝟎−𝟑
= 𝟎. 𝟎𝟏𝟐𝟎 𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝐝 𝟎
𝒖
=
𝟗𝟗𝟎. 𝟓 ∗ 𝟎. 𝟎𝟏𝟐𝟎 ∗ 𝟎. 𝟎𝟏𝟔
𝟓. 𝟗𝟕𝟕𝟓 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟑𝟏𝟖. 𝟏𝟓𝟑𝟏< 2100 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
𝑳𝒊 =
𝟑𝟏𝟖. 𝟏𝟓𝟑𝟏 ∗ 𝟑. 𝟖𝟏𝟗 ∗ 𝟎. 𝟎𝟏𝟔
𝟐𝟎
= 𝟎. 𝟗𝟕𝟐𝟎 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓
Let 𝑻 𝒘 =𝑻 𝒇 − 𝟓  𝑻 𝒘 = 𝟒𝟑. 𝟑 − 𝟓 = 𝟑𝟖. 𝟑 𝒄  Dynamic Viscosity=6.5643∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟑𝟏𝟖. 𝟏𝟓𝟑𝟏 ∗ 𝟑. 𝟖𝟏𝟗 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟓.𝟗𝟕𝟕𝟓∗𝟏𝟎−𝟒
𝟔.𝟓𝟔𝟒𝟑∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟕. 𝟏𝟎𝟕𝟑
𝒉 𝟎 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 =
𝟕.𝟏𝟎𝟕𝟑∗𝟎.𝟔𝟑𝟔𝟒𝟒
𝟎.𝟎𝟏𝟔
= 𝟐𝟖𝟐. 𝟕𝟏𝟎𝟔 𝐰/𝐦 𝟐
∗ 𝐜
Q=m`*Cp*(To-Ti)
Q=𝟗𝟗𝟎. 𝟓 ∗ 𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
∗ 𝟒. 𝟎𝟔𝟕𝟕 ∗ 𝟏𝟎 𝟑
∗ (𝟒𝟓. 𝟖 − 𝟒𝟎. 𝟖) = 279.7979 w
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
279.7979=(3.14∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 (𝟒𝟑. 𝟑 − 𝑻 𝒘
𝟏
)𝟐𝟖𝟐. 𝟕𝟏𝟎𝟔) ∗ 𝟒
𝑻 𝒘
𝟏
= 𝟐𝟖. 𝟓𝟗  Dynamic Viscosity=𝟖. 𝟎𝟏𝟒𝟑 ∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟑𝟏𝟖. 𝟏𝟓𝟑𝟏 ∗ 𝟑. 𝟖𝟏𝟗 ∗
𝟎.𝟎𝟏𝟔
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟓.𝟗𝟕𝟕𝟓∗𝟏𝟎−𝟒
𝟖.𝟎𝟏𝟒𝟑∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟔. 𝟗𝟏𝟏𝟓
𝒉 𝒐 =
𝟔. 𝟗𝟏𝟏𝟓 ∗ 𝟎. 𝟔𝟑𝟔𝟒𝟒
𝟎. 𝟎𝟏𝟔
= 𝟐𝟕𝟒. 𝟗𝟐𝟐𝟐 𝐰/𝐦 𝟐
∗ 𝐜

43. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 43 | P a g e
Counter flow;; cold
𝑸 𝐜𝐨𝐥𝐝 = 50 l/h  𝑸 𝐜𝐨𝐥𝐝 = 𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
m3/s
𝑻 𝒇 =
𝑻 𝟏 + 𝑻 𝟑
𝟐
=
𝟐𝟗. 𝟕 + 𝟐𝟓. 𝟔
𝟐
= 𝟐𝟕. 𝟔𝟓𝒄 𝟎
𝒗 =
𝑸
𝑨
𝒗 =
𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
𝟑. 𝟏𝟒 ∗ (𝟎. 𝟎𝟏𝟒) 𝟐
= 𝟎. 𝟎𝟐𝟐𝟔𝒎/𝒔
𝑹 𝒆 =
𝝆𝒗 𝒅 𝟎
𝒖
=
𝟗𝟗𝟓. 𝟖𝟕 ∗ 𝟎. 𝟎𝟐𝟐𝟔 ∗ 𝟎. 𝟎𝟏𝟒
𝟖. 𝟏𝟖𝟏𝟔 ∗ 𝟏𝟎−𝟒
𝑹 𝒆 = 𝟑𝟖𝟓. 𝟏𝟐𝟒𝟑< 2100
𝑳𝒊 =
𝟑𝟖𝟓. 𝟏𝟐𝟒𝟑 ∗ 𝟓. 𝟒𝟎𝟎𝟔𝟑 ∗ 𝟎. 𝟎𝟏𝟒
𝟐𝟎
= 𝟏. 𝟒𝟓𝟓𝟗
𝟏. 𝟒𝟓𝟓𝟗 > 𝑳(𝟎. 𝟑𝟑𝒎)
.`.Developed & 𝒍𝒂𝒎𝒊𝒏𝒂𝒓 flow
Let 𝑻 𝒘 =𝑻 𝒇 + 𝟓  𝑻 𝒘 = 𝟐𝟕. 𝟔𝟓 + 𝟓 = 𝟑𝟐. 𝟔𝟓 𝒄  Dynamic Viscosity=7.3505∗ 𝟏𝟎−𝟒
𝐍𝐮 = 𝟏. 𝟖𝟔 (𝟑𝟖𝟓. 𝟏𝟐𝟒𝟑 ∗ 𝟓. 𝟒𝟎𝟎𝟔𝟑 ∗
𝟎.𝟎𝟏𝟒
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟖.𝟏𝟖𝟏𝟔∗𝟏𝟎−𝟒
𝟕.𝟑𝟓𝟎𝟓∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟖. 𝟑𝟔𝟑𝟖
𝒉 𝟎 =
𝑵𝒖∗𝒌
𝒅 𝟎
 𝒉 =
𝟖.𝟑𝟔𝟑𝟖∗𝟎.𝟔𝟏𝟔𝟏𝟐
𝟎.𝟎𝟏𝟒
= 𝟑𝟔𝟖. 𝟎𝟕𝟖𝟗 𝐰/𝐦 𝟐
∗ 𝐜
Q=m`*Cp*(To-Ti)
Q=𝟗𝟗𝟓. 𝟖𝟕 ∗ 𝟏. 𝟑𝟖𝟖𝟗 ∗ 𝟏𝟎−𝟓
∗ 𝟒. 𝟎𝟕𝟎𝟗 ∗ 𝟏𝟎 𝟑
∗ (𝟐𝟗. 𝟕 − 𝟐𝟓. 𝟔) = 𝟐𝟑𝟎. 𝟖𝟓𝟗𝟔 𝒘
Q=(3.14∗ 𝒅 𝟎 ∗ 𝒍 (𝑻 𝒇 − 𝑻 𝒘)𝒉 𝟎) ∗ 𝟒
230.8596/4 =3.14∗ 𝟎. 𝟎𝟏𝟒 ∗ 𝟎. 𝟑𝟑𝟓 (𝑻 𝒘
𝟏
− 𝟐𝟕. 𝟔𝟓) ∗ 𝟑𝟔𝟖. 𝟎𝟕𝟖𝟗
𝑻 𝒘
𝟏
= 𝟑𝟖. 𝟐𝟗𝟕  Dynamic Viscosity=𝟔. 𝟓𝟔𝟒𝟔 ∗ 𝟏𝟎−𝟒
𝐮 = 𝟏. 𝟖𝟔 (𝟑𝟖𝟓. 𝟏𝟐𝟒𝟑 ∗ 𝟓. 𝟒𝟎𝟎𝟔𝟑 ∗
𝟎.𝟎𝟏𝟒
𝟎.𝟑𝟑𝟓
)
𝟏/𝟑
(
𝟖.𝟏𝟖𝟏𝟔∗𝟏𝟎−𝟒
𝟔.𝟓𝟔𝟒𝟔∗𝟏𝟎−𝟒
)
𝟎⋅𝟏𝟒
 𝐍𝐮 = 𝟖. 𝟒𝟗𝟕𝟐
𝒉𝒊
𝟏
=
𝟖. 𝟒𝟗𝟕𝟐 ∗ 𝟎. 𝟔𝟏𝟔𝟏𝟐
𝟎. 𝟎𝟏𝟒
= 𝟑𝟕𝟑. 𝟗𝟒𝟗𝟔 𝐰/𝐦 𝟐
∗ 𝐜
∆𝟏 = 𝐓𝟓 − 𝐓𝟑  ∆𝟏 = 𝟒𝟓. 𝟖 − 𝟐𝟗. 𝟕 = 𝟏𝟔. 𝟏
∆𝟐 = 𝐓𝟕 − 𝐓𝟏 ∆𝟐 = 𝟒𝟎. 𝟖 − 𝟐𝟓. 𝟔 = 𝟏𝟓. 𝟐

44. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 44 | P a g e
∆𝑻 𝑳𝑴𝑻𝑫 =
∆ 𝟐−∆ 𝟏
𝒍𝒏(
∆ 𝟐
∆ 𝟏
)
∆𝑻 𝑳𝑴𝑻𝑫 =
𝟏𝟓.𝟐−𝟏𝟔.𝟏
𝒍𝒏(
𝟏𝟓.𝟐
𝟏𝟔.𝟏
)
= 15.65
𝑼 𝒐 =
𝟏
𝟏
𝒉 𝒊
𝒓 𝒐
𝒓 𝒊
+
𝒓 𝒐(𝐥𝐧(
𝒓 𝒐
𝒓 𝒊
))
𝒌
+
𝟏
𝒉 𝒐

