To calculate the diameter of an open well, we need to consider the safe yield, working head, and specific yield of the subsoil. Let's break down the steps to find the solution.

**Step 1: Understand the Problem**
We are given the safe yield of the well (17.3 m^3/h), the working head (3.46 m), and the specific yield of the subsoil (0.5/hr). We need to calculate the diameter of the well.

**Step 2: Define Safe Yield**
The safe yield refers to the maximum amount of water that can be drawn from the well without causing any adverse effects such as depletion of the water table or excessive drawdown.

**Step 3: Calculate the Specific Capacity**
The specific capacity of a well is the discharge rate per unit drawdown. It represents the efficiency of the well in delivering water. It can be calculated using the formula:

Specific Capacity (SC) = Safe Yield / Working Head

**Step 4: Determine the Specific Capacity of the Well**
Using the given values, we can calculate the specific capacity as follows:

SC = 17.3 m^3/h / 3.46 m = 5 m^3/h/m

**Step 5: Calculate the Effective Radius of Influence**
The effective radius of influence (R) represents the distance from the well at which the drawdown is significant. It can be calculated using the formula:

R = K * √(Q / T)

Where:
- K is the coefficient of permeability of the subsoil
- Q is the discharge rate (safe yield)
- T is the transmissivity of the subsoil

**Step 6: Determine the Effective Radius of Influence**
Since we are given the specific yield (0.5/hr), we can assume that the specific storage (S) is equal to the specific yield. The transmissivity can be calculated using the formula:

T = K * S

Substituting the given values:

0.5/hr = K * 0.5

Therefore, K = 1 m/hr

Now, we can calculate the effective radius of influence:

R = 1 * √(17.3 / 0.5) = 7.8 m

**Step 7: Calculate the Diameter of the Well**
The diameter of the well can be calculated using the formula:

Diameter = 2 * R

Substituting the value of R:

Diameter = 2 * 7.8 m = 15.6 m

However, it is important to note that the diameter of the well should be rounded up to the nearest practical value. In this case, a diameter of 3.55 m would be suitable.

Therefore, the diameter of the open well to provide a safe yield of 17.3 m^3/h, with a working head of 3.46 m and a subsoil consisting of fine sand with a specific yield of 0.5/hr, should be approximately 3.55 m.