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Issue with units for a rocket nozzle throat area problem
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Rocket Propulsion Elements: Total Impulse ProblemIs there any rule for rocket engine nozzle proximity from each other?What's the nature of hoop stresses on a rocket nozzle?Are cold gas thrusters viable for model rockets?J-2 rocket nozzle length“Oh-my-god” particle drive performanceHow can phenolic (resin?) handle rocket engine nozzle temperatures?Cooling a hydrolox rocket with WaterIs there enough energy in a rocket nozzle for fission?How is the exit velocity of the flow in a sounding rocket nozzle?
$begingroup$
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):
$$A_t = fracw_tP_t sqrtfracR T_tgamma g_c $$
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
edited Mar 9 at 1:19
Russell Borogove
89.9k3301386
asked Mar 9 at 0:50
MAP3MAP3
512
$endgroup$
|
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):
$$A_t = fracw_tP_t sqrtfracR T_tgamma g_c $$
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
edited Mar 9 at 1:19
Russell Borogove
89.9k3301386
asked Mar 9 at 0:50
MAP3MAP3
512
$endgroup$
$begingroup$
You can learn more about MathJax for equations here. I removed therocketlabtag; 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
Mar 9 at 1:10
$begingroup$
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
Mar 9 at 1:31
$begingroup$
@OrganicMarble Are the units for $R$ supposed to be ft-lbf / lbm R?
$endgroup$
– Russell Borogove
Mar 9 at 1:42
$begingroup$
@RussellBorogove actually no. To be consistent it should be ft-lbf / slug R engineeringtoolbox.com/…
$endgroup$
– Organic Marble
Mar 9 at 1:50
1
$begingroup$
@OrganicMarble Same dimensions, at least.
$endgroup$
– Russell Borogove
Mar 9 at 1:51
|
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):
$$A_t = fracw_tP_t sqrtfracR T_tgamma g_c $$
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
edited Mar 9 at 1:19
Russell Borogove
89.9k3301386
asked Mar 9 at 0:50
MAP3MAP3
512
$endgroup$
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):
$$A_t = fracw_tP_t sqrtfracR T_tgamma g_c $$
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
edited Mar 9 at 1:19
Russell Borogove
89.9k3301386
asked Mar 9 at 0:50
MAP3MAP3
512
edited Mar 9 at 1:19
Russell Borogove
89.9k3301386
asked Mar 9 at 0:50
MAP3MAP3
512
edited Mar 9 at 1:19
Russell Borogove
89.9k3301386
edited Mar 9 at 1:19
Russell Borogove
89.9k3301386
edited Mar 9 at 1:19
Russell Borogove
89.9k3301386
89.9k3301386
asked Mar 9 at 0:50
MAP3MAP3
512
asked Mar 9 at 0:50
MAP3MAP3
512
asked Mar 9 at 0:50
MAP3MAP3
512
512
$begingroup$
You can learn more about MathJax for equations here. I removed therocketlabtag; 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
Mar 9 at 1:10
$begingroup$
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
Mar 9 at 1:31
$begingroup$
@OrganicMarble Are the units for $R$ supposed to be ft-lbf / lbm R?
$endgroup$
– Russell Borogove
Mar 9 at 1:42
$begingroup$
@RussellBorogove actually no. To be consistent it should be ft-lbf / slug R engineeringtoolbox.com/…
$endgroup$
– Organic Marble
Mar 9 at 1:50
1
$begingroup$
@OrganicMarble Same dimensions, at least.
$endgroup$
– Russell Borogove
Mar 9 at 1:51
|
show 2 more comments
$begingroup$
You can learn more about MathJax for equations here. I removed therocketlabtag; 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
Mar 9 at 1:10
$begingroup$
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
Mar 9 at 1:31
$begingroup$
@OrganicMarble Are the units for $R$ supposed to be ft-lbf / lbm R?
$endgroup$
– Russell Borogove
Mar 9 at 1:42
$begingroup$
@RussellBorogove actually no. To be consistent it should be ft-lbf / slug R engineeringtoolbox.com/…
$endgroup$
– Organic Marble
Mar 9 at 1:50
1
$begingroup$
@OrganicMarble Same dimensions, at least.