𝑼 𝒐 =
𝟏
𝟏
𝟑𝟕𝟑.𝟗𝟒𝟗𝟔
∗
𝟎.𝟎𝟖
𝟎.𝟎𝟕
+
(𝟎.𝟎𝟖∗𝐥 𝐧(
𝟎.𝟎𝟖
𝟎.𝟎𝟕))
𝟑𝟒𝟗
+
𝟏
𝟐𝟕𝟒.𝟗𝟐𝟐𝟐
−→ 𝑼 𝒐 = 𝟏𝟒𝟖. 𝟕
𝑸 = 𝑼 𝒐 𝑨 𝒐∆𝑻 𝑳𝑴𝑻𝑫 ∗ 𝑭
𝑸 = 𝟏𝟒𝟖. 𝟕 ∗ 𝝅 ∗ 𝟎. 𝟎𝟏𝟔 ∗ 𝟎. 𝟑𝟑𝟓 ∗ 𝟏𝟓. 𝟔𝟓 ∗ 𝑭
𝑸 = 𝟑𝟗. 𝟏 𝑭 From fig at R=0.8 &p=0.12F=1
𝑸 = 𝟑𝟗. 𝟏 𝒘
PARALLEL FLOW
Q hot Q cold T1 T3 T5 T7
300
300 28.6 23.8 46 42.9
150 29.8 24.4 46 44.2
50 32.2 24.6 46.1 25.2
150
300
28.1 SAIF
ALDIN ALI 24.2 46 20.8
150 29.5 24.2 46 42.7
50 31.7 24.6 46.2 44.5
50
300 27 23.6 45.8 35.9
150 28 23.8 46.2 38.8
50 30 24.1 46 41.9COUNTER FLOW
Q hot Q cold T1 T3 T5 T7
300
300 25.4 27.5 46 42.4
150 25.5 28.8 46.1 44.1
50 26.2 31.2 46.2 45.2
150
300 25.4 27.3 46 40.3
150 29.5 24.2 46.1 42.7
50 26.1 30.8 46 43.8
50
300 25.1 26 46 35.3
150
25.1 SAIF
ALDIN ALI 27.2 45.8 37.8

45. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 45 | P a g e
50 25.6 29.7 45.8 40.8
COUNTER FLOW
Q hot Q cold q LMTD
300 SAIF ALDIN
ALI
300 83.96 17.7
150
65.7594 SAIF ALDIN ALI
17.942
50 57.0542 16.92
150
300 70.049 16.728
150
60.34 SAIF ALDIN ALI
17.185
50 82.1 16.418
50
300 50.05 14.55
150 64.67 15.462
50 39.1 15.65

46. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 46 | P a g e
SAIF ALDIN ALI
6. DISCUSSION
PARALLEL FLOW
Q hot Q cold q LMTD
300
300
40.9 SAIF ALDIN ALI
SAIF ALDIN ALI 17.96
150 66.775 16.523
50 55.76 16.8951
150
300 68.99 16.8
150 93.857 17.14
50 52.7 16.818
50
300 53.63 14.55
150 50 15.9
50 39.16 16.208

47. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 47 | P a g e
0
10
20
30
40
50
60
70
80
90
100
0 50 100 150 200 250 300 350
q(w)
Q hot(l/h)
PARALLEL FLOW
300 150 50
SAIFALDINALI
0
10
20
30
40
50
60
70
80
90
0 50 100 150 200 250 300 350
q(w)
Q hot(l/h)
COUNTER FLOW
300 con 150 con 50 con
SAIFALDINALI

48. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 48 | P a g e
0
10
20
30
40
50
60
70
80
90
100
0 50 100 150 200 250 300 350
q(w)
Q hot(l/h)
PARALLEL&COUNTER FLOW
300 150 50 300 con 150 con 50 con
SAIFALDINALI

49. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 49 | P a g e
0
10
20
30
40
50
60
70
80
90
0 2 4 6 8 10 12 14 16 18 20
q(w)
LMTD
COUNTER FLOW
Linear (300 con) Linear (150 con) Linear (50 con)
SAIFALDINALISAIFALDINALI
0
10
20
30
40
50
60
70
80
90
100
0 2 4 6 8 10 12 14 16 18 20
q(w)
LMTD
PARALLEL
Linear (300) Linear (150) Linear (50)
SAIFALDINALI
SAIFALDINALI

50. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
20/4/2019 50 | P a g e
0
10
20
30
40
50
60
70
80
90
100
0 2 4 6 8 10 12 14 16 18 20
q(w)
LMTD
PARALLEL&COUNTER FLOW
Linear (300) Linear (150) Linear (50)
Linear (300 con) Linear (150 con) Linear (50 con)
SAIFALDINALI