PROBLEM 3.127

As shown in Fig. P3.127, a well-insulated tank fitted with an electrical resistor of negligible mass

holds 2 kg of nitrogen (N2), initially at 300 K, 1 bar. Over a period of 10 minutes, electricity is

provided to the resistor at a constant voltage of 120 volts and with a constant current of 1

ampere. Assuming ideal gas behavior, determine the nitrogen’s final temperature, in K, and the

final pressure, in bar.

KNOWN: Data are provided for nitrogen contained in a well-insulated tank fitted with and

electrical resistor. Voltage is applied and a current flow for 10 minutes.

FIND: Determine the final temperature and pressure.

SCHEMATICA ND GIVEN DATA:

ANALYSIS: Reducing the energy balance; ΔKE + ΔPE + ΔU = Q Welec. Thus

ΔU = Welec → n(

where n = m/M = (2 kg)/(28.01 lg/kmol) = 0.0714 kmol denotes the amount of N2 on a molar

basis. Solving for

(*)

The value = 6229 kJ/kmol can be read from Table A-23. We will next evaluate the work and

calculate the value of . Then, we can return to Table A-23 and interpolate to find T2.

The rate of energy transfer by work (magnitude) in watts due to electric current flow through the

resistance is

Rate of energy transfer in = (voltage)(amperage) = (120 volts)(1 ampere) = 120 watts

Thus

-120 watts = - 0.12 kW

Since the voltage and current are constant, the power is constant and the total amount of energy

transfer by work for the 10 minute period is

Nitrogen, N2

m = 2 kg

T1 = 300 K

p1 = 1 bar

120 volt

1 ampere

Δt = 10 minutes

ENGINEERING MODEL: (1) The

nitrogen in the tank is the closed system.

(2) The tank is well-insulated, so we

assume that =0 (3) The resistor has

negligible mass. (4) The nitrogen can be

modeled as an ideal gas. (4) Kinetic and

potential energy effects are negligible.

PROBLEM 3.127 (CONTINUED)

Welec =

= (-0.12 kW)(10 min)

= -72 kJ

Inserting values in (*)

(6229 kJ/kmol) – (-72 kJ)/(0.0714 kmol) = 7237.4 kmol

From Table A-23; T2 = 348.4 K

Since the volume is constant, the final pressure is

= V =

→ p2 = (T2/T1)p1 = (348.4/300)(1 bar) = 1.16 bar

PROBLEM 3.131

PROBLEM 3.131 (CONTINUED)

PROBLEM 3.134