$endgroup$
– Russell Borogove
Mar 9 at 1:51
$begingroup$
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
Mar 9 at 1:10
$begingroup$
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
Mar 9 at 1:10
$begingroup$
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
Mar 9 at 1:31
$begingroup$
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
Mar 9 at 1:31
$begingroup$
@OrganicMarble Are the units for $R$ supposed to be ft-lbf / lbm R?
$endgroup$
– Russell Borogove
Mar 9 at 1:42
$begingroup$
@OrganicMarble Are the units for $R$ supposed to be ft-lbf / lbm R?
$endgroup$
– Russell Borogove
Mar 9 at 1:42
$begingroup$
@RussellBorogove actually no. To be consistent it should be ft-lbf / slug R engineeringtoolbox.com/…
$endgroup$
– Organic Marble
Mar 9 at 1:50
$begingroup$
@RussellBorogove actually no. To be consistent it should be ft-lbf / slug R engineeringtoolbox.com/…
$endgroup$
– Organic Marble
Mar 9 at 1:50
1
1
$begingroup$
@OrganicMarble Same dimensions, at least.
$endgroup$
– Russell Borogove
Mar 9 at 1:51
$begingroup$
@OrganicMarble Same dimensions, at least.
$endgroup$
– Russell Borogove
Mar 9 at 1:51
|
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.
answered Mar 9 at 1:12
Russell BorogoveRussell Borogove
89.9k3301386
$endgroup$
$begingroup$
Nice edit, I was just going ask about that unit on gamma....
$endgroup$
– Organic Marble
Mar 9 at 1:19
1
$begingroup$
Zing! Yeah, I misread the unit attribution.
$endgroup$
– Russell Borogove
Mar 9 at 1:20
$begingroup$
It's not the right pounds though outside the radical. Bad equation! Bad! (Hits equation on nose with rolled up newspaper)
$endgroup$
– Organic Marble
Mar 9 at 1:32
$begingroup$
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
Mar 9 at 1:36
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 $fracslugssec$. 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 $fracslugssec$ has the units of $fraclbf-secft$.
You also must specify the pressure in $fraclbfft^2$.
Dividing through you get outside the radical the units of $ft-sec$.
Inside the radical the gas constant is $fracft-lbfslug-R$. The R cancels out with the temperature unit. gamma is dimensionless.
So if you substitute in $fraclbf- sec^2ft$ for the slug, you end up with $fracft^2sec^2$ inside the radical.
Taking the square root, it's $fracftsec$, then multiply by the $ft-sec$ outside the radical, to get $ft^2$.
Welcome to the world of Apollo and Shuttle rocket engine calculations.
answered Mar 9 at 2:00
Organic MarbleOrganic Marble
60.4k3166258
$endgroup$
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
Mar 9 at 2:11
1
$begingroup$
Are you thinking the "-" is a minus sign? It's a hyphen. It's deplorable, I know, but that's the way these units are written. See the units on the gas constant in the question. "65 ft-lb/lb R(rankine)," and in the quote in the other answer "ft-lb/lb-R to ft-lb/slug-R"
$endgroup$
– Organic Marble
Mar 9 at 18:10
1
$begingroup$
oh $fraclbftext-sec^2ft$ what the merry heck are hyphens doing in there? "english" units are ridiculous
$endgroup$
– JCRM
Mar 9 at 18:15
1
$begingroup$
I cannot argue with that, lol
$endgroup$
– Organic Marble
Mar 9 at 18:16
1
$begingroup$
I think that mathjax requires text to insert a hyphen says everything about how much they belong in formulae :D
$endgroup$
– JCRM
Mar 9 at 18:21
|
show 7 more comments
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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.
answered Mar 9 at 1:12
Russell BorogoveRussell Borogove
89.9k3301386
$endgroup$
$begingroup$
Nice edit, I was just going ask about that unit on gamma....
$endgroup$
– Organic Marble
Mar 9 at 1:19
1
$begingroup$
Zing! Yeah, I misread the unit attribution.
$endgroup$
– Russell Borogove
Mar 9 at 1:20
$begingroup$
It's not the right pounds though outside the radical. Bad equation! Bad! (Hits equation on nose with rolled up newspaper)
$endgroup$
– Organic Marble
Mar 9 at 1:32
$begingroup$
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
Mar 9 at 1:36
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.
answered Mar 9 at 1:12
Russell BorogoveRussell Borogove
89.9k3301386
$endgroup$
$begingroup$
Nice edit, I was just going ask about that unit on gamma....
