91Ó°ÊÓ

First, calculate the initial and the final kinetic energy of air using the formula \( KE = 0.5 * m * V^2 \), where m is the mass flow rate and V is the velocity. In this case, \( m = 1 \, \mathrm{kg/s} \), \( V_{initial} = 75 \, \mathrm{m/s} \), and \( V_{final} = 125 \, \mathrm{m/s} \). Calculate for both initial and final states.

Next, calculate the change in kinetic energy \( \Delta KE = KE_{final} - KE_{initial}\).

We know that the heat transfer from the compressor to the cooling water is 18 kJ/kg of air which is equivalent to -18 kJ/s for an air flow of 1 kg/s. The first law of thermodynamics for open systems states that \( \Delta H + \Delta KE = Q - W \), where Q is the heat transfer, W is the power required, and ΔH is the change in enthalpy. Here, W is what we want to find, Q is given (-18 kJ/s), and we have already calculated ΔKE. Thus, we can solve for \( \Delta H = Q - \Delta KE - W \).

Since we know that the enthalpy of a gas at pressure P and temperature T is given by \( H = C_p * T \), where \( C_p \) is the specific heat capacity at constant pressure. The change in enthalpy \( \Delta H = C_p * T_2 - C_p * T_1 \), where \( T_1 \) is the initial temperature and \( T_2 \) is the final temperature. Plugging \( \Delta H \) into the earlier equation gives \(C_p * T_2 - C_p * T_1 = Q - \Delta KE - W \). Solving for W gives \( W = C_p * T_2 - C_p * T_1 - Q + \Delta KE\). Note that for air, \( C_p = 1.005 \, \mathrm{kJ/(kg.K)} \). The initial temperature \( T_1 \) is not given, but we know from standard conditions that it's 298 K. The final temperature \( T_2 = 345 \, \mathrm{K} \). Now we can calculate W.