Question

Air enters a hair dryer at \(22^\circ {C}\) and 100 kPa with a velocity of \(3.7 \, {m/s}\), and leaves the dryer at \(83^\circ {C}\) and 100 kPa with a velocity of \(9.1 \, {m/s}\). The exit area of the dryer is \(18.7 \, {cm}^2\), and the ambient temperature is \(22^\circ {C}\). The air is an ideal gas with gas constant \(R = 0.287 \, {kJ/kg-K}\) and isobaric specific heat \(c_p = 1.005 \, {kJ/kg-K}\).
If the change in potential energy is neglected, the second law efficiency (in %) of the dryer is .......... {(rounded off to one decimal place).}

Hint:

Second law efficiency compares the actual energy gain with the maximum possible (exergy) gain.
For devices like dryers, apply both sensible heat and kinetic energy contributions to actual gain.

Answer 1

Step 1: Given values:
Inlet conditions: \(T_1 = 22^\circ C = 295 \, K\), \(V_1 = 3.7 \, {m/s}\)
Exit conditions: \(T_2 = 83^\circ C = 356 \, K\), \(V_2 = 9.1 \, {m/s}\)
\(A_2 = 18.7 \, {cm}^2 = 18.7 \times 10^{-4} \, {m}^2\)
\(c_p = 1.005 \, {kJ/kg-K}, R = 0.287 \, {kJ/kg-K}, T_0 = 295 \, K\)
Step 2: Calculate actual change in energy per unit mass:
\[ \Delta h + \Delta KE = c_p(T_2 - T_1) + \frac{V_2^2 - V_1^2}{2 \times 1000} \] \[ = 1.005(356 - 295) + \frac{(9.1)^2 - (3.7)^2}{2 \times 1000} = 1.005(61) + \frac{82.81 - 13.69}{2000} = 61.305 + 0.0346 \approx 61.34 \, {kJ/kg} \] Step 3: Calculate maximum possible work (exergy change):
\[ \Delta ex = c_p(T_2 - T_0) - T_0 \cdot R \cdot \ln\left(\frac{T_2}{T_0}\right) \] \[ = 1.005(356 - 295) - 295 \cdot 0.287 \cdot \ln\left(\frac{356}{295}\right) = 61.305 - 295 \cdot 0.287 \cdot \ln(1.2068) = 61.305 - 295 \cdot 0.287 \cdot 0.188 = 61.305 - 15.88 \approx 45.42 \, {kJ/kg} \] Step 4: Second law efficiency:
\[ \eta_{II} = \frac{{Actual Energy Increase}}{{Maximum Available Energy}} = \frac{61.34}{45.42} \approx 0.910 \] \[ \eta_{II} = \frac{45.42}{61.34} \times 100 \approx 9.0 % \]