Heat Transfer Question - Calling Current Students

[Cattlegod]

Here is the problem I'm trying to figure out - What is the equation to figure out how long it takes a constant heat source to heat up a volume of air 5 degrees?

I never had thermo or any heat transfer classes in school, so I'm wondering if someone here can shed some light. Assume air density is at 0 feet above sea level and 80% humidity.

Also, the heat source I have is power consumption is 1050 KW - I need to figure out how to turn this into the heat input required in the equation.

Thanks in advance!!

 

[rgwalt]

Here is the problem I'm trying to figure out - What is the equation to figure out how long it takes a constant heat source to heat up a volume of air 5 degrees?

I never had thermo or any heat transfer classes in school, so I'm wondering if someone here can shed some light. Assume air density is at 0 feet above sea level and 80% humidity.

Also, the heat source I have is power consumption is 1050 KW - I need to figure out how to turn this into the heat input required in the equation.

Thanks in advance!!

This really isn't a heat transfer question... it is an unsteady state energy balance question.

V*d*Cp*dT/dt = 1050 KW

where V is the volume of air, d is the density of air, Cp is the heat capacity of the air, T is temperature, t is time. V*d*Cp*dT gives you units of energy. When you devide this by time, you get units of power. There is your unsteady state energy balance. Multiple both sides by dt and integrate to get V*d*Cp*deltaT = 1050 KW * (t1 - t0). t0 is your starting time (zero), t1 is your ending time (the answer to your question). deltaT is 5 degrees. Make sure you convert/cancel your units!

 

[Cattlegod]

This really isn't a heat transfer question... it is an unsteady state energy balance question.

V*d*Cp*dT/dt = 1050 KW

where V is the volume of air, d is the density of air, Cp is the heat capacity of the air, T is temperature, t is time. V*d*Cp*dT gives you units of energy. When you devide this by time, you get units of power. There is your unsteady state energy balance. Multiple both sides by dt and integrate to get V*d*Cp*deltaT = 1050 KW * (t1 - t0). t0 is your starting time (zero), t1 is your ending time (the answer to your question). deltaT is 5 degrees. Make sure you convert/cancel your units!

Thanks - Assume air density is 1.1644 kg/m^3 and deltaT = 3c V = 71 358 M^3

so

71358*1.1644*3*Cp=1050t
t= 237*Cp --- minutes, seconds?

Any idea how to calculate heat capacity of air at 80% humidity? Also, what unit is time in? Minutes?

 

[Cattlegod]

Using wolfram alpha I get heat capacity at .7244 - is this right?

http://www.wolframalpha.com/input/?...ir.EL---.**HeatCapacityMoistAir.RH---&x=6&y=7

 

[Cattlegod]

OK, here is where I'm at:

Assumptions:
Air density at 86f = 1.1644 kg/m^3
V = 71 358 M^3
deltaT = 3c
Humidity = 80%
Heat Capacity = .7244 jk/(kgK)


71358*1.1644*3*.7244=1050t
t = 171 m^3 * C * kg * kj / kW * M^3 * kg * K
t= 171 jC/W*K

Just trying to cancel out the last bit

 

[Fenixgoon]

Thanks - Assume air density is 1.1644 kg/m^3 and deltaT = 3c V = 71 358 M^3

so

71358*1.1644*3*Cp=1050t
t= 237*Cp --- minutes, seconds?

Any idea how to calculate heat capacity of air at 80% humidity? Also, what unit is time in? Minutes?

you need to understand your units before you start running numbers. 1 watt = 1 joule/sec = 1 N*m/s = kg*m*m/s^2 / s. kW is part of MKS unit set (meters, kilograms, *seconds*) so your unit of time must be seconds.

secondly, you can find the heat capacity of moist air in any steam table. otherwise, you could just assume that the heat capacity is (80% water + 20% air), which is not actually true since you have a partially saturated vapor (but it will be much higher than dry air).

in this problem, you care about the *change* in temperature, not the absolute value. going from 0-1C is the same as going from 272-273K.

you really need to *learn* these things.. you can only get so far by blindly plugging and chugging.

 

[Cattlegod]

Thanks for the help flexigoon. I'm a computer engineer by study, so most of this is new to me (hence why I'm turning to ATOT for its vast experience). I'm trying to solve a work problem - we have consultants working on the problem now, but I find the problem curious and am trying to do a simple calculation that I can wrap my head around vs. the super complex thing the consultants are likely to provide. Here is what I have so far, I'm just trying to figure out how to convert C/K:

Assumptions:
Air density at 86f = 1.1644 kg/m^3
V = 71 358 M^3
deltaT = 3c
Humidity = 80%
Heat Capacity = .7244 kj/(kgK) ( per wolfram alpha)


71358*1.1644*3*.7244=1050t
t = 171 m^3 * C * kg * kj / kW * M^3 * kg * K
t= 171 jC/W*K
t=171 s*C/K
t = 37 seconds????

 

[dullard]

t=171 s*C/K

One degree Celsius change is equal to one degree Kelvin Change. Thus, t=171 s. Assuming you got your numbers correctly (I didn't bother to look them up).

And where do you get a 1050 kW heat source? Plenty of things are that powerful, but it just isn't something you'd commonly see in normal daily life.

Finally, that is an AVERAGE of 3°C gained. With such intense heat and with such a large volume, you'll really have temperature zones that are hot and others that are cool. I can't imagine the size of the fan you'd need to make those gradients be negligible.

Oh and, why was it 5°C in your OP and 3°C in your math?