Answer: Plot of
vs
would be linear with a slope of 4.
Explanation:
Given
Equation is ![y=10^{4x}](https://tex.z-dn.net/?f=y%3D10%5E%7B4x%7D)
Taking log both sides
![\Rightarrow \log y=4x\log (10)\\\Rightarrow \log y=4x](https://tex.z-dn.net/?f=%5CRightarrow%20%5Clog%20y%3D4x%5Clog%20%2810%29%5C%5C%5CRightarrow%20%5Clog%20y%3D4x)
It resembles with linear equation ![y=mx+c](https://tex.z-dn.net/?f=y%3Dmx%2Bc)
Here, slope of
vs
is 4.
Answer:
Carbon dioxide temperature at exit is 317.69 K
Carbon dioxide flow rate at heater exit is 20.25 m³/s
Explanation:
Detailed steps are attached below.
Answer:
Ig =7.2 +j9.599
Explanation: Check the attachment
Answer:
T = 858.25 s
Explanation:
Given data:
Reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3, c 550 J/kg K, k 48 W/m K),
initial uniform temperature ( Ti ) = 200 c
Final temperature = 550 c
convection coefficient = 250 w/m^2 k
products combustion temp = 800 c
calculate how long the plate should be left in the furnace ( to attain 550 c )
first calculate/determine the Fourier series Number ( Fo )
![\frac{T_{0}-T_{x} }{T_{1}-T_{x} } = C_{1} e^{(-0.4888^{2}*Fo )}](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7B0%7D-T_%7Bx%7D%20%20%7D%7BT_%7B1%7D-T_%7Bx%7D%20%20%7D%20%3D%20C_%7B1%7D%20e%5E%7B%28-0.4888%5E%7B2%7D%2AFo%20%29%7D)
= 0.4167 = ![1.0396e^{-0.4888*Fo}](https://tex.z-dn.net/?f=1.0396e%5E%7B-0.4888%2AFo%7D)
therefore Fo = 3.8264
Now determine how long the plate should be left in the furnace
Fo = ![(\frac{k}{pc_{p} } ) ( \frac{t}{(L/2)^2} )](https://tex.z-dn.net/?f=%28%5Cfrac%7Bk%7D%7Bpc_%7Bp%7D%20%7D%20%29%20%28%20%5Cfrac%7Bt%7D%7B%28L%2F2%29%5E2%7D%20%29)
k = 48
p = 7830
L = 0.1
Input the values into the relation and make t subject of the formula
hence t = 858.25 s