Chapter 22, Problem 6P

(a)

To determine

The number of liters of fuel gasoline engine consumes in $1.00\text{h}$ of operation if the heat of combustion of the fuel is equal to $4.03\times {10}^7\text{J}/\text{L}$.

Answer to Problem 6P

The number of liters of fuel gasoline engine consumes in $1.00\text{h}$ of operation if the heat of combustion of the fuel is equal to $4.03\times {10}^7\text{J}/\text{L}$ is $29.4\text{L}/\text{h}$.

Explanation of Solution

Write the expression for the input energy of the gasoline for a particular time interval.

  $E_{\text{in}}=E_{\text{rev}}{\omega }_{\text{op}}\Delta t$                                                                                                          (I)

Here, $E_{\text{in}}$ is the input energy of the gasoline for a particular time interval, $E_{\text{rev}}$ is the input energy for each revolution of the crankshaft, ${\omega }_{\text{op}}$ is the operating frequency of the gasoline engine and $\Delta t$ is the time interval.

Conclusion:

It is given that, operating frequency of the engine is $2.50\times {10}^3\text{rev}/\text{min}$, energy required per revolution is $7.89\times {10}^3\text{J}$ and energy exhausted for each revolution is $4.58\times {10}^3\text{J}$.

Substitute $2.50\times {10}^3\text{rev}/\text{min}$ for ${\omega }_{\text{op}}$, $7.89\times {10}^3\text{J}$ for $E_{\text{rev}}$ and $1\text{h}$ for $\Delta t$ in equation (I) to get input energy in each hour.

  $\begin{array}{c}{E}_{\text{in}}=\left(7.89\times {10}^3\text{J}\right)\left(2.50\times {10}^3\text{rev}/\text{min}\right)\left(1\text{h}\times \frac{\text{60}\text{min}}{1\text{h}}\right)\\ =1.18\times {10}^9\text{J}/\text{h}\end{array}$

It is given that heat of combustion of the fuel is equal to $4.03\times {10}^7\text{J}/\text{L}$.

Calculate the number of liters of fuel consumed in $1.00\text{h}$ of operation if the heat of combustion of the fuel is equal to $4.03\times {10}^7\text{J}/\text{L}$.

  $\begin{array}{c}{V}_h=\left(1.18\times {10}^9\text{J}/\text{h}\right)\left(\frac{1\text{L}}{4.03\times {10}^7\text{J}}\right)\\ =29.4\text{L}/\text{h}\end{array}$

Here, $V_h$ is the liters of fuel gasoline engine consumes in $1.00\text{h}$ of operation.

Therefore, the liters of fuel gasoline engine consumes in $1.00\text{h}$ of operation if the heat of combustion of the fuel is equal to $4.03\times {10}^7\text{J}/\text{L}$ is $29.4\text{L}/\text{h}$.

(b)

To determine

The mechanical power output of the engine.

Answer to Problem 6P

The mechanical power output of the engine is $185\text{hp}$.

Explanation of Solution

Write the expression for the energy taken from the hot reservoir using first law of thermodynamics.

  $Q_h=W_{\text{eng}}+\left|Q_c\right|$                                                                                                         (II)

Here, $Q_h$ is the amount of thermal energy taken from the hot reservoir, $W_{\text{eng}}$ is the work done by the engine and $\left|Q_c\right|$ is the thermal energy exhausted at cold reservoir.

Divide above equation by time for continuous-transfer process.

  $\frac{Q_h}{\Delta t}=\frac{W_{\text{eng}}}{\Delta t}+\frac{\left|Q_c\right|}{\Delta t}$                                                                                                      (III)

Here, $\Delta t$ is the time interval.

Write the expression for the power output.

  $P_{\text{eng}}=\frac{W_{\text{eng}}}{\Delta t}$                                                                                                             (IV)

Here, $P_{\text{eng}}$ is the power output of the engine.

Substitute $P_{\text{eng}}$ for $\frac{W_{\text{eng}}}{\Delta t}$ in equation (III) and rewrite the equation for $P_{\text{eng}}$ .

