Some Applications

3. Some Applications

Using Eq. (90), the availability of the hot gases within a boiler can be mapped if the local state data is available. Locations where the availability loss rate is large can be identified. At steady state, neglecting both the kinetic and potential energies, e = h and Eq. (106) simplifies to the form r r

(107) In the absence of internal temperature gradients, for a reversible process

ρ v ⋅∇−∇=∇⋅ ( hTs 0 ) (( λ ∇ T )( 1 − T 0 / )) T − ′′′ − ′′′ w ˙ ˙ i

ρ rr v ⋅∇ = ′′′ ψ w opt , i.e., (108) the local value of ψ is a measure of reversible work for a steady state adiabatic system con-

taining negligible kinetic and potential energy. For an isothermal turbine performing reversible work, Eq. (90) may be expressed as ρ∂a/∂t + rr ρ

(109) where a = u – Ts, and g = h – Ts. Therefore, changes in the Helmholtz and Gibbs functions are rr

v ⋅∇ = − ′′′ g w ,

measures of work in reversible systems. In a closed system ρ v ⋅∇ g = 0 so that ∂a/∂t = w ˙ ′′′ / ρ = w ˙ which is the work rate per unit mass. Since, the system is internally reversible, i.e., there

are no internal gradients, i.e., A 2 –A 1 = –W.

On the other hand in a steady open reversible system ρ r

v ⋅ ∇ = − ′′′ ( g ) w ˙ . (1 1 0) The advective term is absent in nonflow systems and, consequently, Eq. (106) simplifies to the

form

(111) For instance, if a large pot of coffee is cooled such that it is internally isothermal at all instants,

ρ∂ ( eTs − 0 )/ ∂ t =∇⋅∇ ( λ T ( 1 − T 0 / )) T − ′′′ − ′′′ w ˙ ˙ i .

this relation can be expressed as

ρ∂ ∂ u / tTs − 0 ∂∂ / t = ρ c ( 1 − T 0 /) TT ∂ / ∂ t = − ′′′ ˙ i ,

n. Example 14

A turbine blade of length L is subjected to hot gases at a temperature T ∞ . The blade surface temperature T w is maintained by wall cooling. The blade can be simulated as a flat plate subjected to laminar flow of hot gases at T ∞ so that a boundary layer δ(x) grows over the plate as shown in Figure 22 . At steady state, the temperature profile

within the boundary layer is given by the expression Θ=1–2

3 ξ –+ 2 ξ 4 – ξ , where (A) ξ = y/δ(x),

(B) where

δ(x) = C x 1/2 , C is a constant and Θ = (T – T ∞ )/(T w –T ∞ ).

(C) Using the differential form of the generalized availability balance, obtain an expres-

sion for the availability loss rate per unit volume at the wall. Obtain an expression for the availability loss rate per unit volume in the free stream.

If T ∞ = 1300 K, T w = 900 K, the thermal conductivity –6 kW m –2 λ = 70×10 –1 K , and

δ(x) = 0.005 m, determine the availability loss rate per unit volume at the wall. What is the availability transfer rate associated with the heat flow at the wall?

Solution For a steady non–work–producing two–dimensional process, the simplified differen- tial form of availability balance equation is

At the wall v x =v y = 0 so that

T  ∂ y  At the blade wall T = T w so that

Using Eqs. (B) and (C) ( ∂T/∂y) x = (dT/d ξ)(∂ξ/∂y) x =d Θ/dξ (T w –T ∞ )/ δ(x).

(G) At the wall

( ∂T/∂y) x | y=0 =d Θ/dξ | ξ=0 (T w –T ∞ )/ δ(x). (H) From Eq. (A),

d Θ/dξ = –2 + 6ξ 2 –4 3 ξ , and

(I)

T(y)

Figure 22: Laminar flow over a flat plate.

d Θ/dξ(ξ=0) = –2. Using this result in Eq. (H),

( ∂T/∂y) x | y=0 = –2 (T w –T ∞ )/ δ(x). (J) Similarly, ( ∂ 2 T/ ∂y 2 ) = (d/d ξ)((dΘ/dξ)(T –T )/ δ(x)) = (d 2 Θ/dξ x 2 w ∞ )(T w –T )/( δ (x)) 2 ∞ .

(K) Since d 2 Θ/dξ 2 ( ξ=0) = 0, ( ∂ 2 T/ ∂y 2 ) x | y=0 = 0.

