Calculate Thermal Power Output of Electric Furnace

  • Thread starter Thread starter Windseaker
  • Start date Start date
  • Tags Tags
    Power Thermal
AI Thread Summary
The discussion revolves around calculating the thermal power output of an electric furnace with specific parameters, including a heating element drawing 65 amps on a 240V circuit and an air flow rate of 1200 CFM. The initial calculations suggest a thermal output of 8890 Btu/hr based on a delta T assumption of 72 degrees, but there is confusion regarding the incoming air temperature, which is stated as 65 degrees. Participants highlight discrepancies in the temperature readings, questioning the validity of the assumed delta T. A corrected calculation indicates a thermal power output of 52,165 Btu/hr based on the efficiency and power input. Overall, there is a consensus that the temperature values need clarification for accurate calculations.
Windseaker
Messages
45
Reaction score
0
1. Heating element in an electric furnace draws 65amps on a 240v single phase circuit witha .98 ele.to thermal eff. The volume flow rate of air across the element is 1200CFM. the incoming air temp to the element is 65degree and the element temp is 75degree.
Whats the thermal power output in Btu/hr.?



2. Q=1.08xCFMxdeltaT and eff=Power(out)/Power(in)



3. Q=1.08(1200CFM)(72-65), Q= 9072 Btu/hr
Power(out)=(eff)(Power(in) P(out)= (.98)(9072) = 8890btu/hr

The problem is delta T , I am assuming 72degree the temp in but it says incoming temp is 65degree?? I am missing somthing here either the real delta T or Power In--somthinks not right??
 
Physics news on Phys.org
Windseaker said:
1. Heating element in an electric furnace draws 65amps on a 240v single phase circuit witha .98 ele.to thermal eff. The volume flow rate of air across the element is 1200CFM. the incoming air temp to the element is 65degree and the element temp is 75degree.
Whats the thermal power output in Btu/hr.?
2. Q=1.08xCFMxdeltaT and eff=Power(out)/Power(in)
3. Q=1.08(1200CFM)(72-65), Q= 9072 Btu/hr
Power(out)=(eff)(Power(in) P(out)= (.98)(9072) = 8890btu/hr

The problem is delta T , I am assuming 72degree the temp in but it says incoming temp is 65degree?? I am missing somthing here either the real delta T or Power In--somthinks not right??

The thermal power output is \eta \times P_{in} = .98 * 65 * 240 = 15.288 Kw = 52165 Btu/Hr.

This question does not make much sense. It is obvious that the temperature of the heating element cannot be 75 degrees F.

AM
 

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Calculating Efficiency and Thermal Source Rate in a Steam-Electric Power Plant
Replies
4
Views
3K

Hot Threads

Recent Insights

Back
Top