Thermodynamics and Fluid Mechanics (Essay Sample)

UNIVERSITY OF SUNDERLAND
DEPARTMENT OF COMPUTING, ENGINEERING AND TECHNOLGY
EAT106 – THERMODYNAMICS AND FLUID MECHANICS
REFERRED COURSEWORK 2013-2014
NAME:
DATE:
Question 1
Water at 50 degrees Celsius flows at a mass flow rate of 20 kg/s in a 200 mm diameter pipeline.
* Find the density and dynamic viscosity of the water at this temperature
letbe the dynamic viscosity
(kg/ms)= {[2.414 x 10-5)] x[10(247k/(temp-140k))]}
={[2.414 x 10^-5)] x [10(247k/(50 + 273 -140))]}
=0.00002414 x 101.4
=0.000605914kg/ms
 =6. 091 x 10^ -4 kg/ms
Density=988.1kg/m3
(Source Engineeringtoolbox.com, 2014)
* Calculate the velocity of the water.
Q=AV
V=Q/A
Volume flow rate= {(mass low rate)/ density}
Q=20/988.1
= 0.0202m3/s
A=r2
= x 0.1 x 0.1
=0.031m2
V=0.0202/0.031
=0.652m/s
* Hence, calculate the Reynolds number Re.
Re=density x velocity x diameter)
(Dynamic viscosity)
Re=xVx D

=988.1x 0.652 x 0.2
6. 091 x 10^ -4
Re=21255.94
* Is the flow laminar or turbulent?
The flow is turbulent
Because the Reynolds number is greater than 2000
Question 2
Water flows in a circular pipe. At one section the diameter of the pipe is 0.3 m, the static pressure is 260 kPa, the velocity is 3 m/s, and the elevation is 10 m above ground level. The elevation at a section downstream is at ground level and the pipe diameter is 0.15 m.
Find the gauge pressure at the downstream section.
Let upward section be section 1
Pressure at 1= 260Kpa =260 x 103 N/m2
Area 1 = (/4) x (D2)
= (/4) x (0.3 x 0.3)
=0.07071m2
Let downstream section be section 2
Pressure at 2= P
Area 2= (/4) x (0.15 x 0.15)
=0.0176
Equation of flow rate, Q= AV
Q is constant
AV1=AV2
3 x 0.07071= V2 x 0.0176
V2=12.052m/s
Bernoulli equation
(Source Princeton.edu, 2014)
{[260 x 103)/ 999 x 9.81]+ [(32)/ 2 x 9.81] + 10}={ [P/999 x 9.81]+ [12.052x12.052)/ 2x9.81] + 0.075}
26.5+0.459+10=7.478+ P/999 x 9.81
P=29.179 x 999 x 9.81
P=285, 959.74 N/m2
P= 285.959 Kpa
Question 3
A 40 m long horizontal pipe is connected to a water tank at one end and discharges freely into the atmosphere at the other. For the first 25 m of its length from the tank, the pipe is 150 mm diameter. Its diameter is then suddenly enlarged to 300 mm, as shown. The water level in the tank is 8 m above the centre of the pipe. If the Darcy-Weisbach friction factor in both pipes is 0.04, calculate the exit velocity V2 of the water and the volumetric flow rate in m3/s.
(Source Nptel.ac.in, 2014)
Using line XX as the datum
Using pipe1 and the tank
Bernoulli equation
P1=P2=0
Vt for tank=0
Pt/g + V2t/2g + Zt=P1/2g + {V12/2 x g} +Z1+hf
(The crossed values mean they are zero)
8+0.075=V21 /2x9.81 +0.15+ (0.04x25xV2)/ (0.15x2x9.81)
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