Water flows through a tube of 3 {cm} internal diameter and length 20 {m}. The outside surface of the tube is heated electrically so that it is subjected to uniform heat flux circumferentially and axially. The mean inlet and exit temperatures of the water are 10^{\circ} {C} and 70^{\circ} {C}, respectively. The mass flow rate of the water is 720 {kg} {/} {h}. Disregard the thermal resistance of the tube wall. The internal heat transfer coefficient is 1697 {W} {/} {m}^{2} \cdot {K} . Take specific heat {C}_{ {p}} of water as 4.179 {kJ}{/} {kg} \cdot {K} . The inner surface temperature at the exit section of the tube is { }^{\circ} {C} (round off to one decimal place).

GATE Mechanical Engineering (ME) 2020 Shift 2 Solved Paper

Question : 60 of 65

Marks: +1, -0

Water flows through a tube of

3 {cm}

internal diameter and length

20 {m}

. The outside surface of the tube is heated electrically so that it is subjected to uniform heat flux circumferentially and axially. The mean inlet and exit temperatures of the water are

10^{∘} {C}

and

70^{∘} {C}

, respectively. The mass flow rate of the water is

720 {kg} {/} {h}

. Disregard the thermal resistance of the tube wall. The internal heat transfer coefficient is

1697 {W} {/} {m}^{2} ⋅ {K} .

Take specific heat

{C}_{ {p}}

of water as

4.179 {kJ}{/} {kg} ⋅ {K} .

The inner surface temperature at the exit section of the tube is

{ }^{∘} {C}

(round off to one decimal place).

Your Answer:

Validate

Solution:

Dinner = 3 cm, L = 20 m

\;\text " Net heat flux "\;=\;{\;\text " Total heat gain by water "\;}/{\;\text " Total surface area "\;}

\;\; =\;{˙{m C}_{p}(T_{0}-T_{i})}/{π D_{i} L}

=\;{720 × 4.179 × 10^{3} × 60}/{3600 × π × 0.03 × 20}

=26604.340 \;{ {W}}/{ {m}^{2}}

Heat flux at the exist

=26604.34 \;{ {w}}/{ {m}^{2}}

{q}{''}= {h}[ {T}_{\;\text "exit "\;}- {T}_{0}]

26604.34=1697 ×(T_{\;\text "exit "\;}-70)

The inner surface temperature at the exist section of the tube in

=\;{26604.34}/{1697}+70

= 85.67°C

Go to Question:

Prev Question

Next Question