$endgroup$
– Organic Marble
Mar 9 at 1:19
1
$begingroup$
Zing! Yeah, I misread the unit attribution.
$endgroup$
– Russell Borogove
Mar 9 at 1:20
$begingroup$
It's not the right pounds though outside the radical. Bad equation! Bad! (Hits equation on nose with rolled up newspaper)
$endgroup$
– Organic Marble
Mar 9 at 1:32
$begingroup$
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
Mar 9 at 1:36
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.
answered Mar 9 at 1:12
Russell BorogoveRussell Borogove
89.9k3301386
$endgroup$
$R$ 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.
answered Mar 9 at 1:12
Russell BorogoveRussell Borogove
89.9k3301386
edited Mar 9 at 2:08
answered Mar 9 at 1:12
Russell BorogoveRussell Borogove
89.9k3301386
answered Mar 9 at 1:12
Russell BorogoveRussell Borogove
89.9k3301386
answered Mar 9 at 1:12
Russell BorogoveRussell Borogove
89.9k3301386
89.9k3301386
$begingroup$
Nice edit, I was just going ask about that unit on gamma....
$endgroup$
– Organic Marble
Mar 9 at 1:19
1
$begingroup$
Zing! Yeah, I misread the unit attribution.
$endgroup$
– Russell Borogove
Mar 9 at 1:20
$begingroup$
It's not the right pounds though outside the radical. Bad equation! Bad! (Hits equation on nose with rolled up newspaper)
$endgroup$
– Organic Marble
Mar 9 at 1:32
$begingroup$
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
Mar 9 at 1:36
add a comment |
$begingroup$
Nice edit, I was just going ask about that unit on gamma....
$endgroup$
– Organic Marble
Mar 9 at 1:19
1
$begingroup$
Zing! Yeah, I misread the unit attribution.
$endgroup$
– Russell Borogove
Mar 9 at 1:20
$begingroup$
It's not the right pounds though outside the radical. Bad equation! Bad! (Hits equation on nose with rolled up newspaper)
$endgroup$
– Organic Marble
Mar 9 at 1:32
$begingroup$
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
Mar 9 at 1:36
$begingroup$
Nice edit, I was just going ask about that unit on gamma....
$endgroup$
– Organic Marble
Mar 9 at 1:19
$begingroup$
Nice edit, I was just going ask about that unit on gamma....
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– Organic Marble
Mar 9 at 1:19
1
1
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Zing! Yeah, I misread the unit attribution.
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– Russell Borogove
Mar 9 at 1:20
$begingroup$
Zing! Yeah, I misread the unit attribution.
$endgroup$
– Russell Borogove
Mar 9 at 1:20
$begingroup$
It's not the right pounds though outside the radical. Bad equation! Bad! (Hits equation on nose with rolled up newspaper)
$endgroup$
– Organic Marble
Mar 9 at 1:32
$begingroup$
It's not the right pounds though outside the radical. Bad equation! Bad! (Hits equation on nose with rolled up newspaper)
$endgroup$
– Organic Marble
Mar 9 at 1:32
$begingroup$
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
Mar 9 at 1:36
$begingroup$
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
Mar 9 at 1:36
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 $fracslugssec$. 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 $fracslugssec$ has the units of $fraclbf-secft$.
You also must specify the pressure in $fraclbfft^2$.
Dividing through you get outside the radical the units of $ft-sec$.
Inside the radical the gas constant is $fracft-lbfslug-R$. The R cancels out with the temperature unit. gamma is dimensionless.
So if you substitute in $fraclbf- sec^2ft$ for the slug, you end up with $fracft^2sec^2$ inside the radical.
Taking the square root, it's $fracftsec$, then multiply by the $ft-sec$ outside the radical, to get $ft^2$.
