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ladessa [460]
3 years ago
7

A closed, rigid tank fitted with a paddle wheel contains 2 kg of air, initially at 300 K. During an interval of 5 minutes, the p

addle wheel transfers energy to the air at a rate of 1 kW. During the interval, the air also receives energy by heat transfer at a rate of .5 kW. These are the only energy transfers. Assuming the ideal gas model for the air, and no overall changes in kinetic or potential energy, determine the final temperature of the air, in K.

Physics
1 answer:
anzhelika [568]3 years ago
3 0

Answer:

The final temperature of the air is T_2= 605 K

Explanation:

We can start by doing an energy balance for the closed system

\Delta KE+\Delta PE+ \Delta U = Q - W

where

\Delta KE = the change in kinetic energy.

\Delta PE = the change in potential energy.

\Delta U = the total internal energy change in a system.

Q = the heat transferred to the system.

W = the work done by the system.

We know that there are no changes in kinetic or potential energy, so \Delta KE = 0 and \Delta PE=0

and our energy balance equation is \Delta U = Q - W

We also know that the paddle-wheel transfers energy to the air at a rate of 1 kW and the system receives energy by heat transfer at a rate of 0.5 kW, for 5 minutes.

We use this information to calculate the total internal energy change \Delta U=W+Q using the energy balance equation.

We convert the interval of time to seconds t = 5 \:min = 300\:s

\Delta \dot{U}=\dot{W}+ \dot{Q}\\=\Delta U=(W+ Q)\cdot t

\Delta U=(1 \:kW+0.5\:kW)\cdot 300\:s\\\Delta U=450 \:kJ

We can use the change in specific internal energy \Delta U = m(u_2-u_1) to find the final temperature of the air.

We are given that T_1=300 \:K and the air can be describe by ideal gas model, so we can use the ideal gas tables for air to determine the initial specific internal energy u_1

u_1=214.07\:\frac{kJ}{kg}

Next, we will calculate the final specific internal energy u_2

\Delta U = m(u_2-u_1)\\\frac{\Delta U}{m} =u_2-u_1

\frac{\Delta U}{m} =u_2-u_1\\u_2=u_1+\frac{\Delta U}{m}

u_2=214.07 \:\frac{kJ}{kg} +\frac{450 \:kJ}{2 \:kg}\\u_2= 439.07 \:\frac{kJ}{kg}

With the value u_2=439.07 \:\frac{kJ}{kg} and the ideal gas tables for air we make a regression between the values u = 434.78 \:\frac{kJ}{kg},T=600 \:K and u = 442.42 \:\frac{kJ}{kg}, T=610 \:K and we find that the final temperature T_2 is 605 K.

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