Answer:
I = 12.706 Amps
Explanation:
Given:
- The mass of saturated R-134a m = 12 kg
- The initial Conditions
P_1 = 240 KPa
Saturated Vapor
- The final Conditions
P_2 = 240 KPa
T_2 = 70° C
- The amount of Heat transferred Q_in = 300 KJ
- The voltage of current source V = 110 V
- The current supplied by the source = I
- The time duration Δt = 6 min
Find:
Determine the current supplied I.
Solution:
- Look-up enthalpies h_1 and h_2 at both states using Tables A-11 and A-13.
P_1 = 240 KPa
Saturated Vapor ----------> h_1 = h_g = 247.32 KJ/kg
P_2 = 240 KPa
T_2 = 70° C ----------> h_2 = 314.53 KJ/kg
- Using First Thermodynamic Law, set up an energy balance:
E_in - E_out = ΔE_system
Q_in + W_electric,in - W_out = Δ U
Q_in + V*I*Δt - W_out = Δ U
Q_in + V*I*Δt = Δ H
Q_in + V*I*Δt = m*( h_2 - h_1 )
- Make the current I the subject of the expression above:
V*I*Δt = m*( h_2 - h_1 ) - Q_in
I = [ m*( h_2 - h_1 ) - Q_in ] / V*Δt
- Plug in the values in the expression derived above and evaluate current of source I:
I = [ 12*( 314.53 - 247.32 ) - 300 ]*1000 / 110*6*60
I = 12.706 Amps
- The required source of current is I = 12.706 Amps.
