Issue with units for a rocket nozzle throat area problem
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I'm working through How to Design, Build, and Test Small Liquid-Fuel Rocket Engines. The guide explains calculating a rocket nozzle throat area using Eq. (7):
$$A_t = frac{w_t}{P_t} sqrt{frac{R T_t}{gamma g_c}} $$
where $w_t$ is flow measured in lb/s, $P_t$ is pressure measured in psi, $R$ is the specific gas constant for gaseous oxygen and hydrocarbon fuel which equals 65 ft-lb/lb R(rankine), $T_t$ is temperature of the chamber measured in rankine, $gamma$ being the ratio of gas specific heats, and the gravitational constant $g_c$ measured in ft/s².
My issue is for the worked out problem on the guide, they give an example here but the answer is in in² instead of ft². I'm not sure how they got inches instead of feet because they didn't do any kind of conversion that I can tell. I tried following the units to see how they got inches but I'm not sure. Shouldn't the answer be in feet, and if not how is it inches?
rockets nozzle rocket-equation
New contributor
$endgroup$
|
show 2 more comments
$begingroup$
I'm working through How to Design, Build, and Test Small Liquid-Fuel Rocket Engines. The guide explains calculating a rocket nozzle throat area using Eq. (7):
$$A_t = frac{w_t}{P_t} sqrt{frac{R T_t}{gamma g_c}} $$
where $w_t$ is flow measured in lb/s, $P_t$ is pressure measured in psi, $R$ is the specific gas constant for gaseous oxygen and hydrocarbon fuel which equals 65 ft-lb/lb R(rankine), $T_t$ is temperature of the chamber measured in rankine, $gamma$ being the ratio of gas specific heats, and the gravitational constant $g_c$ measured in ft/s².
My issue is for the worked out problem on the guide, they give an example here but the answer is in in² instead of ft². I'm not sure how they got inches instead of feet because they didn't do any kind of conversion that I can tell. I tried following the units to see how they got inches but I'm not sure. Shouldn't the answer be in feet, and if not how is it inches?
rockets nozzle rocket-equation
New contributor
$endgroup$
$begingroup$
You can learn more about MathJax for equations here. I removed therocketlab
tag; after reading that the address and original text was from circa 1967, it doesn't seem to be related to the same company that the tag refers to.
$endgroup$
– uhoh
3 hours ago
$begingroup$
This is really a no-good equation because it has pounds mass (in the flowrate) over pounds force (in the press) outside the radical. It should be the mass flowrate in every engineer's favorite unit, slugs/sec. See Sutton p. 61 pyrobin.com/files/Rocket%20Propulsion%20Elements.pdf
$endgroup$
– Organic Marble
3 hours ago
$begingroup$
@OrganicMarble Are the units for $R$ supposed to be ft-lbf / lbm R?
$endgroup$
– Russell Borogove
2 hours ago
$begingroup$
@RussellBorogove actually no. To be consistent it should be ft-lbf / slug R engineeringtoolbox.com/…
$endgroup$
– Organic Marble
2 hours ago
1
$begingroup$
@OrganicMarble Same dimensions, at least.
$endgroup$
– Russell Borogove
2 hours ago
|
show 2 more comments
$begingroup$
I'm working through How to Design, Build, and Test Small Liquid-Fuel Rocket Engines. The guide explains calculating a rocket nozzle throat area using Eq. (7):
$$A_t = frac{w_t}{P_t} sqrt{frac{R T_t}{gamma g_c}} $$
where $w_t$ is flow measured in lb/s, $P_t$ is pressure measured in psi, $R$ is the specific gas constant for gaseous oxygen and hydrocarbon fuel which equals 65 ft-lb/lb R(rankine), $T_t$ is temperature of the chamber measured in rankine, $gamma$ being the ratio of gas specific heats, and the gravitational constant $g_c$ measured in ft/s².
