Give step-by-step solution with explanation and final answer:
Act like a helpful tutor and exlain me :
As shown in Figure
1
,
a mechanical drive has a coupling made of two concentric cylinders with

Question

Give step-by-step solution with explanation and final answer:
Act like a helpful tutor and exlain me :
As shown in Figure 
1
,
 ﻿a mechanical drive has a coupling made of two concentric cylinders with the space between cylinders filled with a type of oil. The inner cylinder has a radius of 
1
5
 ﻿mm and is attached to a shaft rotating at an angular velocity 
(
 
ω
 
)
 ﻿of 
9
5
5
0
 ﻿rpm 
.
 ﻿The gap between two cylinders is much sinaller than the inner cylinder diameter with a minimum practical gap of 
0
.
2
2
 ﻿mm 
.
 ﻿The drive is required to provide 
2
0
 ﻿W of power. Determine the viscosity of the oil that would safisfy the requirement for this mechanical drive. The length of the cylinder is 
2
5
 ﻿mm 
,
 ﻿and outer cylinder is rotating at an angular velocity equal or greater than 
9
0
0
0
 ﻿rpm 
.

VIPSolutions · Accepted Answer

Let's solve the problem step by step:

---

## **Step 1: Write Down the Given Data**

- **Inner cylinder radius ($ r_i $)**: 15 mm = 0.015 m  
- **Gap between cylinders ($ h $)**: 0.22 mm = 0.00022 m  
- **Length of cylinder ($ L $)**: 25 mm = 0.025 m  
- **Inner cylinder angular velocity ($ \omega_i $)**: 9550 rpm  
- **Outer cylinder angular velocity ($ \omega_o $)**: ≥ 9000 rpm  
- **Required mechanical power ($ P $)**: 20 W

---

## **Step 2: Convert Angular Velocities to SI Units**

$$
\omega = 2\pi N
$$
where $ N $ is in revolutions per second (rps).  
$ 1 $ rpm = $ \frac{2\pi}{60} $ rad/s.

- **Inner cylinder**:  
  $$
  \omega_i = 9550 \text{ rpm} \times \frac{2\pi}{60} \approx 1000 \text{ rad/s}
  $$
- **Outer cylinder**:  
  $$
  \omega_o = 9000 \text{ rpm} \times \frac{2\pi}{60} \approx 942.5 \text{ rad/s}
  $$

---

## **Step 3: Find the Relative Surface Velocity ($ v $)**

The velocity difference between the cylinders at the surface is:
$$
v = r_i \omega_i - r_i \omega_o = r_i (\omega_i - \omega_o)
$$
$$
v = 0.015 \times (1000 - 942.5) = 0.015 \times 57.5 = 0.8625 \text{ m/s}
$$

---

## **Step 4: Shear Force and Power in a Viscous Coupling**

The viscous force on the cylinder surface is:
$$
F = \tau \cdot A
$$
where
- $ \tau = \mu \frac{v}{h} $ (shear stress)
- $ A = 2\pi r_i L $ (area in contact)

So,
$$
F = \mu \frac{v}{h} \cdot 2\pi r_i L
$$

The power required to overcome this force at surface velocity $ v $ is:
$$
P = F \cdot v = \left[\mu \frac{v}{h} \cdot 2\pi r_i L\right] \cdot v = \mu \frac{v^2}{h} 2\pi r_i L
$$

---

## **Step 5: Solve for the Required Viscosity ($ \mu $)**

Rearrange for $ \mu $:
$$
P = \mu \frac{v^2}{h} 2\pi r_i L
$$
$$
\mu = \frac{P h}{2\pi r_i L v^2}
$$

---

## **Step 6: Substitute All Values**

Plug in the numbers:
- $ P = 20 $ W
- $ h = 0.00022 $ m
- $ r_i = 0.015 $ m
- $ L = 0.025 $ m
- $ v = 0.8625 $ m/s

$$
\mu = \frac{20 \times 0.00022}{2\pi \times 0.015 \times 0.025 \times (0.8625)^2}
$$

First, calculate the denominator:
- $ 2\pi \times 0.015 \times 0.025 = 0.002356 $
- $ (0.8625)^2 = 0.7449 $
- $ 0.002356 \times 0.7449 = 0.001755 $

Now the numerator:
- $ 20 \times 0.00022 = 0.0044 $

$$
\mu = \frac{0.0044}{0.001755} \approx 2.51 \text{ Pa}\cdot\text{s}
$$

---

## **Step 7: Final Answer**

$$
\boxed{\mu \approx 2.5 \ \text{Pa}\cdot\text{s}}
$$

---

### **Summary**

- The oil must have a **viscosity of approximately $ 2.5 \ \text{Pa}\cdot\text{s} $** for the mechanical drive to deliver the required 20 W, given the above dimensions and speeds.
- All values are rounded to two significant digits for practical engineering use.

If you have follow-up questions or need clarification on any step, feel free to ask!

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