Question

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?

Answer 1

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.

Answer 2

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)