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3 years ago

Select the correct answer.

Engineering

1 answer:

babunello [35]3 years ago

3 0

Answer:

The information he needs next is;

B. Volume of paint needed per unit area

Explanation:

The operation Andy wants to perform = To apply paint on the walls of the house

The information Andy knows = The area of all walls in the house

The total volume of paint needed = The total area of the walls × The volume of paint needed for each unit area

Therefore, the information required is the volume of paint needed per unit area.

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As the temperature of a thermal radiator is increased Group of answer choices the object appears redder. the object appears blue

Elanso [62]

Answer:

As the temperature of a thermal radiator is increased

Group of answer choices

the object appears redder.
the object appears bluer.
the object emits more power for the same area.
the object emits less power for the same area.
the object expands to keep the same power per units area.

<em>When the temperature of a thermal radiator increases ;</em>

<em>the object emits more power for the same are</em>
<em>the object becomes bluer</em>

Explanation:

Thermal radiation involves the transfer of heat between molecules of two substances without direct contact with each other. When a body is heated to a given temperature it begins to emit light which is transferred to nearby objects as thermal radiation. The medium through which the heat is transferred could be liquid, solid, or in a vacuum.

<h3>How temperature affects thermal radiation.</h3>

Temperature determines the amount of heat that is been radiated from a body. An increase in temperature would increase the thermal radiation of the body. The increase in the heat radiation results to increase in the thermal energy of the body. Also when a body is heated it tends to be bluer than a cool object, this is caused by the rapid movement of the molecules.

Therefore When the temperature of a thermal radiator increases ;

the object emits more power for the same are
the object becomes bluer

4 0

4 years ago

An isentropic steam turbine processes 2 kg/s of steam at 3 MPa, which is exhausted at50 kPa and 100C. Five percent of this flow

borishaifa [10]

Answer:

2285kw

Explanation:

since it is an isentropic process, we can conclude that it is a reversible adiabatic process. Hence the energy must be conserve i.e the total inflow of energy must be equal to the total outflow of energy.

Mathematically,

$\\ E_{inflow} = E_{outflow}$

Note: from the question we have only one source of inflow and two source of outflow (the exhaust at a pressure of 50kpa and the feedwater at a pressure of 5ookpa). Also the power produce is another source of outgoing energy $\\ E_{inflow} = m_{1} h_{1}$ .

$\\$

$E_{outflow} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$

$\\$

Where $m_{1} h_{1}$ are the mass flow rate and the enthalpies at the inlet at a pressure of 3Mpa $\\$ ,

$m_{2} h_{2}$ are the mass flow rate and the enthalpies at the outlet 2 where we have a pressure of 500kpa respectively. $\\$ ,

and $m_{3} h_{3}$ are the mass flow rate and the enthalpies at the outlet 3 where we have a pressure of 50kpa respectively. $\\$ ,

We can now express write out the required equation by substituting the new expression for the energies $\\$

$m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$ $\\$

from the above equation, the unknown are the enthalpy values and the mass flow rate. $\\$

first let us determine the enthalpy values at the inlet and the out let using the Superheated water table. $\\$

It is more convenient to start from outlet 3 were we have a temperature $100^{0}C$ and pressure value of (50kpa or 0.05Mpa ). using double interpolation method on the superheated water table to determine the enthalpy value with careful calculation we have $\\$

$h_{3}$ = 2682.4 KJ/KG , at this point also from the table the entropy value , $s_{3}$ value is 7.6953 KJ/Kg.K. $\\$

Next we determine the enthalphy value at outlet 2. But in this case, we don't have a temperature value, hence we use the entrophy value since the entropy is constant at all inlet and outlet. $\\$

So, from the superheated water table again, at a pressure of 500kpa (0.5Mpa) and entropy value of 7.6953 KJ/Kg.K with careful interpolation we arrive at a enthalpy value of 3206.5KJ/Kg. $\\$

Finally for inlet one at a pressure of 3Mpa, interpolting with an entropy value of 7.6953KJ/Kg.K we arrive at enthalpy value of 3851.2KJ/Kg. $\\$

Now we determine the mass flow rate at each inlet and outlet. since mass must also be balance, i.e $m_{1} = m_{2} + m_{3}$ $\\$