Answer:
fire brick / common brick : 1218 °Ccommon brick / magnesia : 1019 °Cmagnesia / steel : 90.06 °Cheat loss: 4644 kJ/m^2/hExplanation:
The thermal resistance (R) of a layer of thickness d given in °C·m²·h/kJ is ...
R = d/k
so the thermal resistances of the layers of furnace wall are ...
R₁ = 0.200/4 = 0.05 °C·m²·h/kJ
R₂ = 0.120 2.8 = 3/70 °C·m²·h/kJ
R₃ = 0.05/0.25 = 0.2 °C·m²·h/kJ
R₄ = 0.003/240 = 1.25×10⁻⁵ °C·m²·h/kJ
So, the total thermal resistance is ...
R₁ +R₂ +R₃ +R₄ = R ≈ 0.29286 °C·m²·h/kJ
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The rate of heat loss is ΔT/R = (1450 -90)/0.29286 = 4643.70 kJ/(m²·h)
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The temperature drops across the various layers will be found by multiplying this heat rate by the thermal resistance for the layer:
fire brick: (4543.79 kJ/(m²·h))(0.05 °C·m²·h/kJ) = 232 °C
so, the fire brick interface temperature at the common brick is ...
1450 -232 = 1218 °C
For the next layers, the interface temperatures are ...
common brick to magnesia = 1218 °C - (3/70)(4643.7) = 1019 °C
magnesia to steel = 1019 °C -0.2(4643.7) = 90.06 °C
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Comment on temperatures
Most temperatures are rounded to the nearest degree. We wanted to show the small temperature drop across the steel plate, so we showed the inside boundary temperature to enough digits to give the idea of the magnitude of that.
As steam is slowly injected into a turbine, the angular acceleration of the rotor is observed to increase linearly with the time t. Know that the rotor starts from rest at t = 0 and that after 10 s the rotor has completed 20 revolutions.Choose the correct equations of motion for the rotor. (You must provide an answer before moving on to the next part.)
a) a = 2kt, w = 3krº, and 8 = 4kr
b) a = {kt, w = ke?, and 0 = }ke?
c) a = kr?, w = jke', and 0 = krº
d) a = kt, w = jke?, and 0 kr
Answer:
α = kt
ω = [tex]\frac{kt^2}{2}[/tex]
θ = [tex]\frac{kt^3}{6}[/tex]
Explanation:
given data
Initial velocity ω = 0
time t = 10 s
Number of revolutions = 20
solution
we will take here first α = kt .....................1
and as α = [tex]\frac{d\omega}{dt}[/tex]
so that
[tex]\frac{d\omega}{dt}[/tex] = kt ..................2
now we will integrate it then we will get
∫dω = [tex]\int_{0}^{t} kt\ dt[/tex] .......................3
so
ω = [tex]\frac{kt^2}{2}[/tex]
and
ω = [tex]\frac{d\theta}{dt}[/tex] ..............4
so that
[tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{kt^2}{2}[/tex]
now we will integrate it then we will get
∫dθ = [tex]\int_{0}^{t}\frac{kt^2}{2} \ dt[/tex] ...............5
solve it and we get
θ = [tex]\frac{kt^3}{6}[/tex]
One kg of an idea gas is contained in one side of a well-insulated vessel at 800 kPa. The other side of the vessel is under vacuum. The two sides are separated by a piston that is initially held in place by the pins. The pins are removed and the gas suddenly expands until it hits the stops. What happens to the internal energy of the gas?
a. internal energy goes up
b. internal energy goes down
c. internal energy stays the same
d. we need to know the volumes to make the calculation
Answer:
Option C = internal energy stays the same.
Explanation:
The internal energy will remain the same or unchanged because this question has to do with a concept in physics or classical chemistry (in thermodynamics) known as Free expansion.
So, the internal energy will be equals to the multiplication of the change in temperature, the heat capacity (keeping volume constant) and the number of moles. And in free expansion the internal energy is ZERO/UNCHANGED.
Where, the internal energy, ∆U = 0 =quantity of heat, q - work,w.
The amount of heat,q = Work,w.
In the concept of free expansion the only thing that changes is the volume.
10 kg/s Propane at 10 bar and 20 C is directed to an adiabatic rigid mixer and is mixed with 20 kg/s Propane at 10 bar and 40 C. What is the final volumetric flow rate in (m3/s) of the resulting mixture.
Answer:
The final volumetric flow rate will be "76.4 m³/s".
Explanation:
The given values are:
[tex]\dot{m_{1}}=10 \ kg/s[/tex]
[tex]\dot{m_{2}}=20 \ Kg/s[/tex]
[tex]T_{1}=293 \ K[/tex]
[tex]T_{2}=313 \ K[/tex]
[tex]P_{1}=P_{2}=P_{3}=10 \ bar[/tex]
As we know,
⇒ [tex]E_{in}=E_{out}[/tex]
[tex]\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}[/tex]
[tex]e_{1}\dot{v_{1}}h_{1}+e_{2}\dot{v_{2}}h_{2}=e_{3}\dot{v_{3}}h_{3}[/tex]
[tex]\frac{P_{1}}{RP_{1}}\dot{v_{1}} \ C_{p}T_{1}+ \frac{P_{2}}{RP_{2}}\dot{v_{2}} \ C_{p}T_{1}=\frac{P_{3}}{RP_{3}}\dot{v_{3}} \ C_{p}T_{3}[/tex]
⇒ [tex]\dot{v_{3}}=\dot{v_{1}}+\dot{v_{2}}[/tex]
[tex]=\frac{\dot{m_{1}}}{e_{1}}+\frac{\dot{m_{2}}}{e_{2}}[/tex]
On substituting the values, we get
[tex]=\frac{10}{10\times 10^5}\times 8314\times 293+\frac{20\times 8314\times 313}{10\times 10^5}[/tex]
[tex]=76.4 \ m^3/s[/tex]