Question

please answer as soon as possible!

2) Predict the boiling heat-transfer coefficient for water under pressure boiling at 250°F for a horizontal surface of 1/16-in-thick stainless steel having a k of 9.4 btu/h ft °F. The heating medium on the other side of this surface is a hot fluid at 290°F having an h of 275 btu/h ft2 °F. Use the simplified equations (20 points). Hint: You should start by guessing a value for Tw! Question 2 Solution: Tsat = 250 F Tw stainless steel Plate Thickness: 1/16" T 290 F h, 275 Btu/h ft? F

Answer 1

Given figure, we have:

Let the area of heat transfer perpendicular to the direction of heat transfer be 'A'

At steady state, we can write:

h: A (290-T ) = kA = hwater-A (TO-250)

waterLw

Putting required values, we get

275(290-Ti)-9A (Ti )-heat" (Tu-250 0.0052ater (7w 250)

So, we can write

( ) 275 (290-T) = 9.4 0.0052

275 (290-ті) = 1807.69 (Ti-Tu)

0.152 (290-Tİ) = (Tİ-Tu)

...............................................................(1)

ー44 . 117 + 1.1527-T,/

Also, we can write

Ti-Tu: hwater (Tay-250) 9.4 0.0052

, 1807.69 ( TI-T,o) hu'a ter (Tu-250)

...........................................................................(2)

1807.69 (T-Tu) T - 250 hwater

As we know that

h_{boiling water} = 1057 BTU / hr.ft2.F.........................................................(3)

If we assume

T_w = 280 ⁰F

So, as per above calculation, value of 'h' is not matching with equation 3

If we assume

T_w = 280 ⁰F

So, as per above calculation, value of 'h' is not matching with equation 3

If we assume

T_w = 265 ⁰F

So, as per above calculation, value of 'h' is not matching with equation 3

If we assume

T_w = 260 ⁰F

So, as per above calculation, value of 'h' is not matching with equation 3

If we assume

T_w = 257.465 ⁰F

Enter the value of Tw 257.465 As per equation 1 T1 261.7899 hwater 1047.305

So, as per above calculation, value of 'h' is matching with equation 3.

Hence,

T_w = 257.465⁰F

T₁ = 261.7899⁰F

h_water = 1047.305 (BTU / hr.ft2.F)