Heat Transfer Test 4

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Heat Transfer Test 4

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Weekly Quiz Competition

Question 1
1 / -0

In a particular heat exchanger, identical fluids flow through the tubes at the same mass flow rate. Calculate the effectiveness of the heat exchanger if the hot fluid entry and exit temperatures are 70°C and 40°C and cold fluid enters at 27°C and leaves at 57°C.
_______

A
0.65-0.75

B
0.654-0.751

C
0.657-0.756

D
0.659-0.752

Solution

Concept:
Hot fluid exit temperature is lower than the cold fluid entry temperature.
Therefore, it is a counter flow heat exchanger.
Effectiveness = Actual heat transfer/Maximum possible heat transfer
Calculation:
\(\varepsilon = \frac{{{C_h}\left( {{T_{hi}} - {T_{ho}}} \right)}}{{{C_{min}}{\rm{\Delta }}{T_{max}}}}\)
\(\varepsilon = \frac{{{T_{hi}} - {T_{ho}}}}{{{\rm{\Delta }}{T_{max}}}} = \frac{{{T_{co}} - {T_{ci}}}}{{{\rm{\Delta }}{T_{max}}}}\)
ΔT_max = T_hi - T_ci
\(\therefore \varepsilon = \frac{{30}}{{70 - 27}}\)
∴ ε = 0.697
Question 2
1 / -0

Hot gases enter a heat exchanger at 200°C and leave at 150°C. The cold air enters at 40°C and leaves at 140°C. The capacity ratio of the heat exchanger will be:

A
0.40

B
0.45

C
0.50

D
0.52

Solution

Concept:
The heat capacity of the fluid is the product of mass flow rate and the specific heat of the fluid.
The capacity ratio in the heat exchanger is the ratio of minimum heat capacity to the maximum heat capacity
Capacity Ratio \(R = \frac{{{C_{min}}}}{{{C_{max}}}}\)
In a heat exchanger
Heat released by hot gas = Heat gained by cold gas
ṁ_h C_ph (T_hi - T_he) = ṁ_c C_pc (T_ce - T_ci)
Calculation:
Given, T_hi = 200°C, T_he = 150°C, T_ce = 140°C, T_ci = 40°C
Now,
ṁ_h C_ph (200 - 150) = ṁ_c C_pc (140 - 40)
ṁ_h C_ph × 50 = ṁ_c C_pc × 100
clearly cold fluid capacity is more compared to hot fluid
\(R = \frac{{50}}{{100}}= 0.5\)
Question 3
1 / -0

In the counter-current heat exchanger, the ratio of heat capacity rates of cold and hot fluid is 1. If the effectiveness of the heat exchanger is 0.6, the number of transfer units are ____.

A
1.5-1.5

B
1.54-1.51

C
1.57-1.56

D
1.59-1.52

Solution

Concept:
Effectiveness of counter-current flow with fluids having equal heat capacities is given as
\(\epsilon = \frac{{NTU}}{{1 + NTU}}\)
Calculation:
\(\epsilon = \frac{{NTU}}{{1 + NTU}}\)
\(0.6 = \frac{{NTU}}{{1 + NTU}}\)
∴ NTU = 1.5
Question 4
1 / -0

Steam enters the condenser at 287 K and leaves at 295 K. The condenser temperature is maintained at 303 K and the surface area of the tubes is 50 m². If the overall heat transfer coefficient is 2200 W/m²K, then the rate of condensation of steam (kg/min) ____ (At 303 K, h = 2453 kJ/kg)

A
30-32

B
304-321

C
307-326

D
309-322

Solution

Concept:
Condenser can be treated as counter-flow or counter-current heat exchanger.
Calculate the logarithmic mean temperature difference
\(\Delta {T_{lm}} = \;\frac{{\Delta {T_1} - \;\Delta {T_2}}}{{\ln \left( {\frac{{\Delta {T_1}}}{{\Delta {T_2}}}} \right)}}\)
Using heat transfer, Q = hA∆T_lm
And equating this with heat lost from steam we can calculate the mass condensed per min.
Calculation:
ΔT₁ = 30 – 14 = 16°C
ΔT₂ = 30 – 22 = 8°C
\(\Delta {T_{lm}} = \;\frac{{\Delta {T_1} - \;\Delta {T_2}}}{{\ln \left( {\frac{{\Delta {T_1}}}{{\Delta {T_2}}}} \right)}}\)
ΔT_lm= 11.54° C
Q = hA ΔT_lm
Q = 2200 × 50 × 11.54
Q = 12.69.4 kW
Now,
Q = ṁh
\(\dot m = \frac{{1269.4}}{{2453}} = 0.52\;kg/s\)
∴ ṁ = 31.049 kg/min