Steam with a mass flow rate of $6000 \, \text{kg/h}$ enters a turbine. The inlet area is $1.5 \, \text{m}^2$, and the outlet area is $0.55 \, \text{m}^2$. The specific volume at the inlet is $0.30 \, \text{m}^3/\text{kg}$, and the outlet velocity is $80 \, \text{m/s}$. Find the velocity of the steam at the inlet of the turbine.

1. Problem Description

Steam with a mass flow rate of $6000 \, \text{kg/h}$ enters a turbine. The inlet area is $1.5 \, \text{m}^2$, and the outlet area is $0.55 \, \text{m}^2$. The specific volume at the inlet is $0.30 \, \text{m}^3/\text{kg}$, and the outlet velocity is $80 \, \text{m/s}$. Find the velocity of the steam at the inlet of the turbine.

2. Solution Steps

First, convert the mass flow rate from kg/h to kg/s:

$$\dot{m} = 6000 \, \frac{\text{kg}}{\text{h}} \times \frac{1 \, \text{h}}{3600 \, \text{s}} = \frac{6000}{3600} \, \frac{\text{kg}}{\text{s}} = \frac{5}{3} \, \frac{\text{kg}}{\text{s}}$$

The mass flow rate is given by

$$\dot{m} = \frac{A \cdot v}{v_s}$$

where $\dot{m}$ is the mass flow rate, $A$ is the cross-sectional area, $v$ is the velocity, and $v_s$ is the specific volume.

At the inlet, we have $A_1 = 1.5 \, \text{m}^2$ and $v_{s1} = 0.30 \, \text{m}^3/\text{kg}$. We want to find $v_1$. We can rearrange the mass flow rate equation to solve for the inlet velocity:

$$v_1 = \frac{\dot{m} \cdot v_{s1}}{A_1}$$

Plugging in the values, we get:

$$v_1 = \frac{\frac{5}{3} \, \frac{\text{kg}}{\text{s}} \cdot 0.30 \, \frac{\text{m}^3}{\text{kg}}}{1.5 \, \text{m}^2} = \frac{\frac{5}{3} \cdot 0.3}{1.5} \, \frac{\text{m}}{\text{s}} = \frac{\frac{1.5}{3}}{1.5} \, \frac{\text{m}}{\text{s}} = \frac{0.5}{1.5} \, \frac{\text{m}}{\text{s}} = \frac{1}{3} \, \frac{\text{m}}{\text{s}} = 0.333 \, \frac{\text{m}}{\text{s}}$$

So, the velocity at the inlet is $0.333 \, \text{m/s}$.

3. Final Answer

The velocity of the steam at the inlet of the turbine is $0.333 \, \text{m/s}$.