My issue is for the worked out problem on the guide, they give an example here but the answer is in in² instead of ft². I'm not sure how they got inches instead of feet because they didn't do any kind of conversion that I can tell. I tried following the units to see how they got inches but I'm not sure. Shouldn't the answer be in feet, and if not how is it inches?
rockets nozzle rocket-equation
New contributor
$endgroup$
I'm working through How to Design, Build, and Test Small Liquid-Fuel Rocket Engines. The guide explains calculating a rocket nozzle throat area using Eq. (7):
$$A_t = frac{w_t}{P_t} sqrt{frac{R T_t}{gamma g_c}} $$
where $w_t$ is flow measured in lb/s, $P_t$ is pressure measured in psi, $R$ is the specific gas constant for gaseous oxygen and hydrocarbon fuel which equals 65 ft-lb/lb R(rankine), $T_t$ is temperature of the chamber measured in rankine, $gamma$ being the ratio of gas specific heats, and the gravitational constant $g_c$ measured in ft/s².
My issue is for the worked out problem on the guide, they give an example here but the answer is in in² instead of ft². I'm not sure how they got inches instead of feet because they didn't do any kind of conversion that I can tell. I tried following the units to see how they got inches but I'm not sure. Shouldn't the answer be in feet, and if not how is it inches?
rockets nozzle rocket-equation
rockets nozzle rocket-equation
New contributor
New contributor
edited 3 hours ago
Russell Borogove
87k3291376
87k3291376
New contributor
asked 3 hours ago
MAP3MAP3
132
132
New contributor
New contributor
$begingroup$
You can learn more about MathJax for equations here. I removed therocketlab
tag; after reading that the address and original text was from circa 1967, it doesn't seem to be related to the same company that the tag refers to.
$endgroup$
– uhoh
3 hours ago
$begingroup$
This is really a no-good equation because it has pounds mass (in the flowrate) over pounds force (in the press) outside the radical. It should be the mass flowrate in every engineer's favorite unit, slugs/sec. See Sutton p. 61 pyrobin.com/files/Rocket%20Propulsion%20Elements.pdf
$endgroup$
– Organic Marble
3 hours ago
$begingroup$
@OrganicMarble Are the units for $R$ supposed to be ft-lbf / lbm R?
$endgroup$
– Russell Borogove
2 hours ago
$begingroup$
@RussellBorogove actually no. To be consistent it should be ft-lbf / slug R engineeringtoolbox.com/…
$endgroup$
– Organic Marble
2 hours ago
1
$begingroup$
@OrganicMarble Same dimensions, at least.
$endgroup$
– Russell Borogove
2 hours ago
|
show 2 more comments
$begingroup$
You can learn more about MathJax for equations here. I removed therocketlab
tag; after reading that the address and original text was from circa 1967, it doesn't seem to be related to the same company that the tag refers to.
$endgroup$
– uhoh
3 hours ago
$begingroup$
This is really a no-good equation because it has pounds mass (in the flowrate) over pounds force (in the press) outside the radical. It should be the mass flowrate in every engineer's favorite unit, slugs/sec. See Sutton p. 61 pyrobin.com/files/Rocket%20Propulsion%20Elements.pdf
$endgroup$
– Organic Marble
3 hours ago
$begingroup$
@OrganicMarble Are the units for $R$ supposed to be ft-lbf / lbm R?
$endgroup$
– Russell Borogove
2 hours ago
$begingroup$
@RussellBorogove actually no. To be consistent it should be ft-lbf / slug R engineeringtoolbox.com/…
$endgroup$
– Organic Marble
2 hours ago
1
$begingroup$
@OrganicMarble Same dimensions, at least.
$endgroup$
– Russell Borogove
2 hours ago
$begingroup$
You can learn more about MathJax for equations here. I removed the
rocketlab
tag; after reading that the address and original text was from circa 1967, it doesn't seem to be related to the same company that the tag refers to.$endgroup$
– uhoh
3 hours ago
$begingroup$
You can learn more about MathJax for equations here. I removed the
rocketlab
tag; after reading that the address and original text was from circa 1967, it doesn't seem to be related to the same company that the tag refers to.$endgroup$
– uhoh
3 hours ago
$begingroup$
This is really a no-good equation because it has pounds mass (in the flowrate) over pounds force (in the press) outside the radical. It should be the mass flowrate in every engineer's favorite unit, slugs/sec. See Sutton p. 61 pyrobin.com/files/Rocket%20Propulsion%20Elements.pdf
$endgroup$
– Organic Marble
3 hours ago
$begingroup$
This is really a no-good equation because it has pounds mass (in the flowrate) over pounds force (in the press) outside the radical. It should be the mass flowrate in every engineer's favorite unit, slugs/sec. See Sutton p. 61 pyrobin.com/files/Rocket%20Propulsion%20Elements.pdf
$endgroup$
– Organic Marble
3 hours ago
$begingroup$
@OrganicMarble Are the units for $R$ supposed to be ft-lbf / lbm R?
