For the two orifices shown in the figure below, determine Y2 such that X2=(3X1)/4.

Two Orifice Analysis

Analysis of Two Identical Orifices

Problem Statement

For the two orifices shown in the figure below, determine Y2 such that X2 = (3X1)/4.

Two orifices with water tanks

Given Data

Height of water in first tank (H1) 2 m
Total height of first tank (Y1) 10 – 2 = 8 m
Height of water in second tank (H2) 10 – Y2 m
Required condition X2 = (3X1)/4
Total height of tanks 10 m
Required Y2 value

1. Establishing the Relationship Between Velocity Coefficients

The coefficient of velocity for the first orifice is defined as:

Cv1 = X1 / √(4Y1H1)

Similarly, for the second orifice:

Cv2 = X2 / √(4Y2H2)

Since the two orifices are identical, their coefficients of velocity must be equal:

Cv1 = Cv2

Therefore:

X1 / √(4Y1H1) = X2 / √(4Y2H2)

2. Substituting Known Values

We know:

  • H1 = 2 m
  • Y1 = 10 – 2 = 8 m
  • H2 = 10 – Y2 m
  • X2 = (3X1)/4

Substituting these values into our equation:

X1 / √(4Y1H1) = X2 / √(4Y2H2)
X1 / √(4 × 8 × 2) = (3X1/4) / √(4 × Y2 × (10-Y2))
X1 / √(64) = (3X1/4) / √(4Y2(10-Y2))
X1 / 8 = (3X1/4) / √(4Y2(10-Y2))

3. Solving for Y2

Rearranging the equation:

(X1/8) × √(4Y2(10-Y2)) = 3X1/4
√(4Y2(10-Y2)) = (3X1/4) ÷ (X1/8)
√(4Y2(10-Y2)) = (3X1/4) × (8/X1)
√(4Y2(10-Y2)) = 3 × 2 = 6
4Y2(10-Y2) = 36
4(10Y2 – Y22) = 36
40Y2 – 4Y22 = 36
-4Y22 + 40Y2 – 36 = 0
-4(Y22 – 10Y2 + 9) = 0
Y22 – 10Y2 + 9 = 0

Using the quadratic formula:

Y2 = (10 ± √(100 – 36))/2 = (10 ± √64)/2 = (10 ± 8)/2
Y2 = 9 or Y2 = 1

Since Y1 = 8m and Y2 = 9m would make Y2 > Y1, which is not feasible in this context, we take Y2 = 1m as our solution.

4. Verification and Physical Interpretation

Let’s verify our solution by checking if X2 = (3X1)/4 when Y2 = 1m:

For Y2 = 1m:

  • H2 = 10 – Y2 = 10 – 1 = 9m

From the velocity coefficient relationship:

X1 / √(4 × 8 × 2) = X2 / √(4 × 1 × 9)
X1 / 8 = X2 / 6
X2 = (6/8) × X1 = (3/4) × X1

This confirms that X2 = (3X1)/4 when Y2 = 1m, validating our solution.

Physically, this means that the distance from the second orifice to the bottom of the second tank should be 1m to satisfy the condition that X2 = (3X1)/4.

Y2 = 1 m

Conclusion

In this analysis of two identical orifices, we determined:

1. Required Condition: For X2 = (3X1)/4, the value of Y2 must be 1m.

2. Solution Process: We established the relationship between velocity coefficients for identical orifices and solved a quadratic equation to find Y2.

3. Solution Analysis: While solving the quadratic equation Y22 – 10Y2 + 9 = 0 yielded two potential solutions (Y2 = 1m or Y2 = 9m), we determined that Y2 = 9m is not feasible since it exceeds Y1 = 8m.

4. Physical Interpretation: The distance from the second orifice to the bottom of the second tank should be 1m, while the height of water above this orifice will be H2 = 9m.

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