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4 years ago

Is a steam power plant a heat engine?

Engineering

1 answer:

Novosadov [1.4K]4 years ago

7 0

Answer:

Yes

Explanation:

Yes steam power plant is heat engine.

As we know that a heat engine takes from source heat and produce some amount of work and reject the heat to the sink.

So the steam power plant fulfill the all requirement like it take heat from boiler and produce some amount work and then reject the heat by using condenser.So we can say that steam power plant is an example of heat engine.

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3 years ago

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3 years ago

The shear flow at the point depends on the value of Q for the portion of the upper flange to the right of the point. Calculate t

Ostrovityanka [42]

Q)The shear flow at a point is given by q = 1 of 2 An I-beam has a flange width b = 200 mm , height h = 200 mm , web thickness tw = 8 mm, and flange thickness t = 12 mm, For the given point The shear flow at the point depends on the value of Q for the portion of the upper flange to the right of the point. Calculate the value of Q. Express your answer with appropriate units to three significant figures.( <u>I have also attached the diagram for better understanding)</u>

Answer:

Q = 56400 mm^3

Explanation:

<u>I have given the explanation in the attached file below.</u>

6 0

3 years ago

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu

Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

Negligible heat loss through the insulation
Steady state system
One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

$k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}$

During steady state

$\frac{dT}{dt}$ = 0 which gives $k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0$

From which we have $\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}$

Considering the boundary condition at x =0 where there is no heat loss

$\frac{dT}{dt}$ = 0 also at the other end of the plane wall we have

$-k\frac{dT }{dx } =$ hc (T - T∞) at point x = L

Integrating the equation we have

$\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}$ from which C₁ is evaluated from the first boundary condition thus

0 = $\frac{q'_{G}}{k} (0)+ C_{1}$ from which C₁ = 0

From the second integration we have

$T = -\frac{q'_{G}}{2k} x^{2} + C_{2}$

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

$-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k} + C_{2}-T∞)$ → C₂ = $q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$

T(x) = $\frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$ and T(x) = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )$

∴ Tmax → when x = 0 = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))$

Substituting the values we get

T = 167 ° C

4 0

4 years ago