# 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?

Jul 11, 2018

I got this

#### Explanation:

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

Mass of water used in one hour $m = 50 \setminus k g$
Rise in temperature $\Delta T = {70}^{\circ} - {20}^{\circ} = {50}^{\circ} C$
Useful heat ${Q}_{u} = m s \Delta T$

$\implies {Q}_{u} = 50 \times \left(4.19 \times {10}^{3}\right) \times 50$
$\implies {Q}_{u} = 10475000 \setminus J$

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

${Q}_{p} = 0.4 \times \left(40 \times {10}^{6}\right)$
$\implies {Q}_{p} = 16000000 \setminus J$

Thermal efficiency of the heat exchanger $\eta = {Q}_{u} / {Q}_{p}$

$\eta = \frac{10475000}{16000000} = 0.65$, 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 ${20}^{o} C$ to ${70}^{o} C$ (1 hr worth):

Specific heat capacity of water = $4.19 \frac{k J}{k {g}^{o} C}$

so 50 kg raised 50 degrees will need

Energy $= 4.19 \frac{k J}{\cancel{k {g}^{o} C}} \cdot 50 \cancel{k g} \cdot {50}^{o} \cancel{C}$

$= 10475 k J$

$= 10.475 M J$ This is the energy/hr transferred to the water

Energy/hr in from gas $= 40 \frac{M J}{\cancel{m}} ^ 3 \cdot 0.4 {\cancel{m}}^{3} = 16 M J$

so thermal efficiency is $\left(\text{heat out")/("heat in}\right)$

$= \frac{10.475 M J}{16 M J}$

$= 0.6546875$

$= 65.47$% (2dp)