# Compression in a heat pump with dissipation loss

I'm reading an introduction into heat pump cycles right now, and have question about the compression part:

The optimum compression would be an adiabatic one, after which you end at a certain temperature T(1) and pressure p(1). If the compression is non-ideal you still want to reach pressure p(1) but now the textbook says that we have a new temperature T(2) which is higher than T(1). I dont get how this can happen?

Following that logic a part of the compression work would lead to an isobaric compression, and increase the temperature only. But how can the pressure stay constant while I increase temperature? If I want to keep the pressure constant while decreasing the volume, I need an equal amount of heat exchange with the surroundings- again cooling the gas itself?

kind regards

• Why don't you just derive the equations for the two situations (i.e., model them) and see how it plays out? Jun 22, 2020 at 15:20
• i would if i knew how. not studying physics, Im just interested in the topic. I have calculated that for an adiabatic compression T(1) / T(0) = p(1) / p(0) but no idea how to calculate a combined sometimes isobaric- sometimes not isobaric equation Jun 22, 2020 at 16:12
• your equation is incorrect for either reversible or irreversible compressions Jun 23, 2020 at 0:37

$$\Delta U=-W$$ or, equivalently, $$\Delta U=-p_1(V_1-V_0)$$where U is the internal energy of the gas. For the case of an ideal gas, this equation reduces to: $$nC_v(T_1-T_0)=p_1\left(\frac{nRT_1}{p_1}-\frac{nRT_0}{p_0}\right)$$where n is the number of moles of gas (which cancels from the equation). So the gas temperature changes as a result of it exchanging energy in the form of work, with the surroundings.
Solving this equation for the ratio of the final temperature to the initial temperature yields: $$\left(\frac{T_1}{T_0}\right)=\frac{1+(\gamma-1)(p_1/p_0)}{\gamma}$$