Calculating the Cheaper Option for Heating Room

  • Thread starter Thread starter pcoppin
  • Start date Start date
  • Tags Tags
    Heating
Click For Summary
SUMMARY

The discussion focuses on calculating the efficiency and cost-effectiveness of heating a room using an oil-filled radiator versus a heat pump. The radiator operates at a surface temperature of 70°C with a thermal conductivity of 80 Wm-1K-1 and a convection coefficient of 9.5 Wm-2K-1. The heat pump has a coefficient of performance (COP) of 9. The rate of heat transfer from the radiator is calculated to be 1131.67 W, which is crucial for determining the work done by the heat pump to achieve the same heating effect.

PREREQUISITES
  • Understanding of thermal conductivity and convection principles
  • Knowledge of the coefficient of performance (COP) in heat pumps
  • Familiarity with heat transfer equations, specifically for convection and radiation
  • Basic electrical energy cost calculations
NEXT STEPS
  • Calculate the work done by the heat pump using the formula: COP = ˙Qh / ˙W
  • Research the cost comparison between operating an oil-filled radiator and a heat pump
  • Explore the impact of different thermal conductivities on heating efficiency
  • Investigate alternative heating methods and their cost-effectiveness
USEFUL FOR

Engineers, HVAC professionals, and homeowners looking to optimize heating solutions and reduce energy costs in residential settings.

pcoppin
Messages
1
Reaction score
0

Homework Statement



An oil-filled radiator is used to maintain the temperature of a room at 21C when the outside temperature is 10C. The radiator has a surface area of 1.5 m2, and a surface temperature of 70C. Given that the radiator is made of a metal with a thermal conductivity 80 Wm-1K-1, thermal convection coefficient of 9.5 Wm-2K-1 and an emissivity of 0.8.

Question 1: If it was decided that a heat pump would be a better wat to heat this room, how much work would be done by the pump with a coefficient of performance of 9?

Question 2: If both the radiator and pump were run by electricity, which would be the cheaper option? Why?

Homework Equations



Currently, the rate at which heat is transferred into the room by convection plus radiation from the radiator is equal to the rate at which heat is conducted through the metal.
The Rate of Transfer is 1131.67 W (not sure if this is relevant)

The Attempt at a Solution



Not sure what to do as only given the coefficient of performance as 9.
 
Physics news on Phys.org
pcoppin said:

Homework Statement



An oil-filled radiator is used to maintain the temperature of a room at 21C when the outside temperature is 10C. The radiator has a surface area of 1.5 m2, and a surface temperature of 70C. Given that the radiator is made of a metal with a thermal conductivity 80 Wm-1K-1, thermal convection coefficient of 9.5 Wm-2K-1 and an emissivity of 0.8.

Question 1: If it was decided that a heat pump would be a better wat to heat this room, how much work would be done by the pump with a coefficient of performance of 9?

Question 2: If both the radiator and pump were run by electricity, which would be the cheaper option? Why?
First, find the rate at which energy is radiated and convected away from the radiator surface. That is the rate of heat output. Then find the rate at which work is done by a heat pump with a COP of 9 that delivers that rate of heat output . Use:

COP = \dot Q_h/\dot W

AM
 
I think there is a formula in calculating the total amount of heat are you using. I don't remember. Where can I saw this formula? My suggestion is trying to apply the formula if you find it.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
15K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K