$endgroup$
– Russell Borogove
2 hours ago
$begingroup$
@OrganicMarble Are the units for $R$ supposed to be ft-lbf / lbm R?
$endgroup$
– Russell Borogove
2 hours ago
$begingroup$
@RussellBorogove actually no. To be consistent it should be ft-lbf / slug R engineeringtoolbox.com/…
$endgroup$
– Organic Marble
2 hours ago
$begingroup$
@RussellBorogove actually no. To be consistent it should be ft-lbf / slug R engineeringtoolbox.com/…
$endgroup$
– Organic Marble
2 hours ago
1
1
$begingroup$
@OrganicMarble Same dimensions, at least.
$endgroup$
– Russell Borogove
2 hours ago
$begingroup$
@OrganicMarble Same dimensions, at least.
$endgroup$
– Russell Borogove
2 hours ago
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
$R$ and $g_c$ each have a foot component in their units; therefore the feet cancel out when $R$ is divided by $g_c$.
$P_t$ is in pounds per square inch; since you're dividing by that, it's effectively units of square inch per pound, which is where the square inches in the answer comes from.
The rest of the unit cancellation is confusing because of the use of pounds as both a unit of mass and a unit of force (weight). This thread on thespacerace.com discusses the different ways of presenting your equation; the money quote is here:
Outside the square root, my equation uses slugs for the mass flow rate, while the other equation uses pounds (i.e. weight). To convert pounds to slugs we must divide by gc outside the square root.
Inside the square root, my equation uses R' equal to 49,720 ft-lb/slug-R, while the other equation uses 1545.32 ft-lb/lb-R. To convert ft-lb/lb-R to ft-lb/slug-R we must multiple by gc inside the square root.
The combination of these two conversion factors is,
1/gc * (gc)^1/2 = 1/gc^1/2
therefore we end up with gc in the demoninator inside the square root.
$endgroup$
$begingroup$
Nice edit, I was just going ask about that unit on gamma....
$endgroup$
– Organic Marble
3 hours ago
1
$begingroup$
Zing! Yeah, I misread the unit attribution.
$endgroup$
– Russell Borogove
3 hours ago
$begingroup$
It's not the right pounds though outside the radical. Bad equation! Bad! (Hits equation on nose with rolled up newspaper)
$endgroup$
– Organic Marble
3 hours ago
$begingroup$
Ugh, flowrate being mass rather than force? Maybe I'll delete everything except the inch/foot portion of my analysis, since this is way out of my wheelhouse.
$endgroup$
– Russell Borogove
3 hours ago
add a comment |
$begingroup$
That equation as you give it with the values you supply is a bad mish-mash of units. It's painful to do this in the English system, but gather your courage, we can get through it.
You must specify the flowrate in $frac{slugs}{sec}$. Yes, slugs, the real engineer's unit of massTM. A slug is 1 $frac {lbf - sec^2} {ft}$ and equates to ~ 32.2 lbm. So flowrate in $frac{slugs}{sec}$ has the units of $frac{lbf-sec}{ft}$.
You also must specify the pressure in $frac{lbf}{ft^2}$.
Dividing through you get outside the radical the units of $ft-sec$.
Inside the radical the gas constant is $frac{ft-lbf}{slug-R}$. The R cancels out with the temperature unit. gamma is dimensionless.
So if you substitute in $frac{lbf- sec^2}{ft}$ for the slug, you end up with $frac{ft^2}{sec^2}$ inside the radical.
Taking the square root, it's $frac{ft}{sec}$, then multiply by the $ft-sec$ outside the radical, to get $ft^2$.
Welcome to the world of Apollo and Shuttle rocket engine calculations.
$endgroup$
$begingroup$
Welcome to the world of Apollo and Shuttle rocket engine calculations and real engineers.