Welcome to the world of Apollo and Shuttle rocket engine calculations.
answered Mar 9 at 2:00
Organic MarbleOrganic Marble
60.4k3166258
$endgroup$
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
Mar 9 at 2:11
1
$begingroup$
Are you thinking the "-" is a minus sign? It's a hyphen. It's deplorable, I know, but that's the way these units are written. See the units on the gas constant in the question. "65 ft-lb/lb R(rankine)," and in the quote in the other answer "ft-lb/lb-R to ft-lb/slug-R"
$endgroup$
– Organic Marble
Mar 9 at 18:10
1
$begingroup$
oh $fraclbftext-sec^2ft$ what the merry heck are hyphens doing in there? "english" units are ridiculous
$endgroup$
– JCRM
Mar 9 at 18:15
1
$begingroup$
I cannot argue with that, lol
$endgroup$
– Organic Marble
Mar 9 at 18:16
1
$begingroup$
I think that mathjax requires text to insert a hyphen says everything about how much they belong in formulae :D
$endgroup$
– JCRM
Mar 9 at 18:21
|
show 7 more comments
$begingroup$
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 $fracslugssec$. 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 $fracslugssec$ has the units of $fraclbf-secft$.
You also must specify the pressure in $fraclbfft^2$.
Dividing through you get outside the radical the units of $ft-sec$.
Inside the radical the gas constant is $fracft-lbfslug-R$. The R cancels out with the temperature unit. gamma is dimensionless.
So if you substitute in $fraclbf- sec^2ft$ for the slug, you end up with $fracft^2sec^2$ inside the radical.
Taking the square root, it's $fracftsec$, then multiply by the $ft-sec$ outside the radical, to get $ft^2$.
Welcome to the world of Apollo and Shuttle rocket engine calculations.
answered Mar 9 at 2:00
Organic MarbleOrganic Marble
60.4k3166258
$endgroup$
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
Mar 9 at 2:11
1
$begingroup$
Are you thinking the "-" is a minus sign? It's a hyphen. It's deplorable, I know, but that's the way these units are written. See the units on the gas constant in the question. "65 ft-lb/lb R(rankine)," and in the quote in the other answer "ft-lb/lb-R to ft-lb/slug-R"
$endgroup$
– Organic Marble
Mar 9 at 18:10
1
$begingroup$
oh $fraclbftext-sec^2ft$ what the merry heck are hyphens doing in there? "english" units are ridiculous
$endgroup$
– JCRM
Mar 9 at 18:15
1
$begingroup$
I cannot argue with that, lol
$endgroup$
– Organic Marble
Mar 9 at 18:16
1
$begingroup$
I think that mathjax requires text to insert a hyphen says everything about how much they belong in formulae :D
$endgroup$
– JCRM
Mar 9 at 18:21
|
show 7 more comments
$begingroup$
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 $fracslugssec$. 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 $fracslugssec$ has the units of $fraclbf-secft$.
You also must specify the pressure in $fraclbfft^2$.
Dividing through you get outside the radical the units of $ft-sec$.
Inside the radical the gas constant is $fracft-lbfslug-R$. The R cancels out with the temperature unit. gamma is dimensionless.
So if you substitute in $fraclbf- sec^2ft$ for the slug, you end up with $fracft^2sec^2$ inside the radical.
Taking the square root, it's $fracftsec$, then multiply by the $ft-sec$ outside the radical, to get $ft^2$.
Welcome to the world of Apollo and Shuttle rocket engine calculations.
answered Mar 9 at 2:00
Organic MarbleOrganic Marble
60.4k3166258
$endgroup$
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 $fracslugssec$. 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 $fracslugssec$ has the units of $fraclbf-secft$.
You also must specify the pressure in $fraclbfft^2$.
Dividing through you get outside the radical the units of $ft-sec$.
Inside the radical the gas constant is $fracft-lbfslug-R$. The R cancels out with the temperature unit. gamma is dimensionless.
So if you substitute in $fraclbf- sec^2ft$ for the slug, you end up with $fracft^2sec^2$ inside the radical.
Taking the square root, it's $fracftsec$, then multiply by the $ft-sec$ outside the radical, to get $ft^2$.
