Solve Otto Cycle Problem: Heat Rejection (300 kW)

[PauloBuzon]

hey my fellow ME please help me with this problem I am stuck at.

1. Homework Statement
An Otto engine has a clearance volume of 7%. It produces 300 kW of power. What is the amount of heat rejected in kW?

Homework Equations

Wnet=Qadded-Qrejected
(v2=Vc)Clearance Volume = (c)Clearance% x (Vd)Volume Displacement (v1-v2)

where r is the compression ratio

The Attempt at a Solution

rk(compression ratio) = (c+1)/c stuck i don't know where to start
then i should be able to solve the problem where Qrejected = Qadded - Work

 

[DrClaude]

Your missing an equation for efficiency.

 

[PauloBuzon]

Your missing an equation for efficiency.

i just edited it

 

[jack action]

Energy balance:
$$\dot{W} = \dot{Q}_{in} - \dot{Q}_{out}$$
Efficiency definition:
$$n_{th} = \frac{\dot{W}}{\dot{Q}_{in}}$$
Otto cycle efficiency:
$$n_{th} = 1- \frac{1}{r^{(\gamma-1)}}$$
Compression ratio definition:
$$r = \frac{V_d + V_{cc}}{V_{cc}}$$
You get 4 equations, 4 unknowns ($\dot{Q}_{in}$, $\dot{Q}_{out}$, $n_{th}$, $r$), so you can resolve them. $V_d$ and $V_{cc}$ are not known but their ratio is given in the problem (and it is all that is really needed):
$$0.07 = \frac{V_{cc}}{V_d}$$
(or it might be $0.07 = \frac{V_{cc}}{V_d + V_{cc}}$; Not clear from the statement -> 7% of what?)

 

[PauloBuzon]

Energy balance:
$$\dot{W} = \dot{Q}_{in} - \dot{Q}_{out}$$
Efficiency definition:
$$n_{th} = \frac{\dot{W}}{\dot{Q}_{in}}$$
Otto cycle efficiency:
$$n_{th} = 1- \frac{1}{r^{(\gamma-1)}}$$
Compression ratio definition:
$$r = \frac{V_d + V_{cc}}{V_{cc}}$$
You get 4 equations, 4 unknowns ($\dot{Q}_{in}$, $\dot{Q}_{out}$, $n_{th}$, $r$), so you can resolve them. $V_d$ and $V_{cc}$ are not known but their ratio is given in the problem (and it is all that is really needed):
$$0.07 = \frac{V_{cc}}{V_d}$$
(or it might be $0.07 = \frac{V_{cc}}{V_d + V_{cc}}$; Not clear from the statement -> 7% of what?)

7% clearance volume , clearance volume = cVd = c(v1-v2) so 0.07 = c(v1-v2)