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Ymorist [56]
3 years ago
7

For laminar flow over a hot flat plate, the local heat transfer coefficient decreases with distance because (select all that are

correct):
A. The velocities near the plate are decreasing.
B. The thickness of the heated region near the plate is increasing.
C. The fluid temperatures near the plate are increasing.
Engineering
1 answer:
kotegsom [21]3 years ago
3 0

Answer:

B. The thickness of the heated region near the plate is increasing.

Explanation:

First we know that, a boundary layer is the layer of fluid in the immediate vicinity of a bounding surface where the effects of viscosity are significant. The fluid is often slower due to the effects of viscosity. Advection i.e the transfer of heat by the flow of liquid becomes less since the flow is slower, thereby the local heat transfer coefficient decreases.

From law of conduction, we observe that heat transfer rate will decrease based on a smaller rate of temperature, the thickness therefore increases while the local heat transfer coefficient decreases with distance.

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Answer: \dot m_{in} = 23.942 \frac{kg}{s}, \dot H_{out} = 39632.62 kW

Explanation:

Since there is no information related to volume flow to and from turbine, let is assume that volume flow at inlet equals to \dot V = 1 \frac{m^{3}}{s}. Turbine is a steady-flow system modelled by using Principle of Mass Conservation and First Law of Thermodynamics:

Principle of Mass Conservation

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First Law of Thermodynamics

- \dot W_{out} + \eta\cdot (\dot m_{in} \dot h_{in} - \dot m_{out} \dot h_{out}) = 0

This 2 x 2 System can be reduced into one equation as follows:

-\dot W_{out} + \eta \cdot \dot m \cdot ( h_{in}- h_{out})=0

The water goes to the turbine as Superheated steam and goes out as saturated vapor or a liquid-vapor mix. Specific volume and specific enthalpy at inflow are required to determine specific enthalpy at outflow and mass flow rate, respectively. Property tables are a practical form to get information:

Inflow (Superheated Steam)

\nu_{in} = 0.041767 \frac{m^{3}}{kg} \\h_{in} = 3399.5 \frac{kJ}{kg}

The mass flow rate can be calculated by using this expression:

\dot m_{in} =\frac{\dot V_{in}}{\nu_{in}}

\dot m_{in} = 23.942 \frac{kg}{s}

Afterwards, the specific enthalpy at outflow is determined by isolating it from energy balance:

h_{out} =h_{in}-\frac{\dot W_{out}}{\eta \cdot \dot m}

h_{out} = 1655.36 \frac{kJ}{kg}

The enthalpy rate at outflow is:

\dot H_{out} = \dot m \cdot h_{out}

\dot H_{out} = 39632.62 kW

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