Welcome to the world of Apollo and Shuttle rocket engine calculations.
answered Mar 9 at 2:00
Organic MarbleOrganic Marble
60.4k3166258
edited Mar 9 at 2:35
answered Mar 9 at 2:00
Organic MarbleOrganic Marble
60.4k3166258
answered Mar 9 at 2:00
Organic MarbleOrganic Marble
60.4k3166258
answered Mar 9 at 2:00
Organic MarbleOrganic Marble
60.4k3166258
60.4k3166258
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
Mar 9 at 2:11
1
$begingroup$
Are you thinking the "-" is a minus sign? It's a hyphen. It's deplorable, I know, but that's the way these units are written. See the units on the gas constant in the question. "65 ft-lb/lb R(rankine)," and in the quote in the other answer "ft-lb/lb-R to ft-lb/slug-R"
$endgroup$
– Organic Marble
Mar 9 at 18:10
1
$begingroup$
oh $fraclbftext-sec^2ft$ what the merry heck are hyphens doing in there? "english" units are ridiculous
$endgroup$
– JCRM
Mar 9 at 18:15
1
$begingroup$
I cannot argue with that, lol
$endgroup$
– Organic Marble
Mar 9 at 18:16
1
$begingroup$
I think that mathjax requires text to insert a hyphen says everything about how much they belong in formulae :D
$endgroup$
– JCRM
Mar 9 at 18:21
|
show 7 more comments
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
Mar 9 at 2:11
1
$begingroup$
Are you thinking the "-" is a minus sign? It's a hyphen. It's deplorable, I know, but that's the way these units are written. See the units on the gas constant in the question. "65 ft-lb/lb R(rankine)," and in the quote in the other answer "ft-lb/lb-R to ft-lb/slug-R"
$endgroup$
– Organic Marble
Mar 9 at 18:10
1
$begingroup$
oh $fraclbftext-sec^2ft$ what the merry heck are hyphens doing in there? "english" units are ridiculous
$endgroup$
– JCRM
Mar 9 at 18:15
1
$begingroup$
I cannot argue with that, lol
$endgroup$
– Organic Marble
Mar 9 at 18:16
1
$begingroup$
I think that mathjax requires text to insert a hyphen says everything about how much they belong in formulae :D
$endgroup$
– JCRM
Mar 9 at 18:21
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
Mar 9 at 2:11
$begingroup$
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
Mar 9 at 2:11
1
1
$begingroup$
Are you thinking the "-" is a minus sign? It's a hyphen. It's deplorable, I know, but that's the way these units are written. See the units on the gas constant in the question. "65 ft-lb/lb R(rankine)," and in the quote in the other answer "ft-lb/lb-R to ft-lb/slug-R"
$endgroup$
– Organic Marble
Mar 9 at 18:10
$begingroup$
Are you thinking the "-" is a minus sign? It's a hyphen. It's deplorable, I know, but that's the way these units are written. See the units on the gas constant in the question. "65 ft-lb/lb R(rankine)," and in the quote in the other answer "ft-lb/lb-R to ft-lb/slug-R"
$endgroup$
– Organic Marble
Mar 9 at 18:10
1
1
$begingroup$
oh $fraclbftext-sec^2ft$ what the merry heck are hyphens doing in there? "english" units are ridiculous
$endgroup$
– JCRM
Mar 9 at 18:15
$begingroup$
oh $fraclbftext-sec^2ft$ what the merry heck are hyphens doing in there? "english" units are ridiculous
$endgroup$
– JCRM
Mar 9 at 18:15
1
1
$begingroup$
I cannot argue with that, lol
$endgroup$
– Organic Marble
Mar 9 at 18:16
$begingroup$
I cannot argue with that, lol
$endgroup$
– Organic Marble
Mar 9 at 18:16
1
1
$begingroup$
I think that mathjax requires text to insert a hyphen says everything about how much they belong in formulae :D
$endgroup$
– JCRM
Mar 9 at 18:21
$begingroup$
I think that mathjax requires text to insert a hyphen says everything about how much they belong in formulae :D
$endgroup$
– JCRM
Mar 9 at 18:21
|
show 7 more comments
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$begingroup$
You can learn more about MathJax for equations here. I removed the
rocketlabtag; 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
Mar 9 at 1:10
$begingroup$
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
Mar 9 at 1:31
$begingroup$
@OrganicMarble Are the units for $R$ supposed to be ft-lbf / lbm R?
$endgroup$
– Russell Borogove
Mar 9 at 1:42
$begingroup$
@RussellBorogove actually no. To be consistent it should be ft-lbf / slug R engineeringtoolbox.com/…
$endgroup$
– Organic Marble
Mar 9 at 1:50
1
$begingroup$
@OrganicMarble Same dimensions, at least.
$endgroup$
– Russell Borogove
Mar 9 at 1:51