Question

Please show all working including formulas. Thanks you in advance

Q12 A 38% efficient power plant puts out 850 MW of electrical power. Cooling towers take away the exhaust heat. (a) If the air temperature is allowed to rise 7.0 Co, estimate what volume of air (km3) is heated per day. Wil the local climate be heated significantly? (b) If the heated air were to form a layer 180 m thick, estimate how large an area it would cover for 24 h of operation. Assume the air has density 1.3 kgm. and has specific heat of about 1.0 kJ /kg.Co at constant pressure.

Answer 1

In this case we have to use the formula for efficiency of heat engine i.e eta = w/Q_supplied (w = Q_supplied - Q_rejected) (or) eta = w/w + Q_rejected w + Q_rejected = w/eta Q_rejected = w/eta - w = w(1/eta - 1) \r\n Q_rejected = w(1/eta - 1) now heat rejected per unit time is given by Q_rejected/t = (w/t)(1/eta - 1) here w/t = power (p) Q_rejected/t = (p)(1/eta - 1) Q_rejected/t = (850 times 10^6)(1/0.38 - 1) = 13.86 times 10^8 watts \r\n now heat energy rejected by the power plant is given by Q_rejected = m c delta T m = Q_rejected/c(delta T) we know that density = mass/volume rho = m/rho v m = rho v therefore rho v = Q_rejected/c(delta T) \r\n at unit time rho v = (Q_rejected/t)/c(delta t) times t (t = 24 hr = 24 times 60 min = 24 times 60 times 60 sec = 86700 sec v = (13.87 times 10^8)(86400)/(1.3)(1 times 10^3)(7) v = 1.3168 times 10^10 m^3/day (b) now, volume v = A times t A = v/t