The air in a 100 m^3 balloon is electrically heated from 10 C to 35 C in 12 minu

Question

The air in a 100 m^3 balloon is electrically heated from 10 C to 35 C in 12 minutes. Assuming that the balloon is well insulated, and atmospheric pressure is 100 kPa, how much electrical power will the resistance healer need to dissipate to heal up the balloon? Air in a closed system undergoes an isothermal process from an initial state at temperature T_1 = 100 C and pressure P_1 = 500 kPa to a final state at pressure P_2 = 100 kPa. If there is 1 kg of air in the system find: a. The initial volume, V_1 b. The Work done, W_1-2 c. The Heat Transfer. Q_1-2 Saturated water vapor is cooled and condensed into saturated liquid water in an isobaric process at a pressure of 100 kPa. Find a. The heat transferred per kg of H_2O, q_1-2 b. The work done per kg of H_2O, w_1-2 Two pressurized containers are connected with a valve. Before the valve is opened, container A contains m_A = 2kg of steam at a pressure of P_A = 1 MPa, and temperature of T_A = 300C, while container B contains m_B = 3 kg of a saturated liquid-vapor mixture at a temperature T_B = 150 C and quality x_B = 0.5. The valve is opened and the two containers are allowed to come to thermal and mechanical equilibrium at a pressure of P_AB = 300 kPa. Find: a. The temperature of the final state, T_AB b. The heat transferred from the tanks. Q_1-2 Copper rods are to be heated in a furnace from an initial temperature of T_1 = 20 C to a final temperature of T_2 = 200 C. The process calls for 60 steel rods per minute to be heated. Each copper rod has a radius of 0.5 cm and a length of 10 cm. What is the rate that heat must be transferred from the furnace to the copper rods? Assume the density and specific heat of the balls arc given to be r = 8900 kg m^3 and C_c = 0.393 kJ/kg degree C.

Explanation / Answer

1) PV = nRT

PV/T = nR = const

P1V1/T1 = P2V2/T2

V1/T1 = V2/T2

100/10 = V2/ 35

V2 = 350 m3

W = P dV = 100 * 250 = 25000 KJ

2)

For air, 1 mol = 0.029kg   

In 1kg of air, n = 1/ 0.029 = 34.48 moles

V = nRT/P = 34.48*8.3*100/ 500 = 57.24 L

b) P1V1 = P2V2

W = nRT ln (P1/P2) = 34.48 *8.3* 373* ln(500/100) = 171800 KJ

c) The heat transfer Q = -W = -171800 K

  

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