(L) Using Eqs. (F), (J), and (L),

(M) The availability loss rate is always a positive quantity. In the free stream, ∂T/∂y = 0,

y ′′′ = = 0 λ (( TT 0 w − T ∞ )/ T w )( / ( ( )) ) 4 δ x .

2 and 2 ∂ T/ ∂y = 0. Therefore for the free stream

i ′′′ =0 (N) Using Eq. (M),

i ′′′ = 70 –6 2 2 2 ×10 –3 × 298 × ((900 – 1300) ÷ 900 ) × (4 ÷ 0.005) = 2637 kW m . For forced convection

δ(x) = C x 2 where C denotes a constant so that δ = C x. Therefore, using Eq. (M), the dimensionless availability loss rate is

0 ) = 4(1– T ∞ /T w ) = 4(1– 1300 ÷ 900) = 0.79. The local availability flow rate into the plate due to the heat transfer is

2 2 i 2 ′′′ (0) C x/( λT

q ˙ ′′ y=0 (1–T 0 /T w )=– λ(∂T/∂y) y=0 (1–T 0 /T w )

λ dΘ/dξ(ξ=0) (T =– w –T ∞ ) (1–T 0 /T w )/ δ(x) = 70

×10 –2 × 2 × (–400) × (1– 298 ÷ 900) ÷ 0.005 = –7.492 kW m . Remarks:

Assuming a 2.7m 2 of total blade area, there is a net 20 kW loss in availability. This trans- fer should be compared with the work gain due to the higher operational temperature. For instance, (1–T 0 /T g ) is the availability transferred from the gas at T g in the absence of

C.V T 0

Hair

Figure 23 Availability analysis for human hair.

cooling. This availability is increased if cooling is used since operating gas temperature is increased to T g ′ = T g + ∆T g and, consequently, the availability increase is

((1–T 0 /T g ′ )–(1–T

0 /T g )) ≈ (∆ T g T 0 /T g ) per unit amount of heat transferred from the hot gases. The availability loss due to the blade cooling is 2A b λ(T w –T g )(1–T 0 /T w )/ δ(x), where

A b denotes the blade area. If the availability gain is to be larger than the loss, then |˙| Q "

g T 0 /T g > (2 λ(T w –T g )(1 – T 0 /T w )/ δ(x)), where |˙| Q is the heat or energy loss from the hot gases per unit blade area.

2 ∆T "

o. Example 15 Determine the irreversibility loss for a human hair of length L. The hair has a surface temperature of T w at one end and an adiabatic tip at the other end, and is exposed to

an ambient temperature T 0 . The literature informs us that the heat transfer rate from the base of a fin of length L,

Q ˙ w = λA f (T w –T 0 ) tanh αL,

where α = (h H C/

λA 1/2

f ) , C denotes the hair circumference, A f is the cross-sectional area of the fin (hair), and h H is the convective heat transfer coefficient. Use the values

ρ = 165 kg m –3 ,

λ = 0.036 W m –1 K , the hair diameter d = 0.1 mm, and h

H = 0.1 W

m –2 K –1 . Note that C/A f = 4/d for a circular cross section. What is the maximum possible power that can be developed using a “hot head”?

Solution Selecting the control volume around a single hair,

I= ˙ Q w (1 – T 0 /T w )–Q 0 (1– T 0 /T 0 )= ˙ Q w (1 – T 0 /T w ), where T w = 310.2 K (average body temperature). Now,

Q ˙ w = λA –2

f (T w –T 0 ) tanh αL, where L = 10 m,

α = (h H C/

1/2 = 333.3 m × 4 ÷ (0.0001 × 0.036)) –1 . Therefore, Q ˙ w = 0.036

λA 1/2

f ) = (0.1

–4 2 ×(π(1×10 –4 ) ÷4) (310.2 – 298) × tanh (333.3×0.1×10 ) = 0.036

×3.1427×0.1 –7 ÷4 × 10 (310.2–298) × 1 = 3.7×10 W

I = 3.45

×10 –8 × (1–298÷310.2) =1.36×10 W

The maximum possible work rate equals the actual work rate plus the irreversibility rate. Since in this case no work is done, the irreversibility is the maximum possible work.

Remarks Shorter hair will have a lower heat loss. In order to reduce this irreversibility, you can couple a Carnot engine to each strand of hair and use the work so obtained to propel yourself! Since the human body is warm, body heat loss through our skin occurs at the rate of about 1 kW. Again, a Carnot engine may be used to extract work. Since the Carnot ef- ficiency will be 1 – 298/310 = 0.039, the power developed would be roughly 3.9 W which is of the order of power of a night lamp.