Testing the mathjax engine
05-08-2016, 08:56 PM (This post was last modified: 01-31-2017 01:37 PM by emece67.)
Post: #1
 emece67 Senior Member Posts: 321 Joined: Feb 2015
Testing the mathjax engine
From $$Q = mC\Delta T$$ you will get $$\Delta T = {Q\over mC}$$, with $$C=4182 {J\over kg \Delta K}$$ being the specific heat of water at 20ºC. With this you get $\Delta T = {25000 J\over 2kg · 4182 {J\over kg \Delta K}} = 2.989 \Delta K$

Then the final temperature will be: $$T_0+\Delta T = 20 ºC + 2.989 \Delta K =$$ Inconsistent units.

César - Information must flow.
02-19-2017, 03:33 PM
Post: #2
 Csaba Tizedes Member Posts: 219 Joined: May 2014
RE: Testing the mathjax engine
$\frac{N_i}{D_i}=\frac{N_{i-1}}{D_{i-1}}+\frac{1}{F_i}=\frac{F_i\cdotN_{i-1}+D_{i-1}}{D_{i-1}\cdotF_i}$
02-19-2017, 03:35 PM
Post: #3
 Csaba Tizedes Member Posts: 219 Joined: May 2014
RE: Testing the mathjax engine
(05-08-2016 08:56 PM)emece67 Wrote:  From $$Q = mC\Delta T$$ you will get $$\Delta T = {Q\over mC}$$, with $$C=4182 {J\over kg \Delta K}$$ being the specific heat of water at 20ºC. With this you get $\Delta T = {25000 J\over 2kg · 4182 {J\over kg \Delta K}} = 2.989 \Delta K$

Then the final temperature will be: $$T_0+\Delta T = 20 ºC + 2.989 \Delta K =$$ Inconsistent units.
\frac{N_i}{D_i}=\frac{N_{i-1}}{D_{i-1}}+\frac{1}{F_i}=\frac{F_i\cdotN_{i-1}+D_{i-1}}{D_{i-1}\cdotF_i}
02-19-2017, 03:40 PM
Post: #4
 Csaba Tizedes Member Posts: 219 Joined: May 2014
RE: Testing the mathjax engine
(05-08-2016 08:56 PM)emece67 Wrote:  From $$Q = mC\Delta T$$ you will get $$\Delta T = {Q\over mC}$$, with $$C=4182 {J\over kg \Delta K}$$ being the specific heat of water at 20ºC. With this you get $\Delta T = {25000 J\over 2kg · 4182 {J\over kg \Delta K}} = 2.989 \Delta K$

Then the final temperature will be: $$T_0+\Delta T = 20 ºC + 2.989 \Delta K =$$ Inconsistent units.

$$\frac{N_i}{D_i}=\frac{N_{i-1}}{D_{i-1}}+\frac{1}{F_i}=\frac{F_i · N_{i-1}+D_{i-1}}{D_{i-1} · F_i}$$

JÍHÁÁÁÁ!
02-19-2017, 08:14 PM (This post was last modified: 02-19-2017 08:14 PM by Luigi Vampa.)
Post: #5
 Luigi Vampa Member Posts: 186 Joined: Dec 2015
RE: Testing the mathjax engine
(05-08-2016 08:56 PM)emece67 Wrote:  [•••]
Then the final temperature will be: $$T_0+\Delta T = 20 ºC + 2.989 \Delta K =$$ Inconsistent units.
I might be wrong, I haven't been involved in thermal issues since my second course in the University.
Kelvin degrees = Centigrade degrees + the absolute zero offset
Offsets are canceled in the Delta, so that there would be no incoherence. The same Delta in Centigrade degrees would have the very same value.
Fahrenheit degrees are another story, you know.

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06-11-2017, 10:56 AM
Post: #6
 Csaba Tizedes Member Posts: 219 Joined: May 2014
RE: Testing the mathjax engine
(02-19-2017 03:40 PM)Csaba Tizedes Wrote:
(05-08-2016 08:56 PM)emece67 Wrote:  From $$Q = mC\Delta T$$ you will get $$\Delta T = {Q\over mC}$$, with $$C=4182 {J\over kg \Delta K}$$ being the specific heat of water at 20ºC. With this you get $\Delta T = {25000 J\over 2kg · 4182 {J\over kg \Delta K}} = 2.989 \Delta K$

Then the final temperature will be: $$T_0+\Delta T = 20 ºC + 2.989 \Delta K =$$ Inconsistent units.

$$\frac{N_i}{D_i}=\frac{N_{i-1}}{D_{i-1}}+\frac{1}{F_i}=\frac{F_i · N_{i-1}+D_{i-1}}{D_{i-1} · F_i}$$

JÍHÁÁÁÁ!

From these equations ($$\beta$$ in radian):
$$R_2 · \beta=2 · r · \pi$$, and

$$(R_2+s) · \beta=2 · R · \pi$$

we get:
$$R_2=\frac{\frac{h}{cos \alpha}}{\frac{R}{R-h_x}-1}$$, and

$$\beta=\frac{2 · (R-h_x) · \pi}{R_2}$$
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