Question

1. gasoline engine

imagine a gasoline engine with a power output of 180 kw and an efficiency of 28%. how much heat is supplied to the heat engine per second? how much heat is discarded by the heat engine per second? draw the erm of the gasoline engine and correctly write the amount of heat supplied, amount discarded, and work output per second.

1. area under the curve

consider the following process for 50 moles of ideal diatomic gas. calculate the work done from a to b using the straight line path show. what are the temperatures at a and b? calculate the heat absorbed in the process.

1. otto engine

consider an otto engine with a compression ratio of r = 5. calculate the efficiency of the heat engine if it is using air as a working substance. make a simple approximate to the molar heat capacity of air by considering that is it made up of 78% nitrogen gas and 21% oxygen gas, and 1% miscellaneous gas. if the heat engine takes in 400 j of heat from a hot reservoir and repeats itself every 3 seconds, what is the power output of the heat engine? the energy extracted from the engine is used to lift 10 kg boxes by a height of 1 m/ how many boxes will be lifted by the heat engine in 1 hr?

Explanation:

Problem 1

Step1: Recall efficiency formula

The efficiency formula of a heat - engine is $\eta=\frac{W}{Q_{h}}$, where $\eta$ is the efficiency, $W$ is the work output per second (power), and $Q_{h}$ is the heat supplied per second. Given $W = 180\times10^{3}\text{ J/s}$ and $\eta=0.28$.
We can solve for $Q_{h}$: $Q_{h}=\frac{W}{\eta}$.
$Q_{h}=\frac{180\times10^{3}}{0.28}\approx6.43\times 10^{5}\text{ J/s}$

Step2: Calculate heat discarded

The heat discarded $Q_{c}$ can be found using the energy - conservation principle $Q_{h}=W + Q_{c}$, so $Q_{c}=Q_{h}-W$.
$Q_{c}=6.43\times 10^{5}-180\times10^{3}=4.63\times 10^{5}\text{ J/s}$

To draw the Energy - Resource - Model (ERM), we have a box representing the heat engine. An arrow pointing into the box labeled $Q_{h}\approx6.43\times 10^{5}\text{ J/s}$, an arrow pointing out of the box labeled $W = 180\times10^{3}\text{ J/s}$, and another arrow pointing out of the box labeled $Q_{c}=4.63\times 10^{5}\text{ J/s}$

Step1: Calculate work done

The work done $W$ in a $p - V$ diagram is given by the area under the curve. For a straight - line path from $a$ to $b$, the work done is the area of the trapezoid.
The formula for the area of a trapezoid is $W=\frac{1}{2}(p_{a}+p_{b})(V_{b}-V_{a})$.
Here, $p_{a}=4.00\times 10^{5}\text{ Pa}$, $p_{b}=2.00\times 10^{5}\text{ Pa}$, $V_{a}=1.0\text{ m}^{3}$, $V_{b}=2.0\text{ m}^{3}$
$W=\frac{1}{2}(4.00\times 10^{5}+2.00\times 10^{5})(2.0 - 1.0)=3.00\times 10^{5}\text{ J}$

Step2: Calculate temperatures

For an ideal gas, $pV = nRT$.
At point $a$: $T_{a}=\frac{p_{a}V_{a}}{nR}$, with $n = 50\text{ mol}$ and $R=8.31\text{ J/(mol}\cdot\text{K)}$
$T_{a}=\frac{4.00\times 10^{5}\times1.0}{50\times8.31}\approx963\text{ K}$
At point $b$: $T_{b}=\frac{p_{b}V_{b}}{nR}=\frac{2.00\times 10^{5}\times2.0}{50\times8.31}\approx963\text{ K}$

Step3: Calculate heat absorbed

For a diatomic ideal gas, $C_{p}=\frac{7}{2}R$. Using the first law of thermodynamics $\Delta U=Q - W$. Since $\Delta U = nC_{v}\Delta T$ and for a diatomic gas $C_{v}=\frac{5}{2}R$, and $\Delta T = 0$ (because $T_{a}=T_{b}$), $\Delta U = 0$. So $Q = W=3.00\times 10^{5}\text{ J}$

Step1: Calculate efficiency of Otto engine

The efficiency of an Otto engine is given by $\eta = 1-\frac{1}{r^{\gamma - 1}}$, where $r = 5$ is the compression ratio.
The molar - heat capacity of air: $C_{v}=0.78\times\frac{5}{2}R+0.21\times\frac{5}{2}R + 0.01\times C_{v_{misc}}\approx\frac{5}{2}R$ (since the contribution of the miscellaneous gas is small), and $\gamma=\frac{C_{p}}{C_{v}}=\frac{\frac{7}{2}R}{\frac{5}{2}R}=1.4$
$\eta = 1-\frac{1}{5^{1.4 - 1}}=1-\frac{1}{5^{0.4}}\approx0.47$

Step2: Calculate power output

The heat input per cycle $Q_{h}=400\text{ J}$ and the cycle time $t = 3\text{ s}$. The work output per cycle $W=\eta Q_{h}=0.47\times400 = 188\text{ J}$
The power output $P=\frac{W}{t}=\frac{188}{3}\approx62.7\text{ W}$

Step3: Calculate number of boxes lifted in 1 hour

The work done in 1 hour ($t = 3600\text{ s}$) is $W_{total}=P\times t=62.7\times3600 = 2.26\times 10^{5}\text{ J}$
The work done in lifting one box of mass $m = 10\text{ kg}$ by a height $h = 1\text{ m}$ is $W_{box}=mgh=10\times9.8\times1 = 98\text{ J}$
The number of boxes $N=\frac{W_{total}}{W_{box}}=\frac{2.26\times 10^{5}}{98}\approx2306$

Answer:

Heat supplied per second: $6.43\times 10^{5}\text{ J/s}$, Heat discarded per second: $4.63\times 10^{5}\text{ J/s}$

Problem 2