A) A gas-fired heat exchanger uses gas with energy value of 40MJ / m^3 at a rate of 0.4 m^3/h and produces 50kg/h of water at a temperature of 70^oC from feed water at 20^oC. The rest of the question is in the picture attached.PLZ help?

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2 Answers
Jul 11, 2018

I got this

Explanation:

Let us calculate amount of heat required to raise the temperature of feed water at 20^@C to water at 70^@C by the gas fired heat exchanger in one hour.

Mass of water used in one hour m=50\ kg
Rise in temperature DeltaT=70^@-20^@=50^@C
Useful heat Q_u=msDeltaT

=>Q_u=50xx(4.19xx10^3)xx50
=>Q_u=10475000\ J

Heat produced by the gas in the heat exchanger in one hour

Q_p=0.4xx(40xx10^6)
=>Q_p=16000000\ J

Thermal efficiency of the heat exchanger eta=Q_u/Q_p

eta=10475000/16000000=0.65, rounded to two decimal places.

Jul 11, 2018

Thermal efficiency = 65.47% (2dp)

Explanation:

Start by working out how much energy is required to heat 50 kg of water from 20^oC to 70^oC (1 hr worth):

Specific heat capacity of water = 4.19(kJ)/(kg^oC)

so 50 kg raised 50 degrees will need

Energy = 4.19(kJ)/cancel(kg^oC) * 50cancel(kg) *50^ocancelC

= 10475 kJ

= 10.475 MJ This is the energy/hr transferred to the water

Energy/hr in from gas = 40(MJ)/cancelm^3 * 0.4 cancelm^3 = 16MJ

so thermal efficiency is ("heat out")/("heat in")

= (10.475 MJ)/(16MJ)

= 0.6546875

= 65.47% (2dp)