Respuesta :
The rate of heat transfer in the heat exchanger is 484 kW
The exit temperature of water is 99.2°C
Explanation:
Given -
Specific heat of oil, Coil = 2.2 KJ/Kg°C
Specific heat of water, Cwater = 4.18 KJ/Kg°C
ΔEsystem = Ein - Eout
Where E is the energy
Here, change in kinetic and potential energy of the fluid stream are negligible.
Therefore,
ΔEsystem = Ein - Eout = 0
Ein = Eout
mh₁ = Qout + mh₂ (since Δke ≅ Δpe ≅ 0)
Qout = mCp (T1 - T2)
The rate of heat transfer from oil -
Q = [m X Coil (Tin - Tout)]
Q = (2 kg/s) (2.2 KJ/Kg°C) (150°C - 40°C)
Q = 484 kW
Heat lost by oil is gained by water.
Therefore, the outlet temperature of the water is
Q = [m X Cwater (Tout - Tin)]
Tout = Tin + Q/ m X Cwater
Tout = 22°C + 484 kJ/s/ (1.5 kg/s) (4.18 kJ/kg°C)
Tout = 99.2°C
a. The rate of heat transfer in the heat exchanger is equal to 484 kW.
b. The exit temperature of water is equal to 80.71°C.
Given the following data:
- Specific heat capacity of oil = 2.20 kJ/kg°C.
- Specific heat capacity of water = 4.18 kJ/kg°C.
- Initial temperature = 40°C.
- Final temperature = 150°C.
- Rate for oil = 2 kg/s
- Temperature in = 22°C
- Rate for water = 1.5 kg/s
a. To determine the rate of heat transfer in the heat exchanger, we would apply the law of conservation of energy:
[tex]\Delta E = E_{in} -E_{out} =0[/tex]
Thus, rate of heat transfer from oil in the heat exchanger is given by the formula:
[tex]Q_o = m_oc_o\theta\\\\Q_o= 2 \times 2.2 \times (150-40)\\\\Q_o = 4.4 \times 110[/tex]
Heat transfer = 484 kW
b. To determine the exit temperature of water:
Heat lost by oil is equal to the heat gained by the water according to the law of conservation of energy:
[tex]Q_o = Q_w\\\\Q_o = m_wc_w(T_{out} - T_{in})\\\\T_{out}=\frac{Q\; +\;T_{in} }{m_wc_w}[/tex]
Substituting the given parameters into the formula, we have;
[tex]T_{out}=\frac{484\; +\;22 }{1.5 \times 4.18}\\\\T_{out}=\frac{506}{6.27}[/tex]
Exit temperature = 80.71°C
Read more on heat transfer here: https://brainly.com/question/12072129
