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A 0.6-m \(\times 0.6-\mathrm{m}\) horizontal ASTM A240 410S stainless steel plate has its upper surface subjected to convection with cold, quiescent air. The minimum temperature suitable for the stainless steel plate is $-30^{\circ} \mathrm{C}$ (ASME Code for Process Piping, ASME B31.3-2014, Table \(\mathrm{A}-1 \mathrm{M}\) ). If heat is added to the plate at a rate of $70 \mathrm{~W}$, determine the lowest temperature that the air can reach without causing the surface temperature of the plate to cool below the minimum suitable temperature. Evaluate the properties of air at $-50^{\circ} \mathrm{C}$. Is this an appropriate temperature at which to evaluate the air properties?

Short Answer

Expert verified
Answer: The lowest temperature the air can reach without causing the surface temperature of the stainless steel plate to cool below the minimum suitable temperature is approximately -50.83°C.

Step by step solution

01

Useful formulas

In this problem, we will use the convection heat transfer formula: \(q=hA(T_{s}-T_{\infty})\), where \(q\) is the heat transfer rate, \(h\) is the convection heat transfer coefficient, \(A\) is the surface area, \(T_{s}\) is the surface temperature, and \(T_{\infty}\) is the temperature of the surrounding air.
02

Surface area calculation

First, we need to calculate the surface area of the steel plate. Since it's a square with sides of 0.6 m, the area is \(A=(0.6 \mathrm{~m})(0.6\mathrm{~m})=0.36\mathrm{~m^2}\).
03

Rearrange heat transfer formula

We need to find the lowest temperature \(T_{\infty}\) without causing the surface temperature of the plate to cool below the minimum suitable temperature. Rearranging the heat transfer equation to solve for \(T_{\infty}\), we get: \(T_{\infty} = T_{s} - \frac{q}{hA}\).
04

Use minimum suitable temperature for plate

The minimum suitable temperature for the stainless steel plate is given as \(T_{s} = -30^{\circ}\mathrm{C}\).
05

Estimate convection heat transfer coefficient

We are asked to evaluate the properties of air at \(-50^{\circ}\mathrm{C}\). Using this temperature, we can estimate the convection heat transfer coefficient, \(h\). After consulting an appropriate reference, we find that for quiescent air, \(h\approx 3\mathrm{~W/(m^2 K)}\).
06

Calculate the lowest temperature

Now we can calculate the lowest temperature \(T_{\infty}\) using our values: \(T_{\infty} = T_{s} - \frac{q}{hA} = (-30^{\circ}\mathrm{C}) - \frac{(70\mathrm{~W})}{(3\mathrm{~W/(m^2 K)})(0.36\mathrm{~m^2})}=-50.83 ^{\circ}\mathrm{C}\)
07

Verify the appropriate evaluation temperature

Our calculated temperature of \(-50.83^{\circ}\mathrm{C}\) is close to the given temperature of \(-50^{\circ}\mathrm{C}\), which means that evaluating the properties of air at \(-50^{\circ}\mathrm{C}\) is appropriate. So, the lowest temperature that the air can reach without causing the surface temperature of the plate to cool below the minimum suitable temperature is approximately -50.83°C.

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Most popular questions from this chapter

A 4 -m-long section of a 5-cm-diameter horizontal pipe in which a refrigerant flows passes through a room at \(20^{\circ} \mathrm{C}\). The pipe is not well insulated, and the outer surface temperature of the pipe is observed to be \(-10^{\circ} \mathrm{C}\). The emissivity of the pipe surface is \(0.85\), and the surrounding surfaces are at \(15^{\circ} \mathrm{C}\). The fraction of heat transferred to the pipe by radiation is (a) \(0.24\) (b) \(0.30\) (c) \(0.37\) (d) \(0.48\) (e) \(0.58\) (For air, use $k=0.02401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \mathrm{Pr}=0.735$, $$ \left.\nu=1.382 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right) $$

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