Work comparison for compression vs. constant-pressure heating: In a single-stage, single-acting reciprocating air compressor without clearance, the work done during isentropic compression from T1 to T2 is __________ the heat required to raise air from T1 to T2 at constant pressure.
Correct Answer: equal to
Introduction / Context:Relating compression work to heating requirements helps build intuition for compressor energy demands and the role of specific heats in ideal analyses.
Given Data / Assumptions:
- Isentropic (adiabatic, reversible) compression from temperature T1 to T2.
- No clearance volume; ideal gas with constant specific heats.
- Comparison is made to constant-pressure heating over the same temperature rise.
Concept / Approach:
For an isentropic compression of an ideal gas, the specific work input equals the change in specific enthalpy because there is no heat transfer and h = cp * T for an ideal gas with constant cp. Thus w_isentropic,in = h2 − h1 = cp * (T2 − T1). At constant pressure, the heat required for the same temperature rise is also q_cp = cp * (T2 − T1). Hence they are equal in magnitude.
Step-by-Step Solution:
For ideal gas with constant cp: h2 − h1 = cp * (T2 − T1).Isentropic compression work input equals enthalpy rise: w_in = h2 − h1.Constant-pressure heating requires q = cp * (T2 − T1).Therefore w_in = q, i.e., they are equal.Verification / Alternative check:
Using the steady-flow energy equation for an adiabatic compressor (neglecting KE/PE changes) also yields w_in = Δh. The same Δh arises for constant-pressure heating by definition of cp.
Why Other Options Are Wrong:
“Less than” or “greater than” contradicts the enthalpy-based equality under the ideal assumptions; “indeterminate” is incorrect because Δh depends only on temperatures for an ideal gas.
Common Pitfalls:
Mixing isentropic with polytropic processes; assuming variable cp without stating it, which would slightly modify numerical equality but not the learning goal.
Final Answer:
equal to