  $\begin{array}{c}\frac{Q_h}{\Delta t}=P_{\text{eng}}+\frac{\left|Q_c\right|}{\Delta t}\\ P_{\text{eng}}=\frac{Q_h}{\Delta t}-\frac{\left|Q_c\right|}{\Delta t}\end{array}$                                                                                                        (V)

Conclusion:

Substitute $7.89\times {10}^3\text{J}/\text{rev}$ for $\frac{Q_h}{\Delta t}$ and $4.58\times {10}^3\text{J}/\text{rev}$ for $\frac{\left|Q_c\right|}{\Delta t}$ in equation (V) to get $P_{\text{eng}}$.

  $\begin{array}{c}{P}_{\text{eng}}=\left(7.89\times {10}^3\text{J}/\text{rev}-4.58\times {10}^3\text{J}/\text{rev}\right)\left(\frac{2500\text{rev}}{1\text{min}}\right)\left(\frac{1\text{min}}{60\text{s}}\right)\\ =1.38\times {10}^5\text{W}\times \frac{1\text{hp}}{746\text{W}}\\ =185\text{hp}\end{array}$

Therefore, the mechanical power output of the engine is $185\text{hp}$.

(c)

To determine

The torque exerted by the crankshaft on the load.

Answer to Problem 6P

The torque exerted by the crankshaft on the load is $527\text{N}\cdot \text{m}$ .

Explanation of Solution

Write the expression for the power output of the engine in terms of the torque exerted by the crankshaft on the load.

  $P_{\text{eng}}=\tau \omega$                                                                                                                 (VI)

Here, $P_{\text{eng}}$ is the power output of the engine, $\tau$ is the torque exerted by the crankshaft on the load and $\omega$ is the angular frequency of rotation.

Rearrange above equation to get $\tau$.

  $\tau =\frac{P_{\text{eng}}}{\omega }$                                                                                                                  (VII)

Conclusion:

Substitute $1.38\times {10}^5\text{W}$ for $P_{\text{eng}}$ and $2.5\times {10}^3\text{rev}/\text{min}$ for $\omega$ in equation (VII) to get $\tau$.

  $\begin{array}{c}\tau =\frac{1.38\times {10}^5\text{W}}{\left(2.5\times {10}^3\text{rev}/\text{min}\times \frac{1\text{min}}{60\text{s}}\right)}\left(\frac{1\text{rev}}{2\pi \text{rad}}\right)\\ =527\text{N}\cdot \text{m}\end{array}$

Therefore, the torque exerted by the crankshaft on the load is $527\text{N}\cdot \text{m}$.

(d)

To determine

The power the exhaust and cooling system must transfer out of the engine.

Answer to Problem 6P

The power the exhaust and cooling system must transfer out of the engine is $1.91\times {10}^5\text{W}$.

Explanation of Solution

Write the expression for the power transfer of exhaust and cooling system.

  $P_{\text{exhaust}}=\frac{\left|Q_c\right|}{\Delta t}$                                                                                                        (VIII)

Here, $P_{\text{exhaust}}$ is the power transfer of exhaust and cooling system.

Conclusion:

It is given that, frequency of revolution of the crankshaft is $2.5\times {10}^3\text{rev}/\text{min}$

Calculate the energy exhausted in $2500\text{rev}$.

  $\begin{array}{c}\left|Q_c\right|=4.58\times {10}^3\text{J}/\text{rev}\times 2500\text{rev}\\ =\text{11450}\text{J}\end{array}$

Substitute $\text{11450}\text{J}$ for $\left|Q_c\right|$ and $1\text{min}$ for $\Delta t$ in equation (VIII) to get $P_{\text{exhaust}}$ .

  $\begin{array}{c}{P}_{\text{exhaust}}=\frac{\text{11450}\text{J}}{1\text{min}\times \frac{60\text{s}}{1\text{min}}}\\ =1.91\times {10}^5\text{W}\end{array}$

Therefore, the power the exhaust and cooling system must transfer out of the engine is $1.91\times {10}^5\text{W}$.