Thermal efficiency of a Diesel engine

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xJJx
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Hi, i’m learning about the thermal efficiency of a diesel cycle at university and I'm trying to understand the equation. The first law of thermodynamics states Wnet + Qnet = 0, but one of the attachments below says Qnet - Wnet = 0 and I don't understand how this is the case. Also, for process 4-1 (isochoric heat rejection), it says that Qout = -mcv(T1-T4); I don't get why the minus is there. Can anyone explain why? Thank you.
 
Attachments
  • Table of related formulas for a Diesel cycle: isentropic compression 1-2 and isentropic expansion 3-4 with pressure, volume and temperature ratios, constant volume heat addition 2-3 with Qin = mCv(T3 - T2), and constant volume heat rejection 4-1 with Qout = mCv(T4 - T1).
    CD24C8EE-AE12-4EAD-8FC6-BFD82C01FC72.jpeg
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  • Slide showing Otto cycle thermal efficiency: eta_th = W_net/Q_in = Q_net/Q_in = (Q_in - Q_out)/Q_in = 1 - Q_out/Q_in, then first law for process 2-3 at constant volume giving Q_net,23 - W_net,23 = delta U_23, and Q_net,23 = delta U_23 = Q_in = mC_v(T_3 - T_2).
    CB371048-E466-4ACA-9939-35A80F9D6409.jpeg
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  • Slide deriving Otto cycle thermal efficiency: first law for closed system process 4-1 at constant V, Qnet,41 - Wnet,41 = ΔU41, then Qout = -mCv(T1 - T4) = mCv(T4 - T1), and eta = 1 - mCv(T4 - T1)/mCv(T3 - T2).
    A07602B3-5479-42D3-ABD5-4696604EE205.jpeg
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xJJx said:
The first law of thermodynamics states Wnet + Qnet = 0
That's not what the first law says. It says that ΔU = Q - W: any change in the internal energy of a system must come from heat entering/leaving the system or work done on/by the system. The convention used here is that positive W is the work done by the system, hence the negative sign.