A gas flows through a one-inlet, one-exit control volume operating at steady state. Heat transfer at the rate takes place only at a location on the boundary where the temperature is Tb. For each of the following cases, determine whether the specific entropy of the gas at the exit is greater than, equal to, or less than the specific entropy of the gas at the inlet: (a) No internal irreversibilities,
(b) no internal irreversibilities,
(c) no internal irreversibilities,
(d) internal irreversibilities,
(e) internal irreversibilities .

(a) S₁-S₂ = 0 ⇒ S₁=S₂ (b) S₁-S₂<0 ⇒ S₁<S₂ (c) S₁-S₂>0 ⇒ S₁>S₂ (d)S₁-S₂>0 ⇒ S₁>S₂

Explanation:

Solution

Now,

From the entropy rate balance at steady state, the equation is stated below:

Qcv =/Tb + m (S₁-S₂) +σcv = 0

ΔS = Qcv/Tb +σcv

So,

S₁-S₂ = Qcv/mTb +σcv/m

(a) No internal irreversibilities and Q cv =0

Now,

If Qcv =0, the and value of σcv = 0

S₁-S₂ = (Qcv/mTb + σcv/m) = 0

hence

S₁-S₂ = 0 which is S₁=S₂

(b) no internal irreversibilities, and Q cv <0

If Q cv <0 the value becomes this,

S₁-S₂ = (Qcv/mTb + σcv/m) <0

So,

S₁-S₂<0 which is S₁<S₂

(c) no internal irreversibilities and Q cv >0

If Q cv >0 the value is

S₁-S₂ = (Qcv/mTb + σcv/m) >0

So,

S₁-S₂>0 which is S₁>S₂

(d) internal irreversibilities and  Qcv ≥ 0

If Q cv ≥0 the value is

S₁-S₂ = (Qcv/mTb + σcv/m) >0

So,

S₁-S₂>0 which is S₁>S₂

Note: option (d) and (e) are the same

answer:

try looking it up that is always

explanation: