As per text from notebook, having difficulty understand how they tranposed the Head to the left side, and D to the right. I can't seem to figure out how the 3/2 fraction beside H got swapped to the power of 2/3. It's been many years since high school, my co worker tried helping me understand but I still don't get the rules (he couldn't really explain how he did it, just did it). The top is my attempt and bottom is his. Thanks for any help
Please help me understand this transposing (in steam engineering textbook)
- Thread starter Girn
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It appears as if you've not included the full and exact problem text, as there's some information missing. I can read between a lines a bit and guess at what it's meant to be, but regardless the problem is very poorly formulated and the person who gave it to you ought to be ashamed of their abysmal notation!
The relevant assumptions I'm making are as follows. Please correct any that are wrong and provide the correct information as needed. D stands for "Discharge" and H stands for "Head". The very first equation that gives you 4.5 m3 / s is meant to be set equal to D (considering they later mysteriously substitute in 4.5 for the value of D; but this equality is never explicitly stated). In addition to these assumptions, the awful notation I've had to decipher suggests that the final term in the equation for D is not H times 3/2, as it appears to be, but instead H to the power of 3/2. A bit of accuracy is lost in the gravity step, as g is not exactly 9.81, nor is sqrt(19.62) exactly 4.429. For the purposes of whatever you're doing, these approximations seem to close enough, although the rounding ought to be noted with an approximately equals symbol (≈). None of this is your problem or your fault, but I just despise poor notation in math problems. Students learn to hate math just fine on their own without any encouragement from their teachers
Anyway, to the task at hand. I trust you follow the steps up to this point:
\(\displaystyle \displaystyle H^{\frac{3}{2}} \approx \dfrac{D}{\frac{2}{3} \cdot 0.59 \cdot 6 \cdot 4.429}\)
At this point, it's simply a matter of raising both sides to the 2/3 power to cancel out the exponent on H. This is because of the rules of exponentshttp://www.mesacc.edu/~scotz47781/mat120/notes/exponents/review/review.html. Namely:
\(\displaystyle \displaystyle \left( H^{\frac{3}{2}} \right)^{\frac{2}{3}} = H^{\frac{3}{2} \cdot \frac{2}{3}} = H^1 = H\)
As I understand your post, your sole confusion seems to be how the fraction magically turned into a power. As it turns out, the answer is, it didn't, it was always a power.
The relevant assumptions I'm making are as follows. Please correct any that are wrong and provide the correct information as needed. D stands for "Discharge" and H stands for "Head". The very first equation that gives you 4.5 m3 / s is meant to be set equal to D (considering they later mysteriously substitute in 4.5 for the value of D; but this equality is never explicitly stated). In addition to these assumptions, the awful notation I've had to decipher suggests that the final term in the equation for D is not H times 3/2, as it appears to be, but instead H to the power of 3/2. A bit of accuracy is lost in the gravity step, as g is not exactly 9.81, nor is sqrt(19.62) exactly 4.429. For the purposes of whatever you're doing, these approximations seem to close enough, although the rounding ought to be noted with an approximately equals symbol (≈). None of this is your problem or your fault, but I just despise poor notation in math problems. Students learn to hate math just fine on their own without any encouragement from their teachers
Anyway, to the task at hand. I trust you follow the steps up to this point:
\(\displaystyle \displaystyle H^{\frac{3}{2}} \approx \dfrac{D}{\frac{2}{3} \cdot 0.59 \cdot 6 \cdot 4.429}\)
At this point, it's simply a matter of raising both sides to the 2/3 power to cancel out the exponent on H. This is because of the rules of exponentshttp://www.mesacc.edu/~scotz47781/mat120/notes/exponents/review/review.html. Namely:
\(\displaystyle \displaystyle \left( H^{\frac{3}{2}} \right)^{\frac{2}{3}} = H^{\frac{3}{2} \cdot \frac{2}{3}} = H^1 = H\)
As I understand your post, your sole confusion seems to be how the fraction magically turned into a power. As it turns out, the answer is, it didn't, it was always a power.
It appears as if you've not included the full and exact problem text, as there's some information missing. I can read between a lines a bit and guess at what it's meant to be, but regardless the problem is very poorly formulated and the person who gave it to you ought to be ashamed of their abysmal notation!
The relevant assumptions I'm making are as follows. Please correct any that are wrong and provide the correct information as needed. D stands for "Discharge" and H stands for "Head". The very first equation that gives you 4.5 m3 / s is meant to be set equal to D (considering they later mysteriously substitute in 4.5 for the value of D; but this equality is never explicitly stated). In addition to these assumptions, the awful notation I've had to decipher suggests that the final term in the equation for D is not H times 3/2, as it appears to be, but instead H to the power of 3/2. A bit of accuracy is lost in the gravity step, as g is not exactly 9.81, nor is sqrt(19.62) exactly 4.429. For the purposes of whatever you're doing, these approximations seem to close enough, although the rounding ought to be noted with an approximately equals symbol (≈). None of this is your problem or your fault, but I just despise poor notation in math problems. Students learn to hate math just fine on their own without any encouragement from their teachers
Anyway, to the task at hand. I trust you follow the steps up to this point:
\(\displaystyle \displaystyle H^{\frac{3}{2}} \approx \dfrac{D}{\frac{2}{3} \cdot 0.59 \cdot 6 \cdot 4.429}\)
At this point, it's simply a matter of raising both sides to the 2/3 power to cancel out the exponent on H. This is because of the rules of exponents. Namely:
\(\displaystyle \displaystyle \left( H^{\frac{3}{2}} \right)^{\frac{2}{3}} = H^{\frac{3}{2} \cdot \frac{2}{3}} = H^1 = H\)
As I understand your post, your sole confusion seems to be how the fraction magically turned into a power. As it turns out, the answer is, it didn't, it was always a power.
Thanks bud that cleared up everything for me. It's from a text book for steam engineering. Ya its written really poorly Lol.