$endgroup$
– uhoh
2 hours ago
$begingroup$
Real engineers go metric, not like those amateurs at NASA in the old days.
$endgroup$
– Russell Borogove
2 hours ago
1
$begingroup$
I was caught in the transition, my books in college had both. Shuttle and Apollo were all English, ISS is metric. Either is OK if you are used to it but metric is a lot more intuitive.
$endgroup$
– Organic Marble
2 hours ago
$begingroup$
I understand how you got ft^2 but in the example problem they got in^2 without doing any kind of feet -> inches conversion that I can tell at least, which is why I was confused.
$endgroup$
– MAP3
2 hours ago
$begingroup$
My eyes glaze over when they divide lbm by lbf and have them cancel out. That's just wrong so you can quit reading right there. It looks superficially ok but it's like cancelling feet with centistokes.
$endgroup$
– Organic Marble
2 hours ago
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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oldest
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oldest
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$begingroup$
$R$ and $g_c$ each have a foot component in their units; therefore the feet cancel out when $R$ is divided by $g_c$.
$P_t$ is in pounds per square inch; since you're dividing by that, it's effectively units of square inch per pound, which is where the square inches in the answer comes from.
The rest of the unit cancellation is confusing because of the use of pounds as both a unit of mass and a unit of force (weight). This thread on thespacerace.com discusses the different ways of presenting your equation; the money quote is here:
Outside the square root, my equation uses slugs for the mass flow rate, while the other equation uses pounds (i.e. weight). To convert pounds to slugs we must divide by gc outside the square root.
Inside the square root, my equation uses R' equal to 49,720 ft-lb/slug-R, while the other equation uses 1545.32 ft-lb/lb-R. To convert ft-lb/lb-R to ft-lb/slug-R we must multiple by gc inside the square root.
The combination of these two conversion factors is,
1/gc * (gc)^1/2 = 1/gc^1/2
therefore we end up with gc in the demoninator inside the square root.
$endgroup$
$begingroup$
Nice edit, I was just going ask about that unit on gamma....
$endgroup$
– Organic Marble
3 hours ago
1
$begingroup$
Zing! Yeah, I misread the unit attribution.
$endgroup$
– Russell Borogove
3 hours ago
$begingroup$
It's not the right pounds though outside the radical. Bad equation! Bad! (Hits equation on nose with rolled up newspaper)
$endgroup$
– Organic Marble
3 hours ago
$begingroup$
Ugh, flowrate being mass rather than force? Maybe I'll delete everything except the inch/foot portion of my analysis, since this is way out of my wheelhouse.
$endgroup$
– Russell Borogove
3 hours ago
add a comment |
$begingroup$
$R$ and $g_c$ each have a foot component in their units; therefore the feet cancel out when $R$ is divided by $g_c$.
$P_t$ is in pounds per square inch; since you're dividing by that, it's effectively units of square inch per pound, which is where the square inches in the answer comes from.
The rest of the unit cancellation is confusing because of the use of pounds as both a unit of mass and a unit of force (weight). This thread on thespacerace.com discusses the different ways of presenting your equation; the money quote is here:
Outside the square root, my equation uses slugs for the mass flow rate, while the other equation uses pounds (i.e. weight). To convert pounds to slugs we must divide by gc outside the square root.
Inside the square root, my equation uses R' equal to 49,720 ft-lb/slug-R, while the other equation uses 1545.32 ft-lb/lb-R. To convert ft-lb/lb-R to ft-lb/slug-R we must multiple by gc inside the square root.
The combination of these two conversion factors is,
1/gc * (gc)^1/2 = 1/gc^1/2
therefore we end up with gc in the demoninator inside the square root.
$endgroup$
$begingroup$
Nice edit, I was just going ask about that unit on gamma....
$endgroup$
– Organic Marble
3 hours ago
1
$begingroup$
Zing! Yeah, I misread the unit attribution.
$endgroup$
– Russell Borogove
3 hours ago
$begingroup$
It's not the right pounds though outside the radical. Bad equation! Bad! (Hits equation on nose with rolled up newspaper)
$endgroup$
– Organic Marble
3 hours ago
$begingroup$
Ugh, flowrate being mass rather than force? Maybe I'll delete everything except the inch/foot portion of my analysis, since this is way out of my wheelhouse.
