Steam enters a ship's turbine with a pressure of $6205 \, kN/m^2$ and a velocity of $30.48 \, m/s$. At the turbine's exit, the steam has a pressure of $9.86 \, kN/m^2$ and a velocity of $274.30 \, m/s$. The turbine's inlet is $3.28 \, m$ higher than the outlet. We need to find the work produced by the turbine if the mass flow rate is $15 \, kg/s$ and the heat loss is $14 \, kW$. The specific internal energy and specific volume are given at the inlet and outlet.

1. Problem Description

Steam enters a ship's turbine with a pressure of $6205 \, kN/m^2$ and a velocity of $30.48 \, m/s$. At the turbine's exit, the steam has a pressure of $9.86 \, kN/m^2$ and a velocity of $274.30 \, m/s$. The turbine's inlet is $3.28 \, m$ higher than the outlet. We need to find the work produced by the turbine if the mass flow rate is $15 \, kg/s$ and the heat loss is $14 \, kW$. The specific internal energy and specific volume are given at the inlet and outlet.

2. Solution Steps

The steady-flow energy equation for a turbine is given by:

$h_1 + \frac{v_1^2}{2} + gz_1 = h_2 + \frac{v_2^2}{2} + gz_2 + w + q$

where:

$h$ is the specific enthalpy ($kJ/kg$)

$v$ is the velocity ($m/s$)

$g$ is the acceleration due to gravity ($9.81 \, m/s^2$)

$z$ is the height ($m$)

$w$ is the specific work done by the turbine ($kJ/kg$)

$q$ is the specific heat transfer ($kJ/kg$)

The specific enthalpy is given by:

$h = u + pv$

where:

$u$ is the specific internal energy ($kJ/kg$)

$p$ is the pressure ($kN/m^2$)

$v$ is the specific volume ($m^3/kg$)

First, we calculate the enthalpies at the inlet and outlet:

$h_1 = u_1 + p_1 v_1 = 3150.3 \, kJ/kg + (6205 \, kN/m^2)(0.05789 \, m^3/kg) = 3150.3 + 359.22 \approx 3509.52 \, kJ/kg$

$h_2 = u_2 + p_2 v_2 = 2211.8 \, kJ/kg + (9.86 \, kN/m^2)(13.36 \, m^3/kg) = 2211.8 + 131.73 \approx 2343.53 \, kJ/kg$

Now, let's calculate the kinetic energy terms:

$\frac{v_1^2}{2} = \frac{(30.48 \, m/s)^2}{2} = \frac{929.03}{2} = 464.52 \, J/kg = 0.46452 \, kJ/kg$

$\frac{v_2^2}{2} = \frac{(274.30 \, m/s)^2}{2} = \frac{75240.49}{2} = 37620.25 \, J/kg = 37.62 \, kJ/kg$

The potential energy term is:

$g(z_1 - z_2) = 9.81 \, m/s^2 \times 3.28 \, m = 32.18 \, J/kg = 0.03218 \, kJ/kg$

The heat loss is given as $14 \, kW$ and the mass flow rate is $15 \, kg/s$, so the specific heat transfer is:

$q = \frac{-14 \, kW}{15 \, kg/s} = -0.933 \, kJ/kg$ (negative because it's a heat loss)

Now, we can plug everything into the steady-flow energy equation:

$3509.52 + 0.46452 + 0.03218 = 2343.53 + 37.62 + w - 0.933$

$3510.0167 = 2380.25 + w$

$w = 3510.0167 - 2380.25 = 1129.77 \, kJ/kg$

Finally, to find the total work produced by the turbine, we multiply the specific work by the mass flow rate:

Total work = $w \times \dot{m} = 1129.77 \, kJ/kg \times 15 \, kg/s = 16946.55 \, kW$

3. Final Answer

The work produced by the turbine is $16946.55 \, kW$.