Heating Load Required

The 350 kw is about right, but I don't see how 240 kw will ever give the desired temperature, unless the whole situation is described differently.

Power = energy / time. For calculating load required we take time as 1 hour. Taking energy constant, if we increase the time, the system is supposed to attain the same energy in the increased time. I am little confused with it.

You have described a once-through situation. There is no such thing as taking more time. However, you could decrease the cfm and still obtain the same temperature off the heater.

I have to keep the cfm fixed. Are you sure that 240 kw will not get me a rise of 140 deg even in 2 hours?

Yes, I'm sure. (If the conditions are as you have described thus far.)

1. what is the cu ft of the room?

2. what is its insulation rating. R= ?

THe room is about 2000 cu ft. Ps. I donot have to maintain the temp in the whole room, just at the inlet. Please tell me that what will be the rating if I have to raise the temp in 2 hours and not 1 hour .

If you have a small portable room heater, a heat gun, or a hair dryer that you can play with, you will see the issues involved. For a given airflow rate, the difference between inlet temp and outlet temp depends only on the power supplied to the heating element.

Leaving any of these units on for 1 minute, 10 minutes, or one hour does not change the outlet temp. In your case the "room" air is completely exchanged in less than half a minute. Obviously, the room inlet temperature will not change over time (other than during the first seconds after the heater is turned on).

You are not raising the temp in either one hour or two hours (or 30 minutes or 6 hours). (In one hour, you are heating 60 x 5500 cf; in two hours you are heating 120 x 5500 cf.)

You say you have a room of 2000 cu ft, and you are blowing 5500 cfm of hot air into that room. That means you are forcing a complete change of air every 22 seconds!

I assume the room has some form of contents (fruit being dried?). Whatever portion of the volume of the room is occupied by contents, can not be occupied by air, so the exchange rate is even faster. If half of the room volume were filled with contents, then the exchange would occur in just under 11 seconds!

If the air is only in the room for 11 to 22 seconds, it will still hold a lot of heat when it exits. If the air exiting the room is not contaminated by water or other volatiles being driven out of the room contents, then you need to recirculate the air so you no longer need to heat it from ambient, but only from the significantly higher exit temperature. If the air exiting the room is contaminated by water or other volatiles, then you need some form of heat exchanger to recover some of the otherwise lost heat. At this power level, heat recovery or recirculation would pay for itself in a rather short time!

This is a simple once-through calculation involving raising 5500cfm of air 115°C, everything else in your problem statement is irrelevant. Since the dwell time is fixed by the fixed cfm it should be obvious that if you lower the energy input into the air while holding the flow rate constant then the amount of heat transferred to the air will go down and so will the final temperature, time is not a factor under your given conditions.

Why not show us your calculation(s) for getting your 350kW and 240kW answers and we will comment on them.

Employing the simple formula from Engineering Tool Box but using ro=0.851kg/m^3 [for 140 dgr.C dry air] we get, indeed, 254 kw.

However, this is required only to heat the air. In order to circulate the air you need to calculate the fan motor required power.

In order to heat 5500 cft in a minute from 25 to 140 dgr.C you need 254 kw.

Theoretically, if you could open the room so that in a minute you could evacuate all the 25 dgr.C. air you would get 140 dgr.C full room.

In order to heat 2000 cft in 2 hours you need 254/120=2.12 kw if the air losses are neglecting. If the losses will be 5500 cfm [complete open room] you'll need all 254 kw all the time.

Dear Mr.immortani,

You have referred that "the Room is completely open" and the power required - 240 KW may be theoritically OK. But the extended time- the radiation losses also will be there, which is UN-AVOIDABLE.

For radiation losses - Several Factors to be accounted, which ultimately results in Heat Energy. For example, the Velocity of Air Movement plays a vital role.

In the absence of correct data it will be difficult to answer/clear your doubt.

DHAYANANDHAN.S

It is an open room - assume what is pushed in goes out and there is no re-circulation of room air in to heater, you will have to use 300kW heater.

Flow = 155.76 CuM/min

Temperature rise = 140 -25 = 115ºC

Density of air at average temperature of 83ºC = 0.98kg/CuM

Specific heat of air = 0.17 kW.min/kg/ºC

Heater rating = (Multiply above) = 298kW

Rounded off = 300kW

Above rating is assumed at condition that any load inside the room is brought from outside for storing is brought at 140ºC.

In case there is a movement of load in and out of room and same is brought in at lower temperature than that of air - additional capacity of heater will have to be added and calculated similar to above.

What a poorly-designed room!