Experiment No. 3-a
DETERMINATION OF THE COEFFICIENT OF DISCHARGE IN AN ORIFICE
(CONSTANT HEAD)
I. DISCUSSION
An orifice is a sudden flow of restriction of short length (zero length for sharp edge orifice).
Orifices are treated as either a sharp edge orifice or short tube orifice. Orifices are primarily used to
control flow or create pressure differential (drop). Orifices may be fixed or variable (valve). Many types of
valves and flow devices can essentially be viewed as orifices. In valves, there can be numerous flow
passages, but usually somewhere in the flow passage is a restriction that controls flow, which is why a
valve often behaves like an orifice.
The theoretical discharge of an orifice is given by Q = AV.
Where;
Q – discharged through an orifice
A – area of the orifice
V – velocity of water flowing through
g – gravitational acceleration
h – head as drawn
II. EXPERIMENT OBJECTIVES
a. To demonstrate the characteristics of flow over an orifice
b. To determine the discharge coefficient of an orifice under constant head.
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III. MATERIALS AND APPARATUS REQUIRED
a. Orifice
b. Orifice Tank
c. Stopwatch
d. Ruler
e. Water Hose
IV. PROCEDURES
1. Select an orifice to be screwed in the orifice tank
2. Supply the orifice with a constant flow of water until the level of water in the tank is constant
also. Use a hose in supplying the tank.
3. Measure the height of water in the tank, h. Measure the projectile distances, x and y.
4. Use a Vernier caliper to determine the diameter of the orifice, Ao.
5. Determine the actual discharge of the orifice using the volumetric method procedure.
𝒗𝒐𝒍𝒖𝒎𝒆
𝑸𝒂 =
𝒕𝒊𝒎𝒆
6. Determine the actual velocity, va using the concept of projectile in dynamics.
Since the orifice is projected horizontally;
1
𝑦 = 𝑎𝑡 2
2
𝑥 = 𝑣𝑎 𝑡
𝒗𝒂 =
𝒙
𝒕
7. Compute for the actual area of flow or the area of the vena contracta.
𝑸𝒂
𝑨𝒂 =
𝒗𝒂
8. Compute for the theoretical velocity, vt ¸ using the equation
𝒗𝒕 = √𝟐𝒈𝒉
9. Determine the theoretical discharge by,
𝑸𝒕 = 𝑨𝒐 𝒗𝒕
10. Compute the coefficient of discharge.
𝑪𝒅 =
𝑸𝒂
𝑸𝒕
𝑪𝒄 =
𝑨𝒂
𝑨𝒐
𝑪𝒗 =
𝒗𝒂
𝒗𝒕
11. Compute for the coefficient of contraction.
12. Compute for the coefficient of velocity.
2
V. LABORATORY DATA
TRIAL
1
2
3
Height of water in the tank, h
0.51
0.57
0.48
x-component of projectile
0.87
0.83
0.85
y-component of projectile
0.45
0.43
0.46
Area of orifice, Ao
6.3617 × 10−5
3.2675 × 10−5
7.0138 × 10−5
Actual Discharge, Qa
1.6667 × 10−4
8.3333 × 10−5
1.6667 × 10−4
Actual velocity, va
2.8722
2.8031
2.776
Area of Vena Contracta, Aa
5.8029 × 10−5
2.9729 × 10−5
6.004 × 10−5
Theoretical velocity, vt
3.1633
3.3442
3.0688
Theoretical Discharge, Qt (m3/s)
2.0124 × 10−4
1.0927 × 10−4
2.1524 × 10−4
Coefficient of Discharge, Cd
0.8282
0.7626
0.7743
Coefficient of Contraction, Cc
0.908
0.8382
0.9046
Coefficient of Velocity, Cv
0.9122
0.9098
0.856
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VI. CALCULATIONS

Trial 1
𝐴𝑜 =
𝜋
(0.009)2 = 6.3617 × 10−5
4
𝑄𝑎 =
0.5⁄
1000 = 1.6667 × 10−4
3
1
𝑦 = 𝑔𝑡 2
2
1
0.45 = (9.81)𝑡 2
2
𝑡 = 0.3029
𝑉𝑎 =
𝑥
0.87
=
= 2.8722
𝑡 0.3029
𝐴𝑎 =
𝑄𝑎 1.6667 × 10−4
=
= 5.8029 × 10−5
𝑉𝑎
2.8722
𝑉𝑡 = √2𝑔ℎ = √2(9.81)(0.51) = 3.1633
𝑄𝑡 = 𝐴𝑜 𝑉𝑡 = 6.3617 × 10−5 (3.1633) = 2.0124 × 10−4
𝐶𝑑 =
𝑄𝑎 1.6667 × 10−4
=
= 0.8282
𝑄𝑡 2.0124 × 10−4
𝐶𝑣 =
𝑉𝑎 2.8722
=
= 0.908
𝑉𝑡 3.1633
𝐶𝑐 =
𝐴𝑎 5.8029 × 10−5
=
= 0.9122
𝐴𝑜 6.3617 × 10−5
4

Trial 2
𝐴𝑜 =
𝜋
(0.00645)2 = 3.2675 × 10−5
4
𝑄𝑎 =
0.5⁄
1000 = 8.3333 × 10−5
6
1
𝑦 = 𝑔𝑡 2
2
1
0.43 = (9.81)𝑡 2
2
𝑡 = 0.2961
𝑉𝑎 =
𝑥
0.83
=
= 2.8031
𝑡 0.2961
𝐴𝑎 =
𝑄𝑎 8.3333 × 10−5
=
= 2.9729 × 10−5
𝑉𝑎
2.8031
𝑉𝑡 = √2𝑔ℎ = √2(9.81)(0.57) = 3.3442
𝑄𝑡 = 𝐴𝑜 𝑉𝑡 = 3.2675 × 10−5 (3.3442) = 1.0927 × 10−4
𝐶𝑑 =
𝑄𝑎 8.3333 × 10−5
=
= 0.7626
𝑄𝑡 1.0927 × 10−4
𝐶𝑣 =
𝑉𝑎 2.8031
=
= 0.8382
𝑉𝑡 3.3442
𝐶𝑐 =
𝐴𝑎 2.9729 × 10−5
=
= 0.9098
𝐴𝑜 3.2675 × 10−5
5

Trial 3
𝐴𝑜 =
𝜋
(0.00945)2 = 7.0138 × 10−5
4
0.5⁄
1000 = 1.6667 × 10−4
𝑄𝑎 =
3
1
𝑦 = 𝑔𝑡 2
2
1
0.46 = (9.81)𝑡 2
2
𝑡 = 0.3062
𝑉𝑎 =
𝑥
0.85
=
= 2.776
𝑡 0.3062
𝐴𝑎 =
𝑄𝑎 1.6667 × 10−4
=
= 6.004 × 10−5
𝑉𝑎
2.776
𝑉𝑡 = √2𝑔ℎ = √2(9.81)(0.48) = 3.0688
𝑄𝑡 = 𝐴𝑜 𝑉𝑡 = 7.0138 × 10−5 (3.0688) = 2.1524 × 10−4
𝐶𝑑 =
𝑄𝑎 1.6667 × 10−4
=
= 0.7743
𝑄𝑡 2.1524 × 10−4
𝐶𝑣 =
𝑉𝑎
2.776
=
= 0.9046
𝑉𝑡 3.0688
𝐶𝑐 =
𝐴𝑎
6.004 × 10−5
=
= 0.856
𝐴𝑜 7.0138 × 10−5
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