If 7.08 m^3 s^-1 of air at 322.15 K(50^∘ C) and approximately atmospheric pressure is preheat

If 7.08 m^3 s^-1 of air at 322.15 K(50^∘ C) and...

Key Concepts

Mass Flow Rate Conversion

When dealing with fluid flow problems, it is essential to understand that the volumetric flow rate (in m³/s) must often be converted to a mass flow rate (in kg/s) using the fluid's density. This conversion is crucial in energy calculations because most thermal energy equations involve the mass of the substance rather than its volume.

Specific Heat Capacity

The specific heat capacity of a substance is the amount of heat required to raise the temperature of a unit mass of the substance by one degree Celsius (or Kelvin). For air at constant pressure, this property is used to calculate the amount of energy needed to achieve a desired temperature change, making it a central concept in thermal energy transfer calculations.

Energy Balance for Heating Processes

An energy balance in a heating process involves accounting for the energy input required to raise the temperature of a flowing fluid. This is commonly done using the equation Q? = m? · cp · ?T, where Q? represents the rate of heat transfer, m? is the mass flow rate, cp is the specific heat capacity, and ?T is the temperature change. This relationship is fundamental in both engineering and thermodynamics for designing heating and cooling systems.

Temperature Difference (?T)

The temperature difference, ?T, is the difference between the final and initial temperatures of the fluid. It directly influences the energy required for the heating process, as a larger temperature increase results in a higher energy demand when other variables are held constant. In heat transfer calculations, accurately determining ?T is essential for ensuring that the calculated rate of heat transfer is correct.

Transcript

00:01 Okay, so first for this question, we need to determine the rate of heat transfer out of the gas, the combustion gas.

00:08 Since the rate of heat was transferred to the air, that means the rate of heat will be transferred out from the combustion gases, okay, according to energy conservation law.

00:20 So therefore, we can apply the energy balance equation here, which is e -d -in equal to e -d -0.

00:24 If we expand it, we have q -dying plus m .1 times h1 is equal to m .1 has h2.

00:33 So qdy here is the rate of heat that was transfer into the combustion gases.

00:41 You guys may wonder, i just said that the rate of heat was transferred out of combustion gases.

00:47 I will show you why it's transferred out, even though i put it in here.

00:53 So m1 dot here is the mass flow rate for the combustion gases, and h1 is the initial anthope for the combustion gases, and h2 here is the final entropy for the combustion gases.

01:08 If we do some arrangement here, eventually we have q.

01:11 .

01:11 Times cp times t2 minus t1.

01:15 So cp is the specific heats for the combustion gases, and t2 is the higher temperature, i'm sorry, no higher temperature, the final temperature, and the t1 is the initial temperature.

01:28 So now we know m1.

01:29 That is equal to 0 .95 kilogram per second and the specific heat for the combustion gases is 1 .1 kilojouper kilogram times degrees celsius.

01:38 We know the initial temperature is 160 degree celsius and the final temperature is 95 degrees celsius.

01:45 So now that's trying to determine the rate of the heat transfer into the combustion gases, which is equal to 0 .95 kilogram per second times 1 .10 kilojou per kilogram times degrees celsius at n times 95 degrees celsius minus 160 degrees celsius and this will give us the rate of heat transfer into the combustion gases is about 67 actually is in negative negative 67 .95 kilowattes okay so since the value is negative, that means it's transferring out of the combustion gases.

02:48 Okay? so that's why i said it was transferring out of combustion gases.

02:53 So if it's transferring out of the combustion gases, that means it's transferred into the air.

02:58 So therefore the rate of the heat transferred into the air, which is q -dair here, should be negative q -in.

03:04 Okay, which is equal to negative 67 .95 kilowatts, and this will give us the positive 67 .95 kilowatts.

03:13 For the air.

03:14 So now let's determine the x temperature for the air, which is the audio temperature of air.

03:21 So first we need to set such equation here, which is q...

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