$endgroup$
– Russell Borogove
3 hours ago
add a comment |
$begingroup$
$R$ and $g_c$ each have a foot component in their units; therefore the feet cancel out when $R$ is divided by $g_c$.
$P_t$ is in pounds per square inch; since you're dividing by that, it's effectively units of square inch per pound, which is where the square inches in the answer comes from.
The rest of the unit cancellation is confusing because of the use of pounds as both a unit of mass and a unit of force (weight). This thread on thespacerace.com discusses the different ways of presenting your equation; the money quote is here:
Outside the square root, my equation uses slugs for the mass flow rate, while the other equation uses pounds (i.e. weight). To convert pounds to slugs we must divide by gc outside the square root.
Inside the square root, my equation uses R' equal to 49,720 ft-lb/slug-R, while the other equation uses 1545.32 ft-lb/lb-R. To convert ft-lb/lb-R to ft-lb/slug-R we must multiple by gc inside the square root.
The combination of these two conversion factors is,
1/gc * (gc)^1/2 = 1/gc^1/2
therefore we end up with gc in the demoninator inside the square root.
$endgroup$
$R$ and $g_c$ each have a foot component in their units; therefore the feet cancel out when $R$ is divided by $g_c$.
$P_t$ is in pounds per square inch; since you're dividing by that, it's effectively units of square inch per pound, which is where the square inches in the answer comes from.
The rest of the unit cancellation is confusing because of the use of pounds as both a unit of mass and a unit of force (weight). This thread on thespacerace.com discusses the different ways of presenting your equation; the money quote is here:
Outside the square root, my equation uses slugs for the mass flow rate, while the other equation uses pounds (i.e. weight). To convert pounds to slugs we must divide by gc outside the square root.
Inside the square root, my equation uses R' equal to 49,720 ft-lb/slug-R, while the other equation uses 1545.32 ft-lb/lb-R. To convert ft-lb/lb-R to ft-lb/slug-R we must multiple by gc inside the square root.
The combination of these two conversion factors is,
1/gc * (gc)^1/2 = 1/gc^1/2
therefore we end up with gc in the demoninator inside the square root.
edited 2 hours ago
answered 3 hours ago
Russell BorogoveRussell Borogove
87k3291376
87k3291376
$begingroup$
Nice edit, I was just going ask about that unit on gamma....
$endgroup$
– Organic Marble
3 hours ago
1
$begingroup$
Zing! Yeah, I misread the unit attribution.
$endgroup$
– Russell Borogove
3 hours ago
$begingroup$
It's not the right pounds though outside the radical. Bad equation! Bad! (Hits equation on nose with rolled up newspaper)
$endgroup$
– Organic Marble
3 hours ago
$begingroup$
Ugh, flowrate being mass rather than force? Maybe I'll delete everything except the inch/foot portion of my analysis, since this is way out of my wheelhouse.
$endgroup$
– Russell Borogove
3 hours ago
add a comment |
$begingroup$
Nice edit, I was just going ask about that unit on gamma....
$endgroup$
– Organic Marble
3 hours ago
1
$begingroup$
Zing! Yeah, I misread the unit attribution.
$endgroup$
– Russell Borogove
3 hours ago
$begingroup$
It's not the right pounds though outside the radical. Bad equation! Bad! (Hits equation on nose with rolled up newspaper)
$endgroup$
– Organic Marble
3 hours ago
$begingroup$
Ugh, flowrate being mass rather than force? Maybe I'll delete everything except the inch/foot portion of my analysis, since this is way out of my wheelhouse.
$endgroup$
– Russell Borogove
3 hours ago
$begingroup$
Nice edit, I was just going ask about that unit on gamma....
$endgroup$
– Organic Marble
3 hours ago
$begingroup$
Nice edit, I was just going ask about that unit on gamma....
$endgroup$
– Organic Marble
3 hours ago
1
1
$begingroup$
Zing! Yeah, I misread the unit attribution.
$endgroup$
– Russell Borogove
3 hours ago
$begingroup$
Zing! Yeah, I misread the unit attribution.
