Cross-Flow Heat Exchanger Experiment

Running HEAD: CROSS FLOW HEAT EXCHANGER 1
Cross-Flow Heat Exchanger Experiment
Name
Student number
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Date

CROSS FLOW HEAT EXCHANGER 2
Assumed atmospheric pressure in the lab (PA) 9.807x104 N/m2
Density of air in the lab at 220c 0.083038 kg/m3
Diameter of heated specimen 0.0124 m
Length of the heated specimen-Exposed 0.1034 m
Time for rod to remain in heating compartment before
measurements
120 S
Thermal conductivity of air 0.0259 J/msoC
Dynamic viscosity of air 18.2 x 10-6 Kg/ms
Specific heat capacity of air 1004.5 J/kgoC
Ambient temperature of air 295.2 K
Mass of the heated specimen 0.152 kg
Calculation of Density of air in the lab at 220c
From the table the provided values are of Density of air in the lab at 200c and Density of air
in the lab at 300c.
Density of air in the lab at 220c will be given as: Density of air in the lab at 200c +{ (Density
of air in the lab at 200c- Density of air in the lab at 300c)/10}*2
=0.08270+{(0.08270-0.08101)/10}*2=0.083038
Question 1
Air valve opening (%) Pt-pu(Δp1)-(Pa) V1(m/s) V(m/s)
30 -6 3.219746 6.439493
V1= √2 ( Pt −Pu ) 287 T 1❑
PA = √2 (−6 ) 287∗295.2
98070 = 3.219746
V=2V1= 3.219746*2= 6.439493
Reporting
Question 2: V and V1 values

CROSS FLOW HEAT EXCHANGER 3
V is the mean velocity while V1 is the upstream velocity. The value of V is twice that of V1
when all the rods are fitted in the test section. This is because when the five rods are fitted,
they block half of the inlet area of the working section. From the combined gas laws, at
constant pressure, the velocity of a gas is inversely proportional to the area of flow. This
means when the inlet area is halved, the velocity will be doubled as shown by the equation
v1A1=v2A2.
Question 3: Surface area of the rod
The surface area of a cylinder is calculated as 2πrh+πr2. Since the cylinder is open, the
formula becomes 2πrh=2π*0.0062*0.1034=0.00403m2. This is the effective area.
Question 4: Graph
Question 5: Reynolds number equation
The equation for Reynolds number is given as Re¿ ρVL
μ
Re¿ 0.083038∗6.439493∗0.0124
18.2 x 10−6 =364.31651
Question 6: Type of flow
Given:
ρ = 0.083038 kg/m3
V =6.439493(m/s)
μ =18.2 x 10-6 Kg/ms
L = 0.0124 m
Where,
ρ = Fluid density
V = Fluid velocity
μ = Fluid viscosity
L = length or diameter of the
fluid.