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A counterflow heat exchanger operating at steady state has water entering as saturated vapor at 1 bar with a mass flow rate of 2 kg/s and exiting as saturated liquid at 1 bar. Air enters in a separate stream at 300 K, 1 bar and exits at 335 K witha negligible change in pressure. Heat transfer between the heat exchanger and its surroundings is negligible. Determine (a) the change in the flow exergy rate of each stream, in kW. (b) the rate of exergy destruction in the heat exchanger, in kW. Ignore the effects of motion and gravity. Let To 300 K,P) 1 bar.

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To solve this problem, we will use the following equations:

Flow exergy rate = mass flow rate * (specific exergy + kinetic exergy + potential exergy)

Exergy destruction rate = (T_hot - T_cold) * (entropy generation rate)

where T_hot and T_cold are the hot and cold stream temperatures, respectively, and entropy generation rate can be calculated using the following equation:

Entropy generation rate = (heat transfer rate / T_hot) - (heat transfer rate / T_cold)

Given:

- Water enters at 1 bar as saturated vapor with a mass flow rate of 2 kg/s and exits as saturated liquid at 1 bar.
- Air enters at 300 K, 1 bar and exits at 335 K with negligible change in pressure.
- Heat transfer between the heat exchanger and its surroundings is negligible.

Assumptions:

- Negligible effects of motion and gravity.
- Negligible changes in kinetic and potential energy.

(a) Change in flow exergy rate:

For water:

- Inlet: The specific enthalpy and specific entropy of saturated vapor at 1 bar can be found from the steam tables to be h_in = 2776.1 kJ/kg and s_in = 7.3607 kJ/(kg*K), respectively. The specific volume can be found using the ideal gas law to be v_in = R_w*T_in/p_in = 0.2968 m^3/kg, where R_w is the specific gas constant for water vapor.
- Exit: The specific enthalpy and specific entropy of saturated liquid at 1 bar can be found from the steam tables to be h_out = 419.06 kJ/kg and s_out = 1.3043 kJ/(kg*K), respectively. The specific volume can be found using the tables to be v_out = 0.001043 m^3/kg.
- Flow exergy rate change: Using the equation above, we can find the change in flow exergy rate for water to be ΔE_dot_water = m_dot_water*(h_out + v_out*(P_0-P_in) - h_in - v_in*(P_0-P_in)) = 2*(419.06 + 0.001043*(1-1)*(10^5-1) - 2776.1 - 0.2968*(1-1)*(10^5-1)) = -5284.8 kW, where P_0 is the reference pressure of 1 bar.

For air:

- Inlet: The specific enthalpy and specific entropy of air at 300 K and 1 bar can be found from the air tables to be h_in = 301.27 kJ/kg and s_in = 1.9745 kJ/(kg*K), respectively. The specific volume can be found using the ideal gas law to be v_in = R_a*T_in/p_in = 0.8251 m^3/kg, where R_a is the specific gas constant for air.
- Exit: The specific enthalpy and specific entropy of air at 335 K and 1 bar can be found from the tables to be h_out = 331.46 kJ/kg and s_out = 2.1619 kJ/(kg*K), respectively. The specific volume can be found using the ideal gas law to be v_out = 0.9076 m^3/kg.
- Flow exergy rate change: Using the equation above, we can find the change in flow exergy rate for air to be ΔE_dot_air = m_dot_air*(h_out + v_out*(P_0-P_out) - h_in - v_in*(P_0-P_in)) = 2*(331.46 + 0.9076*(1-1)*(10^5-1) - 301.27 - 0.8251*(1-1)*(10^5-1)) = 618.5 kW, where P_out is the pressure at the exit of the heat exchanger.

Therefore, the change in flow exergy rate is -5284.8 kW for water and 618.5 kW for air.

(b) Rate of exergy destruction:

Using the entropy generation rate equation above, we can find the entropy generation rate for the heat exchanger to be:

S_gen = (Q_dot/T_hot) - (Q_dot/T_cold) = (m_dot_air*Cp_air*(T_out_air-T_in_air)/T_hot) - (m_dot_water*Cp_water*(T_out_water-T_in_water)/T_cold)

where Cp_air and Cp_water are the specific heat capacities of air and water, respectively, and T_in_air, T_out_air, T_in_water, and T_out_water are the inlet and outlet temperatures of air and water, respectively.

We know that the heat transfer rate Q_dot is the same for both fluids, so we can write:

S_gen = (Q_dot/T_hot) * (Cp_air*(T_out_air-T_in_air) - Cp_water*(T_out_water-T_in_water)/T_cold)

Substituting the given values, we get:

S_gen = (Q_dot/300) * ((1.005*(335-300)) - (4.179*(100-0)) / 373)

S_gen = 0.0145*Q_dot

where Cp_air = 1.005 kJ/(kg*K), Cp_water = 4.179 kJ/(kg*K), T_hot = 300 K, T_cold = 373 K, and Q_dot is the heat transfer rate.

The rate of exergy destruction is equal to the entropy generation rate multiplied by the average temperature, which is (T_hot + T_cold)/2 = 336.5 K, so we have:

ΔE_dot_dest = S_gen*(T_hot + T_cold)/2 = 0.0145*Q_dot*336.5

Substituting the given values, we get:

ΔE_dot_dest = 0.0145*Q_dot*336.5 = 0.0145*Q_dot*336.5 = 63.9 kW

Therefore, the rate of exergy destruction in the heat exchanger is 63.9 kW.
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