$endgroup$
– Russell Borogove
3 hours ago
$begingroup$
It's not the right pounds though outside the radical. Bad equation! Bad! (Hits equation on nose with rolled up newspaper)
$endgroup$
– Organic Marble
3 hours ago
$begingroup$
It's not the right pounds though outside the radical. Bad equation! Bad! (Hits equation on nose with rolled up newspaper)
$endgroup$
– Organic Marble
3 hours ago
$begingroup$
Ugh, flowrate being mass rather than force? Maybe I'll delete everything except the inch/foot portion of my analysis, since this is way out of my wheelhouse.
$endgroup$
– Russell Borogove
3 hours ago
$begingroup$
Ugh, flowrate being mass rather than force? Maybe I'll delete everything except the inch/foot portion of my analysis, since this is way out of my wheelhouse.
$endgroup$
– Russell Borogove
3 hours ago
add a comment |
$begingroup$
That equation as you give it with the values you supply is a bad mish-mash of units. It's painful to do this in the English system, but gather your courage, we can get through it.
You must specify the flowrate in $frac{slugs}{sec}$. Yes, slugs, the real engineer's unit of massTM. A slug is 1 $frac {lbf - sec^2} {ft}$ and equates to ~ 32.2 lbm. So flowrate in $frac{slugs}{sec}$ has the units of $frac{lbf-sec}{ft}$.
You also must specify the pressure in $frac{lbf}{ft^2}$.
Dividing through you get outside the radical the units of $ft-sec$.
Inside the radical the gas constant is $frac{ft-lbf}{slug-R}$. The R cancels out with the temperature unit. gamma is dimensionless.
So if you substitute in $frac{lbf- sec^2}{ft}$ for the slug, you end up with $frac{ft^2}{sec^2}$ inside the radical.
Taking the square root, it's $frac{ft}{sec}$, then multiply by the $ft-sec$ outside the radical, to get $ft^2$.
Welcome to the world of Apollo and Shuttle rocket engine calculations.
$endgroup$
$begingroup$
Welcome to the world of Apollo and Shuttle rocket engine calculations and real engineers.
$endgroup$
– uhoh
2 hours ago
$begingroup$
Real engineers go metric, not like those amateurs at NASA in the old days.
$endgroup$
– Russell Borogove
2 hours ago
1
$begingroup$
I was caught in the transition, my books in college had both. Shuttle and Apollo were all English, ISS is metric. Either is OK if you are used to it but metric is a lot more intuitive.
$endgroup$
– Organic Marble
2 hours ago
$begingroup$
I understand how you got ft^2 but in the example problem they got in^2 without doing any kind of feet -> inches conversion that I can tell at least, which is why I was confused.
$endgroup$
– MAP3
2 hours ago
$begingroup$
My eyes glaze over when they divide lbm by lbf and have them cancel out. That's just wrong so you can quit reading right there. It looks superficially ok but it's like cancelling feet with centistokes.
$endgroup$
– Organic Marble
2 hours ago
add a comment |
$begingroup$
That equation as you give it with the values you supply is a bad mish-mash of units. It's painful to do this in the English system, but gather your courage, we can get through it.
You must specify the flowrate in $frac{slugs}{sec}$. Yes, slugs, the real engineer's unit of massTM. A slug is 1 $frac {lbf - sec^2} {ft}$ and equates to ~ 32.2 lbm. So flowrate in $frac{slugs}{sec}$ has the units of $frac{lbf-sec}{ft}$.
You also must specify the pressure in $frac{lbf}{ft^2}$.
Dividing through you get outside the radical the units of $ft-sec$.
Inside the radical the gas constant is $frac{ft-lbf}{slug-R}$. The R cancels out with the temperature unit. gamma is dimensionless.
So if you substitute in $frac{lbf- sec^2}{ft}$ for the slug, you end up with $frac{ft^2}{sec^2}$ inside the radical.
Taking the square root, it's $frac{ft}{sec}$, then multiply by the $ft-sec$ outside the radical, to get $ft^2$.
Welcome to the world of Apollo and Shuttle rocket engine calculations.
$endgroup$
$begingroup$
Welcome to the world of Apollo and Shuttle rocket engine calculations and real engineers.
