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:

1. Gasoline Engine

Step1: Find heat supplied per second

The efficiency formula is $\eta=\frac{W}{Q_{h}}$, where $\eta$ is efficiency, $W$ is work - output power, and $Q_{h}$ is heat input power. Given $W = 180\ kW=180000\ W$ and $\eta = 0.28$. Rearranging for $Q_{h}$, we get $Q_{h}=\frac{W}{\eta}$.
$Q_{h}=\frac{180000}{0.28}\approx642857.14\ J/s$

Step2: Find heat discarded per second

We know that $W = Q_{h}-Q_{c}$, so $Q_{c}=Q_{h}-W$.
$Q_{c}=642857.14 - 180000=462857.14\ J/s$

For the Energy - Resource - Model (ERM), draw a box representing the engine. An arrow pointing into the box labeled $Q_{h}\approx642857.14\ J/s$, an arrow pointing out of the box labeled $W = 180000\ J/s$, and another arrow pointing out of the box labeled $Q_{c}=462857.14\ J/s$.

Step1: Calculate work done

The work done in a $p - V$ diagram is given by the area under the curve. For a straight - line process, the work done $W$ is the area of the trapezoid. The formula for the area of a trapezoid is $A=\frac{(a + b)h}{2}$, where $a$ and $b$ are the parallel sides and $h$ is the height. Here, $a = 4.00\times10^{5}\ Pa$, $b = 2.00\times10^{5}\ Pa$, and $h=(2.0 - 1.0)\ m^{3}$.
$W=\frac{(4.00\times10^{5}+2.00\times10^{5})\times(2.0 - 1.0)}{2}=3.00\times10^{5}\ J$

Step2: Calculate temperature at a

Using the ideal gas law $pV = nRT$, at point $a$, $p_{a}=4.00\times10^{5}\ Pa$, $V_{a}=1.0\ m^{3}$, and $n = 50\ mol$, $R=8.31\ J/(mol\cdot K)$.
$T_{a}=\frac{p_{a}V_{a}}{nR}=\frac{4.00\times10^{5}\times1.0}{50\times8.31}\approx962.7\ K$

Step3: Calculate temperature at b

At point $b$, $p_{b}=2.00\times10^{5}\ Pa$, $V_{b}=2.0\ m^{3}$, and $n = 50\ mol$, $R = 8.31\ J/(mol\cdot K)$.
$T_{b}=\frac{p_{b}V_{b}}{nR}=\frac{2.00\times10^{5}\times2.0}{50\times8.31}\approx962.7\ K$

Step4: Calculate heat absorbed

For a diatomic gas, $C_{v}=\frac{5}{2}R$ and $C_{p}=\frac{7}{2}R$. Since $\Delta U = nC_{v}\Delta T$ and $\Delta T = 0$ (because $T_{a}=T_{b}$), $\Delta U = 0$. From the first law of thermodynamics $\Delta U=Q - W$, so $Q = W$.
$Q = 3.00\times10^{5}\ J$

Step1: Calculate efficiency

The efficiency of an Otto - cycle engine is given by $\eta=1-\frac{1}{r^{\gamma - 1}}$, where $r$ is the compression ratio and $\gamma=\frac{C_{p}}{C_{v}}$. For air, considering its composition:
The molar heat capacity of $N_{2}$ ($C_{v,N_{2}}=\frac{5}{2}R$, $C_{p,N_{2}}=\frac{7}{2}R$), $O_{2}$ ($C_{v,O_{2}}=\frac{5}{2}R$, $C_{p,O_{2}}=\frac{7}{2}R$).
$C_{v}=0.78\times\frac{5}{2}R+0.21\times\frac{5}{2}R + 0.01\times C_{v,misc}\approx\frac{5}{2}R$, $C_{p}=0.78\times\frac{7}{2}R+0.21\times\frac{7}{2}R+0.01\times C_{p,misc}\approx\frac{7}{2}R$, so $\gamma=\frac{C_{p}}{C_{v}} = 1.4$.
With $r = 5$, $\eta=1-\frac{1}{5^{1.4 - 1}}=1-\frac{1}{5^{0.4}}\approx0.475$

Step2: Calculate power output

The heat input per cycle is $Q_{h}=400\ J$ and the cycle time $t = 3\ s$. The work output per cycle $W=\eta Q_{h}=0.475\times400 = 190\ J$. The power output $P=\frac{W}{t}=\frac{190}{3}\approx63.33\ W$

Step3: Calculate number of boxes lifted in 1 hour

In 1 hour ($t = 3600\ s$), the total work done $W_{total}=P\times t=63.33\times3600 = 228000\ J$.
The work done in lifting one box is $W_{box}=mgh$, where $m = 10\ kg$, $g = 9.8\ m/s^{2}$, and $h = 1\ m$. So $W_{box}=10\times9.8\times1 = 98\ J$.
The number of boxes $N=\frac{W_{total}}{W_{box}}=\frac{228000}{98}\approx2326.53\approx2326$

Answer:

Heat supplied per second: $642857.14\ J/s$, Heat discarded per second: $462857.14\ J/s$

2. Area under the curve