# 2 Questions - Heating room air, and fuel combustion efficiency

• skrelt
In summary, the first question involves calculating the time it takes for a 1.5 kW heater to heat a room with dimensions 8m x 4.5m x 2.7m from 10.0°C to 21.0°C, using the specific heat capacity of air. The second question involves calculating the amount of fuel (in liters) used by an internal combustion engine with a calorific value of 38 MJ/kg, a density of 900 kg/m3, and an efficiency of 25%, running at 30 kW.
skrelt
2 Questions -- Heating room air, and fuel combustion efficiency

## Homework Statement

2 questions. In advance, i needed to translate it so i am sorry if it is states incorrect. I really did my best to do it right.

1.
The density of air at 1 atm and 10.0 ° C is equal to 1.24 kg / m3. (cubic meter)
A room has the following sizes: 8,0 m x 4,5 m x 2,7 m
Calculate how long it takes for a heater of 1.5 kW to heat the air from the room to 21.0 ° C.
Use c = 1005 J / kg / K

2.
An internal combustion engine uses a fuel with calorific value/heating value of 38 MJ / kg.
The density of the fuel is 900 kg / m3. (cubic meter)
Of each MJ heat only 0.25 MJ is converted to mechanical energy.
The efficiency is 25%.
The engine can deliver 30 kW.
Calculate how many liters of fuel per hour the engine used.

## Homework Equations

http://imgur.com/WQx3BqI

these are the once i used. I probly made some/alot mistakes so i don't know if i need more.

## The Attempt at a Solution

1.
8,0m x 4,5m x 2,7m = 97,2m3 (cubic meters)
1 atm and 10.0 ° C = 1,24 kg/m3
Heater 1,5 kW = 1500 W = 1500 J/s
C = 1005 J/kg/K

This is where I am atm. I figured I needed to change the heater to J/s hence you need to know how long it takes. But after I gathered all the data I could find I don’t know where to start with the calculation or what formula to use first.

2.
Usage = 38 MJ/kg = 0,38 x 10^8 J/kg
P = 900 kg/m3 (cubic meters)
Energy usage = x 0.25
30 kw = 30000 J/s
30000 J/s x 3600 = 1,08 x 10^8 J/hour (first I figured I needed to set it in hours)
0,38x10^8 / 1,08x10^8 = 0,3519 J/kg/hour
900x0,3519 = 316,71 m3 (cubic meters)
316,71 m3 = 316710 dm3 = 316710 Liters of fuel.

I doubt there is an engine that uses that much so I think I made some mistakes.

I hope i can get some help here on 1 or maybe even both questions. Thanks in advance. I tried my best on both and these are the answers and solutions i could produce

One problem at a time...

Look up "Specific Heat Capacity" in your textbook.

Write down the equation that relates energy to mass and the temperature rise.

Then calculate the following in this order..

a) how many Kg of air you are heating.
b) What the temperature rise is in Kelvin (ΔT)
c) The number of joules needed to raise that much mass of air by that ΔT.
d) Then it might help to know that 1W = 1J/S. If you know how many joules are required and the rate at which the heater produces them you can calculate how many seconds it will take.

thanks, going to try that now

I'm off to bed as late here.

For Q2

Eff(%) = 100 * Power_out/power_in

rearrange to give

Power_in = 100 * Power_out/Eff

I make the power going into the engine 120kW.

More tomorrow.

.

1. To calculate the time it takes for the heater to heat the air in the room, we can use the formula Q = mcΔT, where Q is the amount of heat transferred, m is the mass of air, c is the specific heat capacity, and ΔT is the change in temperature. In this case, we know that the heater has a power output of 1.5 kW, which means it is transferring 1.5 kJ of heat per second. We also know the specific heat capacity of air is 1005 J/kg/K, and the change in temperature is from 10°C to 21°C.

To find the mass of air, we can use the density formula ρ = m/V, where ρ is the density, m is the mass, and V is the volume. The volume of the room is 97.2 m3, so the mass of air is 97.2 m3 x 1.24 kg/m3 = 120.528 kg.

Now we can plug these values into the formula:
Q = (120.528 kg) x (1005 J/kg/K) x (21°C - 10°C) = 1215936 J
Since the heater is transferring 1.5 kJ of heat per second, it will take 1215936 J / 1500 J/s = 810.624 seconds = 13.51 minutes to heat the air in the room.

2. To calculate the fuel consumption of the engine, we can use the formula P = E/t, where P is the power output, E is the energy consumed, and t is the time. We know that the engine has a power output of 30 kW, and we want to find the energy consumed per hour. We also know that the efficiency is 25%, which means only 25% of the energy consumed is converted to mechanical energy.

To find the energy consumed per hour, we can use the formula E = P x t, where P is the power output and t is the time. We know that the engine can deliver 30 kW, so we can set up the equation as:
E = (30 kW) x (1 hour) = 30 kWh = 30,000 Wh

Since only 25% of the energy is converted to mechanical energy, the total energy consumed is 30,000 Wh / 0.25

## 1. How does heating room air affect fuel combustion efficiency?

Heating room air can have a significant impact on fuel combustion efficiency. When the air in a room is colder, it is more dense and contains more oxygen. This allows for more efficient and complete burning of fuel, resulting in higher combustion efficiency. Conversely, if the room air is warmer, it is less dense and contains less oxygen, which can lead to incomplete combustion and lower efficiency.

## 2. What factors can affect the heating efficiency of a room?

The heating efficiency of a room can be affected by several factors, including the insulation of the room, the type and efficiency of the heating system, the size and layout of the room, and the outdoor temperature. Other factors such as air leaks, ventilation, and the quality of the fuel being used can also impact the efficiency of heating a room.

## 3. How can I improve the heating efficiency of a room?

There are several ways to improve the heating efficiency of a room. These include proper insulation to prevent heat loss, using a high-efficiency heating system, sealing air leaks, and ensuring proper ventilation. Additionally, using a programmable thermostat and setting it to lower temperatures when the room is not in use can also improve heating efficiency.

## 4. Is it more efficient to use a central heating system or a space heater?

The answer to this question depends on several factors, such as the size of the room, the type of heating system, and the cost of fuel. In general, a central heating system is more efficient for larger spaces, while a space heater may be more efficient for smaller spaces. It is important to consider the cost of running each system and to choose the most efficient option for your specific situation.

## 5. How can I calculate the combustion efficiency of my heating system?

The combustion efficiency of a heating system can be calculated by dividing the heat output by the fuel input. The result is then multiplied by 100 to get a percentage. For example, if a heating system produces 90,000 BTUs of heat and consumes 100,000 BTUs of fuel, the combustion efficiency would be (90,000/100,000) x 100 = 90%. It is important to note that this calculation may not take into account other factors that can affect heating efficiency, such as heat loss and air leaks.

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