$endgroup$
– uhoh
2 hours ago
$begingroup$
Real engineers go metric, not like those amateurs at NASA in the old days.
$endgroup$
– Russell Borogove
2 hours ago
1
$begingroup$
I was caught in the transition, my books in college had both. Shuttle and Apollo were all English, ISS is metric. Either is OK if you are used to it but metric is a lot more intuitive.
$endgroup$
– Organic Marble
2 hours ago
$begingroup$
I understand how you got ft^2 but in the example problem they got in^2 without doing any kind of feet -> inches conversion that I can tell at least, which is why I was confused.
$endgroup$
– MAP3
2 hours ago
$begingroup$
My eyes glaze over when they divide lbm by lbf and have them cancel out. That's just wrong so you can quit reading right there. It looks superficially ok but it's like cancelling feet with centistokes.
$endgroup$
– Organic Marble
2 hours ago
add a comment |
$begingroup$
That equation as you give it with the values you supply is a bad mish-mash of units. It's painful to do this in the English system, but gather your courage, we can get through it.
You must specify the flowrate in $frac{slugs}{sec}$. Yes, slugs, the real engineer's unit of massTM. A slug is 1 $frac {lbf - sec^2} {ft}$ and equates to ~ 32.2 lbm. So flowrate in $frac{slugs}{sec}$ has the units of $frac{lbf-sec}{ft}$.
You also must specify the pressure in $frac{lbf}{ft^2}$.
Dividing through you get outside the radical the units of $ft-sec$.
Inside the radical the gas constant is $frac{ft-lbf}{slug-R}$. The R cancels out with the temperature unit. gamma is dimensionless.
So if you substitute in $frac{lbf- sec^2}{ft}$ for the slug, you end up with $frac{ft^2}{sec^2}$ inside the radical.
Taking the square root, it's $frac{ft}{sec}$, then multiply by the $ft-sec$ outside the radical, to get $ft^2$.
Welcome to the world of Apollo and Shuttle rocket engine calculations.
$endgroup$
That equation as you give it with the values you supply is a bad mish-mash of units. It's painful to do this in the English system, but gather your courage, we can get through it.
You must specify the flowrate in $frac{slugs}{sec}$. Yes, slugs, the real engineer's unit of massTM. A slug is 1 $frac {lbf - sec^2} {ft}$ and equates to ~ 32.2 lbm. So flowrate in $frac{slugs}{sec}$ has the units of $frac{lbf-sec}{ft}$.
You also must specify the pressure in $frac{lbf}{ft^2}$.
Dividing through you get outside the radical the units of $ft-sec$.
Inside the radical the gas constant is $frac{ft-lbf}{slug-R}$. The R cancels out with the temperature unit. gamma is dimensionless.
So if you substitute in $frac{lbf- sec^2}{ft}$ for the slug, you end up with $frac{ft^2}{sec^2}$ inside the radical.
Taking the square root, it's $frac{ft}{sec}$, then multiply by the $ft-sec$ outside the radical, to get $ft^2$.
Welcome to the world of Apollo and Shuttle rocket engine calculations.
edited 2 hours ago
answered 2 hours ago
Organic MarbleOrganic Marble
58.1k3159248
58.1k3159248
$begingroup$
Welcome to the world of Apollo and Shuttle rocket engine calculations and real engineers.
$endgroup$
– uhoh
2 hours ago
$begingroup$
Real engineers go metric, not like those amateurs at NASA in the old days.
$endgroup$
– Russell Borogove
2 hours ago
1
$begingroup$
I was caught in the transition, my books in college had both. Shuttle and Apollo were all English, ISS is metric. Either is OK if you are used to it but metric is a lot more intuitive.
$endgroup$
– Organic Marble
2 hours ago
$begingroup$
I understand how you got ft^2 but in the example problem they got in^2 without doing any kind of feet -> inches conversion that I can tell at least, which is why I was confused.
$endgroup$
– MAP3
2 hours ago
$begingroup$
My eyes glaze over when they divide lbm by lbf and have them cancel out. That's just wrong so you can quit reading right there. It looks superficially ok but it's like cancelling feet with centistokes.
$endgroup$
– Organic Marble
2 hours ago
add a comment |
$begingroup$
Welcome to the world of Apollo and Shuttle rocket engine calculations and real engineers.
$endgroup$
– uhoh
2 hours ago
$begingroup$
Real engineers go metric, not like those amateurs at NASA in the old days.
$endgroup$
– Russell Borogove
2 hours ago
1
$begingroup$
I was caught in the transition, my books in college had both. Shuttle and Apollo were all English, ISS is metric. Either is OK if you are used to it but metric is a lot more intuitive.
$endgroup$
– Organic Marble
2 hours ago
$begingroup$
I understand how you got ft^2 but in the example problem they got in^2 without doing any kind of feet -> inches conversion that I can tell at least, which is why I was confused.
$endgroup$
– MAP3
2 hours ago
$begingroup$
My eyes glaze over when they divide lbm by lbf and have them cancel out. That's just wrong so you can quit reading right there. It looks superficially ok but it's like cancelling feet with centistokes.
$endgroup$
– Organic Marble
2 hours ago
$begingroup$
Welcome to the world of Apollo and Shuttle rocket engine calculations and real engineers.
$endgroup$
– uhoh
2 hours ago
$begingroup$
Welcome to the world of Apollo and Shuttle rocket engine calculations and real engineers.
$endgroup$
– uhoh
2 hours ago
$begingroup$
Real engineers go metric, not like those amateurs at NASA in the old days.
$endgroup$
– Russell Borogove
2 hours ago
$begingroup$
Real engineers go metric, not like those amateurs at NASA in the old days.
$endgroup$
– Russell Borogove
2 hours ago
1
1
$begingroup$
I was caught in the transition, my books in college had both. Shuttle and Apollo were all English, ISS is metric. Either is OK if you are used to it but metric is a lot more intuitive.
$endgroup$
– Organic Marble
2 hours ago
$begingroup$
I was caught in the transition, my books in college had both. Shuttle and Apollo were all English, ISS is metric. Either is OK if you are used to it but metric is a lot more intuitive.
$endgroup$
– Organic Marble
2 hours ago
$begingroup$
I understand how you got ft^2 but in the example problem they got in^2 without doing any kind of feet -> inches conversion that I can tell at least, which is why I was confused.
$endgroup$
– MAP3
2 hours ago
$begingroup$
I understand how you got ft^2 but in the example problem they got in^2 without doing any kind of feet -> inches conversion that I can tell at least, which is why I was confused.
$endgroup$
– MAP3
2 hours ago
$begingroup$
My eyes glaze over when they divide lbm by lbf and have them cancel out. That's just wrong so you can quit reading right there. It looks superficially ok but it's like cancelling feet with centistokes.
$endgroup$
– Organic Marble
2 hours ago
$begingroup$
My eyes glaze over when they divide lbm by lbf and have them cancel out. That's just wrong so you can quit reading right there. It looks superficially ok but it's like cancelling feet with centistokes.
$endgroup$
– Organic Marble
2 hours ago
add a comment |
MAP3 is a new contributor. Be nice, and check out our Code of Conduct.
MAP3 is a new contributor. Be nice, and check out our Code of Conduct.
MAP3 is a new contributor. Be nice, and check out our Code of Conduct.
MAP3 is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
You can learn more about MathJax for equations here. I removed the
rocketlab
tag; after reading that the address and original text was from circa 1967, it doesn't seem to be related to the same company that the tag refers to.$endgroup$
– uhoh
3 hours ago
$begingroup$
This is really a no-good equation because it has pounds mass (in the flowrate) over pounds force (in the press) outside the radical. It should be the mass flowrate in every engineer's favorite unit, slugs/sec. See Sutton p. 61 pyrobin.com/files/Rocket%20Propulsion%20Elements.pdf
$endgroup$
– Organic Marble
3 hours ago
$begingroup$
@OrganicMarble Are the units for $R$ supposed to be ft-lbf / lbm R?
$endgroup$
– Russell Borogove
2 hours ago
$begingroup$
@RussellBorogove actually no. To be consistent it should be ft-lbf / slug R engineeringtoolbox.com/…
$endgroup$
– Organic Marble
2 hours ago
1
$begingroup$
@OrganicMarble Same dimensions, at least.
$endgroup$
– Russell Borogove
2 hours ago