Trees grow in bulk in outward direction

A completely reversible heat pump produces heat ata rate of 300 kW to warm a house maintained at 24°C. Theexterior air, which is

Triss [41]

Answer:

Change in entropy S = 0.061

Second law of thermodynamics is satisfied since there is an increase in entropy

Explanation:

Heat Q = 300 kW

T2 = 24°C = 297 K

T1 = 7°C = 280 K

Change in entropy =

S = Q(1/T1 - 1/T2)

= 300(1/280 - 1/297) = 0.061

There is a positive increase in entropy so the second law is satisfied.

6 0

3 years ago

How engineer can find problems and solutions <br><br><br> Give example

Mademuasel [1]

Answer:

They find problems and solutions by working together

Explanation:

4 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

3 years ago

Relay contacts that are defined as being normally open (n.o.) have contacts that are:_____.

Reptile [31]

Relay contacts that are defined as being normally open (n.o.) have contacts that are open only if the relay coil is known to have de-energized.

<h3>What is meant by normally open contacts?</h3>

Normally open (NO) are known to be open if there is no measure of current that is flowing through a given coil but it often close as soon as the coil is said to be energized.

Note that Normally closed (NO) contacts are said to be closed only if the coil is said to be de-energized and open only if the coil is said to carry current or is known to have energized.

The role of relay contact is wide. The Relays are tools that are often used in the work of switching of control circuits and it is one that a person cannot used for power switching that has relatively bigger ampacity.

Therefore, Relay contacts that are defined as being normally open (n.o.) have contacts that are open only if the relay coil is known to have de-energized.

Learn more about Relay contacts from

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#SPJ1

8 0

1 year ago

On-site oil storage containers must be marked "Used Oil."<br><br> True. <br><br> False.

Scrat [10]

Answer:

How must used oil storage containers be marked? Containers and aboveground tanks used to store used oil at generator facilities must be labeled or marked clearly with the words “Used Oil" (40 CFR Section 279.22(c)).

Explanation:

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7 0

3 years ago

Trees grow in bulk in outward direction​

Trees